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2015-07-29T10:38:09.310 | <p>PSOC4 provides a quadrature decoder functionality but I was unable to find specifics on its performance in the datasheets. Sure I know I need to read and utilize the count until the 16 bits of its counter overflow, but that's some 32k per iteration of the main loop of my program. If a precise encoder is attached to a fast motor, that produces lots and lots of pulses in a very short time, and I'm worried the decoder module won't be able to keep up. </p>
<p>So, does anyone know what is the maximum performance of the decoder block - the minimum timings of its input?</p>
| |electrical-engineering|embedded-systems| | <h2>Quadrature decoder</h2>
<p>Looking at the <a href="http://www.cypress.com/file/131666/download">PSoC 4 TCPWM datasheet</a> the maximum operating frequency of the counter block is 48MHz, with a counter resolution of 21ns.</p>
<p>Assuming that you have a high resolution <a href="https://en.wikipedia.org/wiki/Rotary_encoder#Incremental_rotary_encoder">incremental rotary encoder</a> with 10,000 counts per revolution, and spin the rotor at 15,000 rpm, that's 2,500,000 counts per second (2.5MHz), comfortably less than the 48MHz max. sampling rate.</p>
<p>Sadly, it is not clear if the <em>PSoC 4 TCPWM</em> suffers the same 10x sampling requirement as the <a href="http://www.cypress.com/file/135186/download">PSoC QuadDec</a> mentioned in <a href="http://embedded.stackexchange.com/a/85/186">Simon Jenkins' answer</a>. I suspect not, as the datasheet doesn't include the clock signals on the timing diagrams. This implies that <em>PSoC 4 TCPWM</em> is edge driven (limited by the counter clock), while <em>PSoC QuadDec</em> is driven by level at each clock edge and thus limited by the sampling at the clock frequency.</p>
<p>Even if the <em>PSoC 4 TCPWM</em> does require over sampling, then the max 2.5MHz pulse train is still less than both 10x at 48MHz for the <em>PSoC 4 TCPWM</em> and 28MHz for the <em>PSoC QuadDec</em>.</p>
<p>If you only need 2000 PPR at 6000 RPM (as suggested in your comment), then you only need to support 200 KHz (2000*6000/60), so a 2MHz clock should be sufficient.</p>
<h2>Interrupt handling</h2>
<p>Another concern is whether the CPU will be able to cope with the interrupts this could generate.</p>
<p>If we assume that you are running the quadrature right at the limit of the decoder block, i.e. 48MHz with a 16bit counter you will be getting 1465 overflow interrupts per second, so unless you can service your interrupt (including any context switching required) in less than 682 micro seconds, then you are going to struggle.</p>
<p>Luckily, it looks like the the Cortex-M0 processors have very efficient ISR handling <sub>(See <a href="http://community.arm.com/docs/DOC-2607">A Beginner’s Guide on Interrupt Latency - and Interrupt Latency of the ARM® Cortex®-M processors</a>)</sub> so unless your ISR is very inefficient, it should be able to handle this.</p>
<h2>Resolution</h2>
<p>Incidentally, it looks like you will need to use X4 mode if you want to count individual quadrature state transitions. If you use X1 then you only see whole quadrature cycles (4 states), while if you use X2 then you only see half of the states.</p>
| 3967 | How fast movement (cycles/s) can I reliably read with PSoC4 quadrature decoder? |
2015-07-28T22:52:34.127 | <p>For example, if there is many floating point operations, it could "rewire" the area of some of its integer units to handle floating point instructions, and vice versa. Or if there isn't too many calculations, but there is many I/O need, it could parallelize its IO better.</p>
<p>As I know, the FPGAs load their HDL code on boot, but to me it seems not really impossible, to make it possible to reload differents parts of a bigger HDL on the need, partially.</p>
<p>Does a such FPGA already exist?</p>
| |electrical-engineering|embedded-systems| | <p>It seems that such (Xilinx) FPGAs already exist.
<a href="https://en.wikipedia.org/wiki/Reconfigurable_computing#Partial_re-configuration">https://en.wikipedia.org/wiki/Reconfigurable_computing#Partial_re-configuration</a></p>
<p>In partial reconfiguration, one part of the FPGA continues to run while another part is being reconfigured.</p>
| 3973 | Is there an FPGA which could "re-wire" itself on the need? |
2015-07-30T16:40:59.113 | <p>Atmel's ASF (Atmel Software Framework) is periodically updated by Atmel. When an update is available, Atmel Studio notifies you and asks if you want to download the update. If I install the update, this does not update any code in my existing projects.</p>
<p>I want to determine what code changes have been made in the new ASF compared to what I previously used in a project. And ideally I'd like to be able to cherry-pick the changes that I want to incorporate (and exclude those I don't).</p>
<p>The only way I know to do this now is to create new project with the updated ASF (using the ASF Wizard to add all the drivers I used previously) and manually diff the files against my existing project. Then manually copy/paste from the files I want to include.</p>
<p>I thought I heard that Atmel uses an internal git repository for their ASF. Is there a way to get at this directly and view the changes? Or is there a better way?</p>
<p>It would also be nice to view the changes before installing the ASF update.</p>
| |embedded-systems| | <p>I think what you're looking for is the <a href="http://www.atmel.com/webdoc/asf/asf.VersionConversionWizard.html" rel="nofollow noreferrer">ASF Conversion Wizard</a>. According to the linked page, this wizard is supposed to be launched whenever the ASF version changes. If is not launching for you, you can open the ASF Wizard in your existing project, and under the Version tab, you can change the ASF version for your current project. Make sure the "Trigger upgrade of modules" box is checked:</p>
<p><a href="https://i.stack.imgur.com/1YGdJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1YGdJ.png" alt="upgrade screenshot"></a></p>
<p>Image is from <a href="http://www.atmel.com/webdoc/asf/asf.ModuleExplorer.VersionTab.html" rel="nofollow noreferrer">ASF Wizard documentation</a>.</p>
<p>For viewing changes before you upgrade the ASF version, I think the release notes for a given ASF release are your only resource. They're admittedly not very complete, but I can't find any other documentation about them.</p>
<p>I don't know of any way to cherry-pick different sections of an ASF release. I don't think this is a great idea anyway, unless you're willing to wade through a lot of dependencies. Not all components of the ASF are direct windows between your software and the hardware; some of it is middleware that other components of the same ASF release may rely on. By picking and choosing which aspects of the ASF to update, you risks losing APIs, structure definitions, and other code that you may need in a non-obvious way.</p>
<p>Last note: the only ASF git repository I can find is here: <a href="https://spaces.atmel.com/gf/project/asf/scmgit/" rel="nofollow noreferrer">https://spaces.atmel.com/gf/project/asf/scmgit/</a>. Unfortunately, it doesn't appear to have been updated since 2012. I can't find any more up-to-date repositories.</p>
| 3978 | How can I determine the code changes in an Atmel ASF update? |
2015-07-31T13:43:05.453 | <p>In the Linux kernel's dtsi file (am335x-bone-common.dtsi) for the AM335x processor, there is the am33xx_pinmux tree, with values like this:</p>
<pre><code> i2c0_pins: pinmux_i2c0_pins {
pinctrl-single,pins = <
0x188 (PIN_INPUT_PULLUP | MUX_MODE0) /* i2c0_sda.i2c0_sda */
0x18c (PIN_INPUT_PULLUP | MUX_MODE0) /* i2c0_scl.i2c0_scl */
>;
};
</code></pre>
<p>How do I know what values to use for the <code>pinctrl-single,pins</code> entry?</p>
| |electrical-engineering|embedded-systems| | <p>To update your .dts file (that includes the file am335x-bone-common.dtsi) for a pinmux assignment, consult the datasheet to determine which pin(s) are assigned to a function. This is in section 4.3 (broken into subsections per module). Clicking on the hyperlink for a pin (check your package type, ZCE BALL or ZCZ BALL) will take you to the corresponding entry in Table 4-1 (Ball Characteristics). Use Table 4-1 to determine the correct mode to set for that pin.</p>
<p>Then consult the TRM, Table 9-7 (CONTROL_MODULE REGISTERS), on page 1363 for Rev K. The table lists the registers for the default function (likely mode 0). The pin-mux registers start at offset 0x800, and this offset must be subtracted to give the first value in the .dts file. Consult section 9.2.2 (Pad Control Registers) for options for the second value in the .dts file (in addition to the mode setting).</p>
<p>For example, to assign pin C18 of the ZCZ package to SPI1 SCLK, click on the C18 link in table 4-53 of the datasheet (section 4.3.6.5). This takes you to the <code>ECAP0_IN_PWM0_OUT</code> pin in table 4-1. Here you see that a mode value of 4 will select <code>spi1_sclk</code> for that pin.</p>
<p>Then in the TRM, table 9-7 shows <code>conf_ecap0_in_pwm0_out</code> registers start at address 0x964. We must subtract 0x800, so the first value in the .dts is 0x164.</p>
<p>Then we want to select mode 4 (<code>MUXMODE=4</code>). If we also want to enable a pull-up (<code>PULLTYPESEL=1</code>, <code>PULLUDEN=0</code>), then the pad control field is 0x14. The .dts entry would then be:</p>
<pre><code>&am33xx_pinmux {
spi1_pins_s0: spi1_pins_s0 {
pinctrl-single,pins = <
0x164 0x14 /* ecap0_in_pwm0_out.spi1_sclk */
>;
};
};
</code></pre>
| 3980 | In a device tree for AM335x, how do I know what pinctrl values to use? |
2015-08-03T14:48:51.097 | <p>I'm considering using the i.MX 7 processor for a new design. The overall architecture appears to be similar to the i.MX 6, at least from comparing the block diagrams.</p>
<p>What mechanisms exist for communication between the Cortex-M4 core and the Cortex-A7 core(s)? Does the stock Linux kernel provide any such mechanisms (which likely would assume a specific RTOS or firmware running on the M4), or is this custom firmware for both processors in Freescale's offerings? The processors appear to share pretty much everything, so I would imagine you could implement a shared memory buffer with an atomic semaphore, or something along those lines.</p>
<p>Ideally, I would like to have the M4 firmware "take over" some peripherals while the A7 processor is in a low-power mode (i.e. manage periodic communication transfers or some data analysis until something "interesting" happens that requires the A7 to do something). From the block diagrams and the i.MX 6 SoloX reference manual, it appears the hardware doesn't preclude this.</p>
| |design|embedded-systems| | <p>The generic term for multi-core systems that run different kernels (or bare metal code) on different cores is Asymmetric multi-processing (<strong>AMP</strong>). Other related terms are <a href="https://en.wikipedia.org/wiki/Heterogeneous_computing" rel="nofollow noreferrer"><strong>Heterogeneous Computing</strong></a>, which refers to multiple cores of different types in the same system, and Multiprocessor System-on-Chip (<a href="https://en.wikipedia.org/wiki/MPSoC" rel="nofollow noreferrer"><strong>MPSoC</strong></a>).</p>
<p>Freescale happens to have an <a href="http://www.freescale.com/files/32bit/doc/brochure/PWRARBYNDBITSRAS.pdf" rel="nofollow noreferrer">overview document (PDF)</a> discussing some of the terminology and design issues relating to such systems, which may be good background, though it is discussing a different OS (QNX) and product line (PowerPC). A frequently discussed free software combination is Linux on a processor an <a href="http://www.freertos.org/" rel="nofollow noreferrer">FreeRTOS</a> on the microcontroller. Several existing chips both from Freescale and other vendors have architectures similar to the i.MX7, such as TI's <a href="https://en.wikipedia.org/wiki/OMAP#OMAP_4" rel="nofollow noreferrer">OMAP 4/5</a>. There is also a lot of <a href="http://blog.idv-tech.com/2014/02/26/zedboard-linux-freertos-amp-board-bringup-guide/" rel="nofollow noreferrer">discussion of and example code for</a> AMP on Xilinx's dual-core Zynq ARM+FPGA chips.</p>
<p>Since 2012 or so the mainline Linux kernel has had optional support for the <a href="http://lxr.free-electrons.com/source/Documentation/remoteproc.txt" rel="nofollow noreferrer">Remote Processor</a> (remoteproc) and <a href="http://lxr.free-electrons.com/source/Documentation/rpmsg.txt" rel="nofollow noreferrer">Remote Processor Messaging</a> (rpmsg) frameworks. remoteproc is a generic interface for the Linux kernel to control the "remote" core (power on/off, load firmware). rpmsg is the generic interface for sending messages between cores. The linked documentation pages detail their APIs and give more context. A developer would need to write a custom application-specific Linux kernel driver using this framework to expose I/O to userspace, and then need to write, eg, FreeRTOS firmware that "speaks" rpmsg. The rpmsg framework generally seems to use shared memory to actually implement communication between the cores.</p>
<p>If one does not exist already, I assume Freescale will soon release documentation and examples specific to AMP with the i.MX7 and Linux. You might want to keep an eye on the <a href="http://git.freescale.com/git/cgit.cgi/imx/linux-2.6-imx.git/refs/heads" rel="nofollow noreferrer">Freescale Linux git repositories</a> as they implement i.MX7 support; these branches will presumably be merged into mainline in due time.</p>
| 3982 | What IPC mechanisms exist for the i.MX 7 (or i.MX 6) multi-core processors? |
2015-07-29T14:56:20.197 | <p>I have discovered a problem when using the SPI peripheral in slave mode on an Atmel SAM4S. Occasionally words can become bit-shifted. I can force this condition by enabling the peripheral some number of clocks after <code>NSS</code> goes active. I do this by waiting for NSSR, then using a delay loop to wait a specific time before I enable the peripheral. I have confirmed on a scope that when the peripheral is enabled <code>n</code> clocks after <code>NSS</code> goes active, the words are shifted by <code>n</code> bits.</p>
<p>When the peripheral gets in this state, the only way to recover is to perform a reset (i.e. set the <code>SWRST</code> in the control register). Disabling and re-enabling is not sufficient.</p>
<p>This issue has also been observed on a SAM9G45 MPU.</p>
| |electrical-engineering|embedded-systems| | <p>This is due to the design of their SPI peripheral. In slave mode, the shift register is shifted each <code>SPCK</code> (assuming <code>NSS</code> is low and <code>SPIENS</code> is set). There is an internal counter (not shown in the block diagram) for the shift register, and when it reaches the word size, the contents are shifted to <code>SPI_RDR</code> and from <code>SPI_TDR</code>. Thus if you have a word size of 8, and only 7 clocks are present for the first <code>NSS</code>, on the next <code>NSS</code>, the shift register contents will not be transferred until after the first clock cycle.</p>
<p>When <code>NSS</code> goes high, this counter is not reset. Similarly, setting <code>SPIDIS</code> does not reset the counter. Only <code>SWRST</code> resets the counter.</p>
<p>In addition to the method described to force this condition, glitches on <code>SPCK</code> during <code>NSS</code> (such as from EMI) can also result in a bit-shift.</p>
<p><a href="https://i.stack.imgur.com/udIfG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/udIfG.png" alt="SAM4S Slave Mode Block Diagram"></a></p>
| 3984 | Why are words bit-shifted for an Atmel SAM4S SPI slave? |
2015-07-28T18:26:32.947 | <p>For an embedded Linux system, if I have two or more network interfaces, how do I ensure that they always get the same interface names every boot</p>
<p>In other words, I want, for example, eth0 to always map to one physical Ethernet port, eth1 to the next, etc.</p>
<p>My Linux "distribution" is home-grown, and I use devtmpfs for populating /dev. I use busybox for init (and most everything else), along with custom init scripts for system startup and shutdown.</p>
<p>I do not need hotplug facilities of mdev or udev -- I'm referring to "fixed" Ethernet ports.</p>
| |embedded-systems| | <p>You mentioned that you do not <strong>need</strong> udev but if you are not opposed to installing it then you may want to consider using <a href="http://linux.dell.com/biosdevname/" rel="nofollow">biosdevname</a>. There is a great <a href="http://www.freedesktop.org/wiki/Software/systemd/PredictableNetworkInterfaceNames/" rel="nofollow">wiki page</a> on <a href="http://freedesktop.org" rel="nofollow">freedesktop.org</a> that discusses this issue. As noted on that page, if you're using udev version 197 or higher then you don't even need biosdevname because udev will already do what you need.</p>
| 3986 | How do you ensure physical network interfaces always get the same interface name across reboots on an embedded Linux system? |
2015-07-31T16:08:51.560 | <p>I have an Atmel SAM4S MCU with a 12-bit internal ADC. I would like to achieve 16-bit resolution at a sample rate of 500 Hz. The input to the ADC is fairly low-frequency (about 20 Hz max frequency content), and low noise (less than one 12-bit lsb RMS).</p>
<p>How can I achieve 16-bit resolution with this processor's ADC?</p>
| |electrical-engineering|embedded-systems|adc| | <p>You can increase the effective resolution by intentionally oversampling the input signal. Here is brief summary of the idea from an application note on the topic provided by Atmel:</p>
<blockquote>
<p>The theory behind ‘Oversampling and decimation’ is rather complex, but using the method is fairly easy. The technique requires a higher amount of samples. These extra samples can be achieved by oversampling the signal. For each additional bit of resolution, n, the signal must be oversampled four times. Which frequency to sample the input signal with, is given by Equation 3-1. To get the best possible representation of a analog input signal, it is necessary to oversample the signal this much, because a larger amount of samples will give a better representation of the input signal, when averaged.</p>
</blockquote>
<p>The full application note can be obtained here: <a href="http://www.atmel.com/images/doc8003.pdf">http://www.atmel.com/images/doc8003.pdf</a>.</p>
<p>In your specific case, you would need to sample at about 8KHz to get the equivalent of 16 bits of resolution out of your 12-bit ADC.</p>
<p>Here's how I arrived at this number:</p>
<pre><code>f_oversample_rate = f_current_sample_rate * (4 * (desired_bits - current_bits))
f_oversample_rate = 500Hz * (4 * (16 - 12))
f_oversample_rate = 500Hz * (4 * 4)
f_oversample_rate = 500Hz * 16
f_oversample_rate = 8KHz
</code></pre>
| 3989 | How can I increase my processor's ADC resolution? |
2015-08-04T13:38:27.773 | <p>If I have a 10 MHz digital signal, how fast do I need to sample with a logic analyzer to avoid any bit errors? Does the minimum sample rate increase if I want to attach a protocol decoder (e.g. SPI)?</p>
<p>I am aware of the <a href="https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem">Nyquist-Shannon sampling theorem</a> and the fact that a 10 MHz square wave has significant energy at higher harmonics (30 MHz, 50 MHz, etc.). However, a logic analyzer doesn't need to perfectly sample or reproduce a square wave to know if it is a logic 1 or 0. I'm looking for what is practically necessary to avoid bit errors or protocol decoding errors.</p>
<p>I am using the Saleae Logic Pro 16.</p>
| |electrical-engineering|embedded-systems| | <p>The least you can do is given by <a href="https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem">Shannon-Nyquist theorem</a>: to be able to read a signal, you shall sample at least twice its frequency. Here, you can sample at 20MS/s (Mega sample per second). Nevertheless, 2 points for one period is very few, and introduce much jitter on the rebuilt signal.</p>
<p>Your logic analyzer won't share your signal clock. This is a common case in transmission, and I would advise you to do the same as most UART/SPI/I2C/... interface: take 3 points per state. This allow to filter EMC noise using a majority filter, and, when analyzing, allows to see significant skew between your signals. Please note that the clock of a 10MHz SPI bus goes both high and low every 100ns period, so you need to sample at 6 times your SPI frequency (60MS/s).</p>
<p>60MS/s is common for logical analyzers, and this will allow you to detect excessive skew (between your data and your clock) for most protocols (including SPI). Nonetheless, with a non-symmetric clock (with tHIGH != tLOW), you'll need to sample fast enough to sample the shortest part of the clock. If you clock is HIGH for 1µs and LOW for 9µs, you've got a 100kHz signal, but need 3MS/s.</p>
<p>This answer only considers the signal you want to analyze. For EMC issues, we usually use analog scopes with a sampling frequency 10 times higher than the highest frequency in the system, but it's a bit expensive, and off-topic, considering that the question is to decode an SPI bus.</p>
| 3991 | How fast should I sample with a logic analyzer? |
2015-07-28T23:14:09.687 | <p>I am thinking on analog circuits, whose circuit design could be read (or, at least, significantly altered) from software. Do they already exist?</p>
| |electrical-engineering|embedded-systems| | <p>Yes, it's called Field-programmable Analog Array (FPAA):</p>
<p><a href="https://en.wikipedia.org/wiki/Field-programmable_analog_array">https://en.wikipedia.org/wiki/Field-programmable_analog_array</a></p>
<p>They are exactly the analog version of the FPGAs. There are many different subtypes (current or voltage driven, discrete or continous time). They are built up from CABs (configurable analogous blocks), and from interconnects between them.</p>
| 3993 | Do "analog FPGAs" exist? |
2015-08-07T00:45:24.977 | <p>I am currently working on project where the commands for a PC to the MSP430 will be communicated via UART. These commands need to be converted to binary to be able to written to the SPI bus. </p>
<p>For example 'a' from PC is translated to 0x61 on the microcontroller. What is really needed is hex 'a' Thus<br>
ASCII 'a' from PC = 0x61 => need to translate hex 0x0A </p>
<p><strong>Question :</strong> What is the most simplest way to convert ASCII to binary? I am using "C". </p>
| |electrical-engineering|embedded-systems| | <p>The input from the PC is a character, i.e. an ASCII-value. To understand how this is saved, have a look at the <a href="http://www.asciitable.com/" rel="noreferrer">ASCII-table</a>. The numbers '0' to '9' are represented by the ASCII-codes <code>0x30</code> to <code>0x39</code>. The capital letters 'A' to 'Z' are represented by <code>0x41</code> to <code>0x5A</code>, and the small letters 'a' to 'z' by <code>0x61</code> to <code>0x7A</code>.</p>
<p>You could of course use a function like <code>sscanf</code>, as suggested by @JohnO'M. Such functions (<code>printf</code>, <code>scanf</code>, ...) have many features but thus require a lot of power and memory<sup>[Citation needed]</sup>, which we often don't want to spend on a microcontroller. A simple alternative would be to distinguish the 3 possible cases (0-9 or A-F or a-f) and subtract the correct number from the ASCII code to get to the result:</p>
<pre><code>if( (s >= 0x30) && (s <= 0x39)) { // 0-9
x = s - 0x30;
}
else {
if( (s >= 0x41) && (s <= 0x46)) { // A-F
x = s - 0x37;
}
else {
if( (s >= 0x61) && (s <= 0x66)) { // a-f
x = s - 0x57;
}
else { // wrong input
x = 0xFF;
}
}
}
</code></pre>
| 4007 | How to convert ASCII to binary to facilitate a UART to SPI bridge for an MSP430? |
2015-08-07T23:31:06.100 | <p>I am currently working on develop an IOT architecture. The sensing devices are to be integrated with microcontroller. The microcontroller will be integrated to some type of RF communication device. As the community is well aware RF communication protocols that will fit the bill. Once in consideration are Bluetooth smart/BLE, uIP, zigbee and Zwave. I intend to use module primarily to address FCC certification issues. At this point Bluetooth smart/BLE and Zigbee are the prime candidates.</p>
<p><strong>Question:</strong> From an Embedded Systems design standpoint what are the pros and cons of using Bluetooth smart/BLE over Zigbee or vice versa</p>
<p><strong>References:</strong> </p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/UIP_(micro_IP)" rel="nofollow">uIP (micro IP)</a></li>
</ul>
| |electrical-engineering|embedded-systems|iot| | <p>The present technology is all about IOT i. e with millions of devices connected over internet. Zigbee and BLE are two types of wireless communication which is used for interacting between two or more things or devices. Both have their own protocol and both have their merits and demerits.</p>
<p>BLE(Bluetooth low energy) is a wireless personal area network with low power and low cost. It has high dataa rate. It is specially used in Healthcare application like glucose monitoring,blood pressure monitoring of patients which is connected by IOT.</p>
<p>ZIGBEE is a personal are network with a range of about 100m this is not conventional method even though it uses low Power and has low cost because it does not support all types of operating systems (eg. Microsoft, Linux, Android, etc).</p>
<p>BLE is preferred over Zigbee, because it supports all operating systems in general, and mainly because it is supported by mobile. This is a conventional method as we can update the data using mobile sitting anywhere without the physical presence. </p>
| 4009 | What are the pros and cons of using Zibgee over BLE or vice versa for IOT application? |
2015-08-03T23:35:45.480 | <p>I need to configure the clock systems for a MSP430F5528 microcontroller. The current development board has a 24MHZ crystal attach to XT2 pins (XT2IN/XT2OUT). The current design will need to support the follow peripherals.</p>
<ul>
<li>UART</li>
<li>ADC</li>
<li>SPI</li>
<li>Timers</li>
</ul>
<p>Few GPIO's will be need for system power up. The system is will part of a low power application. The power budget has not yet been determined.</p>
<hr>
<p><strong>About the MSP430 clock system</strong></p>
<p>The MSP430 for the most part has three internal clocks</p>
<ol>
<li>MCLK - Master Clock - CPU</li>
<li>SMCLK - Sub Master Clock - fast peripherals </li>
<li>ACLK - Auxiliary Clok - Slow peripherals</li>
</ol>
<p><a href="https://i.stack.imgur.com/wYV4t.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wYV4t.jpg" alt="enter image description here"></a> </p>
<hr>
<p><strong>Question:</strong> What needs to be considered when configuring the clock system base on the above criteria? </p>
| |electrical-engineering|embedded-systems|power| | <p>It depends a bit on the quality you expect from some of the peripherals and what their qualities are at given clock speeds.</p>
<p>Will you want a very high speed SPI interface and can this specific chip support that on the Fast Auxiliary Clock? Or are you going to do 1MHz SPI at most? So also for all the other devices.</p>
