| Author | Message | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 188 Registered: 10-2001 |
Efwin wrote: ""What I'm trying to say is that if you think of the motor as a simple output device, and think of its output in a flexible construction, you might well come to believe that a motor producing F times D in interval Y, will be better able to raise Kinetic Energy if it produces F-(small quantity) times 2D (e.g., 8krpm v. 4Krpm) over the same interval of time. Simply, it's spinning twice as fast while producing nearly the same torque, ergo, more energy within the same discrete interval. " Todd wrote: "This statement doesn't contradict anything I've said. This is equivalent to the "300ft-lb @ 6000 rpm gives twice the acceleration as 300ft-lb @ 3000 rpm if you double the gear ratio" argument. Efwun's comment here is an accurate description of power. I ask: But Efwun is not talking about doubling the gear ratios. He's talking about being in the same gear at 8000 vs. 4000rpm in a car that has a fairly flat torque curve that may peak at 4000 but still have something close to that at 8000. It seems he believes such a car may actually accelerate faster at 8k rpm than 4k rpm. Hmm... | ||
| Todd Wasson (Todd_wasson)
New member Username: Todd_wasson Post Number: 6 Registered: 4-2003 |
"One possible such car is the turbo vw since there was a chart that showed it accelerating faster at 3000rpm than it's stated torque peak of about 1900rpm. Other people report that they've driven cars that seem to accelerate faster up near hp peak than at their torque peak." ------- 1. The torque and power curves of a turbocharged engine change depending on the RPM and acceleration rate of the engine and turbo; they are not fixed curves, but instead can move around considerably. The published torque curve is simply one of the infinite possible curves you can get with a turbocharged engine. Turbocharged engines should not be used to back up any side of this argument. 2. Our rear ends and ears can be very misleading. I've fallen into this trap myself when saying "it seems to be faster." More than once I used a stopwatch afterwards and found out I was wrong. The car was actually slower when it felt faster. It was a combination of extra noise and wishful thinking, nothing more. ------- "Camp B (well, it's basically just EFWUN, ha!) claims: "Remember also that as the output shaft spins, the torque it produces is really F*D, or Work. Therefore, the amount of work possible varies with D, and in any discrete interval of time, that is governed by F*D/Time, or Power." -------- Right. -------- "What I'm trying to say is that if you think of the motor as a simple output device, and think of its output in a flexible construction, you might well come to believe that a motor producing F times D in interval Y, will be better able to raise Kinetic Energy if it produces F-(small quantity) times 2D (e.g., 8krpm v. 4Krpm) over the same interval of time. Simply, it's spinning twice as fast while producing nearly the same torque, ergo, more energy within the same discrete interval. " - Efwun's words, not mine. -------- This statement doesn't contradict anything I've said. This is equivalent to the "300ft-lb @ 6000 rpm gives twice the acceleration as 300ft-lb @ 3000 rpm if you double the gear ratio" argument. Efwun's comment here is an accurate description of power. --------- "Anything in there want to give you any doubt that f=m*a is all you ever need to determine where any possible car accelerates quickest?" --------- You should come to the same conclusions either way, but yes, f=m*a is all you need. Or you could find it with energy, work, and power as well, but the answers must be identical or somebody made a mistake somewhere. I must still be missing something, as I don't see a conflict anywhere. | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 187 Registered: 10-2001 |
Thanks Todd. I think everyone here understands your point about gear ratios. The questions is whether every car, in its own first gear (for example) accelrates quickest at its torque peak (the A camp says yes) or whether there are cars that can accelerate more quickly at some rpm higher than that torque peak. (the B camp says such cars may exist). One possible such car is the turbo vw since there was a chart that showed it accelerating faster at 3000rpm than it's stated torque peak of about 1900rpm. Other people report that they've driven cars that seem to accelerate faster up near hp peak than at their torque peak. Camp B (well, it's basically just EFWUN, ha!) claims: "Remember also that as the output shaft spins, the torque it produces is really F*D, or Work. Therefore, the amount of work possible varies with D, and in any discrete interval of time, that is governed by F*D/Time, or Power. "What I'm trying to say is that if you think of the motor as a simple output device, and think of its output in a flexible construction, you might well come to believe that a motor producing F times D in interval Y, will be better able to raise Kinetic Energy if it produces F-(small quantity) times 2D (e.g., 8krpm v. 4Krpm) over the same interval of time. Simply, it's spinning twice as fast while producing nearly the same torque, ergo, more energy within the same discrete interval. " - Efwun's words, not mine. == Anything in there want to give you any doubt that f=m*a is all you ever need to determine where any possible car accelerates quickest? | ||
| Todd Wasson (Todd_wasson)
New member Username: Todd_wasson Post Number: 5 Registered: 4-2003 |
""Acceleration is simply how much your speed changes every second." Close! Acceleration is the instantaneous change in velocity (dv/dt = a). On a human time scale you should consider that acceleration in 1/100-to-1/10 second is accurate enough for all practical purposes. 1.0 second is not enough resolution to ferret out answers as to where does maximum acceleration actually occur." ------------------ ------------------ Mitch, Did you read the question I was answering? My response was appropriate for the person asking the question who did not know if acceleration referred to percentage change in speed, total change in speed, or what. I imagine my father, in a clever moment, asking exactly this question. I think my explanation was very intuitive and easier to understand than a statement like "acceleration is the derivitive of velocity, which itself is the derivitive of position," or "acceleration is the second derivitive of position", or "acceleration is the instantaneous rate of change of velocity," don't you? I would not explain acceleration that way to my father or anyone else but someone with a calculus background. It's obvious our educational methods are different. Thank you for your recommendations of integration time step sizes. I don't understand why you would think I might need some help there, but your good intentions are appreciated anyway. Todd Wasson http://PerformanceSimulations.com | ||
| Todd Wasson (Todd_wasson)
New member Username: Todd_wasson Post Number: 4 Registered: 4-2003 |
Rich, you wrote: ----------- Todd, thanks for dropping by. If you go down a few posts I think you'll see my attempt at a summary of what these two guys are in conflict about. You are in camp A - what do you think about the idea B? p -Rich --------- I'm not quite sure what camp B is trying to illustrate. All one needs for this discussion is f=ma from what I've read so far. Perhaps I'm missing a subthread here somewhere though. You can look at a physical system in terms of force, mass, and acceleration, or you can look at a system in terms of work, energy, and power, or even momentum and impulse. They all will give the same results in the end (they better!). It's just so much simpler and more intuitive to stick to force, mass, and acceleration. I think most people try to "switch systems," so to speak, because they see the word "power" being used, and being educated, science respecting people suddenly things get overly complicated as work and energy must enter the picture almost by definition. Calculus explanations for intuitively vague ideas quickly pop up, and guys that aren't very interested in math are quickly lost or lose interest. This is not to say those explanations are wrong, just that there are easier ways to look at all this. Fortunately one can gain a solid, intuitive grasp of power and torque with nothing but basic algebra at the most, and a good imagination at the least. So what can we do when the word "power" is being discussed, and we want to stick to examining the situations in the easier to understand concepts of force, mass, and acceleration? Easy. Convert power to torque with the age old equation every racing nut knows: torque = horsepower * 5252 / RPM Of course, torque can be converted to force very easily (divide by the effective wheel radius), and we are now free to discuss both engine torque and power and their effects on your Ferrari while sticking to easier to understand concepts (mass, acceleration, force). From here on out, there is no need to ever use the words "energy," "work," or even "power" again. When the word "power" pops up, we can convert it to torque or force. It's much easier to think about for most people (myself included). Once you imagine the force from xx power in a few different situations enough times, power starts to make some sense intuitively without having to understand the actual definition of power or having a degree in mathematics or physics. (To reiterate, I haven't read through everything in this thread, so perhaps Efwun is trying to illustrate something that I'm not aware of. I am not speaking against what Efwun has said; I have read hardly any of his comments. I browsed through looking for some key points to be made on this topic, and the fellow I mentioned earlier was the first one I saw making them.) Air Resistance: Moving right along, it appears camp B might be thinking that air resistance alters the picture and other things exist that make this more complicated than it really is. It doesn't. You can still look at this with f=ma. At a given speed, you have a given, constant force due to air resistance. Period. This is simply added to (or subtracted from, however you like to think) the force at the wheels (from torque **at the wheels**, not at the engine). What is being argued is where the driving force at the wheels is greatest; peak torque or peak power. Air resistance is a seperate force, and although it does effect the acceleration of course, it does not change the answer to the following question: "At what point does a vehicle accelerate at maximum, peak torque or peak power?" It should make sense that if you knew the force on the car at peak torque and the force at peak power, you would have the answer to this question. Whichever force is greatest wins. The air resistance force is the same for both if you're talking about a car travelling at a constant speed in both situations, so air resistance drops out of the equation entirely when trying to find out where the greatest force is. To be more precise, this is akin to asking: At what point is the most force available at the driving tires? Torque One side appears to be saying that acceleration will be greatest when the engine is at its peak torque. First of all, engine torque alone does not determine acceleration rate. The acceleration of a car can be calculated from the torque at the driving wheels. The torque at the driving wheels is the engine torque multiplied by the gear ratios, not the engine torque all by itself. Of course, in any given gear, the acceleration will be highest at peak engine torque. Camp B is right on this point (if that's what they're saying). So what gives? If you're sitting at the engine's peak torque and you downshift, the engine runs at a higher speed and provides less torque. Does the car accelerate faster or slower? Remember, the engine torque has dropped. The actual