First year college CALCULUS and PHYSICS,  I can teach.   Send PM - Page 1
Forum section: electronics / beginners

author: RJSV
author_rank: Super Contributor
timestamp: December 01, 2023, 08:09:21 am
content: Hey there folks:Students and others involved with basic CALCULUS and PHYSICS are welcome to ask questions. I did quite well in first 2 years in college, including requested tutor one on one, or sometimes a small group of students wanted help in a casual study time meeting in library or school dorms.Topics include some basis / justification for Newton's math development and the kinetic physics he explored. Send an on-topic question, or even sometimes a 'stupid' or seemingly wide question; I'll try to respond!I've also done plenty of mechanical drawings, like old-school draftsmen, having good 3-D perception...(maybe not perfect).

author: liaifat85
author_rank: Regular Contributor
timestamp: December 01, 2023, 10:49:18 am
content: Hope you will find a student here. You can also post your resume in freelancing websites.

author: armandine2
author_rank: Frequent Contributor
timestamp: December 01, 2023, 11:33:01 am
content: the second problem inHagenRadio Frequency Electronics- stumped me

author: RJSV
author_rank: Super Contributor
timestamp: December 01, 2023, 06:01:35 pm
content: I'm only interested in the subject matter, so any EEVBLOG member can ask questions; not interested in paid tutoring.

author: RJSV
author_rank: Super Contributor
timestamp: December 01, 2023, 06:14:12 pm
content: Armendine2:Thanks for question, but that's more advanced engineering, with supporting math.Interesting question though: They give you the answer, then ask for how that is so.Maybe senior college level complex.My emphasis here is mainly about rates, derivatives, and integrals, with relation to gravity fields and static electric field basics. Something like Maxwell's equations I can only follow along.So, I'd call it basic 'Newtonian' math, essentially the first year or so, for any student starting college school in 2022, or sooner.Derivatives, and integrals, and motivation for inventing 'The Calculus', often termed as 'Analytic Geometry'.I'm considering just make a post just covering those aspects, and let folks read and comment if they have questions. For example, the static field equations dealing with forces are identical for GRAVITY or ELECTRICAL.

author: RJSV
author_rank: Super Contributor
timestamp: December 01, 2023, 11:50:37 pm
content: (having some difficulties, uploading pictures).Here is a sample, narrative on differentiation as a calculus function. Suppose your function is X to power of 14. Idea is to apply the differentiation on that equation.Just for sake of brevity let's just mention a rote action or series of memorized actions:First, take the exponent, and bring it down, to be a multiplier now. See diagram showing that, as rote action for now.Then, with the original exponent, subtract one, and put it back (as exponent of equation).Result is your answer, that is, 14X exp13, as shown at bottom of diagram.

author: RJSV
author_rank: Super Contributor
timestamp: December 02, 2023, 12:00:38 am
content: Now then, you maybe have only rote or mechanical feel for this so far. But be aware (I'd say) the functional pairing where a process of INTEGRATION is performed, by moving backwards up the diagram...plus another item.But basically the two functions are opposites paired by the fact you could go 'back and forth' between the two,...each one simply reversed the other, with the caveat that you will need to bring in an unknown CONSTANT, when integrating. That way, your X exp14 function will literally become (X exp 14) +CONSTANT which might be zero.There's some more detail, but it can be pointed out that INTEGRATION has a sort of higher amount of info in the result, and conversely doing a differentiation has a quality of losing or destroying info. Kind of like a status level, for each of the two converse functions.More, more, and more to follow, thanks

author: RJSV
author_rank: Super Contributor
timestamp: December 02, 2023, 12:48:06 am
content: (Sorry if you have to download the pictures ?...seems like the thumbnail pictures would gum up someone's system memory ?).So...that last diagram, and intro is an important aspect of how the so-called 'Integration Solution Tables' have been built up, by many folks, as some integration functions are not intuitive (in structure).But you can observe, and when asked, "What is the integral of 14 X exp13 ?" you can say:"Ahha, I've seen that..." and plus you (should) know to also add a constant term.The constant has to be solved, usually by some other route, try zero and see if your numbers work, in that situation. Like, perhaps you know that the answer, when X is 40 is 570, but your trial run is giving 550; then you know you need a constant of '20' to fully define your answer.Congratulations, you've just solved one of many Integration problems. Other mathematicians have applied this 'backwards process' to build up a so-called Table of Integrals. I think that might be in Wikipedia.To solve your particular integration problem you would scan the tables, for a match, or at least similar solution. And, of course, you will need that nasty constant as well. At least the table saves some of the work.A quick explanation of the constant, is that Differentiating a constant will give you zero as answer. Now since Differentiation is a function involved with 'rate of change' of something, you can accept that a constant, like '20', doesn't change over time or other variables, so the rate is zero.But looking at that the other way, something that has a rate of change, like '20 degrees per fortnight' is going to have a formula like:Degrees = 20 X TYou can differentiate that, getting '20', the rate. But suppose it'sDegrees = (20 X T ) + 81That also will give same answer, rate is 20. But notice that you've lost a portion of your info, as the 'rate' of any constant will just drop out as a zero.

author: RJSV
author_rank: Super Contributor
timestamp: December 02, 2023, 02:19:12 am
content: So now continuing the dialog, on Integrals and on Differentiation, most of the real detailed solutions were worked out long ago; and many engineers never have to deal with 'solving' some complex Integral in their day to day work.It just helps to illustrate the hierarchy relation.You might encounter a very complex function, such as [e to exp jwt] or [e to power exp sin X jwt] etc. No way most folks can (explicitly) solve those signals processing related Integrals at least not directly and on paper. But this whole explanation serves as an intro to the new and somewhat strange concepts such as infinitesemals, and limits.In that intro, important geometry comes into play (that's Newton's Analytic Geometry). Put in simple terms, the area under a curve helps define Integrals. An Integral (in geometry sense) is performed by the addition of all the little pieces, which sounds like common sense.But then, you have to adjust your thinking, as the process involves shrinking the width of those pieces, while simultaneously getting more and more individual additions, along the integral's boundaries (start and stop points along the horizontal axis).Differentiation, or rate of change of some function is fairly intuitive. Newton worked with concepts of LIMITS, that is, for example, when a complicated curve has a slope at some point that isn't precisely apparent, performing the differentiation can give you the precise answer, using the limit concepts. This is without going directly to the curve and measuring but rather by using or transforming the curve's equation, to get to the function defining slope (rate of change relative to horizontal axis, like time).Pretty powerful tools, for those times.

author: Rick Law
author_rank: Super Contributor
timestamp: December 02, 2023, 11:19:28 pm
content: Quote from: RJSV on December 01, 2023, 08:09:21 amHow recent did you finish quite well? I ask because things are changing rather fast these days.If your goal is to have anincome as tutor, SAT used to be the big money maker a few years ago and may still be. Often, you have High School kids looking for tutor a few month before SAT hoping to get a better score for college application or hope for a merit scholarship. That was a rather big market and may still be one. These days, some top universities are eliminating SAT as primary admission so they can get a more diverse student body. Even UC-Berkeley eliminated SAT/ACT as of May 2020. Harvard eliminated SAT/ACT for class of 2027 (4 years means 2023 admission). So, I am not sure SAT is still a good market. It is worth a moment of consideration if income is your goal.

author: RJSV
author_rank: Super Contributor
timestamp: December 03, 2023, 12:28:07 am
content: Thanks, Rick:Turned 70 this year, so...I'm focused on ability to walk, any distance, otherwise I'd be making good freelance dog walking and other animal care services locally (Oakland, Ca).In early 1970's I was getting requests to tutor, from classmates, first couple of years incollege . Physics test scores, like 100, or nearly so; meanwhile math tests near perfect, so there's that.One person commented: "...Rick, you really know your shit...". Regarding the calculus math help I was giving.Newton, himself, I hear wasn't always up to modern comprehension standard, either, lol.I appreciate your comments, as I didn't clarify (otherwise I'd post in 'jobs'.)But please comment on any errors you might see, as I really should dig up a couple books, to boost my memory.

author: Rick Law
author_rank: Super Contributor
timestamp: December 03, 2023, 04:12:35 am
content: Ah... We are neck and neck in age... So we saw similar world in our walks of life.Now back to the main thing: From your reply, it sounds like you are spreading the good by sharing your gift of understanding with others. I admire that.I was a Physics major, left with a masters degree and moved into the business world to make a living. For about 6 years, I volunteered at a parent operated weekend school with (at a high) of just under 400 students. Having the kid attend more schools during weekends on top of "regular school" is typical "tiger mom" behavior, so I've seen a lot of hard working kids and hard pushing parents. That said, quite frankly, since you are good at Physics and Calculus (even if good just for the first two college years), you possess the analytical skills, math skills, and wisdom as much or more than I could muster.One thing I do want to emphasize:the will to understand. When I was in graduate school, I was a TA. So I had my share of teaching Physics and the accompanying math. Many who fails to learnfailed because they were just trying to "get-by"instead of really understand. In my opinion, the three legs are: (1) Basic math skills, (2) Analytical skills, (3) The ability to think abstract and apply the abstract to the physical world. These skills are useful in Engineering, Business world, and even in personal matters. For the failed, in their haze to just "get pass this course", they want to just do the mechanical "find the right formula and plug the numbers in"without really trying to understandthe reasoning behind. Thereby, besides often failed to figure out "the right formula", they also missed the opportunity to learn skills that could help them with almost every aspect of their life. Well, actually, I think you probably already know what I just said in the above paragraph...This is one that you likely don't know. Prior to my retirement, my colleagues were often puzzled that my degree was in Physics instead of some business degree. They ask, and I told them: "The physics I learn, I hardly ever use. The problem solving skills I build in my study of Physics, I use every day."

author: RJSV
author_rank: Super Contributor
timestamp: December 03, 2023, 04:37:29 am
content: Respectfully, no; You are wrong in that last concept. I am a Polymath, which we are required to disclose, (although the meetings are in undisclosed location). Seeing parallels everywhere, regarding re-using one topic or avenue, into another. Even just the pure feelingof discovery translates fast, across boundaries, while explicit concepts stay in the box.Since there's no boss / taskmaster over me, I actually get a little thrill, when some new (mistake) I've made is pointed out.A person with Masters degree in physics could very easily end up in something like, say, Criminal Forensics !======================================Anyway, a couple thoughts, when you mentioned generational things in common, I recalled how disappointed I was / am that Jimi Hendrix died. That's a creative genius right there. He was consistently the highest grossing performer in those times! (I'm into biographies of creative people).

author: Rick Law
author_rank: Super Contributor
timestamp: December 03, 2023, 05:57:55 am
content: Quote from: RJSV on December 03, 2023, 04:37:29 amI believe people with the 3 skills ( Basic math skills, Analytical skills, The ability to think abstract and apply to physical world) can do anything because those skills are not specific to a discipline. Of course to succeed, one needs other attributes such as continuous learning, conscientiousness, stamina, etc. etc. etc.Criminal Forensics is interesting but too close to evil. I often look at air crash investigation shows and wish I had a part in it. That is a bigger problem that presents more challenges and likely with more resources.Anyhow, for people at our age, we are basically in "God's waiting room". The big guy may call us in any time but hopefully not so soon. If it pleases you to do things without a boss or master, by all means. Besides, what you choose to do, if successful,would make the world just a little bit better-- however little that bit is, better is better and it is better to have if just one more person educated just a little more by your effort.

author: MrAl
author_rank: Super Contributor
timestamp: December 03, 2023, 12:08:41 pm
content: Quote from: armandine2 on December 01, 2023, 11:33:01 amHi,This problem from reply #2 can be solved by following the description very carefully, which means calculating the power reaching the load RL (PLoad) with a lossless network inserted as indicated, then calculating the maximum power the generator could ever produce in the load (PMax), then calculating the ratio PLoad/PMax which is the fraction they are asking for.The maximum power the generator could ever produce in the load is obtained from the Maximum Power Transfer Theorem.