<p>It's of course a nice boasting thing to be able to say that everything runs on many-MHz, but if it's not needed it will create more power consumption, more heat and more programming effort for no result. It can be tempting to just do everything at maximum speed and highest accuracy, but in almost all cases a good systems design will start with indexing what are the minimum speed and accuracy for something to work and then taking the appropriate margins. (Caveat: If you may need a higher speed later, it may be advantageous to make the hardware such that scaling up is only a firmware issue).</p>
<p>As for activating the Crystal; I don't actually know this particular chip, but it seems compatible in build-up to the ones that I know that have a configurable clock source, that is selected at power-up, in which case setting the crystal as a source is a simple configuration and possibly the configuration of all the other clocks is too. If that isn't the case and you need to actively switch while already running the core, you need to make 100% sure the crystal is running before you switch over. The datasheet will make very clear mention of the procedures needed for that if this is the case. Usually it's something along the line of: Activate Clock --> Wait for a bit in a register to be set --> Switch over.</p>
<p>With respect to the ADC, that can become a very complicated case. If you need very low sampling jitter or highly accurate timing (for highly accurate control or very (very) high grade Audio) you will quickly get to the domain of using the Crystal as a clock source. Where you'll then also need to make some hardware considerations. Both in the Analogue domain and on the Crystal end, since crystal accuracy is influenced by the loading capacitance (which includes traces and pins).</p>
<p>But those applications are few and far between.</p>
<p>In most cases your timers and your UART will determine the clocks you need, since they are most sensitive to clock drift and/or offset. If you need a timer just for low-speed timing you are better off relying on a watch crystal, since both the crystal and the internal driving circuitry for such low frequency crystals are usually better suited to high accuracy and low drift over long time.</p>
| 4011 | What needs to be considered when configuring the clock system for a MSP430F5xxx micro-controller? |
2015-08-01T13:39:10.427 | <p>I have been working on and off on small embedded systems project on and off. Some of these projects used an ARM Cortex-M4 base processor. In the project folder there is a <strong>startup.s</strong> file. Inside that file I noted the following two command lines.</p>
<pre><code>;******************************************************************************
;
; <o> Stack Size (in Bytes) <0x0-0xFFFFFFFF:8>
;
;******************************************************************************
Stack EQU 0x00000400
;******************************************************************************
;
; <o> Heap Size (in Bytes) <0x0-0xFFFFFFFF:8>
;
;******************************************************************************
Heap EQU 0x00000000
</code></pre>
<hr>
<p>How does one define the size of the <strong>heap</strong> and <strong>stack</strong> for a microcontroller? Is there any specific information in the datasheet to guide to arrive at the correct value? If so, what should one look for in the datasheet?</p>
<hr>
<p><strong>References:</strong></p>
<ul>
<li><a href="http://community.arm.com/docs/DOC-9701" rel="nofollow noreferrer">Decoding the Startup file for ARM Cortex M4</a></li>
<li><a href="https://www.lpcware.com/content/project/lightweight-ip-lwip-networking-stack/configuring-lwip/default-stack/heap-sizes" rel="nofollow noreferrer">Default stack/heap sizes</a></li>
</ul>
| |electrical-engineering|embedded-systems| | <p>In addition to the other answers, I'd like to add that when carving up RAM between stack and heap space, that you also need to consider the space for static non-constant data (e.g. file globals, function statics, and program-wide globals from a C perspective, and probably others for C++).</p>
<h2>How stack/heap allocation works</h2>
<p>It is worth noting that the startup assembly file is one way of defining the region; the toolchain (both your build environment and run-time environment) mostly care about the symbols which define the start of stackspace (used to store the initial stack pointer in the Vector Table) and the start and end of heap space (used by the dynamic memory allocator, typically provided by your libc)</p>
<p>In OP's example, only 2 symbols are defined, a size of stack at 1kiB and a size of heap at 0B. These values get used elsewhere to actually produce the stack and heap spaces</p>
<p>In @Gilles example, the sizes are defined <em>and used</em> in the assembly file to set a stack space starting wherever and lasting the size, identified by the symbol Stack_Mem and sets a label __initial_sp at the end. Likewise for the heap, where the space is the symbol Heap_Mem (0.5kiB in size), but with labels at the beginning and end (__heap_base and __heap_limit).</p>
<p>These get processed by the linker, which won't allocate anything within the stack space and heap space because that memory is occupied (by Stack_Mem and Heap_Mem symbols), but it can place those memories and all of the globals wherever it needs. The labels end up being symbols with no length at the given addresses. The __initial_sp is used directly for the vector table at link time, and the __heap_base and __heap_limit by your runtime code. The actual addresses of the symbols are assigned by the linker based on where it placed them.</p>
<p>As I aluded to above, these symbols don't actually have to come from a startup.s file. They can come from your linker configuration (Scatter Load file in Keil, linkerscript in GNU), and in those you can have finer grained control over placement. For example, you can force the stack to be at the beginning or end of RAM, or keep your globals away from the heap, or whatever you want. You can even specify that the HEAP or STACK just occupy whatever RAM is left over after globals are placed. NOTE though that you have to be careful that adding more static variables that your other memory will decrease.</p>
<p>However, each toolchain is different, and how to write the configuration file and what symbols your dynamic memory allocator will use will have to come from the documentation of your particular environment.</p>
<h2>Stack Sizing</h2>
<p>As to how to determine stack size, many toolchains can give you a maximum stack depth by analyzing the function call trees of your program, IF you don't use recursion or function pointers. If you do use those, estimating a stack size and pre-filling it with cardinal values (perhaps via the entry function before main) and then checking after your program has run for a while where the max depth was (which is where the cardinal values end). If you have fully exercised your program to its limits, you'll know fairly accurately whether you can shrink the stack or, if your program crashes or no cardinal values are left, that you need to increase the stack and try again.</p>
<h2>Heap Sizing</h2>
<p>Determining heap size is a bit more application dependant. If you only do dynamic allocation during startup, you can just add up the space required in your startup code (plus some overhead for memory management). If you have access to the source of your memory manager, you can know exactly what the overhead is, and possibly even write code to walk the memory to give you usage information. For applications that need dynamic runtime memory (e.g. allocating buffers for inbound ethernet frames) the best I can suggest is to carefully hone your stacksize and give the Heap everything that is left over after stack and statics.</p>
<h2>Final note (RTOS)</h2>
<p>OP's question was tagged for bare-metal, but I want to add a note for RTOSes. Often (always?) each task/process/thread (I'll just write task here on out for simplicity) will be assigned a stack size when the task is created, an in addition to task stacks, there will likely be a small OS stack (used for interrupts and such)</p>
<p>The task accounting structures and the stacks have to be allocated from somewhere, and this will often be from the overall heap space of your application. In these instances, your initial stack size often won't matter, because the OS will only use it during initialization. I have seen, for example, specifying ALL remaining space during linking be allocated to the HEAP and placing the initial stack pointer at the end of the heap to grow into the heap, knowing that the OS will allocate starting from the beginning of the heap and will allocate the OS stack just before abandoning the initial_sp stack. Then all of the space is used for allocating task stacks and other dynamically allocated memory.</p>
| 4013 | Defining the heap and stack size for an ARM Cortex-M4 microcontroller? |
2015-07-31T15:28:14.623 | <p>A colleague mentioned that Thumb instructions were significantly slower than ARM instructions (this was for an AT91SAM7S32 processor). Is this true? What are the performance benefits of one instruction set over another?</p>
| |embedded-systems|computer-engineering| | <h2>In General</h2>
<p>THUMB instructions aren't to my knowledge intrinsically slower than ARM instructions, but rather are more limited in capability. If your code only needs the functionality of THUMB instructions, it will occupy less space than ARM, but would be the same number of instructions and, other things being equal, run at the same speed. If your code needs more functionality, than it would require more THUMB instructions than ARM instructions to execute and would take longer, again other things being equal (see below)</p>
<p>THUMB is popular in microcontrollers because of the smaller size instructions for two reasons:</p>
<ol>
<li>Program space is often limited</li>
<li>Many microcontrollers have 16-bit data buses to their internal flash</li>
</ol>
<p>For the second reason, when your code doesn't require functionality from the ARM instruction set, THUMB code actually runs faster. This is because your instruction can be fetched in one I/O cycle from the flash instead of two. Depending on the speed of your flash interface, that second read can incur one or more wait-cycles per instruction where your CPU is simply stalled and cannot do anything.</p>
<p>This becomes less of an issue if you can copy your code to RAM before executing (which I've usually seen as 32-bit for recent ARM microcontrollers), where the only concern is code density. For that, many tools will try to find which representation is more efficient for a given function. If the compiler can produce THUMB code in fewer instructions it will, but if ARM results in fewer instructions you get ARM. This is the default mode for Keil, if I recall correctly.</p>
<h2>Your specific chip</h2>
<p>For your particular chip (AT91SAM7S32) the documentation mentions that the flash controller has a prefetch buffer that can predict accesses to make things more efficient, which might improve ARM instruction execution. However, it also states that the prefetch is a "dual 32-bit" buffer that "optimizes 16-bit accesses" which is most suitable for "running in Thumb mode", which seems to indicate that it isn't intended to speed up ARM instructions, but to allow your core to run faster in THUMB mode.</p>
<p>From the diagrams, it looks like the flash on your chip actually has a 32bit data bus. The prefetcher seems to work by reading a whole 32bits, giving 16 to the CPU (in THUMB mode) and caching the whole 32bits. During the next cycle, when the CPU reads the second 16 bits, this time from cache, the flash controller is reading the next 32bits and caching it. In this way THUMB code can run without more than an initial wait even if the flash speed would be a bit slower than the CPU core speed. Section 19.2.2 "Read Operations" has more details.</p>
<p>Since your flash is a 32bit bus (as near as I can tell), if your CPU and Flash clocks are the same, THUMB will only give you code density over ARM. If you want your CPU core to run faster than Flash (and note, I didn't review all of the timing of this chip; I assume the CPU can run faster because they let you set wait states), than the prefetch gives a speed advantage to THUMB due to the reduction in actual flash accesses. However, that speed advantage is a per-instruction advantage. If the number of THUMB instructions vs ARM instructions is big enough, it will outweigh the per instruction speed resulting in ARM having a faster per-routine speed.</p>
| 4019 | How do Thumb instructions compare to ARM instructions for performance? |
2015-08-16T10:44:26.503 | <p>I notice that quite a number of deep excavation projects adopt double I-section walers to form a box beam since standard rectangular hollow sections don't come with such large sizes. The I-sections are usually factory butt-welded together at the flanges.</p>
<p><a href="https://i.stack.imgur.com/Thkkd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Thkkd.jpg" alt="Braced Excavation: Struts and Walers"></a></p>
<p>How are such members spliced together? Unlike a single I-section, the inner flanges of the double I-section would not be accessible. There would also be problem connecting the webs due to the same accessibility issue.</p>
<p>From the pictures I found on the internet, it isn't clear how this can be done to achieve effective moment transfer.</p>
| |structural-engineering|steel|geotechnical-engineering| | <h3>Bolts</h3>
<p>Flanges can be added to the members as Wasabi mentioned in his answer. Another option would be to cut a small hole in the web of the members to allow for an iron worker to insert his hand to insert a bolt or hold a wrench. The reduction in strength from the access holes is minimal. This type of access is also used in the bolted connections of truss members. With this method, typical flat splice plates can be used.</p>
<h3>Welds</h3>
<p>Single sided complete joint penetration (CJP) welds are not a problem. They are done frequently. The joint will require a backer rod that will be left in place.</p>
<p>In the case of these walers, fatigue won't be an issue, so partial joint penetration (PJP) welds would even work. As long as the splices were located in good locations, the full strength of the member wouldn't be required at the splice location.</p>
<p>Weld examples from AWS:</p>
<p><a href="https://i.stack.imgur.com/SPxEG.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/SPxEG.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/n2GfS.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/n2GfS.jpg" alt="enter image description here" /></a></p>
| 4043 | How can double I-section walers be spliced to ensure effective moment transfer? |
2015-08-16T15:19:12.180 | <p>I am trying to calculate the loads in the suspension components of a typical car double wishbone setup. I have found <a href="http://www.fsae.com/forums/showthread.php?11179-analysis-of-wishbones" rel="nofollow">a thread on the www.fsae.com forum</a>, which seems to be exactly the kind of thing I am looking for. It involves considering the assembly as a 3D truss, and building up matrices, containing</p>
<ol>
<li>the unit vectors and cross products at the mounting points of the struts at the upright</li>
<li>the forces and cross products at the tyre contact patch. Which when the inverse of matrix 1 is multiplied by matrix 2, results in a third matrix containing the loads on the struts.</li>
</ol>
<p>My question is, can someone give me the name of this method of truss analysis, and any textbooks / resources that specifically cover this method?</p>
| |mechanical-engineering|automotive-engineering|structures|statics|mathematics| | <p>Initially after reading your question I thought that it was using the <a href="https://en.wikipedia.org/wiki/Direct_stiffness_method" rel="nofollow">stiffness method</a> but after seeing the image on page 7 of that forum post I don't think that's what it is. This is because the stiffness method relates forces to displacement using the members' stiffness properties. I don't see any of that discussed on the forum.</p>
<p>Instead, it just seems like a force balance in each of the XYZ axes and a moment balance about each of the XYZ axes giving rise to 6 equations. Depending on how your 3D truss is supported you may have 6 or more unknowns. Using matrices is just an efficient way to solve many simultaneous equations. I'm not sure there's a specific name for this given the context of your application. In terms of mathematics it is just linear algebra (i.e. solving a <a href="https://en.wikipedia.org/wiki/System_of_linear_equations#Matrix_equation" rel="nofollow">system of linear equations</a>).</p>
<p>Perhaps with more information about what you're trying to achieve (maybe a diagram of your suspension system) we can better guide you. The assembly can be modeled as a 3D Truss like you've stated. Normally you would come up with a Free Body Diagram (FBD) to figure out how the 3D Truss is supported and loaded. From there you would solve for the reaction forces where it is supported. This looks like where the 6x6 matrix comes into play - you'd use that to solve for the reaction forces of the suspension system on the chassis. Then you could use this to find the internal forces of the suspension members if that's what your ultimate goal is.</p>
| 4046 | Calculating automotive suspension wishbone loads by considering the structure as a 3D Truss |
2015-08-17T12:21:36.487 | <p>I live and work in the UK, a country known for relentless roadworks. </p>
<p>Last week I drove through roadworks on the M1 that lasted for 18 miles, necessitating road users driving at 50 mph. Of course, almost all of this impressive stretch was simply coned off, with active work only present on one or two small sites. Signs indicate that work will continue into next year.</p>
<p>What justification is there for keeping such a large section of road cordoned off for improvements? Is it a safety or economical consideration? It would seem prudent to close off mile sections at a time to increase traffic flow and reduce the number of cones/signs/average speed cameras in use, but perhaps there is an underlying reason I am missing?</p>
| |civil-engineering|highway-engineering|construction-management| | <p>There is some merit in the argument that one mile on one mile off construction would be dangerous. Works vehicles would have to enter and exit every mile from live traffic. They'd go from 50mph to some stupid speed like 2mph and back every mile go get along the works.</p>
<p>I would seriously question though whether it's cheaper per se. It's <strong>easier.</strong> A lot of road jobs are rolled up into large multi year contracts /partnerships where a safety case has to be made. As part of the UK's war on the motorist, Health & Safety plays a disproportionate role in evaluation of long term contract /partnership bids. It can thus be used as a lever against the pencil heads to extract more money from them. It's really a license to print money as in the UK it's politically incorrect to challenge Health & Safety.</p>
<p>You're not the only one to complain about those works. They've been criticised in the House.</p>
| 4053 | What is the justification for working on such large stretches of road? |
2015-08-17T18:21:48.513 | <p>The AISC 360-10 <em>Specification for Structural Steel Buildings</em> gives provisions for calculating the maximum unbraced length of a compression flange that separates yielding moment from lateral torsional buckling (LTB). This formula is (AISC 360-10, Eqn. F2-5):</p>
<p>$$ L_p = 1.76r_y\sqrt{\frac{E}{F_y}} $$</p>
<p>where</p>
<p>$L_p =$ limiting length that separates full yield moment and LTB <br>
$r_y =$ radius of gyration about the $y$-axis <br>
$E =$ Young's modulus <br>
$F_y =$ yield strength of material <br></p>
<p>Assuming that one is using regular structural steel, the Young's modulus of the material is assumed to be the same regardless of steel grade.</p>
<p>This equation works out such that a steel with a <em>lower</em> yield strength may actually be braced at a lesser interval than one with a <em>higher</em> yield strength. In other words, given the same beam size, the material with the higher yield strength buckles first.</p>
<p>I have also found this to be applicable to design using the ASME <em>Boiler & Pressure Vessel</em> code, specifically Division III, Subsection NF for supports. With temperature effects on yield strength and Young's modulus taken into account, it's possible that a member at an elevated temperature may buckle at a longer length than one at room temperature.</p>
<p>This seems counter-intuitive to me. Why would a weaker material exhibit less LTB action with the same given length?</p>
| |structural-engineering|steel|beam| | <p>As discussed in the earlier answer, if we look at the curve of moment capacity versus unbraced length we see three regions of behavior -- yielding, inelastic LTB, and elastic LTB (see Fig. C-F1.1 in the AISC <em>Steel Construction Manual</em>). It is important to note that we only have inelastic LTB due to residual stresses. This is where that $0.7F_yS_x$ term is coming from (residual stresses are assumed to be $0.3F_y$). It is also important to note that the equation for critical stress at elastic LTB is in the form $\alpha + \sqrt{1+\beta}$ and is not a function of yield stress. <em>Alpha</em> is a term for out-of-plane buckling of the compression flange and <em>beta</em> is a term for torsional stiffness.</p>
<p><a href="https://i.stack.imgur.com/7aVZE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7aVZE.png" alt="Moment Capacity"></a><br></p>
<p>So, conceptually, we could look at a curve that disregards residual stresses - meaning we have just yielding and elastic LTB. When we increase $F_y$, the elastic LTB curve remains the same while $M_p$ increases. The result is that we transition to elastic LTB at a smaller unbraced length. One way to think about it is that with an increased $F_y$, it takes more force to yield the member, making it more likely that it will buckle before yielding.</p>
<p><a href="https://i.stack.imgur.com/KWz9w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KWz9w.png" alt="Moment Comparison"></a></p>
| 4058 | Unbraced Length for Lateral Torsional Buckling vs. Yield Strength |
2015-08-18T13:24:24.987 | <p>The <strong>Darcy-Weisbach equation</strong> is used to calculate the frictional pressure losses in pipes transporting incompressible fluids. This equation uses a dimensionless <strong>Darcy friction factor</strong>, also known as the Moody factor, to account for the relative roughness of the pipe surface.</p>
<p>This empirical factor was experimentally determined by Moody and is normally read off of the <a href="http://www.ndlr.ie/artefact/file/download.php?file=13347&view=2200">Moody Chart</a>. However I am implementing the pressure drop calculation in software, so I need a non-graphical way to find the Darcy friction factor.</p>
<p>Equations for calculating the Darcy friction factor under laminar (Re < 2320) and turbulent (Re > 4000) flow are readily available. But I haven't been able to find one that is valid for the transitional region which exists between laminar and turbulent flow (2320 < Re < 4000), also known as the 'critical zone'.</p>
<p>I understand that fluid flow is complex and difficult to predict in this region. However, <strong>is there a commonly used method that provides <em>reasonable estimates</em> for the friction factor in this critical zone?</strong></p>
<p>I have found a method described in a <a href="http://www.chem.mtu.edu/~fmorriso/DataCorrelationForSmoothPipes2013.pdf">student paper</a>, but it hasn't been peer reviewed and is limited to smooth pipes only. I'm looking for something more tried and tested.</p>
<p><strong>If no formula is available, what approach do other engineers usually take to mitigate or solve this problem?</strong> </p>
| |fluid-mechanics|chemical-engineering|pipelines| | <p>As an engineer, sometimes you need an answer, and it might not be the best answer, but you've got to get one.</p>
<p>In this case, there are two ways to handle this. The first would be to <strong>extrapolate</strong> the turbulent flow backwards to the end of the laminar flow. Since the turbulent flow is always higher than the laminar flow, you'll get a higher than real friction flow in your piping, but since you're mainly using this to size your piping, pumps, etc, in the end you'll just wind up <a href="https://en.wikipedia.org/wiki/Overengineering" rel="nofollow">overengineering</a> the pumps, pipes, etc. This isn't a bad thing, but it isn't desired. However, this way you have an answer. If your program states a warning when this happens </p>
<blockquote>
<p>"This is in the turbulent zone, so an accurate estimate is difficult.
Instead a conservative amount is given. A better design would be in
either the laminar or turbulent zones"</p>
</blockquote>
<p>And most engineers would be satisfied. </p>
<p>The second way would be to require the user to be in either the laminar region or the turbulent region. This is how most engineers handle this situation - they <strong>avoid the region altogether.</strong> This way, the system can be accurately and efficiently designed, to avoid overengineering.</p>
| 4063 | Estimating Darcy Friction Factors for Critical Flow |
2015-08-18T14:10:51.443 | <p>I would like to build a <em>little water mill to place under the sink of the kitchen to produce electricity</em>. Considering that the mill will work when the user opens the tap to wash a dish, using a dynamo and a little electric circuit (bridge rectifier and a low pass filter), I would be able to generate a continuous electric signal of about <em>6 V, 0.1 A</em> for a couple of minutes. The idea is to charge a battery, typically an external one used to charge a smartphone, i.e. a battery with a capacity of <em>2500 mAh</em>.</p>
<p><strong>I am wondering if the repeated short sessions of charges will affect the battery in any way</strong>. Obviously, the battery will need a couple of dish-washing sessions to be fully charged, but will these short sessions affect its efficiency, or even alter the charging time ?</p>
| |electrical-engineering|battery| | <p>The slower you charge a Lithium battery the better. They are not keen on very fast charge/discharge fluctuations.</p>
<ul>
<li><p>One reason is that faster charging also results in higher temperatures which stresses a battery and therefore reduces battery life.</p></li>
<li><p>Another reason is that lithium plating might occur, where the transport rate of the Li-ions exceed the rate at which they can be 'stored' in the anode in a process called intercalation. Instead, the lithium deposits as Li metal onto the anode, which can cause short circuit and degrades capacity, since there are less lithium ions to store electrons.