answer is: It depends. In reality your gear ratios are selected so that the answer is going to be "yes" in all likelihood, especially in a sports car. Now what? If Camp B is right on that point, why are folks saying the power peak is where the greatest acceleration will be? The fact of the matter is that the engine torque does not determine the acceleration rate all by itself. When you downshifted, you increased torque multiplication to the driving wheels through the lower gear. I.e., even though the engine makes less torque at this higher speed, there may be more torque at the driving wheels. Result? More torque = more force = greater acceleration. Example: An engine makes peak torque of 350 ft-lb @ 3000 rpm. Peak power is 300 HP @ 5000 rpm. Let's stop right there. Let's get rid of power and just use torque, something that's easier to think about and not such a nebulous concept to most people. At 5000 rpm (our power peak) there is: torque = horsepower * 5252 / RPM torque = 300 * 5252 / 5000 torque = 315 lb-ft Clearly, there is less engine torque at the horsepower peak than there is at the torque peak (350 vs. 315 lb-ft). Ok, now we're cruising along at some speed with our engine running at 3000 rpm, the peak torque. You nail the gas. Let's just say you've got a 10:1 overall gear ratio to keep the math simple. How much torque is at the driving wheels? Driving wheel torque = engine torque * gear ratio Driving wheel torque = 350 * 10 Driving wheel torque = 3500 lb-ft Pay attention here, because we're going to downshift so our engine is running at the power peak (5000rpm) instead of the torque peak (3000rpm). If we downshift, our car is travelling at the same speed. Our lower gear ratio is 5000 / 3000 * 10 (our original gear) =~ 16.7:1 Is our engine torque lower? Yes (300 vs. 350 ft-lb). Will we accelerate more slowly or more quickly? To answer that question, we need to know the torque at the driving wheels with this new gear ratio. Driving wheel torque = engine torque * gear ratio Driving wheel torque = 300 * 16.7 Driving wheel torque = 5010 lb-ft We have substantially more driving wheel torque at the power peak because of the new gear ratio. We are accelerating more quickly now, even though the engine torque is lower. The new gear ratio offset the slight drop in engine torque. Watch this: Something interesting happens when you look at this on either side of the power peak. What if the gear we downshifted into put us just a little bit below the power peak? What if it put us a little bit above? I'll make up some numbers here for power: RPM--------Horsepower 4500------278 5000------300 ---Power peak 5500------278 Remember, to keep things simple we can talk in terms of torque and force instead of power. Can an analysis like this be done through power, energy and work? Yes, but it's abstract to most people and more complicated. The results must be the same anyway. KISS. Below are simply the horsepower figures above run through the equation torque = horsepower * 5252 / RPM RPM--------Torque(lb-ft) 4500------325 5000------315 ---Torque at power peak 5500------265 Note that the torque is dropping off as it goes through the power peak at an ever progressing rate. You're all probably familiar with this from looking at dyno curves. Note also that all of these torques are substantially less than the peak torque of 350 lb-ft. What if the gear we downshifted into put us just a little bit below the power peak, at 4500rpm? In order to have downshifted and landed at 4500 rpm, our gear ratio must have been 4500 / 3000 * 10 (our original gear) = 15:1 What do you predict will happen to our driving wheel torque? We have changed both the engine torque AND the gear ratio! Errr.... Driving wheel torque = engine torque * gear ratio Driving wheel torque = 325 * 15 Driving wheel torque = 4875 lb-ft We are not accelerating as quickly with the new gear that puts our engine speed a little bit below the power peak. Now let's go the other way and see what happens. What if we shift so our RPM is a little bit above the power peak (closer to the redline)? First, what gear ratio would put us at 5500 RPM? In order to have downshifted and landed at 5500 rpm, our gear ratio must have been 5500 / 3000 * 10 (our original gear) = 18.3:1 What's our driving wheel torque? Driving wheel torque = engine torque * gear ratio Driving wheel torque = 265 * 18.3 Driving wheel torque = 4850 lb-ft See this? The driving wheel torque is the same at 4500 rpm as it is at 5500 rpm (it's a little different only because I rounded off some numbers). The power is the same at those two points, and the torque at the driving wheels is too! Wierd, huh? If you try this for every point on any dyno curve, you'll get the most acceleration if you alter the gear ratio to keep the rpm at the power peak. The confusion on this topic comes about because you are changing two things at once: Engine torque and gear ratio. When you are doing this (by changing gears or selecting gear ratios in the first place) it should become obvious after awhile that peak acceleration will occur at a given speed when the gear ratio puts the engine at the power peak. I encourage everyone to grab a dyno chart and try the above calculations until a little light bulb goes off and you have a new understanding of what power really is. Of course, in any of those gears we just examined, the acceleration will be the highest at the torque peak. But then your car is running at a different speed. At that new speed, you could alter the gear ratios to put the engine speed at peak power and gain acceleration all over again. Acceleration is caused by engine torque times gear ratio. If you produce the torque at a very high speed (like an F-1 car), you can use an incredibly high gear ratio. You then get incredibly high power. A 250 lb-ft @ 16,000 rpm engine would produce the same force at the wheels as a 1,000 ft-lb @ 4,000 lb-ft at any given speed because you could gear the thing down by a factor of four. A useful quantity that reflects this is "power." The 250lb-ft@ 16,000rpm produces the same power as the other engine running 1,000 ft-lb @ 4,000 rpm. The acceleration will also be the same if you gear the car accordingly (factor of four)...... Air resistance does not effect any of this. Air resistance would do nothing but effectively subtract 500 lb-ft (or whatever) torque from the driving wheels in all situations. The peak acceleration would still have occured at the power peak if you had control over the gear ratio. Without the math, you could think to yourself "if I pedaled a little less forcefully, but geared the bicycle down a whole bunch at the same time so my feet were spinning faster, there could be more force at the wheel." Hope that helps! Todd Wasson http://PerformanceSimulations.com P.S. Someone mentioned that Newton's laws don't always hold true. That's correct literally, but not practically. Einstein's derivation of E=MC^2 in relativity theory alters Newton's laws, but there is no measurable difference until something is moving a significant percentage of light speed. Down here on Earth outside of particle accelerators, Newton's laws are iron-clad. Cars do NOT break Newton's laws :-) | ||
| Faisal Khan (Tvrfreak)
New member Username: Tvrfreak Post Number: 25 Registered: 3-2003 |
ok, if you want to split hairs, acceleration is the change in velocity per unit of time, but typically the unit of time is seconds, hence the typical unit of metres per SECOND squared. Why do I get the feeling you never had a date to the prom...and you're still angry. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 581 Registered: 4-2002 |
"Acceleration is simply how much your speed changes every second." Close! Acceleration is the instantaneous change in velocity (dv/dt = a). On a human time scale you should consider that acceleration in 1/100-to-1/10 second is accurate enough for all practical purposes. 1.0 second is not enough resolution to ferret out answers as to where does maximum acceleration actually occur. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 580 Registered: 4-2002 |
"Does 2+2 equal 4???? what a thread" Yes, Unless you want to get into abstract algebra where you can define all sorts of numeric sets and operations thereon. One important kind of abstract set is the ring-sum. Take a 32-bit processor for example. the biggest signed number is 2**31-1 and the biggest negative number is -2**31. If you have 2**31-1 and add 1 you get -2**31, this is normally called overflow. Ring-sum math was designed to allow reasoning about these kind of limitations to 'normal' numbers. All computers have such limitations. One could have a ring-sum {0,1,2} and define the operation 'increment': increment(0) = 1, increment(1) = 2, increment(2) = 0. It then follows that addition in this ring-sum is 'modulo' 3. So 2+2 = 1 (4 mod 3 -> 1). However, in NORMAL math 2+2 = 4 is considered a tautology (unwavering truth). | ||
| Faisal Khan (Tvrfreak)
New member Username: Tvrfreak Post Number: 24 Registered: 3-2003 |
Ken (Allyn): as long as you don't change direction, speed and velocity are the same thing. And, moving in a straight line, the biggest change in real speed or velocity will lead to the biggest percentage change. ie. biggest acceleration and biggest percentage change in speed or velocity will occur at the same point. | ||
| Todd Wasson (Todd_wasson)
New member Username: Todd_wasson Post Number: 2 Registered: 4-2003 |
Ken (Allyn), you wrote: "Excuse what may seem a dumb question, but when we say 'maximum acceleration' in the question of where does max. acceleration occur, do we mean as a percentage change; say going from 5mph to 10 mph in one second doubling the speed, or as a raw number like going from 50 MPH to 90 mph in one second to use an absurd example. The first case doubles the speed, but the second added way more actual velocity. Does my question make any sense? I suppose I could think more about it and answer it myself but my mind is boggeling. " (Sorry, I don't know how to turn on quoting in this forum.) Acceleration is simply how much your speed changes every second. For instance, if right now you are speeding up at a rate of 10 mph every second, you could say your acceleration is 10 mph/sec. This would take you from 0-60 in 6 seconds, 30-100 in 7 seconds, etc.. Acceleration (in the English feet/seconds/pounds system of units) is usually described in feet per second per second(ft/sec^2). If your acceleration is 1 ft/sec^2, this means at time = 0, if you're not moving, one second from now you will be moving 1 ft per second, two seconds from now you will going 2 ft per second, etc.. You can compute the acceleration of an object very easily. Simply divide the force acting on the object by the object's mass (weight in pounds divided by 32.16). Whammo. | ||
| Mike B (Srt_mike)
Junior Member Username: Srt_mike Post Number: 146 Registered: 12-2002 |
The personal attacks are a bit much... I personally think EFWUN is mistaken on this issue but he seems like a nice guy and I really enjoyed reading his thoughts on the matter. Thanks EFWUN. I will really enjoy hearing what your physics professor friend has to say about this. I wonder if his "more than meets the eye" will be weight change due to fuel burning, wind resistance, or something else. There are quite obviously a few intellectual type folks in this thread, and intellectuals enjoy learning. Mitch, I respect your knowledge of calculus, but while I took a TON of math in my engineering days, I don't do engineering and I have forgotten much if not most of it. That doesn't mean I dont have an innate understanding of physics however - and I think EFWUN has the same thing. It took me a alot of reading to understand the differences between TQ and HP and how they produce acceleration, so maybe the slamming of EFWUN was a little bit harsh, no? The guy tried to extend an olive branch and you grabbed it and beat him over the head with it. I think you're right too - but that doesn't mean EFWUN isn't a smart fella too. | ||