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 04:48:05 am
content: Here today, I've got an example of how it's useful to understand (fundamentals) of Differentiation in early physics.Mechanical 'laws' of motion say that when you hold a bowling ball in one hand, and a feather in the other hand, evenly, then both will drop to ground the same way, when released at the same time.Now I worked this out in my home, no books and no WIKI...A Quiz question:Please describe the forces on the two objects, as you hold them. The downward force is considered 'static' just simply because there isn't any movement, relative to, say, an Earth frame of coordinates.I'm better off using a baseball, and with the bowling ball (measuring) say 20 times the weight, of the baseball in your other hand. Assume this question wants a narrative answer, rather than strictly numbers.The classic equation, likely co-invented by others besides Newton, classic equation is:F = M ANow, as I will be teaching soon, that equation reflects result of doing TWO separate differentiations. It's a hierarchy where the first differentiation is performed on the position (equation or function of time elapsed ), while the second differentiation gets you to that acceleration equation, above.O.K. then, so let's go ahead and drop both test objects, at same time, watching them drop and hit ground, all the whole staying abreast of each other. Must do this in a near-vacume of course,but say you've got the rocketry...It's easy to point out that the 'heavy' bowling ball will likely need more FORCE from gravity, vs. the 20X lighter baseball. But since the movement or falling behaviours are same you can conclude: The bowling ball needed 20X the force, for this equivalent falling characteristics.Somehow the uniform gravity field is pulling on the two different objects precisely proportional, to each of the weights.I haven't discussed that force yet, but it turns out to be proportional to any two bodies. For the bowling ball that would be:Weight of Earth X weight of bowling ballAnd, of course, same for a second object dropped; that's Weight of Earth X weight of second object...the baseball.(Don't get mixed up, as each falling object is separate from the other, as the gravity force operates on each (ball) completely separately).So, assuming that your static force equation has that multiplication of the two interacting weights (one object, with the Earth, in two parallel trials), you can start to prove or at least verify that formula, which turns out to involve each of the two weights, with some constant in there. It's inescapable, as you will observe that in each separate trial, the different forces work to create identical outcomes, in movement measured relative to elapsed time. The lighter baseball, (as second object), gets a smaller force, meanwhile the heavy bowling ball (again, as second object relative to Earth's mass), will get a larger force...just enough so that the outcome, ACCELERATIONs, and subsequent falling positions is the same.Only conclusion is that ANY weight object will fall just the same as another.You just have to get air friction out of the test, like on the moon.Q.E.D.thanks for reading!

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 05:41:27 am
content: Had to use WIKI for a second, the term for MASS in the English system is 'SLUG' (I thought so), while 'pounds' is literally a force equivalent.In Metric system it's done in a more sensible 'kilograms and Newtons' setup, a somewhat trivial distinction that perhaps causes more arguments, (and physicist's bar-fights) than it should.Stepping on the scales, I cause a force of 192 pounds, or thus about 6 slugs of mass. Right then.In Europe, in Metric system of weights and measures, it's expressed in MASS, first (kilograms) and, rarely, in terms of force on scale springs, in NEWTONS. That's actually more sensible but in the end a trivial distinction, unless you are looking for a...' Physics contest'.But I'm eager to get back to describing the whole hierarchy thing, concerning derivatives and integrals related to basic mechanics of movement.Static electric fields also behave in very similar ways, concerning electric charge amounts between objects, rather than gravity originated forces.Soon, perhaps tomorrow, coming !

author: MrAl
author_rank: Super Contributor
timestamp: December 05, 2023, 01:36:46 pm
content: Hi,'pound' can be a weight or a force, but when it is a force it is often clarified using the name "pound-force". The abbreviation is usually "lbf" for pound force and just "lb" for pound.I suppose that once you introduce in the context the force as a pound-force, you might revert to the simpler name of just 'pounds' once it has been established that you are talking about a force, or just use the abbreviation "lbf". I guess it is up to the author.

author: jpanhalt
author_rank: Super Contributor
timestamp: December 05, 2023, 02:11:40 pm
content: "lbl" is the symbol for poundal. Thus, pound mass and poundal force are as "logical" as kilogram mass and Newton force. I never thought of "lbl" as just a way to distinguish the two.

author: Haenk
author_rank: Super Contributor
timestamp: December 05, 2023, 03:57:45 pm
content: I really think your offer to help is nice.*However* at least we, back then, had formed smaller groups of 4-5 people, for learning, repeating (in my case, early on, constructing) - so an answer was usually availabe within minutes or even immediatly. There was always someone who could teach the others, if the need should be. So the whole group could proceed very fast, and I don't think there ever was a need for remote teaching.

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 04:41:00 pm
content: In high-school physics (US in 1965), one important concept new to the students was the difference between mass and weight.The course employed both metric and customary units.To keep the customary units consistent, we used mass in slugs and weight/force in pounds (avoirdupois).Of course, in metric, mass is in kg and force/weight is in N.We were warned that some engineering used mass in lb av and force/weight in poundals.Similarly, we were warned that some metric engineering used kgf for force/weight (especially pressure in kgf/cm2).Since then, my mnemonic to convert between N and lb is to remember that 1 kg at mean sea level weighs 9.8 N = 2.2 lb.

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 05:11:17 pm
content: Quote from: TimFox on December 05, 2023, 04:41:00 pmI think that slugs and poundals were theoretically used, sometimes, maybe in the past.In my field (chemical engineering), you would find in US Customary Units that mass is measured in lbmand force is measured in lbf. The relationship between them is expressed with the constant gc, which appears in many formulas.

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 05:36:12 pm
content: Ha! See how this works ? Putting a 'diversion' subject, after the more real Calculus musings I had posted, I inadvertantly caused a distraction, so no real substantial criticism got into (yesterday's) activity! Going to remember this one, just in case.I'll supply a diagram, but the idea is to illustrate 3 levels, that being Position, Speed (or velocity), and, up top, Acceleration. In gravity field the position (freely falling object) is proportional to the square of time elapsed.Taking derivative, of position you end up with Speed or Velocity, which is proportional to elapsed time, looking like F (t). Finally acceleration is just a constant, that being a derivative (of the speed equation). That is simply noted, that a constant has no rate of change with respect to time.Please don't get too confused, with notation, because the variable 'X' is used in various contexts, either as an unknown or as a horizontal variable as in ' X axis, Y axis' etc.Hank; I had an advantageous daily schedule in college, because of having the Calculus math class in 'Period 5',...just following lunchtime at period 5, so that several of us students could do informal study group, by showing up to the room a half hour or so early before teacher came in. So, we even used the blackboard in there as we debated, argued and taught each other, (sometimes making friends!).

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 05:59:52 pm
content: Part of the explanation for how acceleration affects movement speed (free fall) is that the speed increase is compounded due to acceleration being constantly acting.So you've got a speed that got built up until now, and also have acceleration due to the downward force exertion persisting. So it makes sense, that velocity or 'speed' keeps building and it's believable that result is proportional to the square of time.In this context, the horizontal axis for function is 'time', and vertical graph axis is F(t).Much similar to various bouts of money inflation, and how it's a compounded effect.

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 06:42:59 pm
content: One of our roommates is Mechanical Engineer; I cruised his bookshelf and grabbed an older physics book to browse, as I usually don't see the kind of presentation that seems 'logical'...(oh well).WOW, that book weighs a TON and the many pages are super thin, packed with text and very few graphics / diagrams. Using net search engines also not so helpful, wanting to get a more general formula. Distractions as most descriptions seem to want to jump around, inserting for example, some quiz questions on things like:What is the power consumption, relative to time, of a freely falling body, vs. TIME ?Stupid smartness, lacking any real 'narrative' descriptions...What I wish to do, short term, is get all the full terms into the Position / Speed / Acceleration equations (for more fully description).

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 06:51:13 pm
content: Quote from: RJSV on December 05, 2023, 06:42:59 pmIn reality there is no power consumption. The power is zero. This is similar to reactive "power" (VAR) in electrical systems, vs real power (watts).In a free falling body, there is an exchange of energy between potential energy and kinetic energy, but if we consider the "system" to be the body in motion, then the total energy of the system remains constant. Power expresses a transfer of energy across a system boundary. Since there is no such transfer of energy, the power is zero.

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 06:59:40 pm
content: (Using diagram in previous post).I've got a pretty good idea of how to present a more full set of those three equations, using pre-existing Position, and Velocity as the starting condition, rather than simply dropping an object from rest. This relates to my comments previous regarding putting (unknown) constants into an equation after Integration. The next diagram reflects this, as the time variable, starts at zero, but the velocity can start at some existing value. So the expression would contain (V0) for example, to delineate a starting velocity, (P0) as starting position, etc.So, for example, the current instantaneous Velocity will be (V0) + AxT rather than having V0 or 'starting' value = 0. The time spent dropping under acceleration then uses (T- T0) as the elapsed time. That idea also applies to any starting position; where you have the usual equation, but provision for having any starting point (or height in this case).For example, a quiz question could say that your object has already been moving (falling), is now at 309 meters high, and right now moving at 19 meters per second. Working those numbers out actually sounds like a more realistic real-world problem.Thanks.

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 07:13:47 pm
content: There are standard equations of rectilinear motion learned in high school physics class. For instance:v = u + ats = ut + ½at²s = ½(u + v)tv² − u² = 2as(u = initial velocity, v = final velocity, a = acceleration, t = elapsed time, s = distance traveled)These equations will apply to a body in free fall, where the acceleration "a" becomes the acceleration due to gravity, "g".You can apply any of these equations according to need.

author: RJSV
author_rank: Super Contributor
timestamp: December 05, 2023, 08:03:48 pm
content: Yes, thanks, that's the sort of thing that I've been thinking.Never really encountered the more general and everyday type problem. For example I'm assuming that the coordinate system (X,y,z) type, would start calculating at 309 meters high but movement (downward falling) would start involving negative values, as object loses position, and, of course the velocity would be non-negative, but result is a loss of height...hinting that it's numerically a 'negative' velocity (meters per second).But, It's close enough that someone more skilled might suggest corrections.As to those constants needed, of unknown value, I recall the general solution calls for having two equations, with two unknowns to be able to solve. You manipulate one equation so that one of the unknowns drops out, when you've combined the two equations.So I'm going to get the fuller set of the three equations up (Position, Velocity, Acceleration.