The following image might visualize the storage process better:
<a href="https://i.stack.imgur.com/tMr9w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tMr9w.png" alt="Li-ion intercalation"></a>
<em>intercalation in cathode. The opposite happens in the anode(electrons are stored during charge and released during discharge)</em>
<a href="http://tymkrs.tumblr.com/post/7846476684/lithium-ion-battery-how-does-it-work" rel="nofollow noreferrer">Source</a></p></li>
</ul>
<p>EDIT: To answer your question directly:no, short charge cycles do not influence battery life as long as you charge slowly and try not to stress it by making full charge swings from 0 to 100%. In fact, ideally, but maybe not practically, you should try to keep it near 50% at all times(at least, this is the rule of thumb used in EVs). </p>
| 4064 | What are the consequencies of intermittently charging a smartphone battery? |
2015-08-19T14:20:49.930 | <p>Does a standard formula exist for calculating the (maximal) deflection of a constrained rectangular plate that is loaded with a evenly distributed pressure?</p>
<p>I found this <a href="http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html" rel="nofollow noreferrer">Useful tables of mechanics</a>, however I'm not sure if it's correct:</p>
<p><a href="https://i.stack.imgur.com/wNnzz.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wNnzz.gif" alt="http://www.roymech.co.uk/images/Plate_12.gif"></a></p>
| |mechanical-engineering|deformation| | <blockquote>
<p>deflection of a constrained rectangular plate</p>
</blockquote>
<p>By this I assume you mean that all edges are fixed.</p>
<p>My general go-to for these types of formulations is <em>Roark's Formulas for Stress and Strain</em>, 7th Edition. These formulations assume a flat plate with straight boundary conditions and constant thickness. Also, Poisson's ratio for the material is assumed to be $\nu = 0.3$.</p>
<p>From Table 11.4 of that text, Case No. 8a, Uniform load over entire plate:</p>
<p>$$y_{max} = \frac{\alpha q b^4}{Et^3}$$</p>
<p>The $\alpha$ value is found by considering the dimensions of the plate:</p>
<p><a href="https://i.stack.imgur.com/UMXEp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UMXEp.png" alt="enter image description here"></a><br>
<a href="https://i.stack.imgur.com/3YXMD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3YXMD.png" alt="enter image description here"></a></p>
<p>Note that this formula is really only strictly valid for materials with $\nu = 0.3$. If the material has a different Poisson's ratio, they may need to be adjusted. <em>Roark's</em> lists the following references next to this formulation that may be of some help if you have a differing $\nu$ value:</p>
<ul>
<li><p>Timoshenko, S., and J. M. Lessells: "Applied Elasticity," Westinghouse Technical Night School Press, 1925.</p></li>
<li><p>Evans, T. H.: Tables of Moments and Deflections for a Rectangular Plate Fixed at All Edges and Carrying a Uniformly Distributed Load, ASME J. Appl. Mech., vol. 6, no. 1, March 1939.</p></li>
<li><p>Timoshenko, S., and S. Woinowsky-Krieger; "Theory of Plates and Shells," 2d ed., McGraw-Hill, 1959.</p></li>
</ul>
<p>I do not have copies of these documents, so I cannot verify their utility for this problem.</p>
| 4077 | Deflection of constrained rectangular plate |
2015-08-19T16:40:53.750 | <p>I am designing a pattern for a sand casting. It has only one critical dimension which is a 4" inner diameter cylinder. If I use a 4" core, will the casting shrink away from the core? In other words will the inner diameter of the cylinder end up being 4.1" or something like that? What kind of shrinkage can I expect for ductile iron?</p>
<p>The cylinder needs to be a sliding fit on a 4" diameter painted, mill finish steel column. The inner diameter of the cylinder should end up being no more than 4.05".</p>
| |manufacturing-engineering|casting| | <p>Ductile iron is actually likely to expand during the casting process, as shown in the table found on the <a href="https://en.wikipedia.org/wiki/Casting_(metalworking)#Solidification_shrinkage">Casting Wikipeda page</a>, which is in part cited from <a href="https://books.google.com/books?id=JVTJi30phCwC&pg=PA66&source=gbs_toc_r&cad=4#v=onepage&q&f=false">this casting text</a> by Stefanescu. However, as the tables show, the range of volume change is fairly large and not one-sided, which is to say the ductile iron may either expand or contract depending on other conditions of the casting process.</p>
<p>However, another concerning aspect is the range of precision that you're looking for in this casting. A tolerance of $^{+0.05"}_{-0}$ is pretty tight for a ductile iron casting. The ISO standard that lists general tolerances for castings (ISO 8062-3) says that for a machine-molded sand casting of ductile iron, the tolerance range on a linear dimension of $100-160mm$ is at least $\pm 0.035"$, and could be up to $\pm0.14"$. For a hand-molded casting, that goes from $\pm0.098"$ to $\pm0.24"$. These are general tolerances, so you could look for tighter tolerances on your critical feature, but I would suggest (if it's possible in your process) to plan on machining out that 4" diameter and set your core diameter up so that you are cleaning up the entire circumference of the bore. Otherwise, I think hitting the tolerance range you are looking for as-cast will be tricky.</p>
| 4081 | How do the dimensions of a ductile iron casting change as it cools? |
2015-08-21T14:19:19.387 | <p>For a National Board Exam Review:</p>
<blockquote>
<p>To what height will a barometer column rise if the atmospheric
conditions are 13.9psia and 68F and the barometer fluid is mercury?</p>
</blockquote>
<p>Answer 28.3in</p>
<p>I have encountered questions with irrelevant variables; I assume temperature is irrelevant here; so I try;</p>
<p>$${ P = \rho gh }$$
$${ 13.9\frac{lb}{in^2} = (13.6)(62.4 \frac{lb}{ft^3})(\frac{1ft}{12in})^3 (32.2\frac{ft}{s^2})(\frac{12in}{ft})h }$$</p>
<p>Im getting h = 0.073...</p>
<p>Why isn't my solution with the right answer?</p>
| |fluid-mechanics| | <p>You were using the wrong unit for density. It must be in slugs/ft<sup>3</sup>, not in lbm/ft<sup>3</sup>. By the way, 32.2 lbm = 1 slug.</p>
| 4090 | Barometer Column with Mercury |
2015-08-21T16:07:18.440 | <p>I have an (AC) power grid topology with generators, consumers, power lines, tranformers, etc. I also have a "snapshot" of the system's state, i.e., I know how much real power the generators feed, how much real power is used by the consumers, I know the reactive power values, and voltages. I also know the parameters of the power grid's infrastructure such as the maximum power of the transformers or the impedance of the power lines.</p>
<p>The problem I am facing now is this: Suppose one generator feeds more real power, and one consumer uses more real power. How can I calculate what the load of the power grid's nodes is? I want to ascertain that, e.g., no transformer or power line is loaded beyond its limit.</p>
<p>I basically know that there is something called "power flow analysis," which I've been reading up. The matrices, Newton-Raphson approach to solving the equations, etc., is not the problem: What I do <em>not</em> know is how to apply the matrices to arrive at an answer to my question. I basically need help to formulate the equations: What are my input variables, what are my results?</p>
| |electrical-engineering|power| | <p>you can use this free software <a href="http://www.interpss.org/" rel="nofollow">Internet technology based Power System Simulator (InterPSS)</a> also there are a lot of better Commercial software.</p>
| 4091 | How to apply power flow analysis? |
2015-08-21T16:19:21.160 | <p>From my understanding, there are two uses of a gearing system: to change the speed of output rotation (trading it with torque), and to change the axis of rotation. Now, in a car, for example, it is necessary to have multiple available gear ratios, to allow for high torque and high acceleration when the car begins to move from stationary, and also high speed when the car is already on the move.</p>
<p>However, I know that some systems still have gears when they only used a fixed gear ratio. For example, in my lab, we work with a robot arm, and my colleagues often talk about the gearing system in the arm. But the arm does not repeatedly change its gear ratio like in a car. So what are these gears actually doing?</p>
<p>If they are to change the speed / torque of the output of the motor, and this is a fixed gear ratio, then why was the arm not designed with a different motor entirely -- one which provides the desired speed / torque properties? My intuition is perhaps that it is easier to mass produce motors that have high speed and low torque, and so it is more economical to buy one of these generic motors and attach it to a gearing system, rather than design a bespoke motor that has very high torque at its output by default....is this correct?</p>
| |mechanical-engineering|motors|gears|torque| | <p>A given mass of copper and iron can only produce so much torque : increasing torque further would require more current which means (a) the increased magnetic flux saturates the iron and (b) the I^2R losses in the copper overheat the motor. So more torque requires more iron and copper; i.e. a bigger, heavier, more expensive motor. </p>
<p>Furthermore, because increased torque comes at the expense of increased I^2R losses, electric motors are most efficient when used at light torque loadings.</p>
<p>There is no equivalent fundamental limitation on speed. Brushed motors have limits from brush wear, friction, heating, and erosion from sparking but newer motors (BLDC etc) have no such considerations up to the speed limits of their bearings, and material strength limits under centrifugal force.</p>
<p>So to use a motor efficiently you will run it fast producing relatively little torque. Then it has torque in reserve when you need it for short term purposes (acceleration etc).</p>
<p>So, if you need high torque and low speed, gearing a high speed motor will usually be more efficient, lower cost, lighter, using less material and thus probably cheaper than a direct drive motor used at low speed. </p>
| 4093 | What is the use of gears in a device with a fixed gear ratio? |
2015-08-21T17:33:55.650 | <p>I have always thought that lateral load is horizontal and not vertical in direction. For example, from this <a href="http://web.mit.edu/course/4/4.441/1_lectures/1_lecture18/1_lecture18.html" rel="nofollow noreferrer">MIT notes</a> and many others:</p>
<blockquote>
<p>The lateral load resisting systems for earthquake loads are similar to
those for wind loads. Both are designed as if they are horizontally
applied to the structural system.</p>
</blockquote>
<p>But the textbook I have (Mechanics of materials, 7th edition, Gere and Goodno, page 412) shows a vertical load on a beam and says it is lateral load. Here is the page:</p>
<p><a href="https://i.stack.imgur.com/DZboI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DZboI.jpg" alt="enter image description here"></a></p>
<p>Is there a convention for loading directions and the authors happens to use different one or should not Q above be called transverse load?</p>
| |structural-engineering| | <p>In the general case, you need to resolve the transverse load into <em>two</em> components, in the planes of the two principal axes of the beam cross-section. Of course when beginning a study of combined loads, one of the two transverse components would probably be zero, which is apparently the case in the quote from the textbook.</p>
<p>In some application areas (e.g. the frames of buildings in civil engineering), the two transverse directions are almost always "lateral" and "vertical", and those names for the directions relate to the origin of the loads - for example wind loads are "lateral" on the sides of the building, but snow loads are "vertical" on the roof.</p>
<p>In other applications, there are no obvious directions to describe the orientation of the principal axes of the beam, and the general term "transverse load" (or "two transverse load components") would be more appropriate.</p>
<p>Without more context from the book, it's not very clear what scenario is being described. If there is an implication that the X-Y plane in the diagrams is horizontal and the Z direction is vertical (if it is relevant at all), then I think using the term "lateral" is sensible.</p>
| 4097 | Standard specification for direction of lateral load for beams |
2015-08-22T02:42:44.827 | <p>For a National Board Exam Review:</p>
<blockquote>
<p>Calculate the Reynold's Number, $Re$ for water at 20 °C flowing in an open
channel. The water flowing at a volumetric rate of 200 gal/sec. The
channel has a height of 4 ft and a width of 8 ft. At this temperature,
water has a kinetic viscosity of 1.104*10^-5 ft^2/s</p>
</blockquote>
<p>The given answer is 600,000. I have tried to solve the problem like this:</p>
<p>$${ D_{eq} = \frac{2*4*8}{4+8} = \frac{16}{3} }$$</p>
<p>$${ V = \frac{Q}{A} = \frac{200 \frac{gal}{sec} * \frac{1ft^3}{7.48gal}}{4.8 ft} = 5.570409982 \frac{ft}{s}}$$</p>
<p>$${ Re = \frac{VD}{\mu_k} = \frac{ 5.570409982 \frac{ft}{s} \times \frac{16}{3} ft }{ 1.104\times 10^-5 \frac{ft^2}{sec} } = 2691019.315 }$$</p>
<p>Am I using the wrong equivalent diameter? Why am I not getting the correct answer?</p>
| |fluid-mechanics|open-channel-flow| | <p>Reynold Number can be calculated using
Re=VD/Vk
Since the Volumetric flow rate is given Q=200 gal/sec and the area (4x8)ft² we can get the Velocity V
Q=AV</p>
<p>Convert 200 gal/sec to ft³/sec
200 gal/sec x 1 ft³/7.48 gal= 23.74 ft³/s</p>
<p>Q=AV
26.74=(4x8)(V)
V=0.8356 ft/sec.</p>
<p>Sing the channel is a rectangle Diameter can be replace by Dh
Dh=4R
where R=Area/wetted Perimeter
therefore Dh=4(A/P)</p>
<p>Calculating for Reynold number</p>
<p>Re=0.8356x4(4x8/2(4+8))/1.104x10^-5
Re=403671.4976</p>
<p>The answer should be 600,000 but it the most nearest answer.</p>
| 4111 | Reynolds Number for a Rectangular Duct |
2015-08-22T04:38:51.010 | <p>The <a href="https://drive.google.com/file/d/0BwUq8KuaDLbTU2Z3YVd1cGo5YUY3WHlybUpISS0tU3pYNlEw/view?usp=sharing" rel="nofollow">Peng Robinson</a> Equation of State gives a better approximation of the critical compressibility factor $Z_c$, where experimental data ranges between $0.24$ and $0.28$ (<a href="http://books.google.com/books?id=2TpCxMgoxoQC&lpg=PA113&dq=peng%20robinson%20critical%20compressibility&pg=PA118#v=onepage&q=peng%20robinson%20critical%20compressibility&f=false" rel="nofollow">Source</a>). Since the calculated value should be $0.307$ for any substance (<a href="http://books.google.com/books?id=f-u-kPUWmrIC&lpg=PA13&ots=DbcNy90Tqi&dq=0.307%20critical%20compressibility&pg=PA13#v=onepage&q=0.307%20critical%20compressibility&f=false" rel="nofollow">Source</a>), I'd expect to obtain a rather close number by re-applying numerically the Equation of State, given a point $(T_c,P_{c})$ on the Clapeyron's Diagram. </p>
<p>However, when I try to solve the Equation of State for $Z_c$ it gives $Z_c=0.321379$ (<a href="https://drive.google.com/file/d/0BwUq8KuaDLbTV1BGZ2hnZ0ZMbGNEazcxYU9wUDFmR0xjOHNZ/view?usp=sharing" rel="nofollow">Mathematica</a>), a solution with discrepancy of $4.6\%$ from the theoretical value.</p>
<p>Also when I use sightly different parameters for this EoS from <a href="http://en.wikipedia.org/wiki/Equation_of_state" rel="nofollow">Wikipedia</a>, the numerical result with the same code is $Z_c=0.311155$. </p>
| |thermodynamics|chemical-engineering| | <p>It is important to understand that the critical point of a pure substance is characterized by certain properties. For example, the first and second partial derivative of the pressure with respect to the molar volume at constant temperature vanish. Secondly, it is important to understand that the critical point given by an equation of state (EoS) is not necessarily the same point as determined by measurements.</p>
<p>To find the critical point of the Peng-Robinson (PR) EoS one can calculate the first and second partial derivative with respect to the molar volume, and look for a given volume/temperature pair where both partial derivatives vanish. Doing this with a solver for non-linear equation systems (e.g. a Levenberg-Marquardt method) I found the critical point of CO2 to be close to $1.0T_c$ and $0.89\rho_c$, giving a critical pressure of $1.0p_c$, which yields a critical compressibility of $Z_c=0.3074$.</p>
<p>To get good results I recommend the following:</p>
<ul>
<li>If you use iterative solvers be sure to check residuals and convergence of the method.</li>
<li>Use the same constants and critical state variables everywhere, as the solution is quite sensitive.</li>
</ul>
| 4113 | Discrepancy of calculated critical compressibility factor in Peng Robinson EoS |
2015-08-22T12:58:26.757 | <p>Recently there has been road construction around on the main road leading into my office building. I observed the construction crew removing asphalt in the form of a trench as showed in pictures below. Later on they filled up with aggregate and poured a layer of black tar and filled with something. The trench has been filled and resurfaced in line with the existing road surface.</p>
<p><a href="https://i.stack.imgur.com/zKb29.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zKb29.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Iir9K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Iir9K.jpg" alt="enter image description here"></a></p>
<p>Attach is the before picture (Satellite) captured from google maps.</p>
<p><a href="https://i.stack.imgur.com/u5cQj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u5cQj.jpg" alt="enter image description here"></a></p>
<p>Attach is the before picture (Street) captured from google maps.</p>
<p><a href="https://i.stack.imgur.com/w7cWX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w7cWX.jpg" alt="Google Maps Street Image"></a></p>
<p><strong>Question:</strong> The rain water is being directed to a near by manhole. What is the engineering explanation for digging the trench later on was poured with liquid tar (asphalt crack filler)? </p>
<hr>
<p><strong>References:</strong> </p>
<ul>
<li><a href="https://www.gov.uk/government/uploads/system/uploads/attachment_data/file/4382/practicalguidetostreetworks.pdf" rel="nofollow noreferrer">Practical Guide to Street Works</a></li>
</ul>
| |civil-engineering| | <p>Given the very little detail you've provided, it's not possible to be sure.</p>
<p>But I don't think it's any more complicated than a simple resurfacing. The top layer has been planed off, and a new top layer laid.</p>
<p>Planing is done when the underlying structure is fine, but there are superficial problems: a maintenance regime of frequent planing and resurfacing can help reduce the whole-life maintenance bill, by fixing problems early before they propagate to the underlying structural layers (layers which are much more expensive to replace).</p>
<p>Wikipedia currently says that this process is also called "pavement milling", "cold planing", "asphalt milling", and "profiling".</p>
<p>One possible reason (of several) why one might just plane the strip alongside the footway, is that traffic can cause indentations, bumps and grooves along the surface, and when that's along the edge of the road, that creates ponding of water after rainfall: instead of the water all draining off, some of it gets trapped in the dents and grooves. This creates a problem for pedestrians and any other kerbside activity getting splashed, and can decrease the lifetime of the surface.</p>
| 4117 | What is the purpose of digging up trench and then filling with something similar to asphalt crack filler? |
2015-08-22T15:52:33.933 | <p>I'm trying to understand the concept of mechanical resonance.<br>
<a href="https://en.wikipedia.org/wiki/Mechanical_resonance" rel="nofollow">https://en.wikipedia.org/wiki/Mechanical_resonance</a></p>
<p>I cannot find much references about how exactly is the phenomenon known. I.e. how was it discovered and how is it measured in contemporary engineering applications?</p>
| |mechanical-engineering| | <p>I don't think you could point to any one person or group and say, "They discovered mechanical resonance." It's something that's always been around and people have used it to their advantage for centuries, if not millennia.</p>
<p>I've done more than a few vibration tests looking for the kind of resonances where the response is detrimental to whatever's being tested in a variety of machines or devices. A typical approach for a resonance search is to use some kind of measured, controlled input, such as a shaker table, and record the motion (usually with an accelerometer) of both the table and the DUT (device under test). The first step is usually a "sine sweep" where the shaker is operating over a range of frequencies, sweeping up and then down (or vice versa) and the data is examined for regions where the response is high relative to the excitation. In the end, what you're trying to identify are those frequencies where the DUT has a sensitivity that are also present in the environment which the device is supposed to operate.</p>
| 4118 | How is Mechanical Resonance detected? |
2015-08-22T23:52:56.703 | <p>I am just posting here to verify that I am properly using a tool that is new to me, a micrometer. Here is what the precision label specifies, basically that it can measure up to an inch and down to 1/1000" (a notch on the round scale).</p>
<p><a href="https://i.stack.imgur.com/qKpGi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qKpGim.jpg" alt="enter image description here"></a></p>
<p>I would like to describe how I was converting the reading on the linear and on the round gauges and you tell me if I did it right. </p>
<p><a href="https://i.stack.imgur.com/IRQMy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IRQMym.jpg" alt="enter image description here"></a></p>
<p>In the above picture, the linear scale is reading 2/10" and the round scale is reading 15/1000". So the total gauge I am measuring is <code>0.2" + 0.015 = 0.215"</code>. I am particularly curious to verify if the round gauge's each notch is indeed 1/1000" and whether I am using the correct numerator in the fraction.</p>
| |measurements|tools| | <p>These kind of micrometers have two scales, a coarse and a fine. You add them together to get the final reading.</p>
<p>The fine scale is etched around the part that turns. With nothing in the microcometer and it closed all the way, the 0 on this scale should line up with the horizontal line on the fixed part to the left. As you turn the handle to open the gap, you will see that the numbers that line up with the fixed line get larger. These numbers are usually directly in the units the micrometer is intended to measure, which looks like mils (1/1000 inch) from your picture.</p>
<p>This all works fine until the gap is opened so that the handle goes thru a full turn. That's where the second more coarse scale comes in. This scale is revealed as the gap is increased. Usually there is one line per handle turn, and these are usually calibrated to come out to a convenient multiple. For example, if each handle turn opens the gap 100 mils, then there is a line every 100 mils on the fixed coarse scale.</p>
<p>You have to know what the coarse scale multiplier is, but that's easy to see by rotating the handle one turn. Let's say the coarse scale is marked every 100 mils and the handle in mils. Then for example, if you can see 3 lines of the coarse scale revealed and the handle scale is at 37, then the measurement is 337 mils, or 0.337 inch.</p>
| 4124 | How to read micrometer? |
2015-08-23T01:59:35.437 | <p>For a National Board Exam Review:</p>
<blockquote>
<p>A floating cylinder 8cm in diameter and weighing 950 grams is placed in
a cylindrical container 20cm in diameter and partially full of water.
The increase in the depth of water in the container due to placing the
float in it is:</p>
</blockquote>
<p>Answer is 3cm</p>
<p>So I try:</p>
<p>$${ P_1 = \frac{0.950kg * 9.81 \frac{m}{s^2}}{ \frac{\pi}{4} * { (20cm * \frac{1m}{100cm}})^2 } = 296.649 \frac{N}{m^2}}$$</p>
<p>$${ \gamma = 9.81 \frac{kN}{m^3} }$$</p>
<p>$${ P_2 = P_{atm} = 101325 \frac{N}{m^2}}$$</p>
<p>$${ P_1 + \gamma h = P_2 }$$</p>
<p>$${h = 10.29m }$$</p>
<p>What is my error? Am I using wrong approach? Wrong formula?</p>
| |fluid-mechanics| | <p>The first part of your maths is OK.</p>
<p>$P_1$ = $296.649$ $Pa$</p>
<p>Now, pressure at the base of a fluid is:</p>
<p>$P$ $=$ $\rho gh$</p>
<p>Rearrange the equation & $h$ $=$ $P/\rho g$</p>
<p>Using your numbers & the fact that the density of water is $1$ $g/cm^3$ or $1000$ $kg/m^3$,</p>
<p>$h$ $=$ $296.649/(1000)(9.81)$ = $0.0302$ $m$ = $3$ $cm$</p>
| 4129 | Height Increase by a Floating Cylinder Over a Larger Cylinder |
2015-08-23T08:47:52.650 | <p>I am trying to understand what the dynamic factor $K_v$ describes. I know that a low $K_v$ factor results in quieter and smoother work of the gear but what exactly does $K_v$ tell me?</p>
| |mechanical-engineering|gears|dynamics| | <p>A <a href="https://www.google.com/search?q=dynamic%20factor%20gear%20agma" rel="nofollow">google search</a> for this actually returns pretty good information.</p>
<ul>
<li><p>"The dynamic factor $K_v$ accounts for internally generated gear tooth loads which are induced by non-uniform meshing action (transmission error) of gear teeth."</p></li>
<li><p>"It provides allowance for overload effects generated by a pair of meshing gears"</p></li>
<li><p>"It is used to account for the effect of tooth spacing and profile errors..."</p></li>
</ul>
<hr>
<p><strong>Summary:</strong> It is a way to account for transmission errors. Transmission errors occur due to:</p>
<ul>
<li>Inaccuracies in the manufacture of the gear (gear teeth are never perfect).</li>
<li>Meshing of the gear teeth during action (vibration due to tooth stiffness during meshing, tooth friction, tooth wear etc.).</li>
</ul>
<p>The gear ratio will fluctuate during operation because of these transmission errors. Therefore, the gears will always experience slight accelerations/decelerations causing dynamic forces - hence it is called a 'dynamic factor'.</p>
<p>There was a change to the way it was defined (reciprocated) by AGMA. The recent standard is that the factor should be above 1. Note that if $K_v = 1$ this represents perfectly accurate gearing. Under 1.1 is "very accurate gearing".</p>
<p>Hope that helps.</p>
| 4133 | Gears - dynamic factor $K_v$ |
2015-08-24T04:37:34.770 | <p>I made a test in Ansys but the result doesn't seem to be the final steady state. The two bodies didn't touch each other. How can I get the right result?</p>
<p><a href="https://i.stack.imgur.com/e809N.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e809Nm.jpg" alt="manual contact"></a>
<a href="https://i.stack.imgur.com/xQjmF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xQjmFm.jpg" alt="boundary condition"></a>
<a href="https://i.stack.imgur.com/ZOgeL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZOgeLm.jpg" alt="result"></a></p>
| |applied-mechanics|simulation|ansys| | <p>Okay so I ended up having a play in ANSYS Mechanical Workbench. I did a mock up of your problem. Here's how I've been able to successfully model it. Note that a lot of my decisions are based on the <a href="http://www.ansys.com/staticassets/ANSYS/staticassets/resourcelibrary/presentation/2014-sd-mechanical-contact-best-practices.pdf" rel="nofollow noreferrer">Contact Best Practices Guide from ANSYS</a>.</p>
<p>Here's some very important points given your problem but also for contact modelling in general and a frictionless contact type:</p>
<ol>
<li><p>As stated by Trevor there's no need for such a large gap initially - you don't need ANSYS to simulate that sort of behaviour because it should be obvious. So move the roller closer to the ring. I ended up moving the roller within 1mm of the ring. Just doing this on its own, however, isn't enough to get it to converge.</p></li>
<li><p>Mesh $\rightarrow$ Advanced $\rightarrow$ Shape Checking $\rightarrow$ Aggressive Mechanical (This is generally a good idea for any contact modelling).</p></li>
<li><p>After selecting the contact type as frictionless look under Advanced $\rightarrow$ Formulation $\rightarrow$ Normal Lagrange. It has highest accuracy of all the methods, can handle large sliding and interference fit. If you use Augmented Lagrange with frictionless it's likely to diverge when the contact is over a small surface area like your problem. On the other hand if you were considering a bonded or no-separation contact type then Augmented Lagrange formulation seems to converge well.</p></li>
<li><p>Frictionless Contact $\rightarrow$ Advanced $\rightarrow$ Interface Treatment $\rightarrow$ <strong>Adjust to Touch</strong> (should be obvious why I chose this).</p></li>
<li><p>Frictionless Contact $\rightarrow$ Advanced $\rightarrow$ Pinball Region $\rightarrow$ Radius. Set the radius to be at least as large as the gap between the surfaces.</p></li>
</ol>
<p>Below was the deflection result (left) and Von-Mises Stress result (right). As you can see the stress distribution is <strong>not</strong> great - around the contact area in the ring there are probably large stress concentrations. You might need a better mesh around the contact area as I spent hardly any time meshing. The main reason for this exercise was to prove that even though there is a gap between the contact surfaces there's appropriate controls in place to model it:</p>
<p><a href="https://i.stack.imgur.com/1GSxM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1GSxM.png" alt="ANSYS deflection and stress"></a></p>
<p>If you're expecting large deflections then you can turn ON "Large Deflection" in Analysis Settings $\rightarrow$ Solver Controls $\rightarrow$ Large Deflection. That might help it converge.</p>
<p>If you're wondering whether you could do this without moving the roller closer to the ring then you could probably play around with the Pinball Radius as well as another option called "Predict For Impact" in the Time Step Controls. I've never used this personally and I'm not sure what effect that would have.</p>
<p>If you have the time, read the <a href="http://www.ansys.com/staticassets/ANSYS/staticassets/resourcelibrary/presentation/2014-sd-mechanical-contact-best-practices.pdf" rel="nofollow noreferrer">best practices</a> for contact modelling.</p>
| 4140 | How to simulate the contact between two distant parts in Ansys workbench？ |
2015-08-24T13:00:57.333 | <p>The UK is about to start trials of <a href="http://www.thetimes.co.uk/tto/life/motoring/article4524398.ece" rel="nofollow">wireless charging for moving cars</a>.</p>
<p>What kind of power transfer efficiency is likely to be achievable in such a system, compared to just plugging in the car directly?</p>
| |power-engineering|inductive-charging| | <p>This is not exactly a comprehensive answer, but I found an interesting article on <a href="http://eandt.theiet.org/news/2014/apr/onroad-charging.cfm" rel="nofollow">The Institution of Engineering and Technology</a>'s website. It is fairly interesting, if not incredibly technically detailed, and I encourage you to check it out for projects to learn more about. I'll talk about the most interesting parts here.</p>
<p>It has many interesting projects listed. Some are for bus routes, and rather than charging along the whole route, they use 'opportunistic charging' to charge at key places (presumably bus stops and traffic lights) to rapidly transfer energy. This is intended to minimize the disruption and possibly cost of construction (imagine having to dig up an entire busy road to add charging coils). This is called "semi-dynamic charging" by Transport Scotland, who are working on one such project.</p>
<p>However, it sounds like in Korea, a more full route system has been operating.</p>
<blockquote>
<p>[T]he Korea Advanced Institute of Science and Technology (KAIST) is running two online electric vehicle (OLEV) buses on a 12km continuous charging route in the city of Gumi. It claims 85 per cent maximum efficiency in power transfer.</p>
</blockquote>
