Wireless power transmission & over unity - Page 1
Forum section: electronics / beginners

author: jsi
author_rank: Regular Contributor
timestamp: May 06, 2016, 04:05:19 pm
content: On another forum that I hang out in, the folks are all excited for the day when power will be transmitted wirelessly to our computing devices. But, I'm all like don't you people know about the inverse square law?! (Intensity = 1/r^2) So in an effort to make my case for why it will never work I put together a spreadsheet to show the decreasing power over distance. The case I was going to make is that for a widget that requires 5 watts of power it would take 100 watts at the transmitter if it was only 4 feet away. So far so good, but then I discovered over unity! If I can power 1 device why not 2? And if 2 work, why not 100? If 1 widget need 5 watts then 100 would need 500 watts. But, I'm only putting 100 watts into the system. Yeah free power!Clearly building a miniature Dyson sphere of widgets around the power source isn't going to give me infinite power so help me understand where I've gone wrong.

author: IanB
author_rank: Super Contributor
timestamp: May 06, 2016, 04:09:30 pm
content: If you put a sphere around the transmitter the total power hitting the sphere will be equal to the power radiated. So where would the extra power come from?

author: jsi
author_rank: Regular Contributor
timestamp: May 06, 2016, 04:53:28 pm
content: Quote from: IanB on May 06, 2016, 04:09:30 pmI don't know where it would come from and I am not suggesting that there is more power. But, for simplicity sake imagine a 5 watt widget at 4 feet from the 100 watt power transmitter and working fine. Now, if I put another widget 180 degrees away from the first at 4 feet it should work as well. Now put 2 more at 90 and 270 degrees. Shouldn't they work too? The next step is to add more evenly spaced widgets until you add the 21st widget (105 watts total) = over unity (NO! it doesn't, but why?)(Here's the practical application: The forum in question is for educators and teachers often keep their classroom computers in carts to charge and protect them from theft. Plugging 30 - 40 laptops into charge is a complete headache. The plugs get yanked and damage the computers or they don't get plugged in at all resulting in a dead battery. The carts are about 4 feet x 2 feet (hence the 4 foot distance in my thought experiment). Being able to wirelessly charge a cart of computers has a lot of appeal to the folks that support and use them.)

author: stmdude
author_rank: Frequent Contributor
timestamp: May 06, 2016, 05:05:16 pm
content: Ok, lets break it down at bit.Lets assume 100W power radiating, and we're measuring from 2 meters away, and the antenna has a perfect spherical radiation-pattern.Inverse square gives us: 1/2^2 = 25 watts.Now, those 25 watts are distributed evenly over a spherical area with a 2 meter radius. I.e, to capture the whole 25 watts, your receiving antenna would have to be a sphere with a radius of 2 meters.So, lets assume your antenna has an area of a more realistic 5 square centimeters. The area of the sphere is (A=4*pi*r^2) 50.27 square meters.5sqcm is 0,0005 sqm, so, the relative size is 0.0005/50.27, which is a whopping 0.001% of the area of the sphere.0.001% of 25 Watts is 0.00025 watts. Thats the amount of energy you could pick up with a 5 sqcm antenna. Double the area of the antenna, and you'd get twice the amount of power.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 06, 2016, 05:05:46 pm
content: Quote from: jsi on May 06, 2016, 04:53:28 pmWhy do you think your device would receive 5 watts of power from 4 feet away? Where is the math to back up that number? Once you see where that number comes from, you'll see why you won't ever get more than 100 watts (hint: the antenna size required to grab 5 watts of radiated power is large enough that you wouldn't be able to physically fit 21 of them around the power source, there wouldn't be room. The antenna size required to grab 5 watts of power from an omni-directional 100 watt transmitter at a distance of 4' would be exactly 1/20th the surface area of a sphere with a 4' radius.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 06, 2016, 05:28:10 pm
content: Quote from: stmdude on May 06, 2016, 05:05:16 pmThat's not how the inverse square law works, it's still 100 watts of total radiated power at any distance (conservation of power and all that...where would the other 75 watts go?). What changes with distance is the surface area of a sphere with X radius. When you move twice as far away, the surface area of the sphere grows by a factor of four, which means your antenna either needs to be four times larger or it will receive 1/4 the power.