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 08:53:59 pm
content: Note that "rectilinear motion" is a posh way of saying "in a straight line". So there is only one distance coordinate, x.In object in free fall is moving in a straight line, and the acceleration due to gravity is constant. So the equations I listed work perfectly for that scenario.If an object is falling, then you simply say that downwards is the positive direction, and then velocity, distance and acceleration are all positive.Let's do a thought experiment. You let go of an object that is 309 meters high, and you wish to know how fast it will be moving after it has fallen to 289 meters?We pick the following equation as suitable:v² − u² = 2asWe set our datum to 309 meters, and when it has reached 289 meters high it has fallen through 20 meters. So distance traveled, s = 20 meters.We let go of the object at at the beginning, so the initial velocity u is 0 m/s.The acceleration due to gravity is a = g = 9.81 m/s².We plug the numbers into the formula:v² − 0² = (2)(9.81)(20)v² = 392.4 m²/s²v = 19.81 m/sSo after the object has fallen 20 m from rest, it's speed will be 19.81 m/s.

author: MrAl
author_rank: Super Contributor
timestamp: December 06, 2023, 01:54:48 pm
content: Quote from: jpanhalt on December 05, 2023, 02:11:40 pmHi,What is "lbl" ?

author: jpanhalt
author_rank: Super Contributor
timestamp: December 06, 2023, 06:55:12 pm
content: "lbl" is the unit for pound plus the "l" from poundal. I pronounce it as pound "ell" or "el" (i.e, L spelled out).Edit: Why is the unit for "pound" "lb" and not pd or something similar? That's because of Latin.

author: TimFox
author_rank: Super Contributor
timestamp: December 06, 2023, 07:36:41 pm
content: Latin "libra" means pound, or scales/balance.A set of Latin currency values is libra, solidus, denarius; whence pounds, shillings, pence abbreviated L, S, and D.

author: newbrain
author_rank: Super Contributor
timestamp: December 06, 2023, 10:34:39 pm
content: Quote from: TimFox on December 06, 2023, 07:36:41 pmAnd those three words gave origin in Italian to:- "Lira" the Italian currency used before Euro (the symbol most commonly used was the same for pound £, though also L. and ₤ were used)- "Soldo" used in the singular for coin, but more commonly in the plural form "Soldi" meaning money.- "Denaro" another word for money

author: IanB
author_rank: Super Contributor
timestamp: December 07, 2023, 02:04:30 am
content: Quote from: jpanhalt on December 05, 2023, 02:11:40 pmQuote from: MrAl on December 06, 2023, 01:54:48 pmQuote from: jpanhalt on December 05, 2023, 02:11:40 pmSorry. Keep seeing this, and the temptation is too hard to resist

author: jpanhalt
author_rank: Super Contributor
timestamp: December 07, 2023, 10:26:26 am
content: Quote from: IanB on December 07, 2023, 02:04:30 amYou need your nystagmus checked. Please correct the sequence to include the post (#32) that follows , "What is "lbl" and tries to dissect its meaning. Perhaps mistakenly, I assumed everyone knew the derivation of "lb", as there are many such examples in English, Pb, Au, K, Na ... are just a few.

author: watchmaker
author_rank: Unknown
timestamp: December 07, 2023, 02:12:59 pm
content: I want to say thank you for the intended purpose of this thread.I am 70 and reigniting my interest in understanding electronics from my 1960s Voc Ed days. It is better than crossword puzzles or golf.In following the Analog Real course (and and Dave's videos and MIT and NI and Analog Wiki) I realized I needed matrix algebra to solve complex ckts. My kids (daughter (structural), her husband (EE/Patent Attn) started me off on the tangent of strengthening my math skills. At the same time I am doing labs, I am using Kahn and Brilliant to relearn (in reality learn) all the stuff I was taught but failed to learn (including Diffi Q).I failed to learn because at the time it was all taught in the abstract; with no relevance to application. All theorems and proofs. This is where STEM is a game changer. It was the reason I left the gifted program in high school to join the Vocational Education program.I fell into the category of those who passed by not understanding but writing enough stuff to finally get an answer.I am amazed at how "kids" are taught math today. I am starting at middle school algebra and matrix operations. Will need to also relearn trig function math. But the focus on graphing is great!To be honest, I hold a Ph.D. in social research which relied on all the higher order statistical analyses. But I NEVER visualized y=x squared. Nor did I use the distributive property to break large numbers into reasonable chunks for earlier multiplication. Sad, I know. But there it is.It will be some time before I get to Diff Calc but I really appreciate the effort to provide math support on this forum. I started but realized I really needed to back up to middle school.It was because of this forum (and Dave's videos) that I found Wulfram (sic?) for solving the more complex linear ckts.So, I would ask that you not let the "perfect be the enemy of the good". At least for me, there is potential value in this thread. And I am old enough to know I am not unique.Hopefully, this post will encourage others to expose their confusion and ask for help. Maybe there is a way to break this thread out so it becomes a sub-forum if demand is demonstrated?Regards,Dewey

author: IanB
author_rank: Super Contributor
timestamp: December 07, 2023, 04:10:12 pm
content: Quote from: jpanhalt on December 07, 2023, 10:26:26 amJoking aside, the symbol for poundal is actually "pdl". So six poundals would be 6 pdl.

author: RJSV
author_rank: Super Contributor
timestamp: December 07, 2023, 08:47:45 pm
content: Thanks, especially Dewey, for your thoughts.One benefit here, no matter how screwed up my personal organization is, has to do with being available for questions and criticism, unlike some more-perfected text book. (Lol) I could end up benefiting from corrective commentary.Right now, I'm still focused on analogies in the basic mechanical physics, and attempting to get that into diagram form,...that being the three hierarchy related items of Position, Speed, and Acceleration.Looks like I might need to just think up a suitable form, meanwhile noting that it's NOT a correctly expressed formula, for the motion physics. Others responding helped a lot with that frustration!So, if the dynamics of the 3 equations, involving differentiation and integrals isn't leading to heavy use, for numerical answers, I'm hoping to at least give a basis for why that understanding is still useful.For example, Newton, and others, used the calculus when describing the planet orbit problem, of whether an orbiting body is 'falling' by the math analysis, along with the analytic geometry involved in taking apart that (orbit) question.Thanks...more soon

author: RJSV
author_rank: Super Contributor
timestamp: December 08, 2023, 06:45:47 pm
content: Now I've gotten things in a better shape, for illustrating the tiers or 'hierarchy' of the three parameters of motion; that being Position, Speed, Acceleration.I'm in deep water a bit, here, as I'm attempting to generalize for illustrating the calculus aspect, and so have tried to get a most general case.

author: RJSV
author_rank: Super Contributor
timestamp: December 08, 2023, 06:52:10 pm
content: Explanation goes like this:Starting at the top equation (actually just a constant to represent gravity field induced acceleration, AND, in this case I will be putting an additional little rocket motor, for an initial segment of time).I figure to replace my example of falling from 309 meters high, to using a backwards vertical notation system, where the start is up somewhere, like 340, and THAT is newly, the place notated as 'zero' position.Moving downward then is a positive direction for position and velocity.Simply, otherwise it's a hassle in the math.

author: TimFox
author_rank: Super Contributor
timestamp: December 08, 2023, 07:09:57 pm
content: An important nit-pick: position, speed, and acceleration are variables, not parameters: they all vary as a function of time in your example.The strength of the gravitational field is a parameter.

author: RJSV
author_rank: Super Contributor
timestamp: December 08, 2023, 07:23:12 pm
content: In this aspect of my presentation, I think ianb has provided best explanations...please see those posts, a bit back.In the illustration here I've added an Xtra feature to try and get the most general case.Shown us a little rocket motor, let's say the object (we want to characterize) is attached to a rocket motor, before the free-fall phase of downward motion (now considered as a positive direction just to simplify, as ianb has suggested).The reason I'm doing this is to get a general equation, rather than just the '1/2 AT squared' term. Either we could have the rocket motor accelerate, say at 5 newtons per second per second, or just have it give a push at the beginning (looks better).That way you've got all the terms expressedIdea is to start the experiment at (t0), maybe move at a constant rate, to (t1) before doing the free-fall.... But that is wrong because gravity is still there in that initial time period.See how messy this gets ? Hopefully I can at least convey what I want, if not in numerically correct fashion. Sooo, even if incorrect, let's ignore gravity acceleration until t1.... (my bad).This approach let's me introduce the full calculus expressions, where each Integration operation introduces an unknown constant.In that sense, your period where there is acceleration would be from t1 to tx time. For that, you would use (tx - t1) as the length of time that acceleration happens.

author: RJSV
author_rank: Super Contributor
timestamp: December 08, 2023, 07:25:32 pm
content: Here is that modified experiment, for figuring the variables.

author: RJSV
author_rank: Super Contributor
timestamp: December 08, 2023, 07:26:38 pm
content: Uh shit, more upload problems...

author: RJSV
author_rank: Super Contributor
timestamp: December 08, 2023, 07:39:35 pm
content: Sorry about the picture upload glitch...I think it's errors in my cellphone memory (heat caused aging).This last diagram for today shows a more complete use of the TIME variable, where each place that uses time is literally a time period, helping to identify which segment of the little experiment is applicable.Whew, this is confusing for sure, but I think close enough that I can get back into the pure math aspects. That is, if you want to take a constant, and integrate that (twice), as a function then you get that stack of three equations, shown a few illustrations back.Again, though, I'm not sure if the absolute position, in a purely free falling body, will need these fully expressed terms. Part of the whole point, here, is to show the reversibility between integrals and derivatives...Clear as mud...(I may have just convinced 27 students to QUIT Engineering, lol)

author: IanB
author_rank: Super Contributor
timestamp: December 09, 2023, 03:55:33 am
content: The two basic equations we have for linear motion are (1) velocity from acceleration:$$\frac{\mathrm{d}v}{\mathrm{d}t}=a$$And (2) distance from velocity:$$\frac{\mathrm{d}s}{\mathrm{d}t}=v$$If we define \$u\$ to be the initial velocity and \$v\$ to be the final velocity, we can integrate the first equation over a particular interval:$$\int_{u}^{v}\mathrm{d}v=\int_{0}^{t}a\,\mathrm{d}t$$If acceleration is constant it can move outside the integral, to give:$$\int_{u}^{v}\mathrm{d}v=a\int_{0}^{t}\mathrm{d}t$$Evaluating the integrals therefore gives:$$v-u=at$$Or$$v=u+at$$Now, we can integrate the distance equation in a similar way:$$\int_{0}^{s}\mathrm{d}s=\int_{0}^{t}v\,\mathrm{d}t$$Which tells us that:$$s=\int_{0}^{t}v\,\mathrm{d}t$$We can take this integral and substitute in the previous expression for \$v\$ in terms of \$u\$, \$a\$ and \$t\$:$$s=\int_{0}^{t}v\,\mathrm{d}t=\int_{0}^{t}(u+at)\,\mathrm{d}t$$Evaluating this latter integral, with the understanding that \$u\$ and \$a\$ are constants, yields:$$s=ut+\textstyle{\frac12} at^2$$This shows how my earlier posts are related to the fundamental derivatives and integrals that express motion in a straight line, and how they simplify when acceleration is constant.

author: watchmaker
author_rank: Unknown
timestamp: December 09, 2023, 08:22:42 pm
content: Quote from: IanB on December 09, 2023, 03:55:33 amTHAT and circumference/area/volume are the only things I remember from calculus. Always thought it was a great "trick". Those guys were pretty damned smart.Regards,Dewey

author: RJSV
author_rank: Super Contributor
timestamp: December 09, 2023, 09:59:57 pm
content: Creating and Solving Integrals, using Analytic Geometry:_______-----------_________-------------________------By dividing up into smaller areas, the classic solution, for obtaining the area under a curve, can be determined, if maybe only approximately.Nice thing is that, by design, those areas can be made smaller and smaller; meanwhile the sheer quantity (of infinitesemals) of the partitioned areas grow (rapidly!).For example, (see diagram): you could have two pieces, or areas, and then go with four pieces. Obviously you also have each piece at half the area as before.

author: IanB
author_rank: Super Contributor
timestamp: December 09, 2023, 10:54:31 pm
content: Quote from: RJSV on December 08, 2023, 07:23:12 pmBut actually, the simple equation can still work, even in some cases that seem complicated.Consider the illustration below. Someone fires a stone from a catapult up onto the roof of a building. If the launch velocity of the stone is 10 m/s, at an angle of 80°, starting 1 m above the ground, and if the building is 4 m high, how long does it take before the stone lands on the roof? Assume that air resistance can be neglected.To work the solution, we can use the equation derived earlier:$$s=ut+\textstyle{\frac12} at^2$$To work out the time of rise and fall, we only need to consider the vertical component of velocity, acceleration and distance, and we can ignore the horizontal component. So first of all, we can find the vertical component of the initial velocity. This is found with trigonometry:$$u_y=u\sin(80^{\circ})=(10)(0.985)=9.85\,\mathrm{m/s}$$Secondly, we have the total distance traveled:$$s_y=4-1=3\,\textrm{m}$$Since we are using up for the positive direction, then the acceleration due to gravity is negative:$$a_y=-g=-9.81\,\textrm{m/s}^2$$Now we can plug the numbers into the formula:$$s_y=u_y t+\textstyle{\frac12} a_y t^2$$ $$3=(9.85)t+\textstyle{\frac12}(-9.81)t^2$$This is a quadratic equation we can solve for \$t\$. We get two solutions:$$t=0.37\,\textrm{s}\;\;\textrm{or}\;\;t=1.63\,\textrm{s}$$The first time will be on the way up, and the second time will be on the way down. So the answer we are looking for is the second one, and the time to land on the roof is1.63 s.