<p>You're question is focused on the pure efficiency of the charging capacity, but I think also interesting to consider (or possibly implied in the question) is for what cases these technological limitations make actual use a practical exercise. On that note, I will add that this technology may prove of more usefulness on highways, where range is of particular concern in the arena of electric vehicles, leading to projects going to great lengths to provide very close charging stations (25 to 50 miles) like <a href="http://www.westcoastgreenhighway.com/electrichighways.htm" rel="nofollow">West Coast Green Highway</a>.</p>
| 4141 | Efficiency of wireless charging for moving cars |
2015-08-24T21:36:03.453 | <p><a href="http://co.routt.co.us/building/handouts/notching.pdf" rel="nofollow">IBC 2308.8.2</a> states a rule of not notching more than 1/6 the depth of a wooden joist (emphasis added):</p>
<blockquote>
<p><strong>Framing details.</strong> Joists shall be supported laterally at the ends and at each support by solid blocking except where the ends of the
joists are nailed to a header, band or rim joist or to an adjoining
stud or by other means. Solid blocking shall not be less than 2 inches
(51 mm) in thickness and the full depth of the joist. Notches on the
ends of joists shall not exceed one-fourth the joist depth. Holes
bored in joists shall not be within 2 inches (51 mm) of the top or
bottom of the joist, and the diameter of any such hole shall not
exceed one-third the depth of the joist. <em>Notches in the top or bottom
of joists shall not exceed one-sixth the depth and shall not be
located in the middle third of the span.</em></p>
</blockquote>
<p>Would it be acceptable to notch a joist more than 1/6th of the depth if adding a steel plate or sistered wood beam (going the full length) or some other method (short of putting a column in the middle)? I need to cut about 2.5" out of a 9.5"-deep beam to get an HVAC duct through.</p>
<p>The joist is between the basement and first floor. Above is, I believe, the living room.</p>
<p>The joist spans 10'.</p>
| |structural-engineering|beam| | <p><strong>The 1/6th rule</strong></p>
<p>IRC allows a notch 1/6th of the depth of a wooden joist (but not within the middle third) on the basis that the joist is sized sufficiently for the stress at midspan, and a 5/6th depth beam in the end thirds is hence sufficient for the lower stress experienced there. Hence a 1/6th depth notch will not affect the design capacity of the joist.</p>
<p><strong>Engineering</strong></p>
<p>In engineering terms it is eminently possible to reinforce a beam such that a deeper notch will not cause the beam to fail. A simple rule applicable in this case is that the maximum stress is proportional to the width of the beam times the depth cubed.</p>
<p>Hence, if you wanted a beam that was 1/2 of the depth of the original beam, then in order to provide the same stress as the original beam would have experienced, you'd need a beam that was 8 times the width!</p>
<p>Looking at your numbers, and trying to minimise the width as much as possible, I would say:</p>
<ul>
<li>The IRC rules suggest you're allowed a beam depth of 5/6 of 9.5" = 7.92".</li>
<li>You want a beam that has a depth of 9.5" - 2.5" = 7"</li>
<li>Hence your depth is reducing by a factor of 7 / 7.92 = 0.884</li>
<li>Hence your breadth needs to increase by a factor of 1 / (0.884^3) = 1.45</li>
</ul>
<p>So attaching a wooden joist to the side which is half as wide as your original beam should be sufficient. Alternatively, by using a steel plate (as you suggested) you would need much less material.</p>
<p>However, there are of course key points that I've missed here. You need to transfer the stress from the original beam into the attachment. You would hence need to design connections to transfer the stress, and over an appropriate length.</p>
<p><strong>The real world</strong></p>
<p>In most jurisdictions it should be possible for engineering calculations to be carried out, proving that the reinforcement/modification is sufficient. Once the change is then approved by the local authority (on the basis of the calculations), work will then be able to go ahead. I would advise you to either contact a local engineer or contact your local building inspector.</p>
<hr>
<p><strong>Joist width vs IRC width</strong></p>
<p>You mentioned in one of your edits (since superseded I believe) that the joist was wider than the minimum required by IRC. Possibly this is because the joist is wider and/or longer than suitable for the minimum width. I would hence advise against thinking "the joist is wider than the minimum width, therefore I can notch deeper".</p>
| 4149 | Is it acceptable, with reinforcement, to notch a floor joist more than 1/6 of the depth? |
2015-08-24T22:47:30.640 | <p>I have seen this somewhere in the wild, perhaps a space-efficient storage system in lieu of two shelving units. I'm trying to reproduce, and failing.</p>
<p>Imagine two stacks of trays which hold things that must not fall out, so the trays must remain horizontal. The trays are all connected to the same track, which loops around, so that the first stack is on "the way down" for the track, and the second stack on "the way up".</p>
<p>Q: how are the trays connected to the track, so that they can move along the track eternally without ever tipping over?</p>
<p>I sort of remember that there is a wheel on each tray that counteracts the rotation coming from the track, but I don't remember how.</p>
| |mechanical-engineering| | <p>Image in my mind: The track is on a wall, with the trays being perpendicular from the wall. The track is rectangular (with rounded corners obviously). </p>
<p><strong>Solution 1:</strong><br>
If the things that the trays are holding are relatively light (to the tray/shelf), or if its feasible to perfectly balance the load on each tray, then you could just use gravity. You could attach the tray to the track carriage (for lack of a better word) via a bearing, and then just ensure that the CoG of the tray+contents is well below the pivot point (where the bearing is).</p>
<p><strong>Solution 2:</strong><br>
Have an additional track/slot that controls the angular position of the tray (by it's position, so it would not just 'follow' the other tracks). It would probably need to be a slot, and the 'track' would probably need to be redesigned as a slotted/indented track as well that was of greater depth than your angular-controlling slot so that the angular-control-slot could pass over the normal slots/track. </p>
<p><a href="https://i.stack.imgur.com/PYmQ0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PYmQ0.jpg" alt="crude hand-drawing"></a></p>
<p>It would be an idea to offset the arm by something other than the depicted offset (let's call it 90 degrees) as then the two tracks would overlap less. </p>
| 4151 | How to keep a tray horizontal when moving around a closed, vertical track? |
2015-08-24T23:58:14.887 | <p>For a National Board Exam Review: </p>
<blockquote>
<p>A 29.53" * 39.37" pressure vessel contains ammonia with f = 0.041. Compute the minimum required discharge capacity of the relief device in $\mathrm{\frac{kg}{hr}}$.</p>
</blockquote>
<p>The solution is shown as:</p>
<p>$$C = fDL = 0.041(\frac{29.53}{39.37})(\frac{39.37}{39.37}) = 0.03075 \mathrm{\frac{kg}{s}} = 110.71 \mathrm{\frac{kg}{hr}}$$</p>
<p>What I don't get is why DIVIDE? Isn't it already length and diameter?</p>
| |fluid-mechanics|refrigeration|pressure-vessel| | <p>By dividing the diameter and length dimensions, which are in inches, by 39.37 you are converting them to metres. The problem you are trying to solve has inputs that are in both English and metric units, but the answer needs to be in metric/SI, so you need to convert everything to SI units prior/during calculation.</p>
<p>1 inch = 2.54 cm = 0.0254 m</p>
<p>1/0.0254 = 39.37</p>
<p>If I was doing the calculation I would have multiplied by 0.0254 to convert inches to metres. It's just how I do things. The person who came up with the answer you're quoting divided by 39.37 instead.</p>
| 4153 | Why divide the dimensions of a pressure vessel by 39.37 when calculating discharge capacity? |
2015-08-25T08:03:14.777 | <p>What are the advantages and disadvantages of bevel gears?</p>
<p>This is what I've already found:</p>
<p><strong>Advantages:</strong></p>
<ul>
<li>possibility to change operation angle</li>
</ul>
<p><strong>Disadvantages:</strong></p>
<ul>
<li><p>must be precisely mounted</p></li>
<li><p>shafts' bearings will be exposed to significant forces</p></li>
<li><p>limited gear ratio < 5</p></li>
</ul>
<p>Do you know any other? Especially advantages.</p>
| |mechanical-engineering|gears| | <p>Another advantage is to counter mechanical thrust which is why the bearings are usually conical</p>
| 4159 | Advantages and disadvantages of bevel gears |
2015-08-25T20:58:12.193 | <p>Is there a method to test for thermal oxide on a painted steel part after the paint has cured? I'd rather the method be nondestructive but it only being destructive in the presence of thermal oxide is also acceptable.</p>
<p>I am specifically looking for laser oxide under 3mils of polyurethane, if that helps.</p>
| |manufacturing-engineering|product-testing| | <p><a href="http://www.materialsviews.com/no-escape-for-rust-non-destructive-detecting-of-corrosion-under-coatings/" rel="nofollow">This article</a> written on May 02 2012 states that no non-destructive methods exist currently, other than a scanning Kelvin probe (which requires long measurement times).</p>
<p>However, it states that <a href="http://onlinelibrary.wiley.com/doi/10.1002/maco.201005918/abstract;jsessionid=EC9DDC474D028522B0836F76B2023710.f01t04" rel="nofollow">research is in progress into using Pulsed Phase Thermography</a>. According to the article it's got teething problems but has proved that it's possible.</p>
| 4172 | What is a method to detect thermal oxide layer after steel is painted? |
2015-08-26T17:21:01.837 | <p>For a given laminate made of composite material (composed of fibers oriented in a carefully designed sequence coated and embedded in a matrix material), where can I begin estimating the stiffness and/or strength of the material? How can I know (within a beginning estimate tolerance) how to design a part out of composite materials?</p>
| |mechanical-engineering|materials|composite| | <p>There are quite a few ways to find out the stiffness and/or strength of a composite material, with a good understanding of the materials that make up the composite material.</p>
<p><strong>1. Pre-engineered composite materials</strong></p>
<p>Several materials, prevalent mainly in the aerospace industry, are pre-engineered and pre-designed. <a href="https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#safe=active&q=gy70+339+prepreg" rel="nofollow noreferrer">GY70/339</a> is a common carbon fiber epoxy used extensively with highly tested properties. The material is a <a href="https://en.wikipedia.org/wiki/Pre-preg" rel="nofollow noreferrer">prepreg</a>, which means the fibers are pre-oriented and the matrix to fiber ratio is carefully controlled in the factory selling the material. The roll must be kept in careful thermal control to prevent the pre-impregnated roll from curing before desired.</p>
<p><strong>2. Simple formula used for rough estimation</strong></p>
<p>From a far enough point, a very simple <a href="https://en.wikipedia.org/wiki/Rule_of_mixtures" rel="nofollow noreferrer">rule-of-mixtures</a> can be used for a lot of situations using composites. This includes when one of the materials is even missing. Unfortunately, the rule of mixtures can only figure out <a href="https://en.wikipedia.org/wiki/Stiffness" rel="nofollow noreferrer">stiffness</a>, not <a href="https://en.wikipedia.org/wiki/Strength_of_materials" rel="nofollow noreferrer">strength</a>. Note the major differences in definition. Most other methods for estimating materials can only measure stiffness, with strength being better defined as a maximum <a href="https://en.wikipedia.org/wiki/Deformation_(mechanics)#Strain" rel="nofollow noreferrer">strain</a>, then an actual strength.</p>
<p><strong>3. More complex empirical formula</strong></p>
<p>Beyond the rule of mixtures, there are more and more complex formulas that can be utilized. For example, the <a href="http://www.doitpoms.ac.uk/tlplib/fibre_composites/stiffness.php" rel="nofollow noreferrer">Halpin-Tsai</a> formula is utilized frequently. Although semi-empirical, it is used in plenty of applications that need good rough estimates to investigate the plausibility of a design. More complex methods can be seen by various sources, but usually this is considered acceptable with follow up material testing. </p>
<p><strong>4. Highly complex formulas which still need testing</strong></p>
<p>State-of-the-art is considered <a href="https://www.library.ucdavis.edu/wp-content/uploads/2017/03/HDBK17-3F.pdf" rel="nofollow noreferrer">Mil Handbook 17F, Section 3, Chapter 5</a>, which starts on page 211 of the reference. Section 5.2 gives estimates for each layer, than 5.3 combines each layer into an orientation based laminate using matrix methods. These are complex but understandable formulas. As you get further into the document, you will notice that while this system is considered good enough for people who can't afford testing, stress concentration factors, fatigue factors, etc are completely unknown - if you need the system to be more reliable testing will be required, or you can use heavy safety factors and use these equations (10:1 is frequently used in pressure vessel applications, with 5:1 allowable for emergency circumstances, such as seismic and wind loading).</p>
<p>Hopefully this will be helpful to people asking "where do I begin" when it comes to composite material design.</p>
| 4182 | How do I calculate an estimate for the properties of a composite material? |
2015-08-26T18:19:54.513 | <p>I am trying to figure out if a motor I have is capable of spinning a wheel. Here is all of the information I have:</p>
<p>MOTOR: 190 g.cm min.</p>
<p>WHEEL: 26 cm diameter, weighs about 59 grams</p>
<p>I imagine that I would have to calculate the linear speed of the wheel in order to convert it to the same units that the motor torque is listed in. However, I don't know if I am doing it correctly or what the next step would be to solve this. </p>
| |mechanical-engineering|torque| | <p>Check out <a href="https://robotics.stackexchange.com/questions/7796/calculating-required-torque/7799#7799">my answer to a similar question at the Robotics SE</a>. You are asking only about a motor and a <strong>wheel</strong>, and you do not give the weight of the vehicle (assuming there is one), so I cannot comment regarding vehicular motion. </p>
<p>Basically, if you assume friction is negligible, the torque a motor outputs is used to accelerate a load. Then you have to ask, "What is my desired acceleration?" A [Honda Civic] can accelerate 0-60mph in about 9 seconds, while a [1968 VW bus] went the same 0-60mph in 37 seconds. Both vehicles eventually get to the same top speed, but higher torque values give you a higher acceleration rate. </p>
<p>You can calculate your own desired acceleration rate by giving a top speed you would like to reach and a time limit in which you would like to achieve that speed. So, for example, if you wanted to get to 10rpm in 4 seconds, you would need:</p>
<p>$$
a = (\mbox{Top speed} - \mbox{Starting speed})/(\mbox{Time limit}) \\
a = (10\mbox{rpm} - 0\mbox{rpm})/(4\mbox{s}) \\
a = 2.5 \frac{\mbox{rpm}}{\mbox{s}} \\
$$</p>
<p>To be used in an equation, you typically have to convert to radians per second, which is easy knowing that $2\pi \mbox{radians} = 1\mbox{revolution}$ and that $1\mbox{min} = 60\mbox{s}$:</p>
<p>$$
\mbox{rpm} = \frac{\mbox{rev}}{\mbox{min}} \\
$$
$$
a = 2.5\frac{\mbox{rpm}}{\mbox{s}} = 2.5\frac{\mbox{rev}}{\mbox{min s}}\\
$$
Convert to $\mbox{rad/s}^2$:
$$
a = 2.5\frac{\mbox{rev}}{\mbox{min s}} * \frac{1\mbox{min}}{60\mbox{s}} * \frac{2\pi\mbox{rad}}{1\mbox{rpm}} \\
\boxed{a = 0.2618 \mbox{rad/s}^2}
$$</p>
<p>You can calculate your own target acceleration using the same method. </p>
<p>Once you have your target acceleration, you can use the torque formula:</p>
<p>$$
\tau = I \alpha \\
$$</p>
<p>Where $\tau$ is the motor torque <strong>in Nm</strong>, $\alpha$ is the acceleration <strong>in $\mbox{rad/s}^2$</strong>, and $I$ is the <a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia" rel="nofollow noreferrer">moment of inertia</a> of your load <strong>in $\mbox{kg m}^2$</strong>. For a uniform disc, the moment of inertia about the center is given by:</p>
<p>$$
I = mr^2/2 \\
$$</p>
<p>where $m$ is the mass <strong>in kg</strong> and $r$ is the wheel <em>radius</em> <strong>in meters</strong>. This is how you would solve for the required (minimum) torque for your target application; as the question has been asked you do not know the acceleration you can achieve, so you rearrange the torque equation to solve for it as follows:</p>
<p>$$
\tau = I \alpha \\
\frac{\tau}{I} = \alpha \\
$$</p>
<p>Substitute in the definition for $I$ and you can determine your acceleration, <strong>after you convert to standard units</strong>:</p>
<p>$$
\tau = 190 \mbox{g cm} \\
\tau = 190 \mbox{g cm} * \frac{1 \mbox{kg}}{1000 \mbox{g}} * \frac{1 \mbox{m}}{100 \mbox{cm}} \\
\tau = 0.019 \mbox{kg cm} \\
$$
Here the torque is tricky - to get to Nm from kg-m you need to multiply by the gravitational constant, $9.81 \mbox{m/s}^2$:</p>
<p>$$
\tau = 0.019 \mbox{kg cm} * 9.81\mbox{m/s}^2 \\
\boxed{\tau = 0.1864 \mbox{Nm}}
$$
Divide the mass by 1000 to get to kg:
$$
m = 59 \mbox{g} * \frac{1 \mbox{kg}}{1000 \mbox{g}} \\
\boxed{m = 0.059 \mbox{kg}} \\
$$
And lastly divide the wheel diameter by 100 to get to meters <strong>and</strong> by 2 to get to a radius:
$$
r = d/2 \\
r = (26 \mbox{cm} * \frac{1 \mbox{m}}{100 \mbox{cm}})/(2) \\
\boxed{r = 0.13 \mbox{m}} \\
$$
Now, plug those values into the rearranged torque equation:
$$
\frac{\tau}{mr^2/2} = \alpha \\
\frac{0.1864 \mbox{Nm}}{(0.059 \mbox{kg} * 0.13^2 \mbox{m}^2)/2} = \alpha \\
\frac{0.1864 \mbox{Nm}}{0.0004986 \mbox{kg m}^2} = \alpha \\
\boxed{374 \mbox{rad/s}^2 = \alpha} \\
\alpha = 374 \mbox{rad/s}^2 * \frac{1 \mbox{rev}}{2\pi \mbox{rad}} * \frac{60 \mbox{s}}{1 \mbox{min}} \\
\boxed{\alpha = 3570 \mbox{rpm/s}} \\
$$</p>
<p>This is a lot, but it's not so unreasonable - <a href="http://machinedesign.com/sensors/basics-motion-control-part-2" rel="nofollow noreferrer">this article about motion control</a> has the figure below: ("back then" - 2000 - people didn't put graphics on the page, it's linked as Figure 2)
<a href="https://i.stack.imgur.com/kGbWV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kGbWV.jpg" alt="Motor ramp"></a></p>
<p>Here you can see the motor accelerates from stopped to 2000 rpm in 0.12 seconds, for an acceleration of 16,667 rpm/s. Your motor is 3,570 rpm/s for your given load. You probably won't see rates quite that high due to friction, but it's a good ballpark. </p>
<p>So, in conclusion, you haven't stated what "adequate" is, but whatever your case, yes this motor will probably be good enough. </p>
<p>PS - again, <a href="https://robotics.stackexchange.com/questions/7796/calculating-required-torque/7799#7799">see my answer about climbing hills</a> for more information on how ramps affect required torque. If you're building a vehicle. </p>
| 4186 | How to calculate if a motor is capable for a load? |
2015-08-27T23:10:45.337 | <p>I just started mechanics of materials, and it seems that my knowledge from statics has gotten rusty over the summer. But there was a new concept introduced in the first example problem today.</p>
<p>There was a horizontal bar connected to a wall (fixed support). I understand that with a fixed support, there are horizontal and vertical reaction forces and a moment as well. However, when the professor cut the horizontal bar to calculate the internal loadings, he drew vectors for vertical and horizontal forces AND a moment. Since I didn't use this approach in statics, I don't get the reasoning behind adding a vector for the moment when the body is cross sectioned. Can someone explain to me why a moment is added at the newly cut end? </p>
| |mechanical-engineering|beam|statics|torque| | <p>It's useful to be clear about the fact that the 'cut' is an imaginary one. Since the portion of the beam no longer attached to the wall (because of our imaginary cut) is not falling or rotating, it must be subjected to the same type of forces/torques as if it were attached directly to its own 'wall'. The 'wall' in this case is actually the part of the beam on the other side of the imaginary cut though. </p>
| 4197 | Why is there a moment at the cross section end when you cut a body? |
2015-08-28T01:19:03.463 | <p>I've been given the following to memorize; But my head is so crammed with many others that I need to find another way; Are these formulas Empirical? I was hoping that if I derive it from a much more primitive form; I maybe able to remember it.</p>
<p>$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{kJ}{kg} }{3117} \frac{kg_{air}}{kg_{fuel}} }$$</p>
<p>$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{kCal}{kg} }{745} \frac{kg_{air}}{kg_{fuel}} }$$</p>
<p>$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{Btu}{lb_m} }{1340} \frac{kg_{air}}{kg_{fuel}} }$$</p>
<p>I know if I'm provided the answer to one of these; I can easily derive the others because I've memorized the conversions. What I don't get is how ${\frac{kJ}{kg}}$ got to ${ \frac{kg_{air}}{kg_{fuel}} }$. I cant figure out where the 3117 came from.</p>
<p>${Q_h}$ is Higher Heating Value, ${ \frac{A}{F}}$ is Air to Fuel ratio</p>
| |combustion| | <blockquote>
<p>Are these formulas Empirical? I was hoping that if I derive it from a much more primitive form</p>
</blockquote>
<p>This is <strong><em>probably an approximation</em></strong>, let's try to calculate stoichiometric air to fuel ratio and heat of combustion for liquid octane as an example:</p>
<p>$$ C_8H_{18} + \lambda(O_2 + 3.76N_2) \rightarrow 8CO_2 + 9H_2O + 3.76\lambda N_2$$</p>
<p>First we calculate heating value, Cengel (2005) defines the heating value as :</p>
<blockquote>
<p>the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. In other words, the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel.</p>
</blockquote>
<p>Making energy balance we can get $Q_h$ around $-47,891 \ \text{kJ/kg fuel}$</p>
<p>Second, we calculate stoichiometric air to fuel $(\lambda = 8 + \frac{18}{4} = 12.5)$:
$$ (\frac{A}{F})_{stoich.} = \frac{12.5(32 + (3.76*28)}{(8*12) + 18} = 15.052 $$</p>
<p>Substituting in your formula:
$$ (\frac{A}{F})_{stoich.} = \frac{|Q_h|}{3117} = \frac{47891}{3117}= 15.36$$</p>
<p>which is slightly higher than the calculated theoretical value. Of course there is no direct relation between heating value of fuel and stoichiometric air to fuel ratio as there are many parameters that could affect that relation, you can derive the air to fuel ratio by making energy balance and substitute with elements' enthalpy and heat of formation but that won't be any easier than simply memorizing the formula you were given.</p>
| 4200 | Derivation for the Formulas of the Approximate Ultimate Analysis of Coal |
2015-08-28T20:52:38.440 | <p>Why do we design cars to turn by turning the front wheel?</p>
<p>Wouldn't it better to keep all wheels straight and turn by rotating the wheels on one side of a car more then the wheels on the opposite side of the car, like tank tracks do? That way you can turn even if you are standing still. Why don't cars do this?</p>
| |automotive-engineering|wheels| | <p>Meet the skid-steer loader:</p>
<p><a href="https://i.stack.imgur.com/m6M4o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6M4o.png" alt="Skid-steer"></a></p>
<p>It's a versatile beast of a machine, packing lots of power, multiple attachments and (most relevantly to the topic at hand) great maneuverability.</p>
<p>It does exactly what you suggested. It has stationary tires, which turn at different angular velocities in order to turn the vehicle, like a tank. It's easy to turn in a full circle without changing the machine's position. This gives it a number of advantages:</p>
<ul>
<li>It can readjust its position when loading and unloading things, if (for example) the forks need to move a few inches to the left or right.</li>
<li>It can move a bucket sideways to smooth out dirt or gravel.</li>
<li>Its turn radius is 0, meaning it can get into tight spaces not designed for vehicles.</li>
</ul>
<p>Compared to a compact track loader (a similar machine that uses tracks instead of tires), the skid steer has <a href="http://foresternetwork.com/daily/construction/construction-equipment/wheels-vs-tracks-the-debate-over-compact-loaders-part-i/" rel="nofollow noreferrer">these advantages and disadvantages</a>:</p>
<ul>
<li>Lower maintenance and repair cost</li>
<li>Lighter and faster</li>
<li>Operates poorly if at all in muddy or snowy conditions</li>
<li>Affects the ground more</li>
</ul>
<p>A car has little to no need for any of these advantages, as they're not used to load, unload or landscape (though attaching a trailer would be much easier), and generally travel on pathways designed for them.</p>
<p>There actually exist accessories for wheeled vehicles that provide mini tracks for each wheel, to be used in mud or snow:</p>
<p><a href="https://i.stack.imgur.com/zC45x.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zC45x.jpg" alt="Truck tracks"></a> </p>
<p>But even so, the basic steering functionality is still by directed turning rather than skid steering. In fact, there are even machines that are designed for tracks that steer in the same manner:</p>
<p><a href="https://i.stack.imgur.com/QOqAX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QOqAX.jpg" alt="Curbing machine"></a></p>
<p>If even some tracked machines steer in the same manner that cars do, instead of by skidding, there must be some advantage, right?</p>
<p>One advantage (the above curbing machine steers conventionally for this reason) is that it's much easier to form a smooth curve. While computer systems could be used to determine each tire's rotation speed to produce a given turn radius, it's much simpler to simply turn the wheels. Also, as pointed out by @alphazero, one is that skidding reduces traction significantly, which can lead to uncontrollable situations at high speed. There's also the matter of wear and tear resulting from the skidding (and protip, try to turn the steering wheel in your car only when moving, if practical).</p>
<p>In conclusion:</p>
<p>Cars generally steer by turning their wheels instead of skidding because few of the advantages of skid steering have much application to a car but the disadvantages range from expensive to complex to disastrous.</p>
| 4209 | Why do cars turn by turning the front wheel? |
2015-08-30T06:12:53.267 | <p>For a National Board Exam Review:</p>
<blockquote>
<p>A uniform area jet travels to $600 \frac{ft}{s}$ and a
$100\frac{ft^3}{s}$. What is the horizontal force acts on the water jet
if it undergoes a $180^\circ$ turn?</p>
</blockquote>
<p>I would normally provide an answer but even review instructor cannot answer question...</p>
<p>So I try:</p>
<p>$${ F_x = \frac{ \gamma Q }{ g } \left(\nu_2 \cos(\theta) - \nu_1\right)}$$
$${ F_x = \frac{ 62.4 \frac{lb_m}{ft^3} ( 100 \frac{ft^3}{s} ) }{ 32.2 \frac{ft}{s^2} } \left(600 \frac{ft}{s^2} \cos(180^\circ)- 0\right) = -116273.2919\,lb_f }$$
Is the negative sign logical? All choices are from the range of ${1.35*10^5}$ to ${ 5.32*10^5 }$ lbf</p>
| |fluid-mechanics| | <p>Try considering the force required to stop the jet, then the same force again to accelerate the jet back to the same velocity, but in the opposite direction.</p>
| 4225 | The Horizontal Force of Water as it turns 180 degrees |
2015-08-30T11:19:18.820 | <p>I want to mount a heat source onto several heat pipes. The heat will be forwarded to coolers (four heat pipes, 200 mm) which are able to distribute up to 120 W without increasing the temperature of the device which should be cooled. The device itself will generate up to 100 W heat. Now I was wondering what would happen to the mounting point: Will the mounting point resp. the clamp used for fixing the heat source on the heat pipes also heat up? If yes, how much?</p>
| |heat-transfer| | <p>Well, how much the clamp heats up is determined by a great number of factors, chiefly (off the top of my head):</p>
<ul>
<li>The properties of the pipe (material properties: density, material
type, size of contact area etc)</li>
<li>How the clamp interfaces with the pipe (e.g. metal on metal, thermal
interface material between clamp and pipe, an insulator between clamp
and pipe). </li>
<li>The properties of the clamp (material properties, initial temperature etc)</li>
</ul>
<p>Regards the resp. abbeviation I think you're using it incorrectly. I'd say "Will the mounting point w.r.t. the clamp....." where w.r.t. means with respect to.</p>
| 4227 | Mounting heat source onto heat pipes |
2015-08-30T17:31:08.970 | <p>I'm an engineer in industrial design and despite that I studied mechanics, kinematics and dynamics, I got difficulties when I try to label or find components which could enable my designs have relative movements, i.e. one part rotates inside other.</p>
<p>I know I can design most of them but I guess that some must be already created and would like to know how to search them, instead designing new ones or asking for help in each specific project.</p>
<p>What are the basics relative movements and components that generates them?</p>
<p><strong>EDIT</strong></p>
<p>I'm trying to find a <strong><em>"movements library"</em></strong>, to be able to find what are the most common movements and what mechanisms or components are used to achieve them. For example, materials libraries are pretty common in many countries, in them you can find a huge list of materials, with their properties, application and providers.</p>
| |mechanical-engineering|design|kinematics|biomechanics| | <p>There is a fascinating historical collection of mechanisms compiled in the late 19th century here: <a href="http://507movements.com/about.html" rel="nofollow">Mechanical Movements</a> and somebody is working through the task of making animations of them for the website.</p>
<p>Manufacturing technology has progressed since then, but the basic ideas are timeless. We used to have a printed copy of this book in our "office library" before the internet existed, as a source of design inspiration.</p>
| 4232 | What are the basic mechanism, components or parts to generate relative movements? |
2015-08-31T13:11:53.067 | <p>To pick a specific example, a flyover that goes over a ground-level road/rail need only be as tall so that the tallest vehicle travelling in the outermost lane can clear it.</p>
<p>Or another, an elevated road/metro rail that runs along a ground-level road need to be as tall as the tallest vehicle.</p>
<p>But why are they made much taller than that?</p>
| |civil-engineering|transportation| | <p>Looking at the the street view provided by the OP I would say that vertical profile constraints are at play here. The ramp needs a certain vertical clearance near its start and near its end. Rather then have the road return to ground level, the designers may have opted to use one vertical curve over the entire ramp that suited their design needs. The consistent vertical curve should ensure a comfortable ride for the user as well as provide plenty of sight distance to ensure they have enough time to brake if need be.</p>
<p>In addition to what has already been said about required clearance for over sized load, there may be a major industry on this route that routinely ships large/tall objects.</p>
<p>Similar to the need for formwork/falsework clearances, there may be an allowance for future maintenance or rehabilitation work to be performed.</p>
<p>Having done more than a few bridge inspection, I can dream and say they left plenty of clearance for an inspector to work under the bridge in a bucket truck over live traffic without the need for a lane closure! Pipe dream that one!</p>
| 4238 | Why are flyovers/metro rail tracks much higher than traffic? |
2015-09-01T20:43:17.637 | <p>I'm encasing the abutment ends of prestressed concrete bridge girders in concrete before the deck pour. (For lateral stability and to prevent the soil from spilling through.)</p>