| TomD (Tifosi)
Advanced Member Username: Tifosi Post Number: 3414 Registered: 9-2001 |
Does 2+2 equal 4????  what a thread | ||
| David Stoeppelwerth (Racerdj)
Junior Member Username: Racerdj Post Number: 108 Registered: 1-2003 |
WOW-I'm worn out from this thread. Reading about the Calculus and Physics equations brings up past history from my Purdue Civil Engineering days. 1972-1976 2 full years of Calculus and Physics. It's amazing how much I have forgotten!! I now remember how nice it was to get those first two years out of the way. I really enjoyed the education not the personal attacks. Remember we are all interested in the greatest car ever made! | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 845 Registered: 10-2001 |
What a thread! Excuse what may seem a dumb question, but when we say 'maximum acceleration' in the question of where does max. acceleration occur, do we mean as a percentage change; say going from 5mph to 10 mph in one second doubling the speed, or as a raw number like going from 50 MPH to 90 mph in one second to use an absurd example. The first case doubles the speed, but the second added way more actual velocity. Does my question make any sense? I suppose I could think more about it and answer it myself but my mind is boggeling. Regardless, I have to side with Mitch in this one. It seems all the bases are covered in F=MA and no matter what units you use for F, M and A, maximum acceleration occurs at maximum force. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 579 Registered: 4-2002 |
The topic was: where does maximum acceleration occur. We have agreed that maximum acceleration occurs in 1st gear, somewhere. At the velocities encountered in 1st gear air resistance is (essentially) neglegable. In engineering schools the teach you to ignore negligeble influences on the first attempts at a solution. Afterwards, you can go back in and add refinements. Neglecting air resistance was clearly specified early and several time farther on in the argument. EFWUN seems to have a poor memory for details that are contrary to his position. "He similarly demonstrated a perhaps more limited understanding of this concept than mine, " More limited than yours--in what way? I showed and computed why power governs the top end, and why TQ governs the low end. And I showed you how to compute acceleration near top end. The fact you cannot understand is not a problem with the math, it is a problem with the way your brain works. "What I find most insulting and offensive about Mr. Alsup�s continued personal attacks is that they are not motivated by honest purpose, but rather by the apparent need to �win� (??) this debate at all costs. " As opposed to you argument "no it isn't, I can't explain why, and I don't understand the math and physics involved, but since Alsup is a bad guy, we just can't let him win". This may work in Law, but it does not win in science. In science, the equations do the talking. BTW I find it personally insulting that EFWUN assumes he is learned, and yet can not perform simple physics problems using simple math. And then after showing the correct derivation of the equations, and showing the correct usage of the math, EFWUN continues to disbelieve just because of our personal relationship. EFWUN--get over it! "Mr. Alsup insists that F=MA is the ONLY inquiry" Misleading at best. Alsup insists that all sets of equations arrive at the same conclusion as does f=M*a. Including W=f*d, and E=1/2*m*v**2. But your math skills are too thin to be able to comprehend. "I admitted that I�ve forgotten calculus (who wouldn�t want to?)" OK, now we are getting somewhere. EFWUN does not understand calculus. But the thread is why EFWUN should have not forgotten calculus. Especially when he presents himself as someone who understand the subject mater of this thread. Calculus is very useful, even to lawyers! "Mr. Alsup has used calculus as a tool for intimidation " Quite unlike EFWUN, who throws around "I presented arguments in front of some appelate court not so long ago" as intimidation. At this point, I knew he was a lawyer, Lawyers like to argue, and Lawyers like to win. However, law school should have taught him that a good lawyer NEVER asks a question that he does not already know the answer. With me its the same, I never get involved (or stay involved) unless I do know the answer. Data from the web substantiates my position, EFWUN has presented not contrary data, but has presented two equations. He purports these arrive at different conclusions. I showed that, indeed, they arrive at the same conclusion. EFWUN is not capable of accepting this analysis because he is a lawyer and has to win. Get over it. "but in reality, the necessary inquiries are simple physical formulas that are derived algebraically from Newtonian Truths." Here, surprisingly, we are in complete agreement. "�If torque creates low velocity acceleration and power overcomes wind resistance, how do we define where the two intersect? Shall we pick an arbitrary number Mr. Alsup?� " We use calculus! and set up the equations of motion with air resistance included. We then take the derivatives of the equations of motion, set the appropriate derivative to zero (finding the maximum point) and arrive at an answer. Simple math. But remember, we have already agreed that maximum acceleration occurs in 1st gear, and that air resistance at legal speeds is essentially neglegable (about 5 HP at 60 MPH) causing less than 1.2% error in our conclusion. And in 1st gear (50 MPH at max RPMs) it is completely neglegible! "As to Mr. Alsup, I�d respectfully ask that you make no further personal attacks, and save your vituperation for some other person. I�d also ask that you not challenge me to any more �interviews at Weehauken� (see, Aaron Burr and Alexander Hamilton), because it is unseemly, and you�ve shown that you�re not that quick on the draw." As you've shown that in matters of science, you ain't too bright, either. "Similarly, as to Mr. Alsup's graphs, they contain values that assume his thesis, and are therefore self-serving." As opposed to you complete lack of data, with a continual stance of "no its not". But My graphs have to assume someting! So I assumed that f=M*a was useful, and the graphs happen from there. This is not subtrafuge, it is straightforward presentation of data in graphical form--the form most capable of conveying difficult to undersand information in a package the not-quite learned can actually understand! In the court of law you may be capable of holding down a job. If someone comes up and offers you a job in engineering--please don't. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 600 Registered: 2-2003 |
On open letter to readers of this thread: I have tired of the aggravation of going back and forth with Mr. Alsup, and would rather summarize my discussion here for general perusal. What I find most insulting and offensive about Mr. Alsup�s continued personal attacks is that they are not motivated by honest purpose, but rather by the apparent need to �win� (??) this debate at all costs. Succinctly, when Mr. Alsup stated that Torque creates low velocity acceleration, while Power is necessary to overcome high speed wind resistance, he stepped right into a pile of the conundrum I�ve insisted exists, and which he has vehemently denied. He similarly demonstrated a perhaps more limited understanding of this concept than mine, (at least I�m admitting that there is a complex interaction going on here), and if his understanding is that truncated, then his name calling, abuse, and threats are really in the service of his ego, rather than honest scientific inquiry. Mr. Alsup insists that F=MA is the ONLY inquiry, and challenged me to �disprove� F=MA with some expletives deleted. Simply, F=MA is the fundamental truth underlying much of physics, and the equation I�ve derived, simply Torque*Distance =F*D=Work, and W=Delta Kinetic Energy, and therefore W/T =Delta Kinetic Energy divided by time, which equals Power (W/t=P) utilizes that fundamental truth, along with necessary others, such as KE=1/2MVsquared. (I refer to my post where in simplified units, we see that at 8,000rpm there are 500units of Power available to create DeltaKE/time, while at 4,000rpm there are 300units to create that DeltaKE/time. ) I admitted that I�ve forgotten calculus (who wouldn�t want to?) and Mr. Alsup seized upon that admission, and attempts to say that this inquiry depends on calculus. It doesn�t! Mr. Alsup has used calculus as a tool for intimidation (because I am loathe to attempt integrations I don�t remember and then need to apologize for �silly� posts as I have done in the past), but in reality, the necessary inquiries are simple physical formulas that are derived algebraically from Newtonian Truths. Mr. Alsup has stated (and the internet backs him up) that he is a computer designer. (And apparently a good one). I have, embarrassingly enough, mentioned my National Merit Scholarship, my failure in the Westinghouse, and my honors from the University of Pennsylvania. (I am sort of ashamed to have posted that stuff, but I felt a childish need for credibility in the face of Mr. Alsup�s abuse). I submit that computer design and electrical engineering have very little relation to a simple physical problem, and Mr. Alsup�s statement about torque creating low velocity acceleration, while power creates high velocity acceleration shows that computer designers don�t necessarily have a lock on simple Newtonian mechanics. I received a reply today from an old physics professor who stated that he believes there is more involved than simple equations of either torque or of power will define. He invited me to visit after his end of semester duties. I fully intend to take several days off, take my wife and drive to Philly (the Rittenhouse is a great hotel, and Penn is beautiful in Spring!) and pursue this to its end. In the meantime, the following conundrum results from Mr. Alsup�s apparent failure of understanding. �If torque creates low velocity acceleration and power overcomes wind resistance, how do we define where the two intersect? Shall we pick an arbitrary number Mr. Alsup?� Obviously not, and I�d submit that the answer is a fairly complex interaction, and that such complex interactions don�t admit of easy solution. As to Mr. Alsup, I�d respectfully ask that you make no further personal attacks, and save your vituperation for some other person. I�d also ask that you not challenge me to any more �interviews at Weehauken� (see, Aaron Burr and Alexander Hamilton), because it is unseemly, and you�ve shown that you�re not that quick on the draw. Similarly, as to Mr. Alsup's graphs, they contain values that assume his thesis, and are therefore self-serving. Indeed, one graph shows a car accelerating at peak G well after peak torque; that graph is dismissed as "incorrect" or alternatively "the torque peak is where I say, not the manufacturer." As I said, I will take this question up with my old Physics prof, who has already said, there are more issues here than either side has presented. | ||
| Mike B (Srt_mike)
Junior Member Username: Srt_mike Post Number: 143 Registered: 12-2002 |