author: stmdude
author_rank: Frequent Contributor
timestamp: May 06, 2016, 05:31:37 pm
content: Quote from: suicidaleggroll on May 06, 2016, 05:28:10 pmDoh, of course...So, basically most of my post holds up. The end result for the 5sqcm antenna would be 0.001% * 100W, also known as, 0.1W.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 06, 2016, 05:33:46 pm
content: Quote from: stmdude on May 06, 2016, 05:31:37 pmCorrect - except that .001% of 100 W is .001 W, not 0.1 WYou did have that part of the calculation correct in your original post though.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: May 06, 2016, 05:34:10 pm
content: Ah... a subject near and dear to my heart. What about the near field, where the inverse square law doesn't hold? Make the near field large enough with clever antennae and coupling....-- and you _still_ won't get more out than you put in. But it's a pretty interesting effect anyhow.

author: IanB
author_rank: Super Contributor
timestamp: May 06, 2016, 05:39:38 pm
content: Quote from: jsi on May 06, 2016, 04:53:28 pmAs others have said above, the antenna has to be big enough to capture the radiation. If you want to capture the whole 100 W your antenna has to be the whole enclosing sphere. If you want to capture 50 W your antenna has to be half the sphere. For 25 W a quarter of the sphere. And so on.If it makes it easier to visualize, consider light being radiated instead of RF power, and consider capturing the light with solar panels on the inside of the sphere.

author: helius
author_rank: Super Contributor
timestamp: May 06, 2016, 06:50:36 pm
content: Quote from: alsetalokin4017 on May 06, 2016, 05:34:10 pmThe assumption underlying the far field and the inverse square law, is that the behavior of the transmitting antenna is not affected by any receivers. So the VSWR and radiation-resistance of the antenna are immanent in the antenna design and position and independent of whatever is used for receivers.In the near field that assumption isn't useful, because significant two-way interaction exists between transmitter and receiver. This is used by all practical wireless power systems like WiTricity, Energous, etc. The VSWR of the transmitter will be different when receiver antennas are present.

author: RIS
author_rank: Regular Contributor
timestamp: May 06, 2016, 07:19:59 pm
content: what would be the theoretical distance if I have the source of 20MV and 1 AMP and I want to receive one watt

author: ocw
author_rank: Regular Contributor
timestamp: May 06, 2016, 07:43:48 pm
content: QuoteYou want to transmit 0.020 watts and receive one watt?Why doesn't everybody take advantage of the incredible power from that transmitter called the sun and put up an antenna called a solar panel?

author: RIS
author_rank: Regular Contributor
timestamp: May 06, 2016, 07:58:06 pm
content: I mean 20000000 volts and 1 amp Im just checking the possibilities.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 06, 2016, 08:32:05 pm
content: Quote from: RIS on May 06, 2016, 07:58:06 pmPrec = Ptrans * A / (4 * pi * R^2)Where:Prec is your received powerPtrans is your transmitted powerA is the cross sectional area of your receive antennaR is the distance between the transmit and receive antennasPrec and Ptrans should be in the same units, and A and R^2 should be in the same units.This is idealized of course, assuming perfectly matched antennas, 100% efficiency, R is sufficiently large that the receive antenna doesn't couple with the transmit antenna and throw off the VSWR, etc.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 06, 2016, 10:07:31 pm
content: If you're just assuming that the field remains constant and any arbitrary number of loads can draw power from that field, then sure, go for it. But something has to deliver all the power to that field in the first place...A transformer does this. When a constant voltage is applied to the primary, a constant flux is induced in the core. As long as the flux linkage is strong, the same flux flows through the secondary, no matter how many turns or windings are present. If you oppose the flux by putting some resistance on the secondaries, current flows. This current is reflected back to the primary, thus all power is accounted for.The same is true at radio frequencies. The coupling is much weaker, but the load of a receiver is nonetheless perceptible to the transmitter.Normally the receivers for a given radio station are very distant, so the change in transmitter power (for a receiver being present or not) is imperceptible. In the present case, this is not true, and the amount of power absorbed by each nearby receiver must be accounted for, either by weakening the field around each receiver (thus reducing the power-per-receiver drawn by tight groupings of them), or by attempting to draw more power from the transmitter itself (namely, changing its impedance; but with a typical fixed-impedance system, this can only reduce power delivered, so it works doubly against you).Tim