author: MrAl
author_rank: Super Contributor
timestamp: December 10, 2023, 02:03:12 pm
content: Quote from: IanB on December 09, 2023, 10:54:31 pmHi,Another thing you reminded me of from Calculus was Arc Length.For a curve like that parabola:y=A*x^2+Bwe have:g=sqrt(1+(dy/dx)^2)s=integrate(g,x,a,b)where 'a' is the starting 'x' position and 'b' is the ending 'x' position, 's' being the total arc length.dy/dx is of course the first derivative.This is interesting because it gives us the entire length of the arc formed from the start to the end.I remember a long time ago there was a Q Basic program that had apes throwing bananas at each other. One was controlled by you and the other by the program. Your goal was to hit the computer ape, and the computer ape goal was to hit you with a banana.The formula used to generate the path of the bananas was of course the parabola and that is the basic projectile motion expression.

author: MrAl
author_rank: Super Contributor
timestamp: December 10, 2023, 02:16:05 pm
content: Quote from: RJSV on December 09, 2023, 09:59:57 pmHi,This is all about shape, and this reminds me of the old adage, "You can't fit a square peg in a round hole".Funny thing is, YOU CAN. All you have to do is make sure the square peg is smaller than the round hole and it fits through, and if the square is the right size, the four corners just barely touch the circles perimeter.Now if we find the area of the square it will be very roughly the same area as the circle, albeit with a somewhat large error. Making the square smaller and increasing the number of squares, the smaller squares can make a better approximation as the smaller ones start to fill in the gaps between the sides of the original big square and the circle perimeter. The smaller you make the squares the more you need to fill in the circle, and the more you use the better the approximation of the sum of squares becomes to the total area of the circle. As the squares approach an infinitesimally small size the number of squares approaches infinity, and we get a perfect fit so that the area of all the squares exactly equals the area of the circle.So who says you can't fit a square peg in a round hole, I don't think Newton ever said thatThis is also one of the features of using the properties of infinity. You can have two quantities that are infinite in nature, yet maintain a finite ratio to each other like 2/3, 9/11, 5/6, etc.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 10, 2023, 05:38:59 pm
content: I warmly recommend usingwxMaxima, a free computer algebra system for Windows, Mac OS, Linux, and other operating systems.I'm very familiar with numerical integration of particle trajectories. I used to do molecular dynamics simulator development using non-quantum mechanical classical potential models and force field models, which model the total force acting on each atom in a simulation, then use discrete time steps and predictor-corrector models to integrate the particle trajectories at very high precision over time.I like to use \$t\$ for time, \$x\$ for position, \$v\$ for velocity, \$a\$ for acceleration, \$j\$ for jerk or jolt, and \$s\$ for snap.This is one of my favourite examples:What kind of acceleration/velocity should a train or elevator have, so that it would feel the smoothest to us humans?We want everything to be standstill (zero) at time \$t = 0\$, and maximum velocity reached at time \$t = T\$ with zero acceleration left, and maximum acceleration \$A\$ at midpoint during acceleration, \$t = T/2\$. Let's start with a generic cubic polynomial for the jerk:$$j(t) = J_3 t^3 + J_2 t^2 + J_1 t + J_0$$This also gives us snap (the rate of change in jerk):$$s(t) = \frac{d j(t)}{d t} = 3 J_3 t^2 + 2 J_2 t + J_1$$and acceleration (jerk being the rate of change in acceleration):$$a(t) = \int_{0}^{t} j(\tau) d\tau = \frac{J_3}{4} t^4 + \frac{J_2}{3} t^3 + \frac{J_1}{2} t^2 + J_0 t$$Note that because these are definite integrals (integrals over a specific interval), there is no integration constants added.The requirements we have are$$\left\lbrace \begin{aligned}j(0) &= 0 \\j(T) &= 0 \\a(0) &= 0 \\a(T/2) &= A \\a(T) &= 0 \\\end{aligned} \right . \quad \iff \quad \left\lbrace \begin{aligned}J_0 &= 0 \\J_3 T^3 + J_2 T^2 + J_1 T + J_0 &= 0 \\0 &= 0 \\\frac{J_3}{64} T^4 + \frac{J_2}{24} T^3 + \frac{J_1}{8} T^2 + \frac{J_0}{2} T &= A \\\frac{J_3}{4} T^4 + \frac{J_2}{3} T^3 + \frac{J_1}{2} T^2 + J_0 T &= 0 \\\end{aligned} \right .$$This is a system of equations we can solve. (If there was no solution, we'd need to bump \$j(t)\$ one degree higher, until a solution exists.)The solution is \$J_0 = 0\$, \$J_1 = 32 A / T^2\$, \$J2 = -96 A / T^3\$, and \$J3 = 64 A / T^4\$:$$\begin{aligned}s(t) &= \frac{192 A}{T^4} t^2 - \frac{192 A}{T^3} t + \frac{32 A}{T^2} \\j(t) &= \frac{64 A}{T^4} t^3 - \frac{96 A}{T^3} t^2 + \frac{32 A}{T^2} t \\a(t) &= \frac{16 A}{T^4} t^4 - \frac{32 A}{T^3} t^3 + \frac{16 A}{T^2} t^2 \\v(t) &= \frac{16 A}{5 T^4} t^5 - \frac{8 A}{T^3} t^4 + \frac{16 A}{3 T^2} t^3 \\x(t) &= \frac{8 A}{15 T^4} t^6 - \frac{8 A}{5 T^3} t^5 + \frac{4 A}{3 T^2} t^4 \\\end{aligned}$$Let's say that the final velocity (at time \$t = T\$) the velocity is \$V\$:$$v(T) = V = \frac{8 A T}{15} \quad \iff \quad A = \frac{15 V}{8 T}$$The distance traveled during acceleration is \$x(T) = 4 A T^2 / 15 = T V / 2\$.Maximum acceleration or deceleration occurs when its derivative (jerk) is zero; at \$t = T/2\$, when \$a(t) = a(T/2) = A\$. (The two other extrema are \$t = 0\$ and \$t = T\$, but \$a(0) = a(T) = 0\$.)People are most likely to stumble when the magnitude of jerk is maximum. The extrema of a continuous differentiable function occur when its derivative is zero, i.e. when snap is zero, \$s(t) = 0\$. \$s(t)\$ is a second degree function, so the roots are easily calculated, and are \$t = T/2 \pm T/\sqrt{12}\$, or \$t \approx 0.2113 T\$ and \$t \approx 0.7887 T\$. There, \$j(t) = \pm16 A / (\sqrt{27} T)\$, i.e. maximum magnitude of jerk is \$16 A / (\sqrt{27} T) \approx 3.0792 A/T \approx 5.7735 V/T^2\$.While the velocity is a fifth-degree polynomial and position six-degree polynomial, finding say "time \$t\$ when the elevator/train has moved distance \$L\$" is relatively simple to donumerically, when \$T\$ and \$A\$ (or \$V\$) are known: the key is realizing that position \$x(t)\$ and velocity \$v(t)\$ are monotonically increasing functions for \$0 \le t \le T\$, so bothNewton's methodand thebisection methodstarting with \$t=0\$ and \$t=T\$willfind a single solution \$x(0) \le L \le x(T)\$ and/or \$v(0) \le V \le v(T)\$.Algebraic solutions may or may not exist, but even when they do, they tend to be annoyingly complicated to calculate.For example, if \$T = 1\$, \$A = 15/4 = 3.75\$, then \$V = 2\$, and \$x(t) = 2 t^6 - 6 t^5 + 5 t^4\$. \$x(0) = 0\$ and \$x(T) = x(1) = 1\$. Then, \$x(1/2) = 5/32\$, \$x(1/4) = 29/2048\$, \$x(3/4) = 1053\$.By Newton's method, \$f(t) = x(t) - L = t^4 ( 2 t^2 - 6 t + 5 ) - L\$, \$d f(t) / d t = f'(t) = v(t) = 12 t^5 - 30 t^4 + 20 t^3\$. Iteration is$$t_{n+1} = t_{n} - \frac{f(t_{n})}{f'(t_{n})} = (10 t^6 - 24 t^5 + 15 t^4 + L) / (12 t^5 - 30 t^4 + 20 t^3)$$ and \$t_0 = 1/2\$ is a good starting point for any valid \$L\$ (\$0 \le L \le 1\$).For example, for \$L = 0.5\$, \$t_1 = 0.84375\$, \$t_2 \approx 0.745861\$, \$t_3 \approx 0.7420736\$, and \$x(0.7420736) \approx 0.500016\$, so one only needs a few iterations to get a pretty good numerical solution. If you clamp each \$t_n\$ to the valid range (\$0 \le t_n \le T\$), it will converge, because both position and velocity are monotonically increasing functions within this range.

author: IanB
author_rank: Super Contributor
timestamp: December 10, 2023, 09:17:52 pm
content: Quote from: Nominal Animal on December 10, 2023, 05:38:59 pmThat's kind of interesting, but is there a practical control system that delivers that profile? In my experience, trains want to accelerate and decelerate as fast as possible to minimize the time between stations, shorten journey times, and allow a higher train frequency. So the typical control scheme applies as much power as possible from a standstill to maximize acceleration, which naturally can produce a significant jerk for passengers. On a multiple unit train with many driven wheels, the practical limit is what G-force can the passengers withstand without complaining when the train starts and stops? (Wheel slip is not a constraint on such a train unless the track is greasy.)A more realistic problem to solve would be whether it is possible to minimize jerk while still obtaining a short journey time between stations? How much longer would the journey time become?