<<from your link. hmm, is this because a car accelerates more at peak torque??>> Nope! It's because first gear may be a 4:1 ratio, and 2nd may be a 2:1 ratio (greatly simplifying here). Most torque curves are not all that peaky - they are usually pretty flat. So, if you are at 5,000RPM in 1st making 250lb-ft but your torque peaked at 3,000RPM where it was at 300lb-ft, you may think "ahh, I should shift!". But, with your 4:1 first year, you're putting 1,000lb-ft of torque to the wheels. If you shift, yes, you will get closer to your torque peak, but at a 2:1 multiplication. Even if your shift puts you right back at 3,000RPM and again you are making 300lb-ft, you are only putting 600lb-ft to the wheels, because the mechanical advantage is alot less. Now, since horsepower is a function of RPM and torque, you can't have a lot of HP without either a) alot of torque high up in the RPM range or b) really high RPM's. Note that big block chevy's would be the former, whereas F1 cars would be the latter. So a high HP engine means it either makes gobs of torque, or tons of RPM, or both. Either way, it means you can stay in the lower gears up until you run into the redline area, before having to shift and lose mechanical advantage. The example of an F1 car was given that it can smoke the tires at 100mph. It also revs to 18,500RPM. Look at a Corvette vs. an F1 car. The Vette makes 350lb-ft, and it's first gear is something like 9:1 (3:1 tranny and 3:1 rear end - approximate #'s). So at it's torque peak of say, 3500RPM, it's making 350lb-ft x 9 or 3,150lb ft of torque to the wheels. But it redlines at 6,000RPM. That F1 engine may only make 250lb-ft, but it can rev to 18,000RPM. So we can gear the F1 car with a 27:1 first gear. Both cars will reach the same speed in first gear (assuming same tire size), but, the F1 car is putting 6,750lb-ft to the wheels. Both cars shift into 2nd and the F1 car has a lower ratio, and again is putting more torque to the wheels, and accelerating alot faster. In this example, the Corvette has alot more torque... BUT.. the F1 car has WAY more horsepower by virtue of a ton more RPM's, which allows you to take advantage of gearing, producing a faster overall car. Does that make sense? | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 181 Registered: 10-2001 |
Todd, thanks for dropping by. If you go down a few posts I think you'll see my attempt at a summary of what these two guys are in conflict about. You are in camp A - what do you think about the idea B? -Rich | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2892 Registered: 6-2001 |
from your link. hmm, is this because a car accelerates more at peak torque?? | ||
| Todd Wasson (Todd_wasson)
New member Username: Todd_wasson Post Number: 1 Registered: 4-2003 |
Somebody popped into sci.physics and asked for some outside input on this. Well, after reading through some of these posts it looks like there are a few of you that have this down pat and really don't need any help on this one! I haven't read everybody's comments and am not really clear on where this big debate currently is, but Mitch Alsup's posts have been spot on (perhaps there are others as well). If you are arguing with him (and those that agree), I strongly urge you to study the graphs and formulae he has provided. I'm not sure really where the current debate is (300 posts is too much to read through), but basically, where power and torque and acceleration are concerned, in a given gear, peak acceleration will be at the torque peak. I won't put out examples because this has been covered numerous times in this thread already. However, as some have pointed out, you've got gears that change the picture a bit. Shifting down (bringing the rpm's above the torque peak) generally increases the torque at the driving wheels (and therefore force and acceleration). Meanwhile, the engine torque of course falls off a bit. The question is, does the torque at the wheels from the gear ratio change increase more than the engine torque (times the old gear ratio) drops after the change? If so, you made a good move by downshifting. Again, I'll skip the examples because several excellent ones have already been given by others illustrating this. Somewhere in this thread is a graph showing wheel thrust vs. rpm throughout each gear, where the intersections of the lines denote the optimum shift points. This is precisely what that illustrates. When all is said and done, peak acceleration will occur at peak power, if you are free to change gears to keep the RPM near the peak continuously. Remember, what makes a car accelerate is not engine torque all by itself, but rather engine torque times the gear ratio. If you can double the gear ratio and only lose 20% of the engine torque because it's spinning so fast, you're much better off. For all practical purposes, "power" gives a quantity that reflects this precisely, hence the inclusion of power in dyno graphs. In any given gear, peak acceleration will occur at peak torque. Which goes to say that in first gear, peak acceleration will be at peak torque, but don't shift to second and third, etc., until *after* peak power. Center the revs around the power peak, not the torque peak, to maximize the area under the power curve as some described. This is where engine torque times gear ratio will be the highest. I.e., the torque at the driving wheels will be greatest.. I.e., where force and acceleration will be the greatest. Launch the car in 1st gear at or above the torque peak, then shift in a way that centers the rpm split to maximize the area under the **power curve** from there on out. This can be seen by grabbing a calculator and looking at some of the excellent Excell graphs that have been posted. For best acceleration, do you want an engine that makes 400 ft-lb @ 3000rpm or 400 ft-lb @ 6000rpm? If you are free to gear the car however you want, I'd take the 400 ft-lb @ 6000 rpm. You can double the axle ratio and get twice the acceleration. Hmmm.... You also have twice the power at that speed... The best way I think for folks to work this out is to get a basic handle on f=ma and try some simple calculations on paper like several in this thread have illustrated. Or you could go nuts and write a program like mine ;-) http://performancesimulations.com http://performancesimulations.com/scnshot4.htm That bottom link has a Ferrari 333SP and an F-50 ;-) This is the simplest layman explanation on this subject I've seen. Don't know that I could write one better myself: http://www.datsuns.com/torquehp.htm Not bringing anything new to the mix, J. Todd Wasson Performance Simulations Drag Racing and Engine Simulation Software http://performancesimulations.com | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 599 Registered: 2-2003 |
Yes, Chris, I'm sorry it's gotten out of hand too. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 598 Registered: 2-2003 |
See, Rich, what I'm trying to say is that just because an equation is derived from F=MA, doesn't mean it is the Same. We derive equations (carefully, to stay within Newton's Laws) so that we can accomodate more variables, such as time, distance and velocity, among others. Therefore, while W*D/T=P=Delta KE/t is derived from F=MA, it isn't the Same as F=MA, and can, in fact, answer more of the questions in this conundrum than can the simple truth that F=MA. I know that some cars accelerate maximally at peak torque, because two of the cars in Alsup's graphs showed that. I also know that other cars don't, from Alsup's graph, and from my own experience in Atlantic. I think this is far more complex than we think, and perhaps we're trying to evolve an understanding of a concept that wasn't clearly thought out before, who knows. I know that I'm tired of being vilified by Alsup, and I'm going to dinner!! Good night. | ||
| chris cummings (Entelechy)
Junior Member Username: Entelechy Post Number: 213 Registered: 5-2001 |
Actually, I fired that off for fun after reading Mitch's first post at the bottom of this thread. In reading further, I have to say I'm pretty impressed with both of you, but sorry to see this seems to have gotten out of hand. Hope you guys can work things out. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 597 Registered: 2-2003 |
Alsup, one last time before I go enjoy some human company. Power equals work/time, which equals work*distance/time. Work equals force times distance, correct? So, if at 4,000rpm,Force is equal to a value of 300, and distance (say the rotation of the crankshaft) is equal to a value of 1, then we have W=300, right? If at 8,000rpm force is equal to 250, and distance (two rotations of the crankshaft) is equal to 2, then W=500, right? Okay, so, now, let's say time equals 1 (just so we don't have any of that pesky calculus). W/t equals power, right, so let's see, Power at 4,000rpm equals 300units, while Power at 8,000rpm equals 500units. Then, let's make this all simple for us ignorant slob bastard National Merit Scholar magna grads from Penn, P equals delta KE over time, right? You agreed, right? Okay, so, at 4,000rpm and time equal to 1, P=Delta KE which equals 300units. At 8,000rpm, P still equals delta KE, right? and let's use the same time unit of 1, just so us simpletons can follow. P=Delta KE which equals 500units. Ooops! There's more of that pesky power to change Kinetic Energy at 8,000rpm than at 4,000rpm!! Conundrums get sticky on your shoes, don't they? | ||
| Tom Bakowsky (Tbakowsky)
Member Username: Tbakowsky Post Number: 354 Registered: 9-2002 |
Aahhh...is this the right room for an arguement? | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 180 Registered: 10-2001 |
Mitch may indeed be correct but as a member of the jury, I must say he is losing me due to his presentation. He becomes less worthy of trust with each post. Why? Because it's been my experience that those who can't state their own views without insulting other people personally are usually not so confident that their own views are indeed correct. They stray from the issue to the people involved in an attempt to make themselves feel like they must be correct because the other person is simply an idiot while they themselves are the most brilliant human to have walked the earth. Therefore, getting back to the issue. Is this it at this point? a) F=M*A is all that needs to be considered. And "F" is equal to torque alone and therefore acceleration is greatest at every engine's quoted output shaft torque peak. (cvt cars excluded). Evidence for position A includes, a bunch of websites, the brain of a hotshot chip designer, much circular mathematics, Newton (whose "laws" don't always hold), and of course Occam's razor. or b) there is a more complicated relationship which includes time into the equation and that what an engine is really doing is creating "work". Work is defined as F*D and power is defined as F*D/time. Since the amount of work possible varies with D, and in any discrete interval of time, that is governed by F*D/Time, or Power. Therefore it is possible that a motor producing F times D in interval Y, will be better able to raise Kinetic Energy if it produces F-(small quantity) times 2D (e.g., 8krpm v. 4Krpm) over the same interval of time. The engine is spinning twice as fast, producing nearly the same torque, ergo, more energy within the same discrete interval. Therefore some cars may accelerate more at some higher rpm than torque peak rpm. Support for B includes a graph from a magazine appearing to show that at least one car accelerates more greatly after the torque peak listed by its manufacturer, the brain of a lawyer, some hotshots from U of Penn, seat of the pants impression in race cars and timed runs in a modern Ferrari. Oh, and Occam's razor again. I believe Mitch thinks he has shown that the equations proposed by EFWUN are actually the same as F=M*A, but Mitch's proof of this relies on something that EFWUN can't accept: namely, in F*D=W, work(W) can be zero even if D is not zero. Mitch uses some fancy calculus that's way over my head to "prove" this fact that appears by plain simple math to be ridiculous. (Here's where EFWUN could throw Occams's razor back at Mitch! I would think infinitessimal periods of time dealt with by differntial calculus have little to do with real world cars, but maybe that's just me, ha!) If that's a fair summary of what's at stake so far, the issue really becomes how can we figure out if A or B is more applicable to the real world. Some would probably rather stick to the math, but real world timed runs in a real car that shows it accelerating more rapidly at or near redline than it was at published torque peak would be good enough for me. If no cars exists, then "it depends" enters the realm of theory and for all practical purposes we can continue to believe simply F=M*A. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 596 Registered: 2-2003 |