author: RIS
author_rank: Regular Contributor
timestamp: May 07, 2016, 12:20:58 pm
content: Tesla and free energy-to my knowledge he has not worked on the free energy,he simply wanted to make wireless transfer of energy regardless of the distance-During this research he encountered a lots of different problems but it is also solved it an unusual way ,such as low coupling transformer.Even I know that this is a bad idea and I am not an electrical engineer ,but with a bit of thinking and obvious assumptions that this transformer needs to hold a vast quantities of electrical power resulting in a huge magnetic field then loosely coupling is not such a bad idea plus standing waves which obviously is not useful for energy transfer but they may be useful for changing the spacetime or the environment thus creating a favorable environment for the transfer of energy therefore the whole idea may not be so crazy or impossible.so combining what is seemingly useless and what is proven from the classical science as useless can give interesting results which can only be proven if we build something like that.furtherDuring his research I do not know when, maybe before, maybe later of wireless energy transfer he might have found something ,something that does not belong to the classical calculations.Now I'll try to explain and I warn you I am not a specialist in this but I hope you will see the same anomalies.as I've written before about the car-input equally output minus losses so generally input and output are not that important and therefore we should pay more attention on the resistance whatever form was mechanical or electrical.so car has a certain resistance and every second we use a certain amount of energy to beat resistance and logical would be to reduce the resistance on the the lowest possible level right,what we did and lower can not beNow we leave the resistance of the environment as it is,Now we take The typical truck which is fifty times heavier than the usual cars so a bit of computation which says the mass, we have increased fifty times but energy consumption is increased only four times I mean even this is nothing unusual only few more axles.But what is unusual is increase of potential kinetic energy,I mean it is not that unusual this is what we can expectbut if I take that kinetic energy and try to cancel resistance Then it gets interesting with the increase of mass and velocity says it would be possible to completely cancel resistance and even go below that which is bloody very strange.generally speaking if we have a mechanism for the return of that energy more mass more speed and resistance is droping.Now the question is whether this rule applies to electricity?Then I remembered him maybe he's not so crazy because it speaks on mass very similar to what I noticed.generally if the system is bigger then it is better.17.40minWell you judge for yourself boys

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 07, 2016, 12:48:57 pm
content: Quote from: RIS on May 07, 2016, 12:20:58 pmThis again? Cars are not over-unity. They never have been and they never will be. That applies to trucks as well, even "heavy" ones.It is NEVER possible to completely "cancel" resistance and "even go below". And kinetic energy can't "cancel resistance" anyway, what does that even mean? That statement makes no logical sense. Again you are confusing ENERGY and POWER, just like you were doing in your previous "cars are over-unity" posts. Assuming you're driving on flat ground (ignoring changes in gravitational potential energy), the mass of the vehicle only matters when you're either converting the potential energy stored in the fuel to kinetic energy (accelerating) or converting kinetic energy to heat (decelerating with the brakes) or back to potential energy (regenerative braking). When the vehicle is at a steady speed, mass plays NO part in the calculations, at all. No matter how heavy the vehicle is, the power required to overcome rolling resistance and move air out of the way is exactly equal to the power required from the powertrain, which, after taking into account engine efficiency, is exactly equal to the rate of fuel consumption. So what if a truck that's 50x heavier than a car only uses 4x more power to maintain speed? That is a meaningless comparison, since mass has nothing at all to do with maintaining speed, it only matters when you're changing speed.

author: RIS
author_rank: Regular Contributor
timestamp: May 07, 2016, 01:49:17 pm
content: what would be more effectively transport 50 tons one ton at a time 50 times ,or fifty tons at once.and if we give equal limitation for both vehicles a thousand stops for regenerative braking to pull that potential kinetic energyat which vehicle you would pulled more energy out compared to the energy or fuel consumed.

author: TimFox
author_rank: Super Contributor
timestamp: May 07, 2016, 02:06:14 pm
content: The practical wireless power transmission with which I am familiar does not use radiation, but inductive coupling to a resonant circuit in the near field (where energy is stored, not radiated) of the transmitting coil or circuit.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 07, 2016, 02:23:46 pm
content: Quote from: RIS on May 07, 2016, 01:49:17 pmThat's logistical efficiency, which is a completely different topic.Quote from: RIS on May 07, 2016, 01:49:17 pmIt's impossible to say. It depends on the aerodynamics of each machine, the thermal efficiency of the engines, and the efficiency of the regenerative braking system. Regardless, neither vehicle would be anywhere close to approaching 100% efficiency.