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 10, 2023, 10:40:39 pm
content: Quote from: IanB on December 10, 2023, 09:17:52 pmKone elevators, for example. It's not exactly this, but it is derived using similar steps.Quote from: IanB on December 10, 2023, 09:17:52 pmEven at modest accelerations and few kilometers/miles between stations, the travel time is limited by the maximum allowed velocity \$V\$.I already calculated that at the end of acceleration, the velocity is \$v(T) = V = 8 A T / 15\$ and distance \$x(T) = 4 A T^2 / 15 = V T / 2\$.(I will name units within brackets [ ] below. For example, 50 [m/s] means 50 meters per second.)Let's say the maximum acceleration \$A\$ is 1 [m/s²]. This corresponds to constant acceleration from 0 to 100 km/h in a bit under 28 seconds. Let's say the maximum velocity \$V\$ is 180 [km/h] = 50 [m/s]. Because \$V = 8 A T / 15\$, \$T = 15 V / (8 A)\$ or 93.75 . During this time, the train has traveled a distance of \$4 A T^2 / 15 = V T / 2\$ or 2343.75 , or less than two and a half kilometers.Because of symmetries, if the train starts decelerating immediately after it reaches the maximum velocity, it will travel twice that distance (4687.5 ) in twice that time (187.5 , or a bit over three minutes). The average velocity is therefore 4687.5/187.5 [m/s] = 25 [m/s], or 90 [km/h].If the train were to use maximum constant acceleration, \$a(t) = A\$, then \$v(T) = A T\$ and \$x(T) = A T^2/2\$; thus \$T = V/A\$, i.e. 50 , and \$x(T)\$ = 1250 . If it travels the middle part at constant maximum velocity, that'd be 4687.5  - 2 × 1250  = 2187.5 . If it then decelerates at the same constant rate as it accelerated, the entire duration for the 4687.5  trip would therefore be 50  + 50  + 2187.5/50  = 143.75 , thus average velocity 4687.5/143.75 [m/s] ≃ 32.61 [m/s] ≃ 117.4 [km/h].In general, the scheme I showed slows down any trip longer than 4.7 km by 187.5  - 143.75  ≃ 44 , compared to constant acceleration and deceleration to/from the maximum velocity, when the maximum velocity is 50 [m/s] = 180 [km/h] and maximum acceleration is 1 [m/s²] (about tenth of a G).However, because of the smoothly varying jerk, the maximum acceleration using this kind of smooth scheme can be higher than in the constant acceleration (or jerkier acceleration) schemes. It matters much more for elevators, though, because there the smooth jerk and continuously varying acceleration is easier on the power chain. You can easily plug in your acceleration and maximum velocity figures, approximating 1  = 3  to the equations above, and experiment.

author: MrAl
author_rank: Super Contributor
timestamp: December 11, 2023, 10:11:30 am
content: Quote from: IanB on December 10, 2023, 09:17:52 pmI have to wonder if there is a more natural limit to the 'jerk' on a train, both starting and stopping.If there is limited available power vs weight (mass) and friction, the train can only take off so fast anyway.Also, I seriously doubt if a train can stop on a dime, so the question is, just how big of coin would it take (that is, what is the shortest time it can stop in, period).Oh yeah that's another thing, friction. I believe friction will bring some exponentials into the solution.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 11, 2023, 11:56:53 am
content: Turbulent fluid or aerodynamic drag is F ∝ v² (drag equation), laminar is F ∝ v (Stokes' law), and sliding friction is independent of velocity.For simulating motion based on discrete time steps and varying forces,velocity Verlet integrationis often used to obtain the Newtonian trajectories. (However, Gear5 (Gear, C. W., Numerical Initial Value Problems in Ordinary Differential Equations, Prentice-Hall, Englewood Cliffs, N. J., 1971) and Martyna-Tuckerman (Martyna and Tuckerman J. Chem Phys. 102 (1995) 8071) integration are more accurate at similar time step sizes, albeit more complex.)It would not be difficult to create a self-contained HTML+CSS+JavaScript page (that one can open locally in a browser, no network connection needed) to say simulate a 2D or 3D box (with boundaries or with periodic boundary conditions) with one or more particles, points or spheres, using Newtonian mechanics and velocity Verlet integration.Simple systems, like say a cannonball or golf ball, can also be solved piecewise analytically. That is, between events like "exits the cannon" or "is hit by a club", and "hits and bounces from the ground/wall", where all parameters are continuous and sufficiently differentiable during the time interval, the trajectory can be calculated analytically for each interval separately. While one could consider this more accurate, in real life the numerical results from sufficiently short time steps are just as accurate. Where this matters, is in space simulations with accelerated timeframes, where otherwise the discrete time steps would be too long, accumulating integration error. (If I recall correctly, even Kerbal Space Program has had issues with exactly this.)But, before one goes that far, one should learn basic linear algebra, especially vector algebra (dot product, cross product, normalization, vector-matrix multiplication), plus if any kind of rotations are involved, either bivector or quaternion rotation (the component-wise math is the same for those), as they can represent all orientations and changes in orientation without issues like gimbal lock that plagues Euler and Tait-Bryan angles. Transformation matrix orthonormalization ishard, but I could write a basic primer on bivectors/quaternions sufficient for geometry processing and 3D projection (including things like realistic attitude thrusters on a spacecraft) if anyone is interested, because I'm extremely familiar with their use. I could perhaps start a new thread about that in the Programming sub-forum, if there is interest?

author: IanB
author_rank: Super Contributor
timestamp: December 11, 2023, 02:54:07 pm
content: I have not heard of Verlet integration before. I will read up on it.I would also note that in my experience, the Gear method (a variable order, variable time step predictor-corrector method) tends to be weak when applied to "complex" systems that do not have smooth, clean, continuous functions with precise derivatives. Any kind of discontinuous behavior will severely disrupt the integrator. Such disruptions may include external disturbances being introduced to the system, or numerical "noise" in the system such as inner sub-problems that have iteration with numerical convergence tolerances. Also, without any external constraints on the solver, the Gear method will try to capture all the frequency dynamics of the system, including very high frequency components that we might not care about and wish to filter out.Often a simpler integrator will be sufficiently accurate and will be more robust. When a system is dominated by external forcing functions, then first order implicit Euler works remarkably well. If a first order integrator is not suitable, such as when the system is dominated by internal dynamics, the second order trapezoidal method can be very effective.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 11, 2023, 07:21:29 pm
content: Quote from: IanB on December 11, 2023, 02:54:07 pmTrue. I've only used it in molecular dynamics simulations, where interactions are definitely such; and in particular, the "bonds" between particles tend to have a harmonic potential energy well near the equilibrium point. That's the kind of interactions it was developed for.The difference betweenleapfrog integrationand velocity Verlet integration is that in the latter, velocity and position are calculated at the same points in time, whereas in the leapfrog method, velocities are updated half a time step offset from positions.The theoretical modeling can feel a bit odd when first reading it, but if you play around how exactly the calculations are done for even a single particle, it really is quite straightforward; so much so that programmers without the basic physics background will likely end up "reinventing" the leapfrog integration on their own (but not understand all the implications and details of such a model). As long as the time steps are kept sufficiently short, both velocity Verlet and leapfrog integration will produce very realistic results.In the mid-1990s, I played quite a few rounds ofScorched Earthwhile having beers with my friends. Good times.There is an entire genre of theseartillery games. Occasionally, I've wondered whether one should reimplement one of those using HTML+JavaScript+Canvas, using WebSockets for networked play and trajectory arbitration.

author: MrAl
author_rank: Super Contributor
timestamp: December 12, 2023, 12:40:49 pm
content: Hi,This turned into an interesting discussion. I meant to show how the calculation for that one question was done but didn't get to it yet as I have a few other things going on right now.On the subject of integration techniques though there has been some more progress.With the advent of multi-core CPU's some 'new' methods have been ushered in that can take advantage of the parallel processing that multi-core CPU's are known for. I say 'new' in quotes because they are somewhat new to me, in that I have never tried using any of these yet. When I first started using numerical methods for these calculations back in the early 1980's there were no multi-core CPU's, or at least none that I could get ahold of at the time. Now they are in every new personal computer.Since they are somewhat specialized I can not say the newer methods will be good for every single application, but I am guessing they will work decently for the more basic electrical circuits. I intend to find out though as I get to checking some of these methods out as they sound very promising. Some of the pseudo-simulations I have done in the past can take a lot of time to complete, especially when there is more than one dimension involved. The evolution of a numerical grid can take a while to calculate as it makes multiple passes. Since you can get CPU's with a huge number of cores now, imagine say a 64 core CPU where each core can handle part of the problem. Although the calculation would not be able to go a full 64 times faster just because there are 64 cores, It would guess it would still be a lot faster. That's if the method was applicable in the first place.The good news is there is more interest in this area now since multiple core CPU's are a mainstay in computers now. That means more research into these techniques.BTW, one of my least favorite methods was the Adams-Bashforth method I originally tried way back then too. When used for some electrical circuit calculations it would seem to add its own unusual harmonics or something so the solutions would not come out smooth as they should but ratther with little 'humps'. It was kind of funny really to see that print out.

author: RJSV
author_rank: Super Contributor
timestamp: December 17, 2023, 07:41:30 pm
content: Oops; Sorry about delays.My plan is, to show a process where successive approximation attempts helped while (Newton and others) brought the Differential Calculus foundation forward ...I say this partially 'tongue in cheek' as it's likely not a perfect historical layout. But looking at graphical solutions helps get a contextual feel for the invention process that, maybe, happened.By putting focus on a graphical solution we need to look at polygonal areas, that gets more specific using trapezoid shaped areas...that's where you have 3 sides at 90 degrees orientation to each other, while a forth side has a slanted line (at top). That's the curve-fitting portion.Meanwhile, the aspects of the curve fitting are brought into an already established domain of mathematics; that was concepts of limits, and infinitesimals. If solutions for the area under any general curve can be had, by applying the therims in a new way, then that's useful.Sorry about delays in posting...work responsibilities have added up, rather than the 2 weeks off, originally planned. Thanks.

author: RJSV
author_rank: Super Contributor
timestamp: December 18, 2023, 05:47:42 am
content: You folks responding lately are AWESUM; makes me wonder, about all the amazing mathematical related bio's of people here.Also browsing WIKI on some more details of the history aspects. The story I had gotten, first year calculus, was, uniquely, presented by a disabled, former race car driver. (Obviously, he had tragically crashed, losing a leg in the process). But this guy knew his calculus and especially the geometric constructions (employed by Newton). The textbook I recall did also have a straightforward 'flow' that had later parts that built upon earlier simplier results. That's the thing that, later, I've noticed right away...a distinct LACK of continuity or progressive building, one thing after another.I think I lucked out, having that good teacher those first couple years.

author: RJSV
author_rank: Super Contributor
timestamp: December 20, 2023, 08:38:36 pm
content: So I have a struggle understanding the exact details, in the leap of concept going from using delta numbers, even if very small, to using the conceptual infinitesimals. It's helpful to sort-of keep also in mind, the dynamics of infinite series, limits, and convergence. (That last one sounds a bit religious, I know...)One unexpected problem is that I had understood a function 'Slope' to be y over X.But it's usually seen as ∆y over ∆x. Or, as I'm thinking to be 'dx/dy' by regular thinking.At any rate, that leap, of concept in applying limits to some formula involving 'delta or ∆' evolves into the familiar dx/dy'.Looking at area under velocity curve, you can see where a similar term, of 1/2 of something squared, is a simple average! That is, in my case where I wish to state everything in terms of a little differential between 'now', and a short time ago ...not the usual '1/2 At squared'.So you would have each equation in those terms. Starting position is simply at t0, and the next position is that plus a differential. In this case it's 'aT', in velocity terms, or more literally would be the starting velocity plus a bit of posituve change, that being the delta T times 'a' or acceleration constant.To get the area under that velocity curve, you simply average and take 1/2 of result. That's where the '1/2' comes from, but then that result gets multiplied again (by the little ∆ time).

author: IanB
author_rank: Super Contributor
timestamp: December 20, 2023, 11:14:05 pm
content: If acceleration is constant then you don't need calculus, simple arithmetic will work.For example, the distance formula becomes:$$s=\left(\frac{u+v}{2}\right)t$$This is "the area under the curve", but if the acceleration is constant then the "curve" is a straight line and we have the simple average of the initial and final speeds.If the acceleration is not constant, then we can let \$t\$ become smaller and smaller until we reach the limit of zero, and then at the limit the formula is again true.Finally, we may recognize that \$v=u+at\$, so now we have:$$s=\left(\frac{u+u+at}{2}\right)t$$Simplified, this becomes:$$s=ut+\textstyle{\frac12} at^2$$

author: RJSV
author_rank: Super Contributor
timestamp: December 21, 2023, 03:16:06 am
content: Thanks. I did just realize, that I incorrectly stated the average (as being the two numbers multiplied), when it should be the SUM, that then gets divided down, by 1/2 to the average.So, in my case it's that average height, of curve or line, multiplied by ∆t.It's hard to visualize, a general example relating just two points, without the whole graph (from 0 origin).