To Mitch Alsup: Please note that you just stepped into a pile of the conundrum you shrilly suggest doesn�t exist. To paraphrase you, torque creates low speed acceleration, power overcomes wind resistance. REALLY?? In so doing, you have indicated that you really don�t understand these concepts, and that all of your intimidating �calculus� is just that, intimidation. Are you used to getting your way by bullying people with whatever you feel might work? Let me run it down for you. I�ve been suggesting that there is an interplay between the maximal torque and maximal power, or Torque*D/time. You have screamed that the ONLY equation is F=MA, without regard for the fact that in actuality, we derive equations to accommodate more variables, so long as those equations adhere to the truth, which is F=MA. In this case, that equation is: P=W/T=delta KE/time. Simple. In truth, there IS an interplay between the point where torque is maximal, and the point where that torque does most work per unit time, or Power. Why not review these concepts overnight, (or whenever) and see if you still feel that your pejorative (yes, DES, there IS a Santa Claus) terms are necessary. Like most people on this forum and in real life, I respond to people who create an aura of apparent authority, (see, e.g., this fellow who thinks you're Hawking, and you respond pompously, "No, but I understand what he was talking about!" sure you do!) and this is why it�s taken me so long to see through your intimidation and obfuscation. In light of your apparent inability to understand the concepts involved, your position is untenable and your argument bankrupt. You are perhaps a computer scientist, but threatening to integrate or differentiate won�t change the apparent truth; you don�t understand the conundrum. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 577 Registered: 4-2002 |
"You know, Alsup, I'm growing tired of your tirades. P =W/T = Delta KE/t can be derived from the universal truth F=MA, you proved it, I proved it in earlier threads. What it says is simple, no matter how you try to complicate it:" So if I have proved it, then why cant we use F=M*a to arrive at the conclusion that it is force that causes accleration! "Power equals Work over time, which equals change in Kinetic Energy over time. " And both lead to the conslusion that max acceleration occurs at peak TQ. It is not the equations that are wrong, It is your conclusion! "Doesn't seem that, however advanced your calculus is, there is any need for intergration to reach the conclusion that Power equals ability to change kinetic energy as a function of time. " That is NOT what I am arguing. I am arguing that Power leads to the same conclusion as does Force! And that conclusion is summed up as F=M*a. If you can't disprove F= M*a then go shut the F*&K up. Can you not use you intellect to see that all equations lead to the same conclusion! Therefore, use the simplest and most direct means (hint F=M*a) to get from one of the things we know (TQ and HP) and use this (F=M*a) to get from 'F' to 'a'. Is this really too complicated for a lawyer that presents arguments to a appelate court? "I've asked you to stop poisoning your posts with personal invective; I've actually never run into a more humorless, egotistical fellow than yourself. " Ever looked in a mirror? "How do you change the output from the torque that overcomes inertia (a form of resistance) to the output from power that overcomes aerodynamic drag (another form of resistance)? Hint, it is the SAME output." But the conditions are different. At one end of the spectrum air resistance is ZERO, at the other end air resistance = HP. But to get acceleration at high velocities, you take the available HP, subtract the air resistance, then convert the remaining HP back to TQ (HP=TQ*RPM/5252 or for EFWUN who doesn't seem to be able to do math TQ=HP/RPM*5252) and then use the residual TQ to compute acceleration. Simple--at least for those who understand Math. "If you'd stop trying to lift your leg on my pants," Ah, wearing panties to court again,are we now? "you'd see that I'm saying there is a complicated interaction here, and that F=MA doesn't provide enough variables (time) (distance) (force) to explain everything." And I have disproven this several times. You don't seem to understand the work proven as used in science do you? It is simple: F=M*a; it is NOT complicated. "That's why we derive equations from this universal truth, to evaluate more complex situations." And use math that allows us to solve these problems without a lawyer misrepresenting my position to a friendly professor who really doesn't want to spend the time telling his friend that he is just plain wrong! "Please stop the tirades, and the personal attacks." I will stop denigrating you when yo start using sensible arguments. You have posited that E=1/2*m*v**2 arrived at a different conclusion than F=M*a, I showd this not to be the case, and indeed, it agrees with F=M*a. Then you shifted gears (so to speak) and posit that W=F*D would reach a different conclusion than F=M*A or E=1/2*M*v**2. I showed that this too arrives at the same conclusion. The denigration is all yours, and you deserve it for not being able to understand math after posing as someone who does. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 576 Registered: 4-2002 |
"Geez Mitch - did you ghostwrite "A Brief History of Time" under the name Stephen Hawking? Impressive :-)" No, but I did understand what Stephan was talking about. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 595 Registered: 2-2003 |
I'm going home to take some aspirin, take my car and pick up my wife for dinner. I'll say it once again, no more personal attacks. I will not tolerate them. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 594 Registered: 2-2003 |
To quote from Mitch Alsup: "Yes, it says that big TQ implies big acceleration at low velocities, and that power is needed to overcome air resistance at high velocity" How do you change the output from the torque that overcomes inertia (a form of resistance) to the output from power that overcomes aerodynamic drag (another form of resistance)? Hint, it is the SAME output. In other words, why is Power necessary to overcome drag, but not to overcome inertia resisting a change in kinetic energy? Hint, they are the same output!! If you'd stop trying to lift your leg on my pants, you'd see that I'm saying there is a complicated interaction here, and that F=MA doesn't provide enough variables (time) (distance) (force) to explain everything. That's why we derive equations from this universal truth, to evaluate more complex situations. Please stop the tirades, and the personal attacks. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 593 Registered: 2-2003 |
You know, Alsup, I'm growing tired of your tirades. P =W/T = Delta KE/t can be derived from the universal truth F=MA, you proved it, I proved it in earlier threads. What it says is simple, no matter how you try to complicate it: Power equals Work over time, which equals change in Kinetic Energy over time. I'm not saying that means that all cars accelerate hardest at peak power, but it does suggest that there is more going on here than your doctrinaire pronouncements. Of course, no one can disprove F=MA, that's not the point. You're spraying spittle, and calling me names won't change anything, other than the possible conclusion that you've hung your ego on the authority of your statements, and however great you are at computer design, this is simple internal combustion, creating a change in the kinetic energy of a vehicle, not EE, and not CE. Doesn't seem that, however advanced your calculus is, there is any need for intergration to reach the conclusion that Power equals ability to change kinetic energy as a function of time. And no, I don't work with computers, I use them to type. Your graphs are lovely; I had a friend at law school who graduated Penn magna many years after I did, (only in some computer field), and he wasn't really very good at anything else, and couldn't relate his computer near genius to anything resembling real life. I'm giving you lots of credit here; not many people would equate Carnegie-Mellon with Penn. I've asked you to stop poisoning your posts with personal invective; I've actually never run into a more humorless, egotistical fellow than yourself. This isn't any fun, and like I said, I've asked you politely to desist. I'd like not to have to ask again. | ||
| chris cummings (Entelechy)
Junior Member Username: Entelechy Post Number: 212 Registered: 5-2001 |
Geez Mitch - did you ghostwrite "A Brief History of Time" under the name Stephen Hawking? Impressive :-) | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 574 Registered: 4-2002 |
"Proof is not bandied about in law," Proof is used in law all the time in a much less rigorous sense than proof in mathematics. Proof in matematics is not used until ther can never be any other conclusion. Proof in law often is shadow of doubt level "May it please the court, I have proven my client is not guilt....." when in fact he may very well be guilty. "at least not at this level, and in fact, " "credibility is the essence of this or any argument;" You lost all credibility when you refused to accept F=M*a Then you posit that E=1/2*m*v**2 would change the conclusion: I showed that this leads to the same conclusion as F=M*a Then you posit that W=F*D would change the conclusion: I showed that this lead to the same conclusion as F=M*a There you have it, gentlepeople of the jury, you can conclude that different equations in physics leads to different conclusions and therfore physics is untrustworthy; or you can, rightfully, conclude that all the equations need to arrive at the same conclusion and, therefore, that physics reamins trustworthy. "your continual denial of things you posted, and re-re-explanation of things you posted that clearly prove me correct" Maybe if you had the computer skills to download a *.jpg, and add pointers and number to explain your lack of understanding, perhaps I could help. But you level of misunderstanding of the underlying physics and math is incredable, possibly unbridgable. And for a lawyer, you don't seem to be able to read all that well, but perhaps this coinsides with you lack of understanding inother areas. "(my thesis, "it depends")" Is wrong--get over it! "sufficiently denigrate your credibility such that I need make no more offensive statements of the character you continue to make. " It is not character that is on trial here, it is physics. Show why F=M*A is wrong, or go shut the F*&K up. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 573 Registered: 4-2002 |