<p>Discussion of intermediate diaphragms for prestressed girders abounds, but I haven't found information on structural behavior of end diaphragms.</p>
<p><strong>For purposes of a consulting engineer</strong></p>
<p>My gut / engineering judgment is telling me the effect is negligible, but I prefer to understand what's actually going on before making simplifying assumptions.</p>
<p><strong>Conceptually</strong> (versus a girder without the diaphragms):</p>
<ol>
<li>Does this change deflections during the deck pour? What about long-term deflections?</li>
<li>Does this change the live load moments &/or shears?</li>
<li>Is there any logic by which the encased length should be a function of girder depth or span length?</li>
</ol>
<p>I've put a sketch below since terminology and typical bridge configurations vary around the world.</p>
<p><a href="https://i.stack.imgur.com/LUcbf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LUcbf.png" alt="Diaphragm"></a></p>
| |bridges|prestressed-concrete|deflection| | <p>I have no resources to point to in regards to this, but I'm fairly certain your gut-feeling is correct. In fact, end diaphragms exist in bridges primarily to hold the soil on the approach. They can also be useful for temporary lateral stability of the beams, but if that were the only reason, temporary metallic bracing would be much easier and cheaper. Lateral instability (buckling) in the final structure is usually not a controlling factor, especially given the (partial) bracing that is the slab.</p>
<p>Mid-span diaphragms help with transverse load distribution and stop the primary beams from "opening" along with the slab deflection. The same applies to some extent to end diaphragms, but to a much lesser extent since the load is transmitted almost immediately to the bearings. The bearings also help to impede the beams from "opening". This is probably why you've found no resources in regards to end diaphragms: no one's really bothered to think too much about them.</p>
<p>Now, answering your specific questions:</p>
<ol>
<li><p>This would change deflections somewhat, since the end diaphragm is going to change load distributions from your slab to your beams. This will be most relevant if your bridge has a small distance between diaphragms (less than twice the distance between primary beams, according to classic slab methods such as Rüsch). If the diaphragms are more spread out, then they will have almost no influence in the load distribution and therefore will not influence deflections. This also applies to long-term deflections.</p>
<p>Long-term deflections are, however, influenced by one more factor, and that is the prestress' differed losses. Over time, the primary beams are going to try to shrink. This is due not only to the concrete's natural shrinkage, but also due to the prestress' compressive creep. If all the beams were exactly the same, the creep and shrinkage should progress similarly along all of them. In that case, the end diaphragm would not have an effect since all the beams would "pull" it the same amount, implying in a simple rigid-body translation, without any deformation on the diaphragm.</p>
<p>That, however, would be in the perfect world, and that ain't ours. Creep and shrinkage are mysterious and unstable, with a lot of scatter. So even exactly similar beams would probably result in different creeps and shrinkages, meaning that the end-diaphragms would be deformed. The diaphragms deformations (which would appear as horizontal shear forces on the diaphragm) will create tensile forces in the primary beams over time and these tensile forces will influence the creep behavior of the beams over time, leading to a recursive effect.</p>
<p>Also, the beams would never be exactly the same because the loading in each beam is different (perhaps the central beams have similar loads, but they will certainly not be similar to those at the transversal extremities of the bridge), which is sufficient to generate (slightly) different creep behaviors in the beams.</p></li>
<li><p>As mentioned in the beginning of item 1, this will change the stresses suffered in the beams since the load distribution from the slab to the beams is modified by the existence of the end diaphragms. If the diaphragms are sufficiently far apart, however, the effect is almost certainly negligible.</p></li>
<li><p>Here I have a bit less to go on. I can say, however, that at the firm where I work we usually embed the primary beam only a few centimeters (3-5 cm, usually) into the end diaphragms. This allows the end diaphragms' crack-control reinforcement to pass without much fuss. The primary beams are designed with all their reinforcement sticking out at both ends in order to anchor into the diaphragms. Obviously, if your beam is unusual or if is a deep beam, the behavior might be different.</p></li>
</ol>
| 4251 | What is the effect of end diaphragms on prestressed girder forces/deflections? |
2015-09-02T10:33:09.343 | <p>I am taking organic solar cell course on Coursera, and I have a question about I-V curve for a certain solar cell. The question asks for the solar cell characteristics such $V_{oc}$, $P_{max}$ and $FF$. I get the $V_{oc}$ correctly but I couldn't calculate the other two correctly as they provide me with an excel sheet of the current and voltage. </p>
<p>I plotted the voltage on the x-axis and current,power on y-axis below. I need to get the correct $I_{max}$ and $V_{max}$ from the graph with (I-V) in blue and (P-V) in orange.</p>
<p>Here is the question from Coursera:</p>
<blockquote>
<p>Download this dataset <a href="https://i.stack.imgur.com/DJw1t.png" rel="nofollow noreferrer">DataSheet</a> (, decimal mark: dataset) and
calculate the maximum power point (Pmax). Give answer in either W or
mW, a valid form of answer would be '94' or '0.094'.</p>
<p>Hint: You may need to do an interpolation to get the correct answer,
if the value you are looking for does not exist.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/DJw1t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DJw1t.png" alt="enter image description here"></a> </p>
| |chemical-engineering|renewable-energy|photovoltaics| | <p>As you have the power-voltage curve, finding the maximum power point is easy: it's the turning point on the power-voltage curve.</p>
<p>The problem is that the absolute values in your graph look completely wrong. If the graph is flipped upside down, and based at zero power, zero current and zero voltage, then the blue curve does indeed look like an I-V curve for a photovoltaic cell, and the orange curve looks like the corresponding power curve.</p>
<p>So it looks like the shape of your curve is right (albeit upside down from the usual presentation), but the absolute values are wrong.</p>
<p>You need to work out what the right absolute numbers are.</p>
<p>You know that the lowest the actual voltage is going to be is zero, and the lowest observed value is currently marked as -10V. The curve is the right shape going left to right (it's upside down, but does not also need reflecting in a vertical axis), so the correction coefficient for voltage is going to be positive. You know that the lowest the actual current and the actual power can be is also zero, and their lowest observed values are marked as 0.1A and about 6.9W respectively - I say <em>lowest</em>, because we need to flip the curve upside down. And because we do, you know the coefficient to correct them will be negative. And you know that $P_{real}=I_{real}V_{real}$.</p>
<p>Given the shape of the curves, it's safe to assume that current and voltage each need a linear transformation (i.e. $V_{real} = aV_{obs}+b$, where you need to find $a$ and $b$)</p>
<p>This gives you a set of simultaneous equations you can solve, to transform observed voltage (current/power) [$V_{obs}, I_{obs}, P_{obs}$] into actual voltage (current/power) [$V_{real}, I_{real}, P_{real}$]. </p>
<p>Write out those simultaneous equations, and solve them.</p>
<p>That's the recipe on how to solve it. You'll need to follow the recipe yourself, though: given that this is coursework, there's no use in me doing it for you. Be sure to go through the recipe yourself, and make sure you understand why each step exists - try to work out how I derived it.</p>
| 4254 | How to get the maximum power point from I-V curve for Organic Photovoltaics (OPV)? |
2015-09-02T19:13:50.997 | <p>Two questions.</p>
<p><strong>Has anyone had experience in retrofitting consumer (locally-available) square aluminum tube as telescoping structural support?</strong> Ergo; with 0.125" wall thickness square tube at an outside dimension (OD) of 1.5", would minor superficial sanding and end beveling of an inner square tube with OD of 1.25" enable a telescoping pair?</p>
<p>Furthermore, <strong>could said pair be capable of implementing a gas spring internally for vertical assist?</strong> See image below, noted: gas spring improperly sketched upside down.</p>
<p>Loads and dimensions can be provided if necessary.</p>
<p><a href="https://i.stack.imgur.com/Y23Hq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y23Hqm.png" alt="enter image description here"></a></p>
<p><strong>ADDITION:</strong>
Is there an implied ideal mount offset for a gas spring in this circumstance relative to the load distribution (please see 2nd attached image below)?</p>
<p><a href="https://i.stack.imgur.com/nmvvO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nmvvO.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|materials|tolerance| | <p>Aluminium rubbing on aluminum generally will wear very rapidly. You generally want some sort of plastic bushing to reduce friction and wear. (Like with luggage handles there's just a little piece of plastic between the tubes) you should also have a bushing for the other end of the joint (the one inside that's not visible.)</p>
<p>You should do a beam analysis to check that the smaller beam won't buckle right at the joint.</p>
<p>The position of the piston shouldn't matter very much, but the minor benefit is probably worth the small effort to offset it.</p>
| 4263 | Assist for vertically telescoping support? |
2015-09-02T19:31:02.100 | <p>I am designing a block and tackle system to haul wood up a ramp onto my porch. The issue I am having is that the rope seems like it will rub against the top of the ramp and cause it to fray. How can I prevent this from happening?</p>
<p><a href="https://i.stack.imgur.com/PkHNo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkHNo.jpg" alt="enter image description here"></a></p>
| |applied-mechanics| | <p>Here is how I would solve this in true bodge-job fashion:</p>
<h3>Materials:</h3>
<ul>
<li>2x bricks, the type with holes in</li>
<li>1x short section of broom-handle</li>
<li>1x short plastic pipe, or toilet-roll tube (bigger than broom handle)</li>
<li>2x pencils, or bits of string, or something</li>
</ul>
<h3>Assembly:</h3>
<ul>
<li>Put the pipe over the broom handle such that it can rotate freely</li>
<li>Support the broom handle in the holes between the two bricks</li>
<li>Optionally drill holes through the ends of the broom handle and put a pencil through it, to stop it sliding sideways and dropping out of the holes. Alternatively put a bit of string through it and tie to the brick, or anything really.</li>
</ul>
<h3>Usage:</h3>
<p>Place the roller at the apex of the hill. The rope will roll over the roller then lift off as the truck gets to the top. The truck will drive over the top of the roller assembly as you pull it onto the flat section.</p>
<p>Jobsagoodun.</p>
| 4265 | Block and Tackle with Ramp - stopping rope drag |
2015-09-03T07:33:44.673 | <p>"Sketch a combinational circuit which takes as input, two 4-bit binary numbers,
A3A2A1A0 and B3B2B1B0, and which outputs the 8-bit product P7P6P5P4P3P2P1P0.
Assume that you have available sixteen 2-input AND gates and three 4-bit adders.
Please explain the reasoning behind your design." </p>
<p>After a bit of fiddling I got close to the final circuit but I found that three 4-bit adders wouldn't be enough. To save time (and my complete lack of being able to explain anything) I found <a href="http://www.adaelis.com/misc/temp/russian_mul/russian_mul.png" rel="nofollow">this</a>, which is almost exactly what I need except it used two 4-bit adders and one 6-bit adder. I want to know if it's even possible to construct a circuit that can multiply two 4-bit numbers using the gates specified above, and if so, how can you do it? </p>
<p>Thanks in advance.</p>
| |electrical-engineering|computer-engineering| | <p>The first step is ANDing all the combinations of inputs which takes all the 16 AND gates available.</p>
<p>Lets call those results $S_{00}$ $S_{01}$ $S_{02}$ $S_{03}$ $S_{10}$ etc.</p>
<p>Is we arrange them in how we need to add them we get:</p>
<p>$$
\begin{matrix}
& & & S_{03}&S_{02}& S_{01} & S_{00} \\
& & S_{13}&S_{12}& S_{11} & S_{10} & \\
& S_{23}&S_{22}& S_{21} & S_{20} & & \\
S_{33}&S_{32}& S_{31} & S_{30} & & \\
\end{matrix}
$$</p>
<p>$S_{00}$ is the low bit of the output.</p>
<p>Then you use 1 4-bit adder to add {0,$S_{03}$, $S_{02}$, $S_{01}$} and {$S_{13}$, $S_{12}$, $S_{11}$, $S_{10}$} Output is {$A_c$, $A_3$, $A_2$, $A_1$, $A_0$}</p>
<p>If we substitute that result in the original table we see that this collapses the top 2 rows:</p>
<p>$$
\begin{matrix}
& A_c & A_3 & A_2 & A_1 & A_0 & S_{00} \\
& S_{23}&S_{22}& S_{21} & S_{20} & & \\
S_{33}&S_{32}& S_{31} & S_{30} & & \\
\end{matrix}
$$</p>
<p>$A_0$ is bit 1 of the output</p>
<p>Then you use the second 4-bit adder to add {$A_c$, $A_3$, $A_2$, $A_1$} and {$S_{23}$, $S_{22}$, $S_{21}$, $S_{20}$} output is {$B_c$, $B_3$, $B_2$, $B_1$, $B_0$}</p>
<p>$$
\begin{matrix}
B_c & B_3 & B_2 & B_1 & B_0 & A_0 & S_{00} \\
S_{33}&S_{32}& S_{31} & S_{30} & & \\
\end{matrix}
$$</p>
<p>$B_0$ is bit 2 of the output</p>
<p>Then you use the third 4-bit adder to add {$B_c$, $B_3$, $B_2$, $B_1$} and {$S_{33}$, $S_{32}$, $S_{31}$, $S_{30}$} output is {$C_c$, $C_3$, $C_2$, $C_1$, $C_0$}</p>
<p>The carry input on all the 4-bit adders is 0.</p>
<p>The total output is then: {$C_c$, $C_3$, $C_2$, $C_1$, $C_0$, $B_0$, $A_0$, $S_{00}$}</p>
| 4270 | How to create a logic circuit that multiplies two 4 bit numbers, using only 16 2-input AND gates and 3 4-bit adders? |
2015-09-03T09:56:25.713 | <p>I'm studying "signals and systems". We can indentify two kinds of signals: power signal and energy signal.</p>
<p>The definitions are:</p>
<p>Energy of a time-continous signal
$$ E_x = \int_{-\infty}^{+\infty}{|x(t)|}^2dt $$</p>
<p>Power of a time-continous signal
$$ P_x = \lim_{Z\rightarrow+\infty}{\frac{1}{2Z}\int_{-Z}^{+Z}{|x(t)|}^2dt}$$</p>
<p>An energy signal satisfies $0<E_x<+\infty$ and $P_x=0$.</p>
<p>A power signal satisfies $0<P_x<+\infty$ and $E_x = +\infty$.</p>
<p>Now, if we have a signum function</p>
<p>$$sgn(t)=\left\{\begin{matrix}
1 & if\;t \ge 0 \\
-1 & if\;t < 0
\end{matrix}\right.$$</p>
<p>and we calculate the formulas, then we obtain:</p>
<p>$$P_x = 0 \;\; E_x=0 $$</p>
<p>My question is, as the title said: the signum function which kind of signal is?</p>
| |telecommunication|power|energy|signal|systems-engineering| | <p>First point is that signum is not continuous with the jump at $t=0$. </p>
<p>Second because of the absolute value in the integral it has the same $E$ and $P$ as a constant signal $1$.</p>
<p>This means that $E_{sgn}=+\infty$ and $P_{sgn}=1$</p>
| 4271 | What kind of signal is the signum function? A power signal or an energy signal? |
2015-09-03T11:09:09.517 | <p>An input device sends a 20ms positive signal to perform an action. The signal is captured using an interrupt triggered on the rising edge of the positive signal. It appears that there might be some electrical noise on the signal line. I suspect this because the interrupt is getting trigger inadvertently. I currently don’t have access to an oscilloscope to verify the signal integrity. I am using MSP430 microcontroller.</p>
<p><strong>Question:</strong> What is the most effective/<strong>simple</strong> way for the micro-controller to capture the 20ms positive signal with out modifying electrical hardware? </p>
| |electrical-engineering|embedded-systems| | <p>With a 20 ms pulse width you really should not be using interrupts directly.</p>
<p>The usual thing is to have some timer generating periodic interrupts at maybe 250Hz or so, then if on two consecutive timer events you read the input as active you can be reasonably sure you have a pulse lasting at least 1/Timer frequency, similar logic can be applied to filtering the falling edge. </p>
| 4275 | How can electrical noise be isolated in an Embedded System? |
2015-09-04T12:45:05.283 | <p>I am trying to identify the section designation so I can determine it's properties.</p>
<p><a href="https://i.stack.imgur.com/W5znc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W5znc.png" alt="enter image description here"></a></p>
<p>I have looked under "C 150 x 100 x 40" but I can't find its properties or what it's correct designation.</p>
<p>What is this section called? It's not just a "channel" is it?</p>
| |civil-engineering|beam|standards| | <p>In the Southern African context this would be defined as a </p>
<p>"cold formed lipped channel" section. I am not sure if it the same across the world though.</p>
<p>Here is one <a href="http://www.macsteel.co.za/products/cold-formed-sections?b=55" rel="nofollow">supplier's link</a></p>
| 5294 | What is the name of this section? |
2015-09-04T15:21:11.230 | <p>Every morning when I get up, I use <a href="http://www.mrcoffee.com/specialty/grinders/automatic-burr-mill-grinder-stainless-steel/BVMC-BMH23.html" rel="nofollow noreferrer">this coffee grinder</a> to grind beans for my morning coffee.
<a href="https://i.stack.imgur.com/bX9v7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bX9v7t.png" alt="grinder"></a></p>
<p>When I remove and subsequently open the top of the bin, I often find that enough static electricity has built up in the grounds/bin that particles of coffee will literally fly out of the bin and onto my hand and the counter, making a mess. Coffee will also be stuck to the sides of the bin to varying degrees, making it necessary to rap the bin on the counter in order to dislodge the particles from the sides. Of course, more then fly out and increase the mess on my counter.</p>
<p>I have made some observations of a few factors that seem to affect this:</p>
<ol>
<li>It happens to some degree with most beans, but I've noticed that the darker the roast, the greater the static charge. It's the worst with French Roast (my favorite), which is also oily since they add oil to the beans in the roasting process.</li>
<li>The coarseness of the grind doesn't seem to matter except that a coarse grind produces heavier particles which don't tend to fly as far.</li>
<li>Cleaning the bin had no observable effect on the static.</li>
<li>Waiting a while after grinding doesn't seem to do much to alleviate the problem. I've ground the beans and waited 10-20 minutes before removing the bin and observed no apparent decrease in the amount of static.</li>
<li>It is worse in winter when the air is dry, but I don't really let my house get too humid.</li>
<li>Google reveals that this problem is not specific to any one brand or model of grinder, but seems to plague most burr grinders. </li>
</ol>
<p>My questions are these:</p>
<ol>
<li>What is the cause of this static charge in the grinding process?</li>
<li>What modifications can the end-user make to either the grinder or the process to prevent static buildup?</li>
<li>How could this be eliminated in the design itself?</li>
</ol>
| |process-engineering|product-engineering|grinding| | <p>Without getting into the why of static and grinding, I think the problem always lies with these plastic bins that the grinds end up in. Glass would be much better.</p>
<p>The simple solution which nobody has mentioned here,
is to quickly dip a spoon in water and briefly stir the beans adding a few drops right before grinding.
Problem solved.</p>
<p>The more interesting question is why does this easy hack work so well?</p>
| 5297 | How to prevent static buildup in an electric burr mill coffee grinder? |
2015-09-04T18:19:43.767 | <p><strong>Question:</strong> How is shelf deflection determined? More specifically, how is the deflection determined when one or more edges of the plate have reinforcing edging applied to them?</p>
<p>I have been able to find <a href="http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html" rel="nofollow noreferrer">plate deflection equations</a> in a number of places, but of the resources I have found, they are limited to plates without any edging at all.</p>
<p>I have found a <a href="http://www.woodbin.com/calcs/sagulator/" rel="nofollow noreferrer">shelf sag calculator</a> which happens to be well known and loved by woodworkers. But their calculator only handles a single side of edging, and they don't provide any explanation of the formulas that are used. And depending upon what deflections I find, I may need to be able to evaluate the deflection in a lattice structure with a supporting skin.</p>
<p>I would assume that the edging can be treated as simple beams supporting the plate, but I don't know how to apply that support to the plate.</p>
<p>Is there a standard set of equations and / or plate geometries that can be applied to estimate what the shelf deflection may end up being? Or is this getting into the non-trivial end of analysis and design?</p>
<hr />
<p><strong>Background:</strong> For a hobby project of mine, we need to build shelving to support a variety of items which cover a range of shapes and sizes. Some items are large and comparatively lightweight (e.g. a student's desk). Others are small, dense, and kind of heavy (e.g. concrete block). And of course, there are some in the middle for both size and either side of the weight spectrum.</p>
<p>We're planning on building several sets of shelving to accommodate the various sizes of things that need to be stored. And while we do have to design for the unknowns of the future, we're still trying to optimize costs through the amount of material used.</p>
| |mechanical-engineering|beam|deflection| | <p>To add to Wasabi's answer - especially dealing with wood's anisotropy, here are a few things to consider:</p>
<p>1.) IF wood is end grain, plywood, oriented strand board, etc then for all design purposes, it is isotropic - as the grain is going in the "up" direction - where no real bending occurs. As a result, you can use the simpler equations.</p>
<p>2.) Under most conditions, as shown by <a href="http://link.springer.com/book/10.1007%2F0-306-48414-5" rel="nofollow">Dr. Jack Vinson</a>, where your longest length is in the primary grain direction (that are difficult to go into without a lot of partial differential equations), and a non-linear term that <em>for the most part</em> can be ignored, a shelf can be "squished" in the grain direction.</p>
<p>For example, let's say that your shelf is "A" wide and "B" long. And the wood has "a" modulus in the "A" direction, and "b" modulus in the "B" direction. If "B" is the grain direction, simply change the dimensions in the deflection calculator to:</p>
<p>"A" wide by (a/b)*"B", with modulus "a".</p>
<p>The calculated deflection, for woodworking purposes, will be within an order of magnitude of the actual deflection.</p>
| 5299 | How is shelf deflection determined? |
2015-09-05T00:53:52.910 | <p>In the rankine cycle, water is first pressurized and then heated to produce steam before it is used to turn a turbine. As far as i know, this is what is considered the best way to extract mechanical work on the turbine, but i want to understand why this is so. How is it that we can extract more work on a turbine from steam than from regular unheated, un pressurized water?</p>
| |thermodynamics| | <p>A good way to understand why the phase change is so important is by looking at the Pressure vs. Volume diagram for the Rankine cycle (path 1-2-3-4-1):</p>
<p><a href="https://i.stack.imgur.com/Jhxl0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Jhxl0.png" alt="enter image description here"></a></p>
<p>Remember that the work output of any thermodynamic cycle is the integral of its P-V diagram (area under the curve). If the water in the Rankine cycle were not heated to steam, its change in volume would be very small, both in the pumping/heating processes and in the expansion/cooling processes. And, if its volume didn't change significantly, its P-V curve would have almost no area. Thus, the cycle would produce much less work output than if the volume changed significantly and produced a P-V diagram with a large cyclic integral. The dashed line (path 1-2-2'-3'-1) represents what the path might look like if plain water were used.</p>
<p>Many early heat engines used a gas-phase-only working fluid; it was not until Rankine or DeLaval came along (can't remember who), and discovered the benefits of using a phase change in the cycle.</p>
| 5307 | Why use steam in the rankine cycle? |
2015-09-05T01:11:32.173 | <p>In my admittedly naive understanding of the rankine cycle, steam is condensed back into water after passing through the turbine. Why not reuse the steam instead? It would seem to require less effort to reuse heated steam than to intentionally condense the water only to reheat it once again.</p>
<p>I'm sure i'm misunderstanding something about the process. Why do we have to condense the water at all?</p>
| |thermodynamics| | <p>It's for almost the same reason as my answer to <a href="https://engineering.stackexchange.com/questions/5307/why-use-steam-in-the-rankine-cycle">your other question</a> on the Rankine cycle: The work output of the cycle increases as the integral (area) of the P-V diagram (or T-S) diagram is increased. Condensing the steam to liquid water results in a huge change in specific volume, which makes the integral of the P-V diagram much larger than if no phase change occured.</p>
<p>It does seem like energy is being wasted in the condensation process, but think of how much energy it would take to compress the working fluid to the boiler inlet pressure if it were a gas. (It would be much more energy than if we just had to pump a liquid to the same pressure).</p>
| 5308 | Why do we need to condense steam in the rankine cycle? |
2015-09-06T09:01:27.770 | <p>When a rectangular piece of plastic (say a credit card) experiences an increasing compression force, the plastic eventually buckles. Is it possible to find the equation of the plastic's cross section once it has buckled elastically? </p>
<p>What I have in mind is two fingers squeezing a plastic card. The fingers are $x$ cm apart, and the card itself has dimensions $l\times b$ cm and negligible thickness. Suppose $x \leq l$ (so that the card is compressed in a direction parallel to its original length $l$. Can we find an equation for the card's deflection $y$ in terms of $x,h,b$ and (I guess) the elastic modulus? </p>
<p><a href="https://i.stack.imgur.com/1YQ4v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1YQ4v.png" alt="enter image description here"></a></p>
<p>Does the <a href="https://www.google.com.au/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB4QFjAAahUKEwiMoMTj4ePHAhXEN6YKHd5kAuQ&url=http%3A%2F%2Fweb.aeromech.usyd.edu.au%2FAMME2301%2FDocuments%2FChapter09.pdf&usg=AFQjCNE9E-ZyR6-WqQ7N_CkzOeYt6hDNJw&sig2=aLYWveV2N_tMTB0Fyl9ulQ" rel="nofollow noreferrer">Euler column buckling result</a> work here? That is, can we expect a solution of the form $y=A\sin(\pi z/x)$? If yes, for small deflections is it safe to assume that the arc length of the curve is equal to $l$? </p>
<p>I'm nervous about using this, because I'm assuming negligible thickness. Do I have to allow the card to have non-zero thickness? Can I then use this to find the value of $A$? </p>
| |statics| | <p>The axial compression of the member should be negligible compared to the sideways deflection. This means that it doesn't depend on the cross section of the card. This is <em>exactly</em> Euler buckling.</p>
<p>The equations for buckling are usually used to determine the load at which buckling will occur. This is where the cross section matters. Once buckling happens, the shape is only defined by the deflection. The length along the member doesn't change (appreciably).</p>
<p>All of this changes once enough force is applied to move into plastic behavior and hinges form.</p>
| 5322 | The Equation of a buckled card |
2015-09-06T13:50:21.160 | <p>Something hit me during my last flight: our ears feel that the pressure around us changes quickly during liftoff and landing, they hurt more and more until we make them pop. However, the cabin must maintain a level of oxygen higher than outside the aircraft because it isn't dense enough at ~10km to breathe normally. </p>
<p>Does that mean that the aircraft merely takes air from the outside, heats it up (it's around -40°C outside at 10km) and adds oxygen to it before blowing it inside? How else? </p>
| |pressure|aircraft-design| | <p>There's a difference between being pressurized to an absolute value and being pressurized to a relative value. </p>
<p>For instance, if the cabin immediately pressurized to the same atmospheric pressure as your departure terminal, then your ears would not pop on ascent, but they would have to pop at some point on the descent unless the elevation of the landing terminal were the same as the departure. </p>
<p>This would also mean that the aircraft structure would have to be thick enough to withstand the full <em>differential</em> pressure between up to sea level and cruising altitude. </p>
<p>If instead the cabin only pressurized to, say, 7psi (about 50kPa?) above whatever the outside pressure is, then there's no need for an elaborate destination pressure matching scheme and the aircraft walls can be made thinner, thus lighter and more fuel efficient. </p>
<p>Relative pressurization means the absolute pressure changes with altitude, so your ears pop.</p>
| 5325 | If aircraft are pressurised, why do our ears pop during liftoff and landing? |
2015-09-07T07:13:54.313 | <p>I often see specifications for refurbishing works on gas pipelines calling for blanking spades, to separate the area worked on from the rest of the gas system. This is in addition to closing valves.</p>
<p>Googling showed me some pictures of how blanking spades look like, but I don't yet understand the process of installing them. The most straightforward way would be to mount them between flanges, after a section of pipe has been valved off and flushed. How are blanking spades actually used?</p>
| |pipelines|safety| | <p>You are correct, blanks are installed at a pipe flange where two runs connect. Typically you break the joint and pry the flange apart to install the blank, but this can cause problems if/when the prying action torques the next joint in the piping system and damages the gasket. I've heard of a person replacing every gasket in a large system because every time a joint was pried apart it ruined the adjacent gasket. </p>
| 5334 | How are blanking spades entered into gas pipelines? |
2015-09-08T03:39:59.077 | <p>I'm currently working on problems for my mechanics of material course, and I just learned how to calculate the shear stress on a bolt holding stacked plates. That whether it be a single or double shear, I have to account for every shear plane. However, what happens for the bearing stress? I know that it depends on the area of contact, which is diameter * thickness of the plate. If there were three plates (double shear) and I looked at only the portion of the bolt in each plate, would the bearing stress not differ? Do I have to look at the entire bolt at once to calculate the bearing stress?</p>
| |stresses| | <p>I think answer to the question <em>would the bearing stress differ</em> is ==></p>
<p><strong>YES!</strong> </p>
<p>bearing stress <em>would</em> vary along the bolt in different sections in plates...</p>
<p>and you <strong>cannot</strong> look at the entire bolt at once to calculate the bearing stress </p>
<p>here is a pictorial explanation why do we need to consider it in different sections & how to calculate bearing stresses (if you want to).</p>
<p>let us assume that we have a 3 plate arrangement like shown in pic.