author: MrAl
author_rank: Super Contributor
timestamp: December 22, 2023, 01:18:56 pm
content: Quote from: RJSV on December 20, 2023, 08:38:36 pmHi,If you have a little trouble thinking about how something like delta T turns into just T in the limit, you can think in terms of number progressions and that allows you to see the mechanics behind it.For example, here is a number progression:1, 2, 3, 4, 5, ...you have no problem seeing where that is going, as well as:5, 4, 3, 2, 1, ...it's just getting smaller.Also:2, 4, 6, 8, ... or 8, 6, 4, 2, ...where you might stop at zero with that last one.Now how about this one:1.1, 1.01, 1.001, 1.0001, 1.00001, ...You can see that as we progress, the number is becoming closer and closer to '1'. That's the heart of it.The derivative is often defined as:dy/dx=[f(x+h)-f(x)]/h as h approaches zero,and since 'h' is the increment it might be written with h=dx as [f(x+dx)-f(x)]/h as dx goes toward zero.So we really have this:delta_y=f(x+dx)-f(x)anddx=delta_x.If we look at a simple example of y=x^2 and we want the derivative, we can look at the progression of numbersas we make Dx (delta_x) smaller and smaller. Starting with 0.1 and x=3, we get:Dy=f(x+Dx)-f(x)=(3.1)^2-3^2=0.61, and 0.61/0.1=6.1and now let Dx go smaller to 0.01 instead of 0.1:Dy=f(x+Dx)-f(x)=(3.01)^2-3^2=0.0601, and that divided by 0.01=6.01then Dx=0.001:Dy=f(x+Dx)-f(x)=(3.001)^2-3^2=0.006001, and that divided by 0.001=6.001then Dx=0.0001:Dy=f(x+Dx)-f(x)=(3.0001)^2-3^2=0.00060001, and that divided by 0.0001=6.0001It becomes kind of obvious that this is heading toward 6.00000000000...A check of course shows that the exact derivative is 6, but doing this sequence of numerical calculations we can start to get a feel for how Dy/Dx becomes closer and closer to the exact value.You can do this with almost any of these problems. You just have to watch out for numerical problems because calculators and computers alike always have limited precision in number representation.For the corresponding integration, you can let Dx get smaller and smaller and see how the summation gets closer and closer to the exact value.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 22, 2023, 06:53:56 pm
content: The very simple notion that 0.999... = 1, while initially odd/unbelievable, contains several useful realisations related to limits, calculus, and even geometry.

author: RJSV
author_rank: Super Contributor
timestamp: December 24, 2023, 08:24:43 pm
content: Attached diagram shows what I've been (trying) to think about...in between the winter-like trips, and other cold-weathrr hassles.With a more simple function, like in basic physics the speed (or Velocity) of a free falling object, in steady gravity field, speed is going to be, simply the field constant times the elapsed time.Off to the left, I've shown included is a more complicated curve of a function in general...you can surmise that the area under curve in the simple line case, will be easy to calculate, usung a center point or average value.That's going to be (y1 + y0) /2. I think that (Newton) perhaps started thinking about applying limits and convergence ideas, to resolve any general problem (of area under curves). Equivalent in my example, to using rectangles, of diminishing size, each iteration starting with polygons that are 4 sided, (with 3 sides oriented at 90 to each other.

author: RJSV
author_rank: Super Contributor
timestamp: December 24, 2023, 08:36:42 pm
content: So if we are going to talk about comfort, vs Intimidation, in mathematics, that can be a difficult aspect, (especially with school tests and looming homework assignments). It always helps myself, to remember what (the great one), Einstein remarked regarding math and algebra:" ...couldn't factor a basic ALGEBRA equation, to save my life ...". (or something similar to that saying).Math can be that way, harshly. I know I get intimidated around the concepts used in Partial Derivative relations. Best sometimes to just plunge forward anyway, and make good use of any lecture or video materials as they become available.

author: RJSV
author_rank: Super Contributor
timestamp: December 24, 2023, 08:58:28 pm
content: Mr. AI:Your explanation looks to relate fairly directly, and I was also thinking about the parabolic function, (y = X squared). That's what I've been thinking, of how to relate directly the approximate (closer and closer), vs the result when infinite terms are used. I remember how the derivative is obtained, at least by ROTE or procedural memory. Just trying to connect the dots, as to how that rote method, for obtaining a formula as a derivative of (another function derivative).Thanks for that recent explanation, (I'm wading thru that right now).- - - Rick B.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 25, 2023, 04:47:40 am
content: If you have a curve \$f(x)\$ you want to integrate from \$x_\min\$ to \$x_\max\$:The simplest isRiemann sumusing rectangles, also known as themidpoint rule.You sample the curve, modeling the curve as a sequence of rectangles, with the height of each rectangle determined at its midpoint by the curve.Mathematically this is equivalent to $$\sum_{i=1}^{N} f(x_i) w_i$$where \$i = 1 \dots N\$ denotes each interval or rectangle, \$x_i\$ its midpoint, and \$w_i\$ its width. Note that \$w_1 = ( x_1 - x_\min ) / 2\$, \$w_2 = (x_2 - x_\min) / 2\$, \$w_{N-1} = (x_\max - x_{N-1})/2\$, \$w_N = ( x_\max - x_N) / 2\$, and \$w_k = \frac{x_{k+1} - x_{k-1}}{2}\$ for \$k = 3 \dots N-2\$.If instead of midpoint, you pick the minimum or maximum value of the function \$f\$ within that interval, you get an upper or lowerDarboux sumor integral, respectively.If you know the value of the function at each point \$x_i\$, instead of a rectangle, you can approximate the area in that interval using a rectangle and a triangle. This is known as theTrapezoidal rule. Mathematically this is equivalent to $$\sum_{i=1}^{N-1} \frac{x_{i+1} - x_i}{2} \left( f(x_{i+1}) - f(x_i) \right)$$If you use \$N\$ regular intervals (\$N+1\$ values for \$x_i\$), then the approximate area under the curve is $$\frac{x_\max - x_\min}{2 N} \left( f(x_\min) + f(x_\max) + 2 \sum_{i=1}^{N-1} f(x_\min + i x_\Delta) \right), \quad \text{where } x_\Delta = \frac{x_\max - x_\min}{N}$$Better approximations can be calculated by fitting a curve to each interval, for example using the next neighboring values of \$f\$ (i.e., a total of four samples of \$f\$ per interval) and a cubic curve. Computationally, however, the trapezoidal rule is faster to calculate, so it is typically recommended to simply use the trapezoidal rule and more intervals (larger \$N\$).For annoyingly behaving functions, it can be even more useful split the intervals (recursively) for each interval where the function changes a lot. Since the intervals are still regular, you end up only having division/multiplication by the number you split the interval into. Splitting into power of two (2, 4, 8, 16, 32, 64, \$2^N\$) is extremely fast and efficient (with minimal effects wrt. rounding et cetera) when using standard IEEE 754 Binary32/Binary64 (float/double) floating-point numbers, as most computers nowadays do, so the numerical integration problem actually simplifies to detecting whether a function "behaves nicely" (is nearly linear enough) within a specific interval or not, in an efficient manner.For high-dimensional functions and functions that can be easily numerically calculated but whose shape or behaviour is unknown,Monte Carlo integrationcan be very effective. It means you sample the function value at \$N\$ uniform random values of \$\mathbf{x}\$ (being a vector, \$(x_1, \dots, x_D)\$, in Cartesian coordinates here). If \$V\$ is the hypervolume of the bounds of \$\mathbf{x}\$ (\$V = x_\max - x_\min\$ for the one-dimensional case), then the approximate integral is$$\frac{V}{N} \sum_{i=1}^{N} f\left(\mathbf{x}_i\right)$$i.e. the range/area/volume/hypervolume of the region of \$\mathbf{x}\$ you sampled in, multiplied by the average value of \$f\$ thus sampled.I know I should add a small illustration for each, but it's lazy holidays and I cannot be arsed to. The linked Wikipedia pages have the illustrations anyway.

author: IanB
author_rank: Super Contributor
timestamp: December 25, 2023, 09:52:40 am
content: Quote from: RJSV on December 24, 2023, 08:24:43 pmSeems like you might be driving towards theFundamental Theorem of CalculusThere's a lot of material there and you will need to work through it carefully.

author: MrAl
author_rank: Super Contributor
timestamp: December 25, 2023, 03:54:38 pm
content: Quote from: RJSV on December 24, 2023, 08:58:28 pmHello again,In these approximations, as the increment used to form the straight line used for the approximation gets smaller, the straight line approximates the curve better so the approximation gets more accurate. The increment is just the difference between two 'x' values used to calculate the function values which are two points on the curve itself. Once we let the increment get infinitesimally small, the calculation for the derivative equals the tangent to the curve, and the calculation for the integral equals the area under the curve.(See diagram)For integration of the top curve, it's a straight line so the approximation is exact. The can be found using the area of a triangle and the area of a rectangle added together. The average area is what we are after, so we could just use the two end points and the width, and average the area using that. that works because the average area is the area within the rectangle formed with a rectangle which has height that is halfway between the two 'y' end point values. That height times the width gives us the area, and that is exact because the curve is a straight line.For the bottom curve, do the same thing, but it becomes an approximation because we lose some of the area as shown above each straight line in red+violet for the blue points and in just violet for the green points. Notice though that the error is less for the green points because they are closer together, while the error for the blue points is much greater. That's because the straight line used in the approximation better approximates the curve for points closer together. That means the closer we make the points, the better the approximation to the exact integral.For the derivative, we can start with the point-slope form of a line in analytic geometry. That is:y=m*x+bwherey is the 'curve',m is the slope (derivative),x is the independent variable (along the horizontal), andb is the y intercept.We don't care about the y intercept because when we do subtractions that goes away, so we can focus on:y=m*xNow solve for 'm' the slope and we get:m=y/xand that is the derivative of a straight line.We do not have straight lines very often, so we have to look at curves. To do that we need again two points instead of just one.We can start with the two-point form of a line from geometry again:y=x*(y2-y1)/(x2-x1)+band drop 'b' again and get:y=x*(y2-y1)/(x2-x1)and here y1 and y2 are the function values and x1 and x2 are just the 'x' values again.Now since the y1, y2, x1, and x2 part is the slope again, we solve for that again and get:y/x=(y2-y1)/(x2-x1)Now when we look at the denominator, we can see that we have the difference in x's so we can call that Dx:y/x=(y2-y1)/Dxand when we look at the numerator, we see that is the difference in y's so we can call that Dy:y/x=Dy/Dxso what we have here is the derivative.Now back to:y/x=(y2-y1)/DxAs we make Dx smaller (which is x2-x1) the approximation to the curve between y2 and y1 gets more accurate. This means that the derivative gets more accurate. As we let Dx get very very small, Dy/Dx gets very very accurate, and if we look at the progression of numbers as before, we will see it get closer and closer to the exact derivative.It's actually in the limit as Dx goes toward zero that we see an exact derivative come out of it.The progression of numbers would again look something like:6.1, 6.01, 6.001, etc., and in the limit it would become 6 exactly (for that particular derivative at that particular point).Now in the above, we think of Dx as:Dx=x2-x1but really we can use an increment to represent x2 because it is close to x1. This means:x2=x1+hand since y2 and y1 are the functions of x1 and x2, we can rewrite the above:y1=f(x)y2=f(x+h)x1=xx2=x+hand then:Dy/Dx=[f(x+h)-f(x)]/[x1+h-x1]which of course is:Dy/Dx=[f(x+h)-f(x)]/hwith 'h' the increment.and that is the one-sided derivative.For a better approximation, we can use both a positive and a negative increment:Dy/Dx=[f(x+h)-f(x-h)]/(h+h)and this gives us a slightly better approximation to the derivative.So really the main point is that as 'h' gets smaller, the two functions approximate the curve better because the straight line formed by the two functions becomes equal to the tangent of the curve, which is the derivative.Note this is similar to how the integration worked also. The straight line used in the approximation gets closer and closer to the tangent of the curve.