"I figured that if I left you a way out of our dilemna with your ego intact, you might understand that there can be no equivocation as to the answer to this question and that is simply "it depends." " It does not depend! F=M*a implies that max 'a' occurs at max 'F' there is no other way to interpret this equation (by intellegent people). "Remember the old saw, torque equals low elapsed times, horsepower equals high mph (in the quarter mile)." Yes, it says that big TQ implies big acceleration at low velocities, and that power is needed to overcome air resistance at high velocity. However, we are discussing where does maximum acceleration occur a) without air resistance, b) in cars with gears. We have agreed that 1st gear accelerates faster than higher gears. So where in 1st gear doe peak acceleration occur? A: at max TQ. "Remember also that as the output shaft spins, the torque it produces is really F*D, or Work. Therefore, the amount of work possible varies with D, and in any discrete interval of time, that is governed by F*D/Time, or Power." I showed that his set of equations can be converted into the F=M*a equations. You have never found a way to get around that issue. Because you can't! Its the Law (Newton's)! Are you too stupid to understand that? Or are you too ignorant to be able to believe in freshman level math? "I believe there is a complicated interaction here that makes this more difficult than F=MA. " Then you are wrong. And after being told you are wrong many time, shown how you are wrong, you continue to believe; make me think you aren't that good a lawyer to begin with. Until you can find some way to disprove F=M*a you will stay wrong. Idiot. "This is a complex issue, not in the physics, but in figuring out what the interplay of force and energy really is." F=M*a F is related by multiplication to TQ M is fixed a is whatver the equation causes it to be based on given F and given M How hard it that? As hard as you head? Are yo so thick that you cannot see that you can never escape from F=M*a. It is the simplest form relating force, mass, and acceleration. Why don't you get you so called professor friend to figure out what is wrong with F=M*a? | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 572 Registered: 4-2002 |
"Well Mike, U Penn just said that my derivations, which yielded the equation: P= W/T = delta KE/time were completely valid, and agreed with the derivations. " But did you bother to ask him what is wrong with my use of those equations. Either my use of the F=M*a equation is wrong, or you are wrong. It is simple as that. "Simply put, you stated that you would disprove my equation, and then you publicly proved it correct. Do not now deny that. " I proved your equations sits in the family of equations that connect accleration, velocity, energy, power, and work. As such, that equations PROVES that F=M*a can be used to compute acceleration from a force (TQ). Peak Force results in peak acceleration (mass being constant). What you don't seem to be able to get though your thick skull is that the rest of the world used F=M*a to determine acceleration given a force. You continue to be so pig-headed to avoid actually understand the physics and math involved. The use of a so called physics professor that cannot find the error in my equations (if there is one) denigrates not me but him. You ignorant bastard. "Your statement that you "argue in front of the Supreme Court everyday" is beneath contempt. We all believe our jobs are important, trust me, neither of us are of that elevated a stature. Careers rise and fall in state court all the time, responsibility to your shareholders or officers is not "The Supreme Court." " There are hundreds of lawyers that argue infront of the highest courts in the land. There are at last cound about a dozen CPU architects currently practicing. Why is that? Because its such and easy field? "As I said, I started on this site as a means to enjoy pleasant discussion of my hobby with others who enjoy the same hobby and passion. I do not enjoy being endlessly vilified by you, and I don't enjoy you trying to wiggle out of proving my equations correct, or posting a graph proving my thesis, "it depends" correct. " In order for you to achieve this position, you have to FIND something WRONG with the F=M*a derivation. You have not found anything wrong with F=M*a, nor has your so called professor. Why don't you have your so called professor come on to this board under an untraceble name and argue those subjects that it is so clear you are incapable of arguing? "Further, no amount of calculus can turn F*D into zero, where D is not equal to Zero and F is not equal to Zero. " This is a definition of someone who does not understand calculus. Taking a position of authority and not understanding the subject mater is grounds for calling you an idiot. And you are an idiot. But I digress: dt in calculus is the representation of a period of time so small that from a human persepctive, no time has past. It is used to find the slope of a curve at a given point. It requires the knowledge of how 'LIMIT(f(x))as t->0' works--another thing you aparently do not understand. You silly little ignorant slob. "I fully understand these concepts, " It is obvious that you do not, and cannot without years of succesful additonal schooling. "that doesn't license you to respond by further denigrating me." But your being stupid and failing to accept the inevitable loss in theis discussion does. F=M*a. Big force causes big acceleration; little force causes little acceleration. Get it--or do you want to remain stupid? "I do not enjoy being endlessly vilified by you," Then achieve the understanding that F=M*A works for this problem. "and I don't enjoy you trying to wiggle out of proving my equations correct," I have not wiggled out at all. I am still here fighting for math and physics, are area where 'it depends' is NOT an acceptable answer. F=M*a. "or posting a graph proving my thesis, "it depends" correct. " I disagree with you conclusion here. But since you seem to be stuck to only words, and cannot express yourself with graphs, formulae, and an understanding of the math, then you cannot point out the (so called) error in my 'multiple' derivations showing that all forms: F=M*a E=1/2*m*v**2 W=F*D all lead to the same conclusion: max acceleration occurs at max TQ in any given gear. It doens't matter to me how stupid you are, until you get someone who can demonstrate what is wrong with F=M*a, OR you take a new position, your positioin remains cannon fodder. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 589 Registered: 2-2003 |
Well Mike, U Penn just said that my derivations, which yielded the equation: P= W/T = delta KE/time were completely valid, and agreed with the derivations. I think the real answer is "it depends", and because we all shift at redline, and that usually drops the motor to revs that are still well above the torque peak I think we need to concede that there are other variables entering the mix, and acceleration isn't necessarily always hardest at peak torque. Remember the old saw, torque equals low elapsed times, horsepower equals high mph (in the quarter mile). Remember also that as the output shaft spins, the torque it produces is really F*D, or Work. Therefore, the amount of work possible varies with D, and in any discrete interval of time, that is governed by F*D/Time, or Power. What I'm trying to say is that if you think of the motor as a simple output device, and think of its output in a flexible construction, you might well come to believe that a motor producing F times D in interval Y, will be better able to raise Kinetic Energy if it produces F-(small quantity) times 2D (e.g., 8krpm v. 4Krpm) over the same interval of time. Simply, it's spinning twice as fast while producing nearly the same torque, ergo, more energy within the same discrete interval. I believe there is a complicated interaction here that makes this more difficult than F=MA. Incidentally, I've found physics departments hydebound as well, the same profs argued with me that all cars should (in theory) stop in the same distance. No equation explains this multi-dimensional conundrum fully; that's why I'd like Mitch Alsup to let this rest at "it depends", that and the fact that I haven't enjoyed his personal attacks, nor other things I'll leave rest here. Further, I think that graph is accurate, and the fact that it speaks of acceleration increasing as power ramps up, doesn't necessarily mean it's wrong. Finally, Mitch Alsup proved my equations correct, and should not now deny them. This is a complex issue, not in the physics, but in figuring out what the interplay of force and energy really is. Like I said, I'd like to enjoy the forum for discussions of Ferraris, not defending myself from personal attacks. | ||
| Mike B (Srt_mike)
Junior Member Username: Srt_mike Post Number: 135 Registered: 12-2002 |
EFWUN, Mitch, I really enjoyed this thread - not because of the bickering, but because both of you gents are obviously professionals in your field. It was sort of like watching Jeff Gordon and Michael Schumacher compete in a Rally race... not something they make a living at, but they are both pro's and it was interesting to behold. Now, having said that, I am STILL interested in the ultimate conclusion here. It seems the point of debate is that the graph shows a VW accelerating hardest around 3,200RPM or so, yet the stated torque peak for this vehicle is 1,900RPM or so. This deviation is what you, EFWUN, are saying validates your point. Mitch is saying obviously the graph is wrong. I am not trying to take sides, but I agree with Mitch. I dont have the level of understanding he does of this subject (nor do I need to), but it's something I investigated alot a while back, even going to my alma mater and talking to the head of the physics dept about - and he explained to my why I was wrong in thinking a car accelerated harder at any other time than peak TQ. If that graph is correct, then one of two things is going on... either peak TQ was actually produced at 3,200RPM (or whatever exact RPM it was), or the graph is wrong. Mitch provided a plausible argument as to why the graph is at it is - the turbo kicking in. EFWUN I don't think you will concede that the graph is either wrong or that peak TQ did occur at 3,200RPM. But what I would be interested in is what is the deal with the dept. of physics at UPenn? They are saying acceleration isn't hardest at peak TQ? What is their reasoning? Can we put Mitch in touch with the professor in question and solve this and maybe get a public answer posted? For no reason other than to see where they are coming from - because if indeed some professor at UPenn is saying that, it goes against my understanding of this completely and I would like to hear their explanation. Also, being a stubborn prick that I am, I know that acceleration is hardest at peak TQ, and I would like my belief vindicated That is NO disrespect to you EFWUN, but a few people seem wishy-washy on the veracity of that, but I know there is no deviation from it, and I hate to be anything less than right when I know I'm right. I think this thread has great value. P.S. Mitch, do you know Jeff Ellison at AMD? I made a killing a couple of years back - AMD's stock was down around $11 and I bought a ton against everyone's warnings. I know a fair bit about computers at the component level and I was sure the Athlon would be a hit. It was. The stock went up to around $85 I believe and then split and went to around $65. My best stock success ever! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 586 Registered: 2-2003 |