<a href="https://i.stack.imgur.com/W3kr1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/W3kr1.png" alt="enter image description here"></a></p>
<p>where this blue thing on the right is called <strong>CLEVIS</strong> and on the LHS we have a plate.</p>
<p>Looking at the cross sectional view let us say:</p>
<p><a href="https://i.stack.imgur.com/q8TbY.png" rel="noreferrer"><img src="https://i.stack.imgur.com/q8TbY.png" alt="enter image description here"></a> </p>
<p>that</p>
<p>This connection consists of a flat bar A (which i will refer here as just plate), a clevis C, and a bolt B that passes through holes in the plate and clevis. Under the action of the tensile loads P, the plate and clevis will press against the bolt in bearing, and contact stresses, called bearing stresses, will be developed.</p>
<p><strong>Bearing stresses being contact stresses depend upon the contact between two surfaces so the answer to * Do I have to look at the entire bolt at once to calculate the bearing stress * is answered here</strong> and the answer is <strong>NO</strong></p>
<p><em>you have to look at each contact area not just because you want to calculate but more because looking at the entire bolt at once to calculate the bearing stress is</em> CONCEPTUALLY WRONG!!!</p>
<p>&
NOW, assume that clevis has upper and lower plate of different thickness... even if it doesn't look like... </p>
<p>but let's just say it for a moment.</p>
<p>these two plates of CLEVIS (upper and lower) are under forces say: <span class="math-container">$P_1$</span> and <span class="math-container">$P_3$</span> (respectively).</p>
<p>and using equations of static equilibrium: <span class="math-container">$P_1$</span> + <span class="math-container">$P_3$</span> = <span class="math-container">$P$</span></p>
<p>The <strong>Bearing</strong> stresses exerted by the clevis against the bolt appear on the left-hand side of the free-body diagram (see 1 and 3) . </p>
<p>&</p>
<p>The <strong>Bearing</strong> stresses from the plate appear on the right-hand side (see 2).</p>
<p>Here, we make an assumption i.e.:</p>
<p>Since, The actual distribution of the bearing stresses is difficult to determine, so we customarily assume that the stresses are uniformly distributed.</p>
<p>Based upon the assumption of uniform distribution, we can calculate an average bearing stress <span class="math-container">$\sigma_{b}$</span> by dividing the total bearing force <span class="math-container">$F_b$</span> by the bearing area <span class="math-container">$A_b$</span> , since stress is Force divided by acting area.</p>
<p>The projected area <span class="math-container">$A_b$</span> on which they act is a rectangle having a height equal to the thickness of chosen plate (upper or lower) of the clevis and a width equal to the diameter of the bolt.</p>
<p><a href="https://i.stack.imgur.com/bwEKl.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bwEKl.png" alt="enter image description here"></a></p>
<p>SO,following would be the EXPRESSIONS for bearing stresses at different contacts:</p>
<p><span class="math-container">$plate_1$</span> (of clevis) <span class="math-container">\begin{align} \begin{pmatrix} \sigma_1 = \dfrac{P_1}{d * t_1} \end{pmatrix} \end{align}</span></p>
<p><span class="math-container">$plate A$</span> <span class="math-container">\begin{align} \begin{pmatrix} \sigma_2 = \dfrac{P}{d * t_2} \end{pmatrix} \end{align}</span> </p>
<p><span class="math-container">$plate_3$</span> (of clevis) <span class="math-container">\begin{align} \begin{pmatrix} \sigma_3 = \dfrac{P_3}{d * t_3} \end{pmatrix} \end{align}</span> </p>
<p>and until and unless <span class="math-container">$t_1$</span> == <span class="math-container">$t_3$</span> bearing stresses would differ </p>
| 5344 | Does the bearing stress vary along the bolt? |
2015-09-08T06:48:31.593 | <p>I have a system with a motor, some masses, damping and springs. One of the masses is loaded with an unknown external force F. I want to estimate the force F using data of the motor (mainly current). There are some unknown disturbances, like temperature and friction in the system. Can this be done or are there better ways to estimate the unknown force? There will be no feedback loop. The available data will look be a time series of the current acting on the motor.</p>
| |mechanical-engineering|control-engineering|sensors|kalman-filters| | <p>First, start with a precise mathematical description of your system. Second, analyze it observability. If the system is observable, of course you can estimate the (augmented) state vector (along with the load as an unknown parameter). In practice, there are always some factors that are often a subject to experimental estimation rather than system identification. These are, for instance, the efficiency map of the motor, liquid friction coefficient (may be estimated from the step response), motor inertia etc. I do not know which exactly feedback loop you are taking about, but the mechanical torque in electrical machines can be estimated from the current, that is true.</p>
| 5345 | Can a load be estimated using a Kalman filter? |
2015-09-09T02:34:25.843 | <p>I know lasers can cut or create holes on some hard materials such as diamond and metals. Is there any similar concept in tunnel engineering?</p>
<p>If not, what are the difficulties?</p>
| |materials|lasers|tunnels| | <p>Barring other difficulties, it is unclear what advantages laser cutting would have in such an application. Lasers are <a href="https://engineering.stackexchange.com/questions/2991/why-are-lasers-used-for-concentrated-light-applications-instead-of-incoherent-li">useful in drilling/cutting of materials</a> because of their high precision and ability to make very small cuts. Digging a tunnel is not a precision activity though. </p>
<p>Even if a tunnel drilling machine were set up to use lasers, the cut between the bore and the tunnel wall would have a very small separation (~1 mm), and a massive amount of equipment would still be needed to remove the rest of the material from the bore. </p>
| 5355 | Is it possible to use lasers to build tunnels instead of mechanical digging machines? |
2015-09-09T08:08:40.533 | <h2>Short form: How would one calculate the perfect shape of a range hood?</h2>
<p>See drawing below. Will the curve of the profile of the range hood be exponential? Quadratic? What if the range is enclosed from three sides that create a box open only from the front - what curve should one choose? Sigmoid? Cubic? Some sort of chain curve? How does one calculate the parameters of the curve?</p>
<p>How does one select the minimal air flow necessary to suck the steam in?</p>
<h1>Long form:</h1>
<p>I'll start with a drawing.</p>
<p><img src="https://i.stack.imgur.com/WgRDm.png" width="200" height="340" alt="schematic drawing of range hood" /></p>
<h1>EDIT</h1>
<p>I have added the sections "Fan throughput" and "Transition from laminar flow to turbulent flow".</p>
<h2>Dimensions</h2>
<p>Assume the range is a 60x60 cm square that produces steam (or 60 cm diameter disc if that's easier). The range hood begins 70 cm above that and is 110 cm high, terminating in a 7cm diameter pipe (which I might have to replace for something wider, but that's what I'm working with right now). This pipe is 80 cm high and terminates in a fan. I assume having the fan in a pulling configuration is better than having it push into the pipe. In addition it'll be quieter (which I would prefer). The fan has a constant throughput and blows into open air.</p>
<p>There is a wall behind the range. There may be walls on the sides but not the front which is where you access the pots. I assume that's better than just having the wall behind the range. As a bonus it will minimize splatters around the kitchen. The walls may also only go part way from the range hood towards the stove if that's optimal.
The hood may be wider than the range, but with side walls that might not be necessary. It may not protrude towards the cook.</p>
<h2>Fan throughput</h2>
<p>I have tried to calculate the necessary fan throughput by calculating the expansion rate of heated steam during normal use (one large and one medium heater turned on, a total of roughly $4000\mathrm{W}$).</p>
<p>I am not sure which of my simplifying assumptions are too strong.</p>
<p>Assume the volume above the range (60 cm * 60 cm * 70 cm) is filled with steam at 100 degrees Celsius, normal atmospheric pressure, and is heated uniformly by a 4000W heating element with no heating loss.</p>
<p>Normally, the excess volume would spill in all directions - to the top and to the sides. We want to suck enough air into the top that the whole excess is taken in, so that the sides have negative pressure, i.e. air is coming into the volume from the sides, not out of it, and only escaping through the top.</p>
<p><img src="https://i.stack.imgur.com/C8FLX.png" width="300" height="200" alt="Steam volume" /></p>
<p>Let us calculate the volume gained over one second:</p>
<ol>
<li>(starting temperature) $20 ^\circ \mathrm{C} = 293 \mathrm{K}$</li>
<li>(specific heat of steam between $20$ and $100 ^\circ \mathrm{C}$) between $1.86$ and $1.89$ $\mathrm{J/(gK)}$, let's settle for roughly $c_P = 1.88 \mathrm{J/(gK)}$.</li>
<li>(density of superheated steam at $100 ^\circ \mathrm{C}$) $\rho=597.42868305543 \mathrm{g/m^3}$</li>
<li>(mass of steam) $m = \rho*0.6*0.6*0.7 \mathrm{m^3} = \rho * 0.252 \mathrm{m^3}$</li>
<li>(power of heater) $H = 1000 \mathrm{W}$</li>
<li>(energy increase) $\Delta U = H * 1\mathrm{s} = 1000 \mathrm{J}$</li>
<li>(temperature increase) $\Delta T = \Delta U/(c_P*m)$</li>
<li>(volume increase) $\Delta V=(nR/P) \Delta T$</li>
<li>(gas constant) $R = 8.3144621 \mathrm{J/(K*mol)}$</li>
<li>(pressure) $P = 101.3 \mathrm{kPa}$</li>
<li>(amount of substance) $n = m/m_m \mathrm{mol}$</li>
<li>(molar mass of water) $m_m = 18.02 \mathrm{g/mol}$</li>
</ol>
<p>Plugging everything in:</p>
<p>$$
\begin{align}
\text{from 8:} \\
\Delta V & =(nR/P) \Delta T \\
\\
\text{from 7:} \\
\Delta V & =(nR/P) \Delta U/(c_P*m) \\
\\
\text{simplify:} \\
\Delta V & =nR \Delta U/(P*c_P*m) \\
\\
\text{from 11:} \\
\Delta V & =(m/m_m)R \Delta U/(P*c_P*m) \\
\\
\text{simplify:} \\
\Delta V & =R \Delta U/(P*c_P*m_m) \\
\end{align}
$$</p>
<p>Plugging everything into Google gives me the answer:</p>
<p><code>(((8.3144 joule) / (kelvin * mol)) * (1000 joule)) / (101.3 kilopascal * 1.88 * (joule / (gram * kelvin)) * 18.02 * (gram / mol)) =
0.00242275129 m^3</code></p>
<p>So $1000 \mathrm{W}$ gives us $0.002 \mathrm{m^3/s}$, and $4000\mathrm{W}$ gives us $0.008 \mathrm{m^3/s} = 16.95 \mathrm{CFM} = 8 \, \mathrm{liters/s}$.</p>
<p>This is the minimum according to the above model. However, the volume in question isn't made out of jelly, and so a lot more suction will be necessary to make sure steam doesn't escape out the sides. My gut feeling tells me around three times more, or <strong>24 liters per second</strong>, will be adequate <strong>according to this model</strong> (but below we reduce this to 5 liters per second in a realistic scenario).</p>
<p>We can compute at what air throughput turbulent flow will begin for our range hood of 60x60cm.</p>
<p>The dynamic viscosity of steam at 100 degrees celsius = 212 deg K can be read off for <a href="http://www.engineeringtoolbox.com/steam-viscosity-d_770.html" rel="nofollow noreferrer">this table</a> by using <a href="http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html" rel="nofollow noreferrer">this table</a> to get the absolute atmospheric pressure of $14.7 \, \mathrm{psia}$.</p>
<p>(dynamic viscosity of steam) $\mu = 0.013 \mathrm{centiPoise}$</p>
<p>The kinematic viscosity of a fluid <a href="http://www.engineeringtoolbox.com/dynamic-absolute-kinematic-viscosity-d_412.html" rel="nofollow noreferrer">can be calculated</a> using $\nu = \mu / \rho$ where $\rho$ is the density of steam.</p>
<p>(density of steam as above) $\rho=597.42868305543 \mathrm{g/m^3}$</p>
<p>(kinematic viscosity of steam) $\nu = 2.18 * 10^{-5} \mathrm{m^2/s}$</p>
<p>We want to calculate the Reynolds number and keep it under $2300$ for it to indicate laminar flow. Above $4000$ it indicates turbulent flow.</p>
<p>$Re = Q D_H / (\nu A)$</p>
<p>where</p>
<p>$A$ is the cross-sectional area of the range hood's opening: $A = 60\mathrm{cm} * 60\mathrm{cm} = 0.36 \mathrm{m^2}$,</p>
<p>$Q$ is the flow rate: $Q = 24 \, \mathrm{liters/s} = 0.024 \mathrm{m^3/s}$,</p>
<p>$D_H$ is the hydraulic diameter for a square duct, which is just the length of the side: $D_H = 0.6 \mathrm{m}$</p>
<p>Plugging everything in:</p>
<p>$$
\begin{align}
Re & = 0.024 * 0.6 / (2.18 * 10^{-5} * 0.36) \\
Re & = 183.48 \\
\end{align}
$$</p>
<p>Therefore we are perfectly within laminar flow.</p>
<p>For a circular pipe at $24 \, \mathrm{liters/s}$, we can calculate the minimum radius $r$ to be below turbulent flow thus:</p>
<p>$$
\begin{align}
2300 & > 0.024 2r / (2.18 * 10^{-5} * \pi r^2) \\
2300 * 2.18 * 10^{-5} * \pi / (0.024 * 2) & > 1/r \\
(2300 * 2.18 * 10^{-5} * \pi / (0.024 * 2))^{-1} & < r \\
0.30 & < r \\
\end{align}
$$</p>
<p>Therefore, we would need roughly a half-meter pipe otherwise we are dealing with slightly turbulent flow.</p>
<p>However, in the real world, we are dealing with air that only contains a portion of steam. The steam buildup will be an order of magnitude less than 8 liters per second (imagine how much water you'd have to put in a pot!). This makes me think that <strong>realistically we only need a fan rated about 5 liters per second</strong>. On the other hand dry air is only two-thirds as viscous as steam, which will mean turbulence is easier. All in all, with those adjustments compared to our model we can multiply the $Re$ calculated above by 0.3, or in other words, we only need <strong>a pipe of 18 centimeters diameter</strong> for perfectly laminar flow. A pipe of 7 centimeters diameter which I have will cause turbulent flow.</p>
<h2>Transition from laminar flow to turbulent flow</h2>
<p>I understand because we are transitioning from laminar to turbulent flow, the correct profile curve will minimize pockets of air blocked by turbulence, which end up "choking" the flow, further reducing the laminar core and causing even more turbulence.</p>
<h2>Range hood profile curve</h2>
<p>My experience from acoustics, and horns specifically, gives me the hunch that if the range hood terminated in mid air, the perfect profile would be an exponential curve. However I appreciate that with a horn speaker the pressure changes on the order of 15 Hz or higher, while here we're purely dealing with constant pressure. Also in this case the range would not terminate in open space but in a confined box (cylinder) that's open on one side (say an arc of 90 degrees). Would the best curve then be a sigmoid? Some sort of chain curve? A cubic?</p>
<p><em>How does one calculate the right kind of curve for the profile connecting the semi-closed box and the pipe inlet?</em></p>
<p>I assume that the case where everything has a round cross-section is easier than when it's square, but can then be generalized using a little common sense.</p>
<h2>Airflow</h2>
<p>I assume there is some minimal air flow that is necessary to suck in the steam. How does one calculate that? I guess this has to do with pressure building up where air flows from the kitchen to the range top; my guess is that you could calculate the pressure of the air flowing past the range, and if that pressure is high enough, the <strong>flow</strong> will expand to cover the whole available space (i.e. all the way down to the range top). I understand that's a <strong>fluid</strong> <strong>dynamics</strong> question, but I know almost nothing on the subject.</p>
<h2>Impedance</h2>
<p>Is there a good approach to this problem (the curve profile as well as the that uses the concept of impedance? I understand impedance based approaches only really deliver answers for AC signals, not DC, but perhaps one could calculate the perfect transformer as a limit as frequency tends to 0, or use an approximation (e.g. assume frequency is 0.1 Hz)?</p>
<h2>Background</h2>
<p>My range hood doesn't really work at all. That is, it's so inefficient that putting my hand just under the hood I don't feel any draft at all. Having looked at it, there are many, many things wrong with it that I won't bother to list - it's just bad, leaks air everywhere, etc. The market doesn't seem to provide any sensible solutions either. Not even the professional offerings seem to make sense from an aerodynamics perspective. That made me wonder what the perfect range hood <em>would</em> be, and I don't know how to approach the problem.</p>
<p>Note: I have posted this on the Physics StackExchange previously, but it was pointed out to me that the question was off topic for that SE, and I was asked to move it to Engineering. It has been put on hold there.</p>
| |fluid-mechanics|hvac|airflow| | <p>Making a duct that flares along an exponential curve to 2 feet (60cm) in diameter would create a larger duct that won't solve your problem.</p>
<p>It would be an expensive duct to make (think <a href="http://www.all-brass-instruments.com/uploads/brass/20/200400529718-7.jpg" rel="nofollow">tuba bell</a>), and if you focus on the flare instead of the overall dimensions, then the duct might come down too low and you won't have enough space. </p>
<p>Keep in mind that a flared duct doesn't do anything to move air - that's all done with a fan. The duct also isn't going to generate noise - that's also all done by the fan. If you want a quiet, high-flow hood, get a quiet, high-flow fan. I searched online for "low noise high flow kitchen hood fan" and I found <a href="http://www.broan.com/products/series/broan-evolution-qp4-series-1783e9c0-5b48-4785-b04b-d4d01fad493e" rel="nofollow">this fan</a> that advertises <0.3 sones during normal operation, with a boost mode of 630 CFM. 630CFM means that all of the air in a 18x18 foot room gets removed every 4 minutes.</p>
<p>It appears to me that you should ignore the hood shape and focus on the fan. At your desired air speeds, for your desired application, the shape of the hood is going to make virtually no difference at all. That is why there is <a href="https://www.google.com/search?q=kitchen+hood&biw=1920&bih=955&tbm=isch&source=lnms&sa=X" rel="nofollow">such variety</a> in kitchen hoods. </p>
<h2>:EDIT:</h2>
<p>The same conclusion, this time with numbers!</p>
<p>Consider evaluating the <a href="https://en.wikipedia.org/wiki/Reynolds_number" rel="nofollow">Reynold's number</a>, which "is defined as the ratio of momentum forces to viscous forces and ... also used to characterize different flow regimes within a similar fluid, such as laminar or turbulent flow." </p>
<p>For <a href="https://en.wikipedia.org/wiki/Reynolds_number#Flow_in_pipe" rel="nofollow">flow in a pipe</a>, the Reynolds number can be calculated with:</p>
<p>$$
\mbox{Re} = \frac{Q D_H}{\nu A} \\
$$</p>
<p>where $Q$ is the volumetric flow rate in $m^3/s$, $D_H$ is the hydraulic diameter (which is just the diameter for a <a href="https://en.wikipedia.org/wiki/Hydraulic_diameter" rel="nofollow">circular tube</a>), $\nu$ is the kinematic viscosity in $m^2/s$, $A$ is the cross-sectional area of the pipe in $m^2$, and $\mbox{Re}$ is the Reynolds number, a unitless number that indicates laminar flow for values under approximately 2300, turbulent flow for values over 4000, and transitional flow between the two. </p>
<p>SO, for air, here are the values I used in the equation:</p>
<ol>
<li>Going off the grid in your question (thanks for that!), if the bell end flare is 60cm wide at 6 grids wide, I estimated your duct to be about 1 grid wide, so a diameter about 10cm or 0.01m. </li>
<li>This diameter gives the cross-sectional area $A = \pi r^2$, or $A = pi (0.05^2)m^2$, so $A = 0.00785m^2$. </li>
<li>Your stated flow of $50 L/s = 0.050 m^3/s$. </li>
<li>A <a href="https://en.wikipedia.org/wiki/Viscosity#Air" rel="nofollow">kinematic viscosity</a> of $\nu = 1.48 \times 10^{-5} m^2/s$. </li>
</ol>
<p>That's it! So, plugging everything in:</p>
<p>$$
\mbox{Re} = \frac{Q D_H}{\nu A} \\
\mbox{Re} = \frac{(0.050 m^3/s)(0.01m)}{(1.48 \times 10^{-5} m^2/s) (0.00785m^2)} \\
\mbox{Re} = \frac{5 \times 10^{-3}}{1.16 \times 10^{-7}} \\
$$
$$
\boxed{\mbox{Re} = 43,015} \\
$$</p>
<p>So, where a Reynolds number over 4,000 indicates turbulent flow, at your duct size and your flow rate, you are at 43,000, so an order of magnitude past the threshold for turbulent flow. </p>
<p>This demonstrates and hopefully reinforces my earlier statement, "the shape of the hood is going to make virtually no difference at all," because again, at your flow rate, you're already in turbulent flow. </p>
<p>Since you can't do anything about the flow noise due to turbulence, the best thing you can do for noise reduction is to look at a fan that's designed for low noise emission. This is typically done by increasing the blade surface area while reducing the speed, because one of the largest factors for fan noise is the tip velocity on the blades. See <a href="http://www.moorefans.com/pdfs/NOTES4_1.PDF" rel="nofollow">this guide</a> for more information. </p>
| 5357 | Perfectly aerodynamic kitchen range hood |
2015-09-09T08:09:39.283 | <p>I am trying to manipulate some data that were obtained from a three point load experiment on a porcine tibia.</p>
<p>I would like to obtain a stress-strain curve for the various specimens; however, the only data that was given from the Instron three-point loading set up was Load (N) vs Deformation/Extension (mm).</p>
<p>My question is: <strong>how do you convert the Load vs Deformation curve to a Stress-Strain curve</strong>?</p>
| |mechanical-engineering|materials|stresses|biomedical-engineering|biomechanics| | <p>It's fairly complicated. I found an article in The Bone Journal <a href="http://www.thebonejournal.com/article/S8756-3282(98)00076-3/fulltext?refuid=S8756-3282(10)01400-6&refissn=8756-3282">Comparison of three-point bending test and peripheral quantitative computed tomography analysis in the evaluation of the strength of mouse femur and tibia</a> where they used tomogrphy to calculate inertia. However if you don't have CT you can try the hard way.</p>
<ol>
<li>You'll have to calculate bending moment M. It should be simple.</li>
<li>Locate Neutral Axis. You can assume it as a simple shape but probably you have a complicated shape and have to calculate it.</li>
<li>Calculate Area Moment of Inertia. Again if you assume simple shape it's easy else complicated.</li>
<li>Formula for bending stress is $\sigma = -\frac{M\cdot y }{I}$ which is called <a href="http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/flexure-formula">flexure formula</a> M is moment, y is distance from neutral axis and I is area moment of inertia.</li>
<li>Strain is $\varepsilon = \frac{y}{\rho }$ where $\rho$ is radius of curvature. A simple presantation is <a href="https://www.google.com.tr/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCEQFjAAahUKEwjM4Kqn0OnHAhXnpnIKHWGUBKU&url=http%3A%2F%2Fengr.bd.psu.edu%2Frxm61%2F213%2FCh6_sec3_4.ppt&usg=AFQjCNHJlOFr9JLexGpzpSwVC6aofMQOeg&sig2=_R_C6lMrIKcaiyVYC1beSw">here (PDF)</a></li>
</ol>
<p>This <a href="http://academic.uprm.edu/pcaceres/Courses/INME4011/MD-3A.pdf">presentation (PDF)</a> and <a href="http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?topic=me&chap_sec=04.1&page=theory">this course notes</a> are also informative.</p>
| 5358 | Convert Load Deformation (Extension) Data to Stress-Strain Curve |
2015-09-10T09:37:59.240 | <p>How can I compute the Frequency Response Function (FRF) if I use two three-axis MEMS accelerometer to measure the input excitation and output response?</p>
<p>I have read from some books that the FRF can be computed as the ratio of response to excitation, e.g., $ H(s) = x(s)/F(s)$, where $s$ is the Laplace variable. or a similar way via Fourier transform.</p>
<p>However, those books have not mentioned how to compute FRF when it comes to three-axis accelerometer. As there are three channels of data, <em>i.e.</em>, X, Y, Z, should I compute the FRF for each channel separately, or I need to perform some kind of combination of these three axes before computing the FRF?</p>
| |civil-engineering|structural-engineering|vibration|modal-analysis| | <p>The body of your question (the title is somewhat different) asks about computing the frequency response function (FRF) between a multi-input, multi-output (MIMO) system. First off, the numerical transfer function/FRF between two time domain signals is usually calculated by calculating the cross power spectrum between the two signals and dividing it by the power spectrum of the input
$$
FRF_{A\rightarrow B}=\frac{S_{AB}(f)}{S_{AA}(f)}.
$$
National Instruments <a href="http://www.engr.sjsu.edu/bjfurman/courses/ME120/me120pdf/FFT_tutorial_NI.pdf" rel="nofollow">Application Note 41</a> has a helpful introduction to this type of analysis. It is important to also calculate the coherence of the input and output signals so that you have an idea of how accurate the estimated transfer function is. </p>
<p>To answer your question about how to deal with a MIMO system; usually people think about it as a matrix of transfer functions. In your case there will be 9 independent transfer functions:
$$
\begin{pmatrix}
FRF_{x\rightarrow x} & FRF_{x\rightarrow y} & FRF_{x\rightarrow z} \\
FRF_{y\rightarrow x} & FRF_{y\rightarrow y} & FRF_{y\rightarrow z} \\
FRF_{z\rightarrow x} & FRF_{z\rightarrow y} & FRF_{z\rightarrow z}
\end{pmatrix}
$$
Understanding how any excitation shows up at the output is then just a matter of multiplying the x, y, and z inputs (expressed in the frequency domain) by the transfer function matrix.</p>
| 5365 | How do I conduct an experimental modal analysis with a three-axis accelerometer? |
2015-09-13T14:35:18.607 | <p>Let us assume you have a large forging press or drophammer, and drawings for a set of smaller parts that all need to be forged from the same material. Would it be feasible to construct a forging die so that all the parts could be forged from a single billet and then separated as a machining step? Or is this more expensive than making a forging die for each part and forging them separately from individual, smaller billets?</p>
<p>For the purposes of this question, you can assume that the parts all have a linear geometry -- say, if you wanted to forge a crankshaft and camshaft set for an engine in a single operation.</p>
| |manufacturing-engineering|forging| | <p>I've previously worked in the forging industry and you can certainly forge multiple parts in a single stroke of a large press but there are problems. The obvious ones are the size of the available press and the economics of the operation. However, even from a metallurgical point of view, forging a complex part like a crankshaft in one stroke would induce more plastic deformation than the material could take, at least at room temperature. The shaft would then have to be hot-forged to maintain the ductility of the material during deformation and the temperature required is critical. I can't imagine that you could do a crankshaft and a camshaft at the same time. And apart from that, complex parts are usually forged in stages, each stage taking the component closer to its final shape. Again, I can't see this being done for a crankshaft and a camshaft in a single die.</p>
<p>But I could be out-of-date...</p>
| 5396 | Is it feasible to forge multiple, smaller parts in the same forging operation? |
2015-09-14T18:46:12.063 | <p>Just a short question: Is there any usefulness in doing block diagram reduction piecewise?</p>
<p>The reason I am asking is that I find it much (!) easier to just find the final $\frac{output}{input}$ transfer function using mathematics (the brute force method), but my professor does it stepwise. </p>
<p>If I am to do it stepwise, I still feel like I have to do it mathematically, rendering the stepwise reduction pretty useless as far as I can tell. </p>
<p>I realize this seems like a banal question, but I cannot find anything answering this question is my controls book.</p>
| |control-engineering|control-theory| | <p>When you are solving a problem in the real world, there are 2 main important requirements</p>
<ol>
<li>Get the correct answer. </li>
<li>Solve it using sound methods. </li>
</ol>
<p>There might be a half-dozen different ways to solve a given problem, so long as you arrive at the correct answer and don't pick a method involving rain dances or Satanic rituals, you are free to pick the method that you find easiest.</p>
<p>However, the requirements in academia are often different. There is a good chance that your professor has a reason for using the method he does. If he asks you to solve a problem step-wise, you should do what he asks. </p>
<p>It's likely that, at a certain point, doing it directly rather than piecemeal becomes overly complex and involves massive equations. At this point, knowing how to do it step-wise becomes necessary, and your professor is preparing you for this. But the only way to know for sure is to ask him. </p>
<p>As I said, beyond this class, the only thing that will matter is getting the correct answer and being able to explain how you arrived at that answer, so if you get the same answer through both methods, both are equally valid. </p>
| 5408 | Block Diagram Reduction: Is it necessary to do it stepwise? |
2015-09-14T21:09:45.450 | <p>I want to simulate a turbine in Fluent. For this reason, I studied some papers. In these papers, Turbulence is modeled using the realizable k–e model and Non-equilibrium wall functions are used in the near wall flow modeling. There are the following sentences in these papers:</p>
<ul>
<li><p>The Boussinesq approach is applied to relate the Reynolds stresses to the mean velocity gradient.</p></li>
<li><p>The Reynolds stress is related to the mean velocity gradient by means of the Boussinesq approach. </p></li>
</ul>
<p>What is the exact meaning of above sentences? And is there a direct option in Fluent?</p>
<p>Is there any adjustment that must be made in Fluent before starting the calculation for this approach?</p>
| |mechanical-engineering|fluid-mechanics|turbines|cfd| | <p>The <strong>basic assumption</strong> of the RANS approach (see the answer by @Algo) is that there is an overall steady solution to the problem. This overall steady solution features small scale oscillations <em>around</em> the mean value. </p>
<ul>
<li>In a first step the Navier Stokes equations (NS) are rewritten with this assumption
$u_i = \bar{u_i} + u_i'$</li>
<li>The second step is to average the NS again </li>
<li>By doing this only on term does not cancel out except: $\frac{\partial }{\partial x_j}(\overline{\rho {u_i}'{u_i}'})$ (this is a 3x3 matrix and it's called the Reynolds Stress Tensor)</li>
</ul>
<p>Since there is no analytical solution for the <em>mean of the product of the fluctuations</em> <strong>an additional assumption</strong> was needed to solve the equation system.