author: RJSV
author_rank: Super Contributor
timestamp: December 25, 2023, 08:21:09 pm
content: Right now there's a 'toss-up' as to which is relevant, (historically in the discovery process), as to whether you show a complete and separate derivation, of each of the derivative, and of the integral. That is, whether the reversability of the two calculus functions can be enough to get your INTEGRAL, after deriving the derivative. That would involve building up a set or table of known integrals...when each derivative is determined.I find that the 'Summary in plain English' I can do very well (eventually, lol), whereas the exact formula manipulations are a (bulky) challenge.Many, many reaaaallly smart doctors and some teachers live in this so-called common sense 'gap' or bubble, often lacking common sense oversight, but meanwhile doing very well in their descriptive math. It takes just a sheer volume of study time!

author: RJSV
author_rank: Super Contributor
timestamp: December 27, 2023, 02:27:16 am
content: Nominal Animal:? Would you be able to to take a couple seconds, to explain something you outlined ?It's; (X1 - X min.) and that was multiplied by two; W1 = 2 (X1 - X min.)I didn't follow that, as to where the '2' came from, in the Rieman method. Plus, actually, not making progress putting that (above formula) in a plain English and graphical context.Help appreciated !Rick

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 27, 2023, 04:56:47 am
content: Quote from: RJSV on December 27, 2023, 02:27:16 amGood catch! It should be *divided* by two, not multiplied by two.Let's consider the case where we have five \$x\$ coordinates, \$x_\min\$, \$x_1 = -1\$, \$x_2 = 0\$, \$x_3 = 1\$, and \$x_\max = 2\$,in the middle of each interval, each rectangle –– except for the first and last one –– so we have five intervals in total.The first interval is from \$x_\min\$ to midway between \$x_\min\$ and \$x_1\$: \$x_\min\$ to \$(x_1 + x_\min) / 2\$.The second one is from midway between \$x_\min\$ and \$x_1\$ to midway between \$x_1\$ and \$x_2\$: \$(x_1 + x_\min)/2\$ to \$(x_2 + x_1)/2\$.The third one is from midway between \$x_1\$ and \$x_2\$ to midway between \$x_2\$ and \$x_3\$: \$(x_2 + x_1)/2\$ to \$(x_3 + x_2)/2\$.The fourth one is from midway between \$x_2\$ and \$x_3\$ to midway between \$x_3\$ and \$x_\max\$: \$(x_3 + x_2)/2\$ to \$(x_\max + x_3)/2\$.The last one is from midway between \$x_3\$ and \$x_\max\$ to $x_\max\$: $(x_3 + x_\max)/2\$ to \$x_\max\$.Their widths are thus$$\begin{aligned}w_1 &= \frac{x_1 + x_\min}{2} - x_\min &= \frac{x_1 - x_\min}{2} \\w_2 &= \frac{x_2 + x_1}{2} - \frac{x_1 + x_\min}{2} &= \frac{x_2 - x_\min}{2} \\w_3 &= \frac{x_3 + x_2}{2} - \frac{x_2 + x_1}{2} &= \frac{x_3 - x_1}{2} \\w_4 &= \frac{x_\max + x_3}{2} - \frac{x_3 + x_2}{2} &= \frac{x_\max - x_2}{2} \\w_5 &= x_\max - \frac{x_\max - x_3}{2} &= \frac{x_\max - x_3}{2} \\\end{aligned}$$The height of the first rectangle is \$f(x_\min)\$, the second rectangle \$f(x_1)\$, the third rectangle \$f(x_2)\$, the fourth rectangle \$f(x_3)\$, and the fifth rectangle \$f(x_\max)\$.Thus, the areas of each rectangle, i.e. the approximate integral over each interval, are$$\begin{aligned}A_1 &= f(x_\min) \frac{x_1 - x_\min}{2} \\A_2 &= f(x_1) \frac{x_2 - x_\min}{2} \\A_3 &= f(x_2) \frac{x_3 - x_1}{2} \\A_4 &= f(x_3) \frac{x_\max - x_2}{2} \\A_5 &= f(x_\max) \frac{x_\max - x_3}{2} \\\end{aligned}$$and the approximate integral is their sum, \$A_1 + A_2 + A_3 + A_4 + A_5\$.Yes, the first and last rectangles are narrower than the others. (I also should have shown how \$w_2\$ and \$w_{N-1}\$ are calculated, but forgot, because I usually use \$x_0 = x_\min\$ and \$x_{N-1} = x_\max\$.)

author: RJSV
author_rank: Super Contributor
timestamp: December 27, 2023, 04:04:21 pm
content: Actually, maybe I 'caught' one small glitch...However, feels like the fictitious Rip Van Winkle story, where, this time, I've woken up and discovered a whole new world (mathematics): it's maybe like a fabric with many new moth holes..., or actually mostly holes with nothing but air, plus a few strands of intact fabric.Just trying to get a more literal grasp of the nomenclature, for labels on repeated divisions (using 'i'). How to work in the time variable 't' in those, as it's differential with respect to 't', and integrals w. respect to 't'...Getting all that coordinated, before even dealing with solving or modifying (the equations).I don't recall an explicit use of 'anti-derivative' anywhere, although certainly it was discussed in classroom...(that was what? 55 years ago!)Anybody see, where I've set my coffee cup ?- - -Rick

author: IanB
author_rank: Super Contributor
timestamp: December 27, 2023, 04:19:14 pm
content: Quote from: RJSV on December 27, 2023, 04:04:21 pmYou would use the anti-derivative when doing symbolic integration. For instance, if you want to integrate \$x^2\$, you will need that the anti-derivative of \$x^2\$ is \$\frac13 x^3\$. That will give you the indefinite integral as:$$\int x^2 \,\textrm{d}x=\textstyle\frac13 x^3 + C$$Or the definite integral as:$$\int_a^b x^2 \, \textrm{d}x=\left[ \textstyle\frac13 x^3 \right]_a^b$$

author: RJSV
author_rank: Super Contributor
timestamp: December 27, 2023, 06:58:26 pm
content: (December 5th post has my diagram, of the three hierarchial related formulas, increased 'order' equation has X squared, mid equation at complexity of 'X', and lastly the third, least complex equation is just a constant.But, again, don't recall anything said about 'anti-derivatives', although that dynamic (of reversal) was certainly a major part of explanation of the history of developing an ever expanding table of integrals (solutions).Also, similar comment for the phrases 'indefinite' and 'definite' that are associated with anti-integration. Instructors maybe used different phrases, in their classroom.

author: IanB
author_rank: Super Contributor
timestamp: December 28, 2023, 01:19:33 am
content: You have a mistake in your formulas from that December 5th diagram that would lose you marks as a student.For example, if the (constant) acceleration is \$a\$, then the velocity is given by$$v=at+C$$where \$C\$ is a constant of integration. \$C\$ can take any value, but is commonly chosen to be \$v_0\$, the initial velocity at time zero. Thus we would have:$$v=v_0+at$$Similar considerations apply when integrating velocity to get position.The terms "definite integral" and "indefinite integral" are, I think, universal in time and place. Perhaps you just don't remember them?The idea of a definite integral would apply to thechangein velocity when integrating the acceleration. Thus, we could say:$$\Delta v=at$$In this case the change in velocity is definitely known, and therefore there is no constant of integration. An indefinite actual velocity became a definite change in velocity.

author: RJSV
author_rank: Super Contributor
timestamp: December 28, 2023, 03:42:52 am
content: Yeah, NO, I'm pretty sure I'd remember that, but remember, classroom tests etc were not well focussed all the way to a numerical answer, such as distance traveled, water amounts leaked, etc. More like proofs and theorms.Quiz questions were more like area under such and such curve, for some of that, but more like show what is needed to calculate that.(Maybe that reflects the traditional 'white collar' engineering in office setting, meanwhile the technicians ran around obtaining the numerical results...in the lab or on factory floor)

author: IanB
author_rank: Super Contributor
timestamp: December 28, 2023, 08:44:33 am
content: Well the area under a curve is an example of a definite integral. For example, if we say that the area under the curve of y = x² between x = 3 and x = 6 is 63, then that is a definite integral. Any time the answer can be stated as an actual number, that is a definite integral. If the answer contains a constant of integration, and we don't know the value of the constant, then that is an indefinite integral.

author: RJSV
author_rank: Super Contributor
timestamp: December 28, 2023, 04:25:20 pm
content: Beginners reading, here's an outline, of what's happened, (in some simple language):It was pointed out that the expression for the speed of an object is:'a X t' plus a constant, and that constant would be your speed at an earlier time, when you started. For example, moving at 4 meters per second, at start of some time interval. That's the constant.Then, the 'a X t' term gives the accelerated motion or what's called free-fall, from the start of the time interval, on to an end (of measured time period). Let's say the object picked up 10 meters per second...that means you end up at speed of '4 + 10' or 14 meters per second.I believe you have similar situation for calculations of distance. That is, integrating the speed term, above, which was 'at + c' would give the equation '1/2 X a t squared' plus 'ct', and then plus another constant, (let's call it c2).That 'c2' comes from integrating as explained by Ian, and that 'ct' portion of your position gives gives a way to account for whatever position is when an interval is started (its speed x time)= position.)Lol, this is making me squint really hard, but simple explanation is that you are accounting for the initial position, and initial velocity, in your measure interval, and then calculating the ending position.An initial position and initial velocity, if there, and plus the changes during the (time) interval gives you the real position, at end.....whew, complicated to explain, in proper frame.!

author: RJSV
author_rank: Super Contributor
timestamp: December 30, 2023, 01:18:44 am
content: Yes, thanks Ian, for the constructive criticism (reply #79 from Nov 28th). Pushing on those comments took a bit of time.But what I get, when attempting to do the integration, for POSITION, is three terms. That was a messed up area, as there were 2 constants, or so it seemed, after integrating from the velocity equation:One of those 'constants' gets re-labled as V0, the initial velocity, for the current measurement interval, while the other constant is simply the initial position. That's nice, as I didn't like the look, with having TWO constants (after integrating the Velocity line equation).I got: (1/2 a X t squared) + (V0 X t) + C...where the 'C' is your initial position, or y0, (assuming its motion along y axis).This should make for a nicer stack of the three equations (Position / Speed / Acceleration) in the derivative / integration heirarchy diagram.Thanks a lot, Ian!

author: RJSV
author_rank: Super Contributor
timestamp: December 30, 2023, 03:51:35 am
content: So in getting down to business, with a new year and all (the holiday parties done), I've come up with a count of 16 places that you folks have responded in detail...usually with extensive formulas illustrated!That's kind of high volume of work, to wade through and figure through (most of them anyway). If I can average approx. one every two days...brings me out, into the light around February ...plus a week or two.That's some (moderately) dense shit, (and intimidating)!But of course I could unpack a good deal of it by way of asking questions, of the handful of maybe four members who posted, over November and December.Ambitious, but if done transparently I might actually be able to learn a thing or two, while other members learn the stuff.