Reply to Mitch Alsup: Despite my well-intentioned apology you continue to denigrate my understanding of physics, and denigrate the entire department of physics at the University of Pennsylvania (your "so called" professor). It is apparent that you have convinced yourself that the graph you posted somehow shows something it doesn't show. That graph, no matter how you now re-explain it and re-produce it, shows that the VW accelerated at peak G well past peak torque. You insist that peak torque occurs well past all published reports. You are wrong. You proved that W/t = P= delta KE/time, and that equation cannot be equivocated by smearing me, my knowledge or understanding, or the professors to whom I deferred. Simply put, you stated that you would disprove my equation, and then you publicly proved it correct. Do not now deny that. Proof is not bandied about in law, at least not at this level, and in fact, credibility is the essence of this or any argument; your continual denial of things you posted, and re-re-explanation of things you posted that clearly prove me correct (my thesis, "it depends") sufficiently denigrate your credibility such that I need make no more offensive statements of the character you continue to make. I apologized as a method of defusing what is turning out to be a really vicious, unpleasant argument. I figured that if I left you a way out of our dilemna with your ego intact, you might understand that there can be no equivocation as to the answer to this question and that is simply "it depends." You have instead thrown this putative olive branch back in my face, and rather than apologize sincerely as did I, you apologize with the caveat that it is okay for you to denigrate me!! Succinctly, designing chips may well give you a facility with math, it does NOT allow you to re-write Newtonian physics, particularly after you yourself prove (inadvertently) that my humbly derived equation is absolutely correct. Further, no amount of calculus can turn F*D into zero, where D is not equal to Zero and F is not equal to Zero. F*D unequivocally equals Work, and therefore, in the nanosecond that Torque (F) causes ANY rotation, you have WORK, plain, simple and unambiguous. I fully understand these concepts, (and I'm certain you do too) but it seems you're unwilling to examine the plain truth, that over a discrete interval of time, torque must be evaluated as Work/time, and that equals Power, unequivocally and by your own proof of my derivation. Therefore, while torque is the force responsible for overcoming resistance or initial inertia, once any work (F*D>0) is accomplished, we must evaluate power output and its ability to increase kinetic energy as a function of time. As I said, I apologized for pushing your buttons, that doesn't license you to respond by further denigrating me. You have posted some real clunkers, including: no wear at 7000rpm (except, of course "ancillary components!"); and your post calling me to the Weehauken interview and then shooting yourself with your own slide rule by proving my equations unequivocally correct; and the graph that clearly showed that while one car accelerated at peak G exactly at peak torque, ANOTHER ONE DID NOT. Your statement that you "argue in front of the Supreme Court everyday" is beneath contempt. We all believe our jobs are important, trust me, neither of us are of that elevated a stature. Careers rise and fall in state court all the time, responsibility to your shareholders or officers is not "The Supreme Court." As I said, I started on this site as a means to enjoy pleasant discussion of my hobby with others who enjoy the same hobby and passion. I do not enjoy being endlessly vilified by you, and I don't enjoy you trying to wiggle out of proving my equations correct, or posting a graph proving my thesis, "it depends" correct. I once again urge you to concede that the answer is "it depends" and apologize for your offensive remarks; further "argument" can have no possible good outcome. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 567 Registered: 4-2002 |
"Initially, let me apologize for things I may have said that pushed the wrong buttons in you. Reviewing your posts on other threads, I see you are (perhaps rightly) used to being an authority, and if I inadvertently turned this into a pissing contest, I apologize. I hope a similar apology will be forthcoming from you, because you�ve said some pretty offensive things." Appology accepted. I will return the favor by appologizing for the irrevrent choice of words I used to describe your lack of understanding of the math and physics involved in this discussion. I do not appologize for calling you stuborn, as you are the third most stuborn person with whom I have ever had a conversation. Indeed, you did end up pushing my buttons, and the major button was taking a position of athority on an area where it was clear from your choice of words you did not really understand the subject matter (on which you chose to present yourself as an authority). This is the calculus line permiating through this thread. In the school I went through (Carnegie-Mellon) we had calculus drummed through our heads until all we could think was calculus. If you ever saw the movie/TV-series "The Paper Chase" simply convert reading 'law stuff' to reading 'math stuff' and this is a good representation of an engineering education at CMU. Math was calculus, Physics IS applied calculus, engineering was the use of the calculus we learned only an hour ago, chemistry degenerated into calculus, even the history course was on some of the side line players in (you guessed it) the creation of calculus. "I recently argued a civil rights case in the Second Circuit Court of Appeals, which you may know is the court directly below the Supreme Court of the United States. (Not ambulance chasing, as you can see) One justice took my adversary to task for miss-stating the holding in an important case, saying �You lost all credibility with me when you did that, and your argument with it.�" I am happy that you find your work so fulfilling. And now I see why you so haplessly used 'prove' for 'demonstrate' or 'show'. In the world of math, physics, and other sciences, 'prove' is ONLY used when no other conclusion can EVER be reached. I understand that in the world of law that the word 'prove' is thrown around with reletive abandon. We are/were arguing a math, physics, science issue, not a law issue, so you need to use the math, physics, science definitions of the words, or your argument will fail. This time it did. If you go back to your so called professor and have him review my facts and figures, he will substntiate my position. And, indeed, a week ago, I suggested that he and I work this out offline, but NOOOOOO, you had to continue fighting for something you simply do not understand. If you want to understand go back and get an education in engineering. Choose a school where the first engineering class utilizes calculus, and one where you cannot graduate unless you really know the stuff. I happen to design computers for a living (CPU chips). I was the chief architect of Motorola's Mc88000 architecture, I was CTO for a medium sized startup company, and now I am back as chief architect of the K9 chip design at AMD. This is the CPU that comes after the 'Hammer' generation of processors. I have to be able to use this kind of math everyday. If you google me, you will find that I have numerous patents, puplications, and am held in high esteem by others who practice my art. In my field, I am arguing in from of the Supreme Court, constantly. If the chip (aka argument) fails, so does the company. Math, physics, automobiles, optics, and astronomy are sciences I understand at doctorate level, and in many circles I am consider an authority figure in these areas. I spend most of my free time adding to my collection of knowledge of these areas. In the future if you start to get into an argument and someone complains that you are not using the correct words (prove, demonstrait, show) to associate your argument to the subject at hand; do yourself a favor, and restudy the subject. Question: At what point does a vehicle accelerate at maximum, peak torque or peak power? Answer: In the absense of: a) air resistance, and b) tractive issues, and c) change in the mass of the car by consuming fuel to perform work; the only answer is the one congruent with F=M*a, and that is at peak TQ. This answer is also computable from W=P*D, and E=1/2*m*v**2, as it must in any complete set of physical system of equations. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 585 Registered: 2-2003 |
Letter to Mitch Alsup: Initially, let me apologize for things I may have said that pushed the wrong buttons in you. Reviewing your posts on other threads, I see you are (perhaps rightly) used to being an authority, and if I inadvertently turned this into a pissing contest, I apologize. I hope a similar apology will be forthcoming from you, because you�ve said some pretty offensive things. I recently argued a civil rights case in the Second Circuit Court of Appeals, which you may know is the court directly below the Supreme Court of the United States. (Not ambulance chasing, as you can see) One justice took my adversary to task for miss-stating the holding in an important case, saying �You lost all credibility with me when you did that, and your argument with it.� My point is simple, when you publish a graph showing a car that doesn�t accelerate hardest at peak torque, and then back track and even try to state that peak torque occurs in another place than that published in several magazines, you lost credibility. That graph alone shows that the answer is �IT DEPENDS.� I made some earlier posts in haste, and without proper thought and evaluation; I publicly owned up to those posts, and said they made me look �silly� and �like I have early-onset Alzheimers!� You remain unrepentant about your earlier posts that make little sense ( I won�t repeat them here). Further, you challenged me to an �interview at Weehauken�, slide rules at high noon. You came back and triumphantly announced that MY derivations and equations were correct! I stated that my equations have the necessary flexibility to delve into this difficult concept; having proven my equations correct (and the University of Pennsylvania physics department concurs), you still won�t allow that there is room to meet in the middle, e.g., Question: At what point does a vehicle accelerate at maximum, peak torque or peak power? Answer: IT DEPENDS. I would like to put this rather vicious argument to rest. I do not enjoy waking up in the morning and looking to see how you may have vilified me overnight, nor do I enjoy digging out from under a morass of rather intellectually dishonest statements. If my actions put your back against a wall and made you act this way, I apologize (see supra.) Having said that, however, it is proven that P=W/t=Delta KE/t. Torque is a twisting force that overcomes resistance, Power is the motor�s ability to accomplish Work (or Delta KE) over time. The acceleration of a vehicle is a complex interaction of these two factors, and the answer to which the question above MUST be �It Depends.� I hope you will join me in putting this dispute to rest with that as the coda. I would like to enjoy this forum for its capacity for pleasant discussion of my hobby, rather than vicious, no-holds barred argument. Sincerely, EFWUN | ||
| PSk (Psk)
Member Username: Psk Post Number: 387 Registered: 11-2002 |
Rob, This has got way out of hand. I think we should put this NEW thread in the shame area. Come on guys lets let this one die, and agree to disagree, or we will have to get your Mum's involved (no disrespect to anybodies Mum intended) Until the day I die I will be comfortable in the knowledge (or belief) that in gear acceleration peaks at peak torque ... but I am 100% comfortable that others have different beliefs. Somebody way clever than me came up with F=m.a and I'm sure if he was wrong somebody would have rained on his parade by now. I'm going for a drive , and my cars acceleration will peak somewhere ... but I'm not sure I care where anymore. Pete | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 177 Registered: 10-2001 |
I must say I'm happy the earlier part of the thread has been archived separately so that I don't have to scroll right on every line to read them (something didn't wrap as it was supposed to: possibly only a problem on macs). Forgetting physics and math for a moment, from a philosophical standpoint, EFWUN's style is much more pleasant and less filled with common argumentative fallacies than Alsup's, though both of you are a bit much at times. As a pragmatist, I don't much take to absolutes and so the answer "it depends" is also more attractive. Even when the science is concerned because we all know that "proof" and "laws" change over time as new perspectives are adopted: ask Newton and Euclid. Rob, one vote for letting it continue. The content - whenever we get there - has the potential to outweigh the bickering. | ||