There were a lot of ideas on how to <em>model</em> or <em>solve</em> this problem. One turned out to be very useful because it predicted the reality very well.</p>
<p>The idea is based on the observation that turbulence basically increases mixing of the fluid. To put it in other words: velocity gradients <em>mix</em> out a lot faster in turbulent flows, as if the viscosity (momentum diffusion) is increased. </p>
<p>So the first step in the solution was to model the unknown term as a kind of additional viscosity term as shown in the answer from @Algo. Here the mean velocity gradient is known and only one unknown remains (after the RANS approach we were faced with a 3x3 matrix).</p>
<p>Turbulence models are then used to calculate $\mu_t$. So the only adjustment you can make is to choose a turbulence model and in some cases tweak its parameters.</p>
| 5410 | What is the Boussinesq approach? |
2015-09-14T21:37:19.430 | <p>While studying some papers, I didn't understand the following sentences:</p>
<blockquote>
<p><code>deltP</code> the pressure drop across the turbine, evaluated using the mass average values at the inlet and outlet sections of the computational domain, respectively.</p>
</blockquote>
<p>What is the exact meaning of the "mass average values"?</p>
<p>And is there such a option of "mass average values" in Fluent? I dont know that how to compute this parameter in Fluent.</p>
| |mechanical-engineering|pressure|cfd| | <p>Averaging is a tricky business (see <a href="https://www.researchgate.net/publication/245354712_Averaging_Nonuniform_Flow_for_a_Purpose" rel="nofollow">Cumsty 2006</a> for an introduction). Basically the problem is, that our thermodynamics usually talk about <strong>a</strong> pressure and <strong>a</strong> temperature etc. But in real life the pressure and temperature are fields (i.e. varying in all three dimensions). So the question is: How to come up with <strong>one</strong> representative value of (for example) the temperature.</p>
<p>Depending on the way the averaged value is used different averaging techniques are suited better.</p>
<p>For example think of the temperature of water in a big tank which is filled via two pipes from two different heat exchangers delivering two different water temperatures. Therefore not every water molecule has the same temperature. If you were supposed to calculate <strong>the</strong> temperature in the tank you would need to come up with a way to calculate a representative temperature. How high would this mixed out temperature be?</p>
<p>You can of course just use the mean of the two pipes. But what if one pipe pumps more water into the tank. The solution to this particular problem is to weigh every pipe-flow with the mass-flow rate. This is called a mass-flow-average.</p>
<p>As a rule of thumb:
To validate numerical simulations the averaging method in CFD and experiment have to be the same. It might be a good idea to use area-averaging to compare basic values as temperature or pressure.
If the averaged values are used to do further calculations use mass-average for all except for static pressure which is always area-averaged. </p>
| 5411 | Understanding pressure drop and mass average values |
2015-09-15T11:50:55.000 | <p>I have gas pipeline with biogas, low pressure - approx 40mbar over atmosphere. I want to remove a small amount of condensate, inside a building, without creating an <a href="http://www.engineeringtoolbox.com/hazardous-areas-classification-d_395.html" rel="nofollow">ex-zone</a>. Small amount: I honestly don't know but it should be next to nothing. But I create a low point and want to be sure I don't block my pipeline.</p>
<p>What is a valve that I can build into the pipeline at the low point, that allows me to ex-filtrate gas without creating an ex-zone?</p>
<p>No need for an actual supplier, but some terms to aid my googling my writing of the RFQ would be great. Plus the technical info to decide if it's the right gadget for my problem!</p>
| |mechanical-engineering|pipelines|safety|valves| | <p>There are <a href="http://mankenberg.de/en/20/prgb:7/product:kondensatableiter-schwimmergesteuert/produkte.html" rel="nofollow">swimmer operated condensate traps</a> (link contains cut view). The operating principle seems fairly simple, a swimmer in the pot opens a valve when enough condensate has accumulated and closes before the pot is empty. The specific product I linked has been accepted as tight enough for not creating an ex-zone, I'm sure there's others.</p>
<p>Note: I don't want to advertise for Mankenberg, it's just the first example I found.</p>
| 5422 | Device to remove condensate, without creating a hazard (EX) zone |
2015-09-15T16:06:10.750 | <p>Consider a centrifugal compressor. The compressor map usually plots the adiabatic head vs volumetric inlet flow for rated conditions of pressure, molecular weight and temperature at a given speed. But when pressure, temperature, molecular weight and speed are fixed, how can the inlet flow be made to vary ? Is it accomplished through the use of inlet vanes ?</p>
| |mechanical-engineering|compressors| | <p>In general every compressor (axial and radial) will deliver a certain flow rate depending on the back-pressure down stream of the compressor. So by changing the pressure difference over the compressor: for example closing a throttle upstream or downstream of the compressor the flow rate will change accordingly. This can also be achieved by changing the <em>geometry</em> of the compressor. One possibility are the mentioned variable inlet guide vanes.</p>
<p>Below is a compressor map taken from <a href="https://en.wikipedia.org/wiki/Centrifugal_compressor" rel="nofollow noreferrer">Wikipedia</a>. As you can see (away from the surge line) every mass flow rate is connected to a pressure ratio. At constant rotor speed, the flow rate changes according to the pressure ratio which is needed.</p>
<p><a href="https://i.stack.imgur.com/stlhG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/stlhG.png" alt="enter image description here"></a></p>
| 5431 | How can flow in a centrifugal compressor at fixed conditions vary? |
2015-09-16T22:53:40.500 | <p>A motor gives its strength in N-cm or oz-in, but how do you know how much you require? I know you can just calculate the minimum torque you need to move the load, but is there a typical amount over that minimum that is standard? Like 20% or reaching a certain acceleration?</p>
<p>What I'm using a motor for is my electrical engineering capstone project. I need to rotate a mirror about the z axis at between 30/60 Hz. The mirror is the only moving component, so there aren't any other mechanical considerations besides that. I don't know the mirror size yet, as I figured I'd buy as good a motor as I can afford and pick a mirror that's small enough to work with it, but I want to know how to figure out approximately what a given torque will get me. I don't want to spend $50 on a motor with something like "100 oz-in min holding torque" and then get it and find out thats only strong enough for a 2"x2" mirror or something.</p>
| |electrical-engineering|stepper-motor| | <p>Without more details about your design, the dynamics of your mechanism, or your calculations, its virtually impossible to address your question with any certainty. </p>
<p>However, i offer some advice that may prove useful to you:</p>
<p>If affordable, I would use double the minimum holding torque needed for the application because i usually find that my calculations don't usually account for all the true dynamics of the mechanism. When starting out with an untested design, it's better to err on the side of caution. Having too much torque is usually better than having too little. </p>
<p>Depending on speed requirements, you may be even able to use gears to increase your torque while using a motor with less torque. I may even recommend experimenting with gear ratios since its usually cheaper to buy a different gear than to buy a different motor. Over time, you may find gears to be the better choice until you find the torque that meets your needs.</p>
<p>But if you're really confident that your calculations truly capture the physics, you might be able to get away with 20%-30% more than the minimum torque. Ultimately, it all boils down to how well you can capture the physics of the mechanism.</p>
| 5451 | How much stronger than the bare minimum should a motor be? |
2015-09-17T21:22:49.660 | <p>I'm working on analysis of a bridge wall-pier for dead and live loads. When modeling the pier in finite element software (LARSA), I'm running up against several modeling questions:</p>
<ol>
<li>Is it valid to model the wall as a single column element? (Going to a plate mesh seems like overkill for a run-of-the-mill highway bridge.)</li>
<li>How should the connectivity be modeled between the girders and the wall pier? Are rigid links appropriate? Joint constraints? Beam elements with really high moments of inertia?</li>
<li>If I wanted to run modal analysis, would this change the modeling considerations?</li>
</ol>
<p><a href="https://i.stack.imgur.com/9GXyA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/9GXyA.png" alt="Wall Pier"></a></p>
| |structural-engineering|bridges|finite-element-method| | <p>I'll go through your questions one by one:</p>
<ol>
<li><p>A simple model like you have shown is fine to get model loads. Once you get those loads, you can then use a more specific technique to design the pier. (strut and tie).</p></li>
<li><p>Rigid links are ok. Just make sure that you are capturing the various induced moments. I usually have rigid links from the bearing locations to the pier nodes and rigid links from the center of girders to the bearings. That way you can see individual bearing axial loads and shears (and set moment/rotation releases properly).</p></li>
<li><p>All of the above should work for modal analysis. In modal analysis you will have to be extra careful that members are drawn along their centerlines so that you can apply mass accurately.</p></li>
</ol>
| 5462 | Simple Finite Element Modeling of Wall Piers |
2015-09-18T15:35:39.757 | <p>I am currently tinkering with typewriters and two things striked me when I started to read about their maintenance:</p>
<ol>
<li>They are supposed to run almost completely dry, except for a very tiny amount of lubricant at very specific points;</li>
<li>Gummy or sticky gunk buildup - usually due to over-oiling + dust buildup + passage of time - is one of the main causes of typewriter malfunctioning.</li>
</ol>
<p>As an avid cyclist I am, I have already noticed some oils have a tencency to get sticky as they dry out and mix with dust (e.g. gearbox oil), while others seem to be very stable and somewhat "repel" dust, mostly automatic transmission fluid - ATF, power steering fluid, or suspension oil. I believe these are somewhat intended to remain fluid and to actively repel dust, which seems to "fall out of the way" instead of get increasingly mixed with the oil and the working surfaces.</p>
<p>I know a lot of typewriter folks use sewing machine oil (Singer, mostly), but I wonder if its characteristics are suitable for this need of not getting gunky as the years pass.</p>
<p>So que questions are:</p>
<ol>
<li>Given the considerations above, would some type of oil or non-oily lubricant be more adequate than sewing-machine oil?</li>
<li>Is the hydraulic fluids hypothesis a good one?</li>
<li>Any other relevant consideration?</li>
</ol>
| |mechanical-engineering|tribology| | <p>The best option would probably be "clock oil" which is intended to be long lasting, non-acidic, and (importantly) non-spreading. If you apply it in small enough quantities, it will stay where you put it, and not move around through capillary action.</p>
<p>Sewing machine oil is probably a reasonable alternative, but the fact that it is formulated not to mark or damage any fabric or thread that it comes in contact with is probably irrelevant. Also it was not meant to be long-lasting. Singer used to recommend that their sewing machines should be oiled daily if they were in continuous use (i.e. in a factory.)</p>
<p>Clock repairers sometimes use transmission fluid to lubricate clock springs (if only because of the expense of using clock oil on big rusty springs in a low-quality clock!) but not for lubricating bearings and pivots.</p>
| 5467 | Which lubricant (oil or not) to use if I want to prevent sticky/gummy buildup? |
2015-09-20T09:37:06.603 | <p>For rectification I am given the following information</p>
<p>$x_e$ = 0.95</p>
<p>$x_f$ = 0.50</p>
<p>$x_b$ = 0.15</p>
<p>$v$ = 8</p>
<p>$q$ = 1.17</p>
<pre><code>**VLE Data**
x y
0.00 0.00
0.10 0.17
0.20 0.31
0.30 0.44
0.40 0.55
0.50 0.645
0.60 0.73
0.70 0.81
0.80 0.88
0.90 0.94
0.95 0.97
1.00 1.00
</code></pre>
<p>The VLE data is derived from a given diagram. There may be some inaccuracies there. However drawing with pen and paper in the diagramm gives me the same error I'll explain later on.</p>
<p>With all this data I developed the McCabe Thiele Diagramm for this problem.</p>
<p>Putting everything together in Excel</p>
<p><a href="https://i.stack.imgur.com/EPMrb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EPMrbm.png" alt="enter image description here"></a></p>
<p>The grey dotted line indicates my $x_b$ and in orange I drew the theoretical stages. My Problem is: Usually you would come up with a whole number of stages, in this case I would get $9\frac{5}{7}$.</p>
<p>There is also an $E_k = 0.7$ given.</p>
<p>$$n=\frac{n_{th}}{E_k} = 13.87$$</p>
<p>My question is: Why do I not get whole numbers? I am under the impression that there always have to be whole numbers. However in Absorption we where supposed to give answers like $n = 1.2$.</p>
<p>I thought maybe the given mole fractions took $E_k$ into account and therefore I would get a whole number after dividing $n_{th}$ by 0.7.</p>
<p>Is this just an error in my VLE Data and actually I would end up with n = 14? 13+1 to be accurate. It's hard to use the given diagramm since it's very small and I couldn't draw accurate enough in it. Also on what stage is the feed? I wouldn't get a whole number the either.</p>
| |thermodynamics| | <p>Well a stage is an equilibrium stage ie. given some feed composition z, temperature and pressure, vapour of composition y goes up and liquid composition x goes down. </p>
<p>Whether you are calculating actual or theoretical stages, you will hardly ever get a whole number. If you look at the graph, last stage (where xb = 0.15), the 5/7 stage probably comes from the (Stage N-1 composition - xb)/(Stage N-1 composition - Stage N composition). You can verify this yourself. Looks about right from eyeballing it.</p>
<p>Another way to see it is if your column had 10 theoretical stages, then xb will be around 0.12 (from the graph). </p>
<p>Ek just accounts for the fact that in an actual column the stages will not be at equilibrium. Meaning on each stage n y will be less than what is predicted on the VLE curve. </p>
<p>To conclude, you didn't do anything wrong. If this were an actual design and you come up with 13.87 actual stages required you would then select 14 stages. Then if you really want to meet xb=0.15 you can do that by playing with the bottom or top temperature.</p>
| 5480 | McCabe Thiele - number of stages |
2015-09-21T11:22:14.297 | <p>A journal I am reading about circular hydraulic jumps provides the equation and I quote:</p>
<blockquote>
<p><span class="math-container">$H$</span> Depth after hydraulic jump, <span class="math-container">$h$</span> depth before hydraulic jump, <span class="math-container">$V$</span> velocity after hydraulic jump, <span class="math-container">$v$</span> velocity before hydraulic jump, <span class="math-container">$R$</span> radius of hydraulic jump.</p>
<p>The condition for conservation of momentum can be written as
<span class="math-container">$$\frac{dp}{dt} = 2\pi R\rho HV^2 - 2\pi R\rho hv^2 = F_1-F_2$$</span>
where
<span class="math-container">$$F_1 = 2\pi R \rho g \int_0^hx\cdot dx = \rho g\pi Rh^2$$</span>
<span class="math-container">$$F_2 = 2\pi R \rho g \int_0^Hx\cdot dx = \rho g\pi RH^2$$</span></p>
</blockquote>
<p>I understand that units for <span class="math-container">$F$</span> and <span class="math-container">$\frac{dp}{dt}$</span> are both <span class="math-container">$kg\ m\ s^{-2}$</span>, and that they are equivalent.</p>
<p>But I don't know how, in this example, <span class="math-container">$F_1$</span> and <span class="math-container">$F_2$</span> are derived i.e. where <span class="math-container">$g$</span> and the integrals come from. Could somebody please explain?</p>
<p>Page 4 from: <a href="http://adatbank.transindex.ro/vendeg/htmlk/pdf5017.pdf" rel="nofollow noreferrer">On the circular hydraulic jump</a></p>
| |mechanical-engineering|fluid-mechanics|mathematics| | <p>$F=\frac{dp}{dt}$ is, in some sense, a more fundamental expression of Newton's law than $F=ma$ because $F=ma$ doesn't allow for situations with changing mass. You can easily derive the second from from the former by using $p=mv$ which gives $F=m\frac{dv}{dt}=ma$. </p>
<p>Your question seems to be more about how to derive the expressions for the force at the boundary of the hydraulic jump. To do this, think about the hydraulic jump as a wall; on one side of the wall the height of the water is $h$ and on the other it is $H$. </p>
<p>Now consider an infinitesimal square in this wall at depth $x$; the pressure exerted on this square is given by
$$
P=\rho\ g\ x.
$$
The <em>force</em> exerted on the square is calculated by multiplying by the surface area of the infinitesimal patch
$$
F_{inf}=\rho\ g\ x\ dx\ dy.
$$
Now you just have to sum up (with integration) all of the patches on the wall. The y direction is easy
$$
\begin{align}
F_{tot}&=\rho\ g\int_{0}^{h}\left(\int_0^{2\pi R}dy\right) xdx\\
&=\rho\ g\ 2\pi R\int_0^{h}x dx\\
&=\pi R\ \rho\ g\ h^2
\end{align}
$$</p>
<p><a href="https://i.stack.imgur.com/nAAyS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nAAyS.png" alt="enter image description here"></a></p>
| 5490 | Equating $\frac{dp}{dt}$ and $F$ at a hydraulic jump? |
2015-09-22T12:05:07.420 | <p>The following scenario:</p>
<p>I am tasked to design a 6.6m high reinforced concrete wall in an industrial building. We have a 10 ton Telehandler (telescopic loader) with 7m maximum reach, loading 1500kg loose material in a scoop.</p>
<p>I need to cater for a "runaway vehicle collision" where:</p>
<p>the vehicle reaches a maximum velocity of 2m/s before it collides with the wall. </p>
<p>I need to find out the worst case collision force on my wall.</p>
<p>Usually I would employ the design methodology for vehicle barriers and parapets in car parks in Eurocode 1 (EN 1991-1-1). But in this case the point of contact is significantly higher than the centroid of the colliding vehicle. </p>
<p>I would assume that the front wheels of the loader leave the ground during the collision, thus there is some sort of energy dissipation there too.</p>
<p>I am pretty sure that the long boom just frightens me and is enough of a distraction to stop me from thinking straight here.</p>
<p>Any help would be greatly appreciated.</p>
<p><a href="https://i.stack.imgur.com/rk8ID.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rk8ID.png" alt="enter image description here"></a></p>
| |civil-engineering|reinforced-concrete| | <p>In a situation like this you will need to resort to energy methods. Depending on how complicated you wanted to get you could do a dynamic calculation also.</p>
<p>I would recommend looking at the guidance in Eurocode 1991-1-7 (Accidental Actions). In particular look at Annex C which describes dynamic design for impact. This essentially involves equating the kinetic energy of the object impacting the wall to the energy absorbed by deformation of the wall. There are a few simple formulas which can be used depending on how you classify the impact.</p>
<p>You will need to consider a variety of loading cases for example: </p>
<ul>
<li>Load at various heights - separate (or possibly simultaneous) impacts
of the loader and load. </li>
<li>Load and loader all impacting as one 'lump'</li>
<li>etc</li>
</ul>
<p>From my experience with impact calculations it is unusual for a structure to be able to withstand any significant impact through elastic deformations alone. In which case the structure will obviously be damaged. In this case you have a few options:</p>
<ul>
<li>Protect the structure from being hit in the first place</li>
<li>Perform a plastic analysis and provide appropriate detailing so that the resulting plastic configuration can be achieved</li>
<li>Allow the structure to be damaged in some controlled way (imagine a fuse or some kind of energy dissipating system)</li>
<li>Accept that whatever is being hit will not exist after the impact and deal with the consequences (structural integrity)</li>
</ul>
<p>In my view it is prudent to consider the consequences of your constraints being exceeded:</p>
<ul>
<li>What if someone is driving too fast? </li>
<li>What if someone is carrying a load which is too heavy?</li>
</ul>
<p>To address your comment below. Clause 4.4 in EN 1991-1-7 states for accidental actions caused by forklift trucks:</p>
<blockquote>
<p>The National Annex may give the value of the equivalent static design
force F. <strong>It is recommended that the value of F is determined according
to advanced impact design for soft impact in accordance with C.2.2</strong>.