author: RJSV
author_rank: Super Contributor
timestamp: December 31, 2023, 10:08:32 am
content: I was saying that there are 16 value-packed responses, and since I can't easily generate a decently formatted TABLE (lame cellphone input), I'll just list 4 entries, for now:(helpful) replies #29, #46, and #47 (by IanB),and reply #50; (Mr.AI).Plus you can see the 'hierarchy' diagram,reply #27. UpSome of difficulty I'm having is simple nomenclature stuff, not too hard, but initially confusing. I was looking for a 'slope' in graphical sense, to be 'y/x' or 'dy/dx' in the geometric sense, but if you use the cannonball parabolic path, (reply #49), can start correcting the nomenclature.The 'Independent variable' is horizontal, that would be the elapsed time (not 'X axis that I usually would consider). We have to ignore air friction, so that horizontal time axis is evenly divided.Next, the ACTUAL use of 'X', is as the dependent variable, for the cannon shot problem, confusing, simply because that 'X' variable is VERTICAL, on the graph, and, is backwards in some gravity problems.(I'm leaving that go for a second, as a cannon ball parabola has both up and down travel acceleration and deceleration).O.k. so, the main differential will actually bedx/dt, as the projectile moves up, and then down. Contributer IanB has this path and distance problem with shorter path on the right half.That gets the first stuff into more coherent status. Now to the integration hierarchy,(reply #24). While I could understand the first integral, taking the acceleration constant and creating the two items of speed, doing that once more seems to then end up with TWO constants; That's been explained now, as you simply need to re-lable your velocity starting point (constant), to 'u'That way, when doing the second integration, (reply #24), you get, in general, THREE terms, which I like to classify as:(Top) (1.) Acceleration Constant.(Middle) (2.). Velocity linear+constant.and finally, (bottom formula) (#3.) The full Squared term, linear term and constant, defining you position.It makes good sense, and a 'neat' form that has the more 'generic' quality in the relations, of the three equations for a 'free fall' type of (cannonball) motion.Nice so far, but needs some minor considerations for the SIGN, positive or negative, of the gravity caused motion. I guess it's really too much to ask, to do these aspects without a quite literal feeling for the actual case or problem posed.Thanks

author: RJSV
author_rank: Super Contributor
timestamp: January 02, 2024, 06:53:06 pm
content: One roommate here wants me to 'derive' the equation for the area of a circle, 'π r exp 2', (the classic Pi R squared).Basic formula isr squared = X squared + Y squared,in various forms, but I think I'd start from a function standpoint: y = f(x).Looking at my included diagram, I've come up with an interesting take on the various option routes for solving the area, that I wish to share:Viewing the included diagram, I considered obtaining an integration solution, for area under a subsection of a full disk, or even by way of Angular function, using infinitesimal angle segments.Thinking, If you do the integration within 1/8th of the disk, that would lead to the complete result, by way of multiplying by 8.Now, when using y= f(X) you could always perform the area calculation arriving at a four sided shape that has a curved top, using X value that ends aligned with 45 degree point...that would be 0.707 X the full x at radius size...0.707 X radius.'Cute' thing is that it's then easy to turn the equation into the area of 1 'pie slice' shape, by simply subtracting the triangle area left, by the integration, that is the trianglular area underneath the pie-shaped area. Nice.However, it's likely easier to do integration over the entire quarter circle, as appears in the first quadrant of disk.

author: IanB
author_rank: Super Contributor
timestamp: January 02, 2024, 09:55:23 pm
content: It's easier to consider thin "shells" around the circumference of the circle and add them all up.For instance, the circumference of a circle is2πr. Then the area of a small "shell" of thicknessdraround the circumference is2πrdrTherefore the area of a circle of radiusR, say, can be found by adding up (integrating) over all the little shells:A= ∫ 2πrdrbetween 0 andR( =πR²)

author: IanB
author_rank: Super Contributor
timestamp: January 03, 2024, 07:26:11 am
content: The area can equivalently be found by integrating in \$xy\$ coordinates, but it's much more complicated, as you can see below.We have for the equation of a circle:$$x^2+y^2=r^2$$We can find the area of the quarter circle in the upper right quadrant with this integral:$$A=\int_0^r y\,\mathrm{d}x$$Substituting for \$y\$:$$A=\int_0^r \sqrt{r^2-x^2}\,\mathrm{d}x$$To integrate this, we can make the substitution \$x=r\sin{t}\$, whereupon \$\mathrm{d}x=r\cos{t}\,\mathrm{d}t\$With this substitution:$$A=\int_0^\frac{\pi}2 \left(\sqrt{r^2-r^2 \sin^2{t}}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$$$A=\int_0^\frac{\pi}2 \left(\sqrt{r^2(1- \sin^2{t})}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$$$A=\int_0^\frac{\pi}2 \left(\sqrt{r^2\cos^2{t}}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$$$A=\int_0^\frac{\pi}2 \left(r\cos{t}\right)\left(r\cos{t}\right)\,\mathrm{d}t$$$$A=r^2 \int_0^\frac{\pi}2 \cos^2{t}\,\mathrm{d}t$$To make the next step we can use the identity \$\cos{2t}=2\cos^2{t}-1\$. Thus:$$A=\frac{r^2}2 \int_0^\frac{\pi}2 \cos{2t}+1\,\mathrm{d}t$$Solving the integral gives:$$A=\frac{r^2}2 \left[\frac12 \sin{2t} + t\right]_0^\frac{\pi}2$$$$A=\frac{r^2}2\left(\frac{\pi}2 \right)=\frac{\pi r^2}4$$Since this is the area of a quarter circle, the area of a whole circle is four times this, or:$$A=\pi r^2$$

author: RJSV
author_rank: Super Contributor
timestamp: January 03, 2024, 06:45:31 pm
content: Thank you, this is very much what I was thinking, (except I'm more interested in getting a 1/8th portion, as illustrated).Doing the 1/8 'pie slice' intersecting the perimeter at 45° (degrees), the triangular area to subtract becomes 0.707 X 0.707, which in perfect numbers will give exactly 1/2 square inches or square centimeters, depending on using a 'unit circle' of radius = 1.The reason I like to play with this alternative is, mainly that I haven't seen that before...just exploring...(out of love, for pure mathematics)!Of course it's much more difficult, but I think using that 'y = f(X)' form is more formal and general. But now, the sort of reverse proof is to differentiate the integral...that is using the 'rote' or memorized way;Taking the exponent (2) down into the multiplying terms, getting '2π' then, and decreasing the actual exponent, to R to the first power, gets your anti-integral, '2πR'.That goes back and forth, if you want, between Integral and accompanying derivative, as I've mentioned being partially using 'rote' (memorized) and the reversal...if there is one, in your bag of tricks (table of solutions).But then, of course, question comes to mind, is there an actual literal method, other than;"I've seen that one, where a constant times a variable ends up as constant times the variable squared".In other words, the 'rote' method, (and it's anti-calculation) seems a little like cheating, with some luck....Is there a more defined way, of building up an Integral like this, rather than the method of: "Saw that solution before" ?Sorry if my English structure is a bit circular (no pun).At any rate, after obtaining the full area, under the curve, from 0 to 45° that triangular region comes out to a half square, of 1 unit by 1 unit (1/2 of that).Thanks

author: IanB
author_rank: Super Contributor
timestamp: January 03, 2024, 07:08:54 pm
content: Quote from: RJSV on January 03, 2024, 06:45:31 pmSurely it's 0.5 x 0.707 x 0.707 = 0.25? (Area of a triangle is half base times height.)QuoteAlthough the triangular bit may be "easy", you still have the circular bit to deal with afterwards, which will use exactly the same method I outlined. So it's really no easier at all.QuoteOf course, there is naturally a literal method. Nothing in mathematics is "rote". It all can (and must) be derived and proved. Any reasonable calculus course will include these derivations and proofs. I learned them when I was 16 and can still broadly remember them (or can recreate them). What I did above was pulling back memories from high school mathematics studies about 45 years ago.You will see that I didn't use a "table of solutions" to do the integration in my post above (though I could have). Rather, I showed the actual details. It was easier to work it out from scratch than to spend time looking for a table of integrals.

author: RJSV
author_rank: Super Contributor
timestamp: January 21, 2024, 01:25:30 am
content: Oops, yeah, I goofed that one up, estimating the area, but off by factor of two! Sorry.I think I got distracted, into thinking about a shortcut to a smaller portion of the basic multiplication tables.One shortcut involves the multiplication of 1 digit, below 10, by another digit above 10.Doing this (see diagram), gives a kind of 'reflection' with the '10' in the middle, of a number line.

author: RJSV
author_rank: Super Contributor
timestamp: January 21, 2024, 01:35:48 am
content: So, it's a simple math related trick, and uses a Quality Factor, usually up as close to 100 'perfect'.Taking 9, 8, 7, and 6, while the lower multiplies are too trivial to deal with),You will have; 9 matched or paired with 11.That, multiplies as '9 X 11' or 99, thus having a Quality Factor of '1', or minus one, technically. But it's the distance from '100'.Doing the '9 X 11' quick multiply, the more accurate version will be, (interestingly);'9.090 909 09...etc.' X 11, to get perfect 100 as result. OR, you could do it as;'9 X 11.111 11 repeating' as giving a perfect result.Doing this, as pairing '8' with a '12' gives slightly off-results, of 96 instead of 100.

author: RJSV
author_rank: Super Contributor
timestamp: January 21, 2024, 01:52:52 am
content: Now you can do this little memory speedup for just a few; '9, 8, 7, and 6, practically. The case for 7 is also interesting; When doing'7 X 14' the Quality Factor is good, at 99, and there is (shown in previous diagram) the wonderous result of '0 .142857142...' as mentioned elsewhere, as a longer pattern of 6 digits that repeats, in the fraction.But also, there is the '7' doubling itself, at least briefly, doing '7, 14, 28, and 57,... ALMOST continuing the simple digit doubling.The pattern occurs, similarly when doing division, say of 1/7... Some of that seems to veer (dangerously ?) close to the simple √2.

author: IanB
author_rank: Super Contributor
timestamp: January 21, 2024, 02:15:06 am
content: It's a common mental multiplication trick.In general, you have:$$(x-a)(x+b)=x^2-ax+bx-ab$$So, if you wanted to multiply 99 by 102, you would have:$$(99)(102)=100^2-100+200-2=10,098$$Taking your example of 8 times 12, we have:$$( 8 )(12)=10^2-20+20-4=96$$More interestingly, 98 times 102 gives:$$(98)(102)=100^2-200+200-4=9,996$$Also, 1/7 being 0.14285714... is not so much close to √2 as toπ. Hence we have:$$\pi\approx\frac{22}{7}$$

author: RJSV
author_rank: Super Contributor
timestamp: January 21, 2024, 05:51:03 pm
content: Hey IanB ! Thanks for reading!You get a more interesting figure, when dividing by 14, instead of 7:1/14 = 0 .071428571...Note that it has the same structure, where the pattern should be doing '56' as the next doubling value,...but, alas, it diverges from that pattern (of new lower value digits in the fractional number).It gets interesting, when trying to compare with 'root two' related values. That would include the various such as '1/√2'; at '.707106...'Maybe this has simpler explanation, but also relates to the use of '22/7' in some calculation shortcuts.Now looking at the extreme ends, you would 'pair' the numbers 2 and 50; (for a perfect quality factor). The next 'pairing', you may notice, would be 3 with 33.333... which is (approx) 17 counting units apart, and each subsequent pair gets a smaller and smaller separation. If you had started on the other end, it would appear that the distance stepped is exactly '1.0' but that quickly shows as false assumption, (witness the '13' got skipped, in the next pair, '7 with 14'.)At any rate, by the time you've tracked each pair, down to (4 with 20), it becomes really too trivial to include in some simple gimmick...it's just apparent on inspection, with '4, 3, and 2' doing the in-head figuring.Playing with the various combos, for illustration with an example (estimate): Suppose someone says that business improved by about 1/9th last year, you immediately can be thinking: 'Uhh, that'd be 11 percent gain,...(or multiply by 1.11 for the gained new figure.)Another quick example, someone saying 'down by 1/16th,...you know it's (about) 6 percent diminished...and about how accurate your rule of thumb is...That is, '6 X 16' being a bit 'low' in this context.