author: RIS
author_rank: Regular Contributor
timestamp: May 07, 2016, 03:04:09 pm
content: Exactly not even close to 100%-This requires an increase in the overall mass and velocity, etc.therefore we have the numbers who simply show a very strange results.but if this behavior is to some extent accurate then all that transfer to the electrical system that is smaller and acceptable then who knows what's going on with this return EMF

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 07, 2016, 03:25:03 pm
content: Quote from: RIS on May 07, 2016, 03:04:09 pmWhat gives you the impression that increasing mass and/or velocity increases efficiency? By your own admission, large trucks use 4x MORE energy to move at the same speed vs smaller, lighter cars. The logistical efficiency is higher since they're able to move more freight in fewer trips, but that's irrelevant to discussions about energy efficiency and over-unity.Quote from: RIS on May 07, 2016, 03:04:09 pmWhat numbers? You haven't provided any numbers. And what strange results?

author: RIS
author_rank: Regular Contributor
timestamp: May 07, 2016, 03:50:03 pm
content: efficiency comes from somewhere that is from real numbers or physicstry to calculate yourself Take some mass speed etc.I do not claim overunity only unusual numbers If you're looking at several different directions

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 07, 2016, 07:13:32 pm
content: Well, this went from "beginner" to "numpty" pretty darn quick...Tim

author: jsi
author_rank: Regular Contributor
timestamp: May 07, 2016, 09:15:16 pm
content: Quote from: IanB on May 06, 2016, 05:39:38 pmBeautiful! Absolutely beautiful!I see the error in my thinking. There isn't x Wattage available at every point in the sphere which is what I was assuming. It seems simple now that I get it, but isn't that how it always is.So what are the chances that practical, affordable wireless power transmission across room size distances will happen? My gut feeling is that the physics of it will stop it from ever being anything more than a novelty.

author: IanB
author_rank: Super Contributor
timestamp: May 07, 2016, 10:04:52 pm
content: Quote from: jsi on May 07, 2016, 09:15:16 pmThe chances are very low indeed. To avoid wasting power in all directions you have to confine it to a beam. This beam has to pass through anything (like people) in its way, but be effectively captured by the receiver. It's very hard to do this. If the beam is microwaves it will cook any people in its path. If the beam is light (a laser), it will fry any people in its path. Basically, making a beam that is easily captured when desired and not captured when not desired is quite tricky.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 07, 2016, 11:10:26 pm
content: You're as well off using broad sources of radiation that are already available.Typical room lighting is on the order of several watts, visible range. It's quite reasonable to draw small amounts of power, microwatts to milliwatts perhaps, for CMOS devices. Indeed, optical wireless devices have been around for some decades: solar powered calculators for instance!Tim

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 08, 2016, 06:03:54 pm
content: Quote from: RIS on May 07, 2016, 03:50:03 pmThen provide some of those "real numbers" that back up what you're trying to say, because at this point I honestly have no idea what your point is.Quote from: RIS on May 07, 2016, 03:50:03 pmI have, many times. Again, what is your point? What are you trying to say?Quote from: RIS on May 07, 2016, 03:50:03 pmThere's nothing "unusual" about the numbers. If you don't understand something, then get over yourself and ask for clarification, but quit trying to imply that physics doesn't add up and everybody else is just as clueless as you are about how energy and power work. I'd rather not have a repeat of the whole over-unity bouncy ball argument you were trying to make in that other thread. Just stop with the vague buzzwords and provide an actual, concrete example, start to finish, of whatever "unusual numbers" you're talking about.

author: aurmer
author_rank: Regular Contributor
timestamp: May 14, 2016, 05:46:21 am
content: There is a lot of discussion (digression?) here!Simply put, if energy is radiated evenly in a circle, the amount of energy you can collect is proportional to your antenna's "tangential surface" relative to the origin of the circle.In other words, if your antenna spans a 15 degree arc of the circle, then the origin "sees" that your antenna (15/360 = 1/24 ~> 4%) occupies 4 percent of the 'space' that it is sending the energy into.Therefore, the theoretical maximum energy you can receive is 4% of the total power being radiated.If you keep squeezing more widgets in the field, their antenna will overlap each other and rob each other of the full 5W they need to run.