Alternatively, it is recommended that F may be taken as 5 W, where W
is the sum of the net weight and hoisting load of a loaded truck (see
EN 1991-1,1 , Table 6.5), applied at a height of 0,75 m above floor
level. However, higher or lower values may be more appropriate in some
cases.</p>
</blockquote>
<p>So one way of approaching the problem would be to apply an equivalent single degree-of-freedom analysis: First obtain the maximum dynamic interaction force from Equation C.1 (substituting the stiffness of your wall for k as specified in clause C.2.2(1) ):
$$ F = v_r \sqrt{k m} $$
Where $v_r$ is the velocity at impact, k is the wall stiffness, and m is the mass of the impacting object.</p>
<p>The time duration of loading can be obtained from equation C.2:
$$ \Delta t = \sqrt{\frac{m}{k}} $$</p>
<p>The static deflection can then be calculated by:
$$ \frac{F}{k} $$
A dynamic load factor assuming a rectangular pulse load (see Fig C.1) can be calculated according to [1]:
$$ DLF = 1 - \cos{2\pi\frac{\Delta t}{T}} $$
Where T is the natural frequency of vibration of the wall. The dynamic displacement is then:
$$ \delta = DLF \frac{F}{k} $$
You will immediately notice that:
$$ 0 \leq DLF \leq 2$$</p>
<p>You can then obtain the desired section forces/moments by using the dynamic displacement in the standard deflection formulas.</p>
<p>The value of the wall stiffness k will depend on the impact location, construction of the wall, support conditions, if the wall remains elastic (if it is not elastic than the above analysis is no longer valid), etc. </p>
<p>The book cited below is quite good for explaining how to perform an equivalent single-degree-of-freedom analysis. </p>
<p>[1]: Biggs, John M. "Introduction to structural dynamics." (1964). (One of the best introductory books on structural dynamics in my opinion - despite its age)</p>
| 5507 | Telehandler wall collision forces |
2015-09-23T03:59:12.370 | <p>I have a very basic understanding of how gears work, and while trying to learn more, I've run into a small point of confusion.</p>
<p>If a 2:1 gear ratio is achieved by having twice as many teeth on one gear than the other, then how are 250:1 ratios achieved? Certainly there aren't gears with 2500 teeth on one gear, and 10 on the other...</p>
<p>For example, here's a very tiny motor with a 250:1 gear ratio: <a href="http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_253534_-1" rel="noreferrer">DC Motor with Gearhead</a> </p>
| |mechanical-engineering|gears| | <p>Planetary Gear Sets can provide a high reduction in a small space. Also, they can be connected in a similar fashion to gear train to further increase the reduction. Also, one benefit is the output shaft is co-axial with the input shaft.</p>
| 5514 | How are high gear ratios achieved in a small space? |
2015-09-23T12:04:21.123 | <p>What would the ideal thickness of the dump trailer bed be, so it can withstand loading rock from a height of 1 meter?</p>
<p>The spacing of ribs is yet to be decided.</p>
<p>The goal is that the trailer be as light as possible, whilst being strong enough to withstand hauling rock every once in a while.</p>
| |mechanical-engineering|metals| | <p>See <a href="https://engineering.stackexchange.com/questions/5494/steel-selection-for-building-a-trailer">this similar-ish question</a>. The ideal thickness will be the thickness >= the thickness required to keep stresses below the yield strength of "the metal." </p>
<ol>
<li>Pick a metal, find its yield strength. </li>
<li>Pick a factor of safety. Use this and the yield strength to calculate an allowable stress. </li>
<li>Since you're dropping a load onto the trailer, find out spring/damping constants for the suspension as the suspension will affect peak loading.</li>
<li>Pick a rib spacing. </li>
<li>Draw a <a href="https://en.wikipedia.org/wiki/Shear_and_moment_diagram" rel="nofollow noreferrer">bending moment diagram</a>, find the peak moment, then use that and your allowable stress to find your thickness.</li>
</ol>
<p>Note that your scenario will be harder to design for (you should have a large factor of safety) because the rocks will not deform to create a true distributed load. Instead you'll have corners concentrating load in a (relatively) few number of places. </p>
| 5518 | What is the ideal metal sheet thickness used in a 1 tonne atv/tractor dump trailer? |
2015-09-23T13:11:17.030 | <p>I'm currently designing a small reinforced concrete bridge on private property. The project is in Scotland. </p>
<p>It's very small scale, a single lane, with a span of around 4.5m. Service life is taken as 120 years.</p>
<p>Using Eurocode design for standard road bridges seems far too over-the-top (there will not be any lorries crossing this bridge for example), so I'm looking for a good load design precedent (with backup documentation). The largest vehicle that may foreseeably cross the bridge would be, say, a 30 tonne concrete mixer. </p>
<p>If there's a good code that relates to, say, loading on temporary works or construction traffic loading then we feel that would be perfect for use on this bridge. Lest we over-engineer the thing and it costs a fortune... </p>
| |civil-engineering|structural-engineering| | <p>I'm going to take a slightly unconventional approach, and suggest you design to an <em>assessment</em> standard; namely BD21/01<a href="http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol3/section4/bd2101.pdf"> [pdf link]</a>.</p>
<p>My reasons are:</p>
<ol>
<li>It's a UK standard, and you're in the UK</li>
<li>It provides loading details for vehicle classes which are lighter than the full 40/44 tonnes.</li>
</ol>
<hr>
<p><em>The rest of this answer probably won't make much sense to those not familiar with BD21. I thought it useful to include this advice, but I don't think a full explanation of all terms and loading to BD21 is appropriate here.</em></p>
<p>I would suggest that seeing as it's a private bridge, the surfacing maintenance will be minimal, and such the design should be for <strong>Poor</strong> surfacing. Clearly the traffic flow is <strong>Low</strong>, so you're looking at Figure 5.4 for your K factor.</p>
<p>As an absolute minimum, I think your bridge needs to be sufficient for all fire engines (i.e. group 1 FE loading to BD21). Beyond that, you've suggested a concrete mixer truck might need to use it for access. A quick bit of googling and I've found <a href="http://www.hanson.co.uk/en/technical-information/truck-information">two</a> different <a href="http://www.smithsconcrete.co.uk/truck.htm">links</a> which suggest a concrete truck might be 26 tonnes or 32-33 tonnes. The difference between 26 tonnes and 40 tonnes loading for a 5m span is a K factor of 0.85 vs 0.87. Given the small difference, and that you wanted to allow the larger concrete trucks on your bridge, I think you should design for the 40 tonnes loading.</p>
| 5523 | Looking for a loading code for use in the design of a small scale private bridge? (4.5m span, one lane) |
2015-09-23T17:32:43.037 | <p>According to <a href="http://rads.stackoverflow.com/amzn/click/0962373109">Hoffman</a>, the standard $k-\epsilon$ turbulence model incorporates a two-layer approach (inner region and outer region formulations to reperesent mixing length) when it comes to regions adjacent to wall surface $(y^+ \approx 30 - 50)$. </p>
<p>My understanding of the $k-\epsilon$ model is that a two additional equations are added to the system of flow governing equations, one equation for kinetic turbulence energy $k$ and the other is for the turbulence energy dissipation rate $\epsilon$, and the closure problem is solved by using dimensionless groups as boundary conditions such as turbulence intensity, turbulence length scale and hydraulic diameter.</p>
<p>My question is why the $k-\epsilon$ model uses zero equations models (algebraic models) for regions adjacent to the wall, what advantages do zero equations models have over $k-\epsilon$ if by definition zero equations models assume that the rate of production of turbulence and the rate of dissipation are approximately equal and they do not include the convection of turbulence?</p>
| |mechanical-engineering|fluid-mechanics|cfd|turbulence| | <p>There are two ways of "<em>wall treatment</em>" both feature two layers but at different $y^+$. The general idea is that the velocity gradients in the <a href="https://en.wikipedia.org/wiki/Boundary_layer" rel="nofollow noreferrer">boundary layer</a> are so high that one would need a very high number of grid cells in order to resolve those gradients. In order to overcome that the flow close to the wall is modeled by one of the following algebraic models (see <a href="http://www.cfd-online.com/Wiki/Two_equation_turbulence_models#Near-wall_treatments" rel="nofollow noreferrer">cfd-online</a>:</p>
<ol>
<li>Low Reynolds number treatment (LRN)</li>
<li>High Reynolds number treatment (HRN)</li>
</ol>
<p>Both <em>treatments</em> are based on the <a href="http://www.cfd-online.com/Wiki/Law_of_the_wall" rel="nofollow noreferrer"><strong>Law of the Wall</strong></a>. Numerous experiments have shown that the boundary layer of a flat plate consists of two distinct regions. From the wall until $y^+ \approx 4$ it is called viscous sublayer with a linear velocity gradient and a logarithmic region from $y^+ \approx 30$. Both layers are joint by a region called buffer-layer.</p>
<p>If one chooses a LRN approach the numerical grid needs to fully resolve the boundary layer in the log-region with (rule of thumb) the first grid-cell at $y^+ =1$. The viscous sub-layer is then modeled algebraically.</p>
<p>In case the HRN approach is chosen the first grid-cell should be in the log-region. Both layers viscous sublayer and log-layer are than modeled algebraically.</p>
<p><a href="https://i.stack.imgur.com/SXibb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SXibb.png" alt="enter image description here"></a><br/> <em>from Turbulence and Transport Phenomena by <a href="https://www.researchgate.net/profile/Kemal_Hanjalic" rel="nofollow noreferrer">Hanjalic</a> and <a href="https://en.wikipedia.org/wiki/Law_of_the_wall" rel="nofollow noreferrer">Wikipedia</a></em></p>
<p>The answer to your question has three parts:</p>
<ol>
<li><p>Due to the high velocity gradients in the boundary layer it is beneficial with respect to the computational costs to model the flow close to the wall.</p></li>
<li><p>The modeling of the boundary layer makes the simulations more robust.</p></li>
<li><p>The error introduced into the solution by algebraically modeling the boundary layer is generally small (obviously depending on the simulation itself). Your mentioned neglect of the convective transport of turbulent properties is correct (Which in certain cases will introduce an error).</p></li>
</ol>
| 5528 | Why does the standard $k-\epsilon$ model use zero equation models? |
2015-09-23T21:22:32.013 | <p>I would like to fold thin metal sheets to a particular angle. </p>
<p>For example: fold a 0.1mm thick steel strip of width 20mm and length 150mm to an angle of 45° along its length (10mm panels on each side of the fold).</p>
<p>I would like to keep the radius of the fold as small as possible, ideally on the order of the thickness or smaller. I would also like to do this on a bench top. I plan to use a range of thicknesses, up to 1mm thickness. But I am flexible on this constraint (but need to use thicknesses <1mm).</p>
<p>Are there any tools/machines which can do this in a repeatable way (ideally with 1-2° variation or less)? Are there any standards for this type of forming?</p>
| |mechanical-engineering|metal-folding| | <blockquote>
<p>I would like to fold thin metal sheets to a particular angle. </p>
</blockquote>
<p>It sounds like you're looking for a <a href="https://www.northerntool.com/shop/tools/product_7825_7825" rel="nofollow noreferrer">sheet metal brake</a>.</p>
<p><a href="https://i.stack.imgur.com/B2zem.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B2zem.png" alt="enter image description here"></a></p>
<p>I'm sure you can find less expensive models elsewhere. Lots of places rent them too.
Heck, with the thicknesses you're using, you can make one with some hardwood and hinges.</p>
| 5531 | Folding of thin gauge metal sheets |
2015-09-24T15:10:36.517 | <p>I am intending to construct a small pressure vessel. It needs to be around 7 centimeters in length and width and around 1-2 centimeters in depth. It must be capable of being pressurised to 300 psi with sufficient space to contain a small amount of liquid (say 5cc) </p>
<p><strong>Constraints</strong></p>
<ul>
<li><p>The body of the vessel must be made out of nickel-plated annealed brass.</p></li>
<li><p>The vessel will have a brass cap, held in place with silver solder. </p></li>
<li><p>The cap must be sufficiently thin to be pierced when it is inserted into a receptacle. </p></li>
</ul>
<p><strong>Is this design feasible?</strong></p>
| |mechanical-engineering|pressure|pressure-vessel| | <p>Sure. <a href="http://rads.stackoverflow.com/amzn/click/B0015V45F4" rel="nofollow">CO2 cartridges</a> are made of <a href="https://www.reddit.com/r/Blacksmith/comments/119rd7/co2_cartridges_as_a_metal_source/" rel="nofollow">steel</a> and hold <a href="https://www.physicsforums.com/threads/12g-co2-cartridge-energy-in-j-please-help.686293/" rel="nofollow">850 psi</a>. You want to use brass, which has <a href="https://en.wikipedia.org/wiki/Ultimate_tensile_strength" rel="nofollow">80% of the yield strength of steel</a> (200MPa vs 250MPa) to hold 35% of that pressure. Not to mention the fact that you specify a 7x7x"1-2"cm vessel, which is 49-98cc, to hold 5cc of liquid, so you're left with up to 93cc of brass to use as the wall thickness, where again, the same dimensions as a CO2 cartridge should suffice due to the difference between yield strength and desired pressure limits. The use of silver solder won't matter because you can use whatever surface area you need to provide enough contact area to provide a solid joint - see the comment above about having plenty of excess volume. </p>
| 5539 | Is this pressure vessel feasible? |
2015-09-25T14:34:11.360 | <p>We have some metal parts in our machines, which are anodised aluminium. Those machines are due to be shipped to the customer in the near future, but I have noticed some pitting or slight rust developing on some of the anodised aluminium surfaces. I have tried cleaning it with 70% IMS, but to no avail (the surface is clean, there is no dirt). This doesn't really affect the proper function of the machine, but for something that is supposed to be brand new and relatively expensive, I would like it to look better. </p>
<p>Here are a couple of photos to illustrate what I mean:</p>
<p><a href="https://i.stack.imgur.com/LAufj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LAufjt.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/jiCME.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jiCMEt.jpg" alt="enter image description here"></a><br>
<sub><em>Click for larger images</em></sub></p>
<p>Can anybody suggest a cleaning method for removing slight surface pitting/corrosion on anodized aluminum parts?</p>
| |materials|chemical-engineering|surface-preparation| | <p>Chuck is right that there could be some galvanic action contributing to your issue. One other possibility is that your aluminum has been contaminated with steel particles at some point. Your second picture does seem to clearly show rust, which wouldn't be happening if the metal was all aluminum. It's less clear what the issue is in the first picture, but it could be smaller pieces of foreign material that interfered with the anodizing coating.</p>
<p>If this is the case, there are a few possible sources. The most likely is that the tooling used for the machining of your parts had been used on steel before. Since aluminum (at least the common alloys and tempers) is a very soft metal, it is easy to embed particles of other metals in the surface. For this reason, many shops will segregate cutters, grinding wheels, files, etc for ferrous and non-ferrous metals. Selection of the material used on the cutting tool, and speeds & feeds also have an impact on the risk of entrapping foreign material. Other possible sources are defects in the aluminum stock, contamination in the anodizing baths, or impact with a steel object.</p>
<p>Based on your picture, I would guess that steel was embedded while the part was machined, and then the moisture in the anodizing process accelerated the rusting. This is problematic not just cosmetically, but also bad for the coating because as the rust progresses, it will expand and separate the anodizing from the base aluminum. The best way to solve this problem is probably to use dedicated tooling for these parts that is only used on non-ferrous materials, so there is minimal risk of getting steel into the aluminum in the first place. This segregation is especially important where aluminum is to be welded, because a small amount of steel in the weld pool can contaminate the weld and cause excessive porosity.</p>
<p>Assuming this is the problem, cleaning it after the fact is fairly difficult. A chemical process won't work without damaging the whole surface, you have to actually remove the offending material or it will rust again. If it's loose enough and the anodizing layer is thin, you may be able to scrape it off with a knife or a needle file. Depending on the function of your part, you might be able to mill or face the surface down with clean tooling and re-anodize. One important thing is to take a big enough cut that you don't get any 'tearing' in the surface. Too thin a cut will tend to rub and push the material in deeper. If that's not possible, the next best thing would probably be to dig out that specific area by drilling or with a clean abrasive. Again, depending on the function of your part, you may then be stuck welding the spot full, re-surfacing, and re-anodizing.</p>
<p>In the end, you may decide that if the issue is only cosmetic, it's better to live with the stain than the cosmetic impact of a repair. Whatever you do, make sure you remove all of the steel, otherwise the rust stain will likely re-appear soon.</p>
| 5553 | Cleaning anodised aluminium |
2015-09-26T22:26:33.333 | <p>Say, in a firearm, after giving off a shot, if the barrel were to be closed off, does there exist a way to store some of it in an external pressure vessel, without requiring a mainly-external source of power and complex machinery?</p>
<p>Forgive the possibly stupid question, I'm a bit of a layman as far as engineering goes.</p>
| |pressure|combustion| | <p>Absolutely. There is a device called a bore evacuator which has been in service for a long time on large caliber guns. Its purpose is to draw combustion gas out through the muzzle after the shot cycle is over, so that said gas (which is usually toxic) doesn't flow backwards out the breech and into the crew compartment (assuming the gun is mounted on a vehicle like a tank or self-propelled howitzer). The picture below shows a bore evacuator on on an Abrams tank. It is the bulge in the middle of the main gun barrel with the words "99 Problems" faintly visible on it.</p>
<p><a href="https://i.stack.imgur.com/Ce71G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ce71G.jpg" alt="enter image description here"></a></p>
<p>The way it works is simple; after the bullet passes the bore evacuator, the high-pressure combustion gas behind the bullet is allowed to flow into the bore evacuator through small holes. After the bullet exits the gun, the pressure in the tube begins to drop, eventually going below the pressure in the evacuator. The gas then begins to flow out of the evacuator into the tube through holes that are slightly angled towards the muzzle. The resulting discharge induces gas flow towards the muzzle and creates a slight vacuum at the breech. When the breech is opened, hardly any combustion gas escapes into the crew compartment due to this forward flow.</p>
<p>The bore evacuator stores the gas for only a few milliseconds, but it should not be difficult to store it longer (a few seconds or minutes) by using check valves. In fact some bore evacuators in service do use check valves at the inlet to increase their effectiveness. I don't know how much gas you need to store, but you may not need to close off the barrel completely; you may be able to harvest enough gas just during the normal shot cycle, before the bullet exits the gun.</p>
<p>There is also a German recoilless rifle called the "Armbrust":</p>
<p><a href="https://i.stack.imgur.com/L5rAd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L5rAd.jpg" alt="enter image description here"></a></p>
<p>It uses captive pistons to completely hold the propellant combustion gas inside the weapon after firing. It's a one-use weapon though, which may not work for what you're doing.</p>
| 5560 | Is it possible to store leftover pressure from burning a propellant in a pressure vessel? |
2015-09-28T08:10:40.533 | <p>I've learned in school that pressure in water changes like</p>
<p>$$p(h) = \rho g h$$</p>
<p>where $h$ is depth in meters, $\rho$ is density (e.g. 1000 $\frac{\text{kg}}{\text{m}^3}$ for water) and $g$ is gravitation acceleration ($\approx 9.81 \frac{\text{m}}{\text{s}^2}$) and $p$ is the pressure in Pascal.</p>
<p>I guess there is no similar law for pressure in earth as it is to different, depending on where you are. But is there a rule of thumb? What do engineers who build tunnels / underground stations do?</p>
| |civil-engineering|pressure| | <p>In simple terms earth pressure is both very similar and very different. </p>
<p>The vertical earth pressure is given by:
Density x height x gravity.
Here the density depends on the material, which varies with type of soil.</p>
<p>The horizontal earth pressure is where it diverges from the simple water model. The percentage of the vertical force applied horizontally depends on the ability of the soil to support and transfer load.
Usually this is a simple coefficient for granular material (around 0.5) and for cohesive takes account of the shear strength.</p>
<p>There are theories, such as silo theory, that reduce the volume of soil acting on an point base on failure planes.</p>
| 5566 | How does pressure change with depth in earth? |
2015-09-28T13:58:53.050 | <p>The recent <a href="http://www.nature.com/news/the-science-behind-the-volkswagen-emissions-scandal-1.18426?WT.mc_id=SFB_NNEWS_1508_RHBox">Volkswagen pollution test cheating scandal</a> has shocked many people with its widespread extent as well as how it was hidden for so many years. </p>
<blockquote>
<p>Last year, the International Council on Clean Transportation (ICCT) in
Washington DC contracted scientists from the Center for Alternative
Fuels Engines and Emissions at West Virginia University in Morgantown
to test emissions from three light-duty diesel vehicles under
more-realistic conditions than are possible in the lab. To do so, the
scientists fitted cars with a portable emissions measurement system to
gather a continuous stream of data over a variety of US road types.</p>
<p>The tests found that the levels of NOx emitted by a Volkswagen Jetta
were 15–35 times greater than dictated by the US standard (31
milligrams per kilometre), depending on road and driving conditions.</p>
</blockquote>
<p>How was this cheating program carried out, and why were the lab tests unable to detect the presence of this cheating program? </p>
| |automotive-engineering|product-testing|air-quality|emissions| | <p>Details have emerged in the German press that indicate this was known about in 2007 (<a href="http://blog.caranddriver.com/report-bosch-warned-vw-about-diesel-emissions-cheating-in-2007/" rel="nofollow">from German newspaper Bild am Sonntag, as translated by Automotive News</a>):</p>
<blockquote>
<p>According to the report, Bosch claims to have supplied diesel
engine-management software to Volkswagen under the impression that it
would be used only in vehicle testing. That software, which was able
to activate emissions-control devices when a testing environment was
detected and deactivate them during normal driving, somehow ended up
in production vehicles. According to Bild am Sonntag, Bosch wrote to
Volkswagen in 2007 warning the automaker that using this software in
publicly sold vehicles was illegal.</p>
<p>...</p>
<p>In a statement released last week, Bosch revealed that it supplied
common-rail fuel-injection systems, as well as supply and dosing
modules for exhaust-gas treatment, on the Volkswagen and Audi models
at the center of the growing emissions-cheating scandal. “As is usual
in the automotive supply industry, Bosch supplies these components to
the automaker’s specifications,” the statement reads. “How these
components are calibrated and integrated into complete vehicle systems
is the responsibility of each automaker.”</p>
<p>...</p>
<p>Automotive News reports that the crisis began when
then–VW brand chief Wolfgang Bernhard and engineer Rudolf Krebs began
developing a new diesel engine for the U.S. market in 2005. Bernhard
and Krebs realized that an AdBlue urea exhaust-treatment system would
be needed to meet U.S. emissions standards, at an estimated cost of
$335 per vehicle. Reportedly, VW finance heads determined this cost
was too high, as a company-wide cost-cutting exercise was then
underway.</p>
</blockquote>
<p>I understand most, if not all, of these vehicles have an electronic anti-skid braking system and/or an electronic stability control for roll, pitch and yaw. As the tests are done on a "rolling road," which only uses the driven axle, the sensors for these must be disabled, as the undriven wheels are stationary.</p>
| 5570 | How was Volkswagen able to trick the lab pollution test? |
2015-09-28T19:33:40.583 | <p>Take a hollow aluminium cylinder with outer radius $r$ and length $h$, capped with two circular endcaps. How thick does the aluminium have to be, that is what is the inner diameter of the cylinder, to withstand 1 atm on the outside and 0 mbar on the inside, without crumpling?</p>
| |materials|metals|geometry|vacuum|buckling| | <p>I found this formula <a href="http://www.eng-tips.com/viewthread.cfm?qid=91966" rel="nofollow">here</a>:</p>
<p>$$p_{crit}=\frac{2\,E\,t}{D}\left(\frac1{(n^2-1)\left(1+\left(\frac{2\,n\,L}{\pi\,D}\right)^2\right)^2}+\frac{t^2}{3(1-\nu^2)D^2}\left(n^2-1+\frac{2\,n^2-1-\nu}{\left(\frac{2\,n\,L}{\pi\,D}\right)^2-1}\right)\right)$$</p>
<p>So you should set $p_{crit}$ to about 3 atm to be safe, then find $E$ and $\nu$ for your aluminum, plug those in along with your diameter, and length. Then plug in a few integer values of $n>1$ to find a thickness $t$ that is strong enough for all values of $n$.</p>
| 5576 | How thick does an aluminium vacuum cylinder have to be? |
2015-09-29T03:05:37.480 | <p>I´m developing a tool to generate an <a href="http://www.openfoam.com/" rel="nofollow noreferrer">openFoam</a> wind tunnel case of a wind turbine. I already have the blade geometry constructed, but I'm really new to CFD and parametric meshing. I found some articles where they mention a C topology to mesh the airfoil and also some software (<a href="http://bladedesigner.sourceforge.net/" rel="nofollow noreferrer">bladedesigner</a>) that mentions a topology named O10H, but none of these has a reference about how to implement these topologies and if there are others. They also do not specify what the best use cases for each one. </p>
<p>Can someone help me to get started making an efficient mesh for my tool? </p>
<p><strong>EDIT:</strong> At the end I want to have a domain like the picture below, with other blade orientation, but first I need to define the mesh topology around the blade.</p>
<p><a href="https://i.stack.imgur.com/Dv73A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dv73A.jpg" alt="enter image description here"></a></p>
<p><strong>My example Blade:</strong></p>
<p><a href="https://i.stack.imgur.com/0kNCh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0kNCh.png" alt="enter image description here"></a></p>
<p>Some of the mesh topologies that I found:</p>
<p><strong>Bladedesigner O10H topology</strong></p>
<p><a href="https://i.stack.imgur.com/B37LM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B37LMm.png" alt="Bladedesigner O10H"></a></p>
<p><strong>Various C topologies</strong></p>
<p><a href="https://i.stack.imgur.com/RW352.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RW352m.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/el5Mr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/el5Mr.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Xwv8o.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xwv8o.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/dFhvz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dFhvz.jpg" alt="enter image description here"></a></p>
| |airflow|cfd|wind-power|meshing|wind-tunnels| | <p>I would recommend using third party software for meshing like gmsh, gambit or fluent where you can import your geometry, edit it, define the topology and get the mesh.
Gmesh is free while the rest are not.
You might be able to find many tutorials on meshing in these softwares on youtube</p>
| 5583 | Efficient 3D mesh topology for CFD simulation of turbine blade |
2015-09-29T13:25:58.347 | <p>My understanding so far is that a jump occurs when the Froude Number $\left(Fr=\frac{v}{\sqrt{2gh}}\right)$ of the radial flow falls with a decreasing velocity due to friction and viscosity; and the jump occurs at the critical point where $Fr=1$. However, I also learnt that the height of the jump is dependant on the difference in $Fr$ before and after the jump which contradicts the former theory where there is no difference.</p>
<p>Could somebody please explain and help correct my understanding?</p>
<p>Sources</p>
<p><a href="http://thatsmaths.com/2014/01/09/white-holes-in-the-kitchen-sink/" rel="nofollow">http://thatsmaths.com/2014/01/09/white-holes-in-the-kitchen-sink/</a>
<a href="https://en.wikipedia.org/wiki/Hydraulic_jump#Tabular_summary_of_the_analytic_conclusions" rel="nofollow">https://en.wikipedia.org/wiki/Hydraulic_jump#Tabular_summary_of_the_analytic_conclusions</a></p>
| |mechanical-engineering|fluid-mechanics|hydraulics| | <p>I am researching on this topic actually. Hydraulic jump is not an easy matter as it seems to be. the location of the jump also depends on the thickness after the jump. if you increase the height after the jump(for example, put a bump). the location of the jump will move upstream. it depends on a lot of parameters, such as the upstream speed, the downstream configuraion, surface tension. etc. with all being said, the true reason behind this phenomenon is still unknown.</p>
| 5591 | What causes circular hydraulic jumps in kitchen sinks? |
2015-09-30T02:11:15.903 | <p>I am printing a wheel with a 2.5mm hole in the middle to accept a shaft of a motor with a set screw. The problem I am having is that the wheel is always wobbly when spinning (sometimes more, sometimes less without changing any parameters). Is there anything I can do to make this print as perfect as possible and prevent the wheel from wobbling?</p>
| |mechanical-engineering|3d-printing|prototyping| | <p>Also, the plastics used in most hobbyist printers are really not well suited to carrying any load and are usually too soft and flexible to machine properly. </p>
<p>However if you can find a piece of tube, either metal or rigid plastic, to fit the axle accurately, design the wheel bore so that the tube can be pressed in. It should be a snug fit. Screws, glue, etc can attach the axle to the tube.</p>
<p>When you say the wheel is "wobbly" do you mean that the bore is not aligned with the tire along the axle axis? Or is the bore offset from the centre somehow? Can you can share your design so that we can understand the problem more clearly?</p>
| 5601 | How do you make a 3D print with an accurately aligned bore hole? |
2015-09-30T10:04:35.117 | <p>Instead of using an alternator to charge the battery, and a starter to start the engine, why not just put one electric motor which is always connected with the engine? When we want to start the motor, we supply electricity to it; and when the engine is started, it spins the motor to generate electricity and charge the battery.</p>
| |mechanical-engineering|electrical-engineering|electric-vehicles| | <p>One problem with generating electrical energy from a car engine is that the RPMs are constantly changing, and consequently the power output of an electrical generator would vary with engine speed. An alternator is very good at responding to these variations; the current in the rotor can be quickly adjusted to change the power output at a given RPM. I don't know how you would do the same with an electric motor as a generator, but I imagine that it would involve rejecting some of the surplus energy as waste, which would be inefficient.</p>
<p>Another thing to consider is that starting an engine and using it to generate electrical energy each require controlling very different amounts of electrical power. When starting the engine, a high amount of electrical power is needed, but only for a few seconds. When using the engine to <em>generate</em> power, typically only a small amount of power is needed, but it has to be generated continuously, as long as the engine is running. This means that the starter motor has to be built to withstand much higher current than the alternator, and the alternator has to be built to operate much longer than the starter.</p>
<p>It all comes down to the fact that the starter and the generator each have different design requirements. You could use one device that does the job of both, but it might be something of a "jack of all trades, master of none". The best solution for the time being seems to be the use of a separate starter and generator. That may change as technology advances though.</p>
| 5603 | Why not use just one electric motor in a car? |
2015-10-01T11:48:03.677 | <p>I am in the process of helping my son design a sort of model of the <a href="http://www.bloodhoundssc.com/project/car" rel="nofollow noreferrer">Bloodhound SSC car</a> for his school project, using balsa wood, model aircraft wheels, etc... It doesn't have to a like of like, but vaguely resemblant so that you can tell what it is just by looking at it. It would also be nice if the car could move and I was thinking of using on this small CO2 gas canisters as a means of propulsion:</p>
<p><a href="https://i.stack.imgur.com/OVvrz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OVvrz.jpg" alt="enter image description here"></a></p>
<p>You can get some which are threaded, as above, or some which are just plain. What I haven't been able to find is a way to quickly release the pressure from the canister to propel the car. <strong>Are there any small on/off manual valves that would fit this type of gas canister?</strong> I haven't been able to find any, the only ones apparently in existence are for fitting onto bike tyres to (re)inflate them.</p>
<p>The only method I have come across is people using some sort of puncturing device (e.g. with a nail) that is generally sprung-loaded to pierce the gas canister when launching the model car. This puncturing device can be made part of the car, but more generally tends to be part of the "launching pad" as it adds weight and unnecessary complexity to the car. I would like to avoid this if I can.</p>
| |valves|propulsion|compressed-gases| | <p>In the end, I decided to use an <a href="http://www.estesrockets.com/rockets/engines/mini/1502-1-4a3-3t" rel="nofollow noreferrer">Estes 1/4A3-3T rocket engine</a> (the smallest one they do):</p>
<p><a href="https://i.stack.imgur.com/Xoy8X.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xoy8X.jpg" alt="enter image description here"></a></p>
<p>We'll see how it goes!!</p>
| 5621 | Valve for mini CO2 gas canister? |
2015-10-01T23:21:58.187 | <p>So, I must do the following conversion from my engineering textbook: </p>
<p>$0.155 \frac{m^2 * ^{\circ} C}{W}$ to $\frac{^{\circ}F * ft^2 * hr}{Btu}$</p>
<p>I can get as far as: </p>
<p>$0.155 \frac{m^2 * ^{\circ} C}{W} * (\frac{1 ft}{0.3048 m})^{2} * \frac{1 W}{3.4123 \frac{Btu}{h}} = .39108 \frac{ft^2 * ^{\circ}C *hr}{Btu}$, but I can't figure out how to do the final conversion properly, which involves converting that from </p>
<p>$.39108 \frac{ft^2 * ^{\circ}C *hr}{Btu}$ to $\frac{^{\circ}F * ft^2 * hr}{Btu}$. </p>
| |heat-transfer| | <p>I suspect you're converting a value of <a href="https://en.wikipedia.org/wiki/R-value_(insulation)" rel="nofollow">thermal resistance</a>, which means that the temperature units represent a temperature <em>difference</em>, rather than specific values of temperature. In this case, the constant 32 drops out of the conversion and you would just just multiply by $\frac{1.8^\circ F}{^\circ C}$. Also, I agree with idkfa that your math is wrong in the conversion of power and area. I get 0.489.</p>
| 5625 | Convert 1 one compound unit to another involving temperature |