| Tom Bakowsky (Tbakowsky)
Member Username: Tbakowsky Post Number: 347 Registered: 9-2002 |
Holy smokes guys!! Is this a mathimatic's club membership quiz? I can't beleve this post has been going on this long. Anyways torque is a turning or twisting force. eg when you turn a cars steering wheel you apply torque to the steering wheel. Engine torque is a rating of the turning force at the engine crankshaft. Eg. when combustion pressure pushes the piston, a strong rotating force is applied to the crank, through the trans and out to the wheels, propelling the car. Horsepower is a measure of an engines ability to perform work. I realize that you guys already know this stuff. I just thought I'd throw it in so you don't have to scroll through 400 posts to fined it. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 566 Registered: 4-2002 |
 I have annotated three points on this graph I want to discuss. point 1: This is the point of maximum HP as given by Road and Track magazine 180 HP @ 5500 RPMs. Using HP = TQ*RPMs/5252 we find that the TQ needed to produce this HP is: Tq(5500) = 171 lb-ft Point 2: This is the point of maximum TQ reported in Road and Track (173 lb-ft at 1950 RPMs). I draw the readers attention to the fact that the TQ at max HP is 99% of the TQ at max TQ. I have drawn a flat line from the point of max HP to the point fo max TQ. Point 3: This point is the time when the turbo has spooled up but the waste gate has not yet opened. The extra boos pressure creates a temporary surge in TQ and acceleration. Had the turbo spooled up earlier, the bump would have been earlier, had it spooled up later, the bump would have been later. So this graph does NOT show the engine delivering peak TQ at the RPMs that peak TQ is supposed to happen! At this point it is clear that the MT graph is showing the turbo lag of this engine and this particular graph needs to be discounted from the discussion. This graph represents an anomoly, and good scientists discuss and dismiss anomolies while discussing their data. A second anomoly is the perfectly flat TQ curve (99% of max TQ at mx HP.) any minor noise on the dyno, and the RPM of peak TQ can move around considerably. In other words, in this kind of engine configuration, the point of max TQ is 'not a single RPM' but a broad range of RPMs. Even the Ferrari engine, we so love, don't deliver this much of their peak TQ at the point of peak RPMs. The F355 engine I showled earlier is delivering only 95% of peak TQ at peak HP. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 565 Registered: 4-2002 |
"Then, you reposted your graph, but smaller, so the "as power ramps up" is less legible, and you make an intellectually dishonest claim that torque peaks at 3,000, despite clear evidence in magazines that it peaks at 1,950rpm. " It is now clear that my posit that peak TQ occurs at 3000 RPMs is and was false. I admit that. The graph shows that the turbo has not kicked in by the time that the TQ RPMs have been passed. My contention (now) is that if the turbo was already kicked in (at the time the enine was at 1950 RPMs) that peak acceleration would have occured at that point (1950 RPMs turbo kicked in). With the turbo not kicked in, the engine did not produce peak TQ in this run at the RPM where peak TQ would have been produced if the turbo was kicked in. However, this particular (of the three graphs) shows that peak acceleration occurs after the turbo kicks in, rather than at the RPMs that peak TQ occurs. My contention is that if the turbo had already kicked in, that peak acceleration would occur at peak TQ. "Alright Mitch, look, you said you'd disprove my equation, and you inadavertently proved it was true." No I said I would disprove (and have) your conclusion. "There is apparently no way to convince you that you've just proven me correct, even trying to mollify you doesn't seem to work." There is aparently no way to convince you that peak acceleration occurs at peak TQ. This is all that I have shown (other than the ability to do math). "W/T=P, which equals Delta KE over time. Plain and simple, proven by both of us, and indicative that power creates delta V over T, or Acceleration." Shown by you, proven by me. Do you know the difference between show, derive, and prove? And P=F*v*a leads directly to the conclusion that peak acceleration occurs at peak TQ. We solved the equations 3 different ways, and arrived at the same conclusion. It is not the equations that are amiss, it is your conclusion that is amiss. "Are you really going to say that W/T=P=delta-KE/t is the SAME as F=MA? " These two equations describe different aspects of the same thing! One (who is skilled in math) can transform either into either. "I suggest we find an amicable way to end this dispute" Then give up, and say: "Peak acceleration occurs at peak TQ". You still never addressed the other web sites and litterature references that agree with me, nor have you addressed why all the other graphs presented agree with me. Nor have you shown the ability to do the math necessary to clarify your point, or to solve the equations. QUESTION: "at what point does a car reach peak acceleration?" ANSWER: at peak TQ in any choosen gear. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 584 Registered: 2-2003 |
Alsup, enough already. Your own posted graph doesn't agree with you, why should I? Are you really going to say that W/T=P=delta-KE/t is the SAME as F=MA? Just because my equation can be derived from the fundamental truth doesn't mean it is the same, and the increased number of variables and the addition of the time factor make it more flexible in describing various scenarios, while still adhering to Newton's fundamental law. I've reached the point where I am satisfied with my equations and argument. The fact that EVERY car does not accelerate hardest at peak torque is proven by your own graph, your own proofs of my equations, and by respected journalists and physics profs. I would like to have an evening with my wife, my Ferrari, and some aspirin for the Alsup headache. I suggest we find an amicable way to end this dispute, and I suggest that amicable end might well be if you admit what all the proofs have shown: QUESTION: "at what point does a car reach peak acceleration?" ANSWER: "IT DEPENDS." | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 583 Registered: 2-2003 |
Alright Mitch, look, you said you'd disprove my equation, and you inadavertently proved it was true. Then, you reposted your graph, but smaller, so the "as power ramps up" is less legible, and you make an intellectually dishonest claim that torque peaks at 3,000, despite clear evidence in magazines that it peaks at 1,950rpm. There is apparently no way to convince you that you've just proven me correct, even trying to mollify you doesn't seem to work. W/T=P, which equals Delta KE over time. Plain and simple, proven by both of us, and indicative that power creates delta V over T, or Acceleration. Why don't you stop now, before you get yourself convicted of the Jon Benet Ramsey murder or something? (Just joking, really, this has to end somewhere. You've proven my equations correct, and I've suggested that the real answer is "it depends" as a respected automotive journalist stated several weeks ago! I suggest we call a halt!) | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 564 Registered: 4-2002 |
"Please stop the nonsense, you set out to disprove my equations, and inadvertently proved them 100% correct." Equations are correct, you conclusions are faulty. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 563 Registered: 4-2002 |
"I'm no longer contending that EVERY car accelerates maximally at peak Power, or even that most cars do. I've proposed an equation that proves with particularity that "it depends!" " So, now, all you are contending is that peak acceleration occurs somewhere/anywhere other than where Alsup contends it occurs? "I've proposed an equation that proves with particularity that "it depends!" " You have proposed an equations that delivers identicle conclusions as the other equations. "it depends" is not on the available list of answers. "Why not admit that I've re-learned my physics" Noting in you posts lindicates that yo have the foggiest notion of how to use the equations of physics, nor the calculus that ties the bunch together. "Is there NO meeting of the minds with you? Will you now continue to argue that I'm completely wrong, even after you inadvertently proved me correct?" I showed that your equations (both sets) lead to the same conclusion as F=M*a. You don't like thie conclusion--thats not my problem. Until you say "yes, indeed, paek acceleration occurs at peak TQ" this argument will continue. BTW I did not disprove your equations, I proved that they exist in the complete system of equations governing acceleration mass, and velocity. I drew a different conclusion that you did from these equations. My conclusion is in agreement with the available litterature. Your is not. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 562 Registered: 4-2002 |
"Alsup, you've posted equations that prove, beyond doubt that: Power = Delta Kinetic Energy over time You've proven that equation can be DERIVED from F=MA, thus proving that it is true." So far so good. "If Power equals the rate of change of kinetic energy," still on target. "and therefore acceleration," Here is where you argument goes wrong. The rate of change in 'KE' is nonlinear with respect to the rate of change in 'a'. dE/dt = P = m*v*a Both the 'v' term and the 'a' term are related by the differential equation: dv/dt = a So if you add a constant amount of power over a period of time, acceleration is NOT constant over that period of time! However, 'v*a' is constant over that period of time. And since we are talking about acceleration, 'v' over that period of time necessarily increases, therefore 'a' has to decrease because 'v*a' is constant. "how can you continue to argue any further?" The real question is how can you? "You've also posted graphs proving that in at least one vehicle, acceleration is maximal past peak torque. How can you argue that "it depends" isn't the correct answer to this conundrum?" You seem to believe I said something I did not. I tried to correct the problem by restating my position, then reposted a graph restating my position, and you still don't understand my position. Perhaps the problem is not in my position? "My equations, (which you've proven obey Newtonian Law), explain a host of things you haven't been able to explain" Like the fact that maximum acceleration occurs at peak TQ. That is the only thing I have tried to prove, and I have been successful using three different computational methods (5 actually if you consider that doing differential equations are different than numerical integration in eXcel). "ALSUP!! You've proven my equations to be true beyond any doubt, how can you now say they aren't true? " Your equations agree with my position! get over it. However, the conclusion you draw from those equations is different than the conclusion I draw from those equations. Litterature and web sites back my position--your mouth backs yours--which do you think stronger? | ||
| Rob Lay (Rob328gts)
Board Administrator Username: Rob328gts Post Number: 4544 Registered: 12-2000 |
Should this be Threads of Flame and Shame now or everything still cool with this thread? There's much valuable information in here, although I haven't been reading much lately. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 582 Registered: 2-2003 |
Alsup, look, you can repost that graph and claim that torque peak occurs wherever you like, the plain fact is that R & T tested the Turbo S, and torque peaks at 1,950rpm and stays flat from there. Peak acceleration occurs at 3,000rpm, indicating that "as power ramps up" acceleration is increasing. Please stop the nonsense, you set out to disprove my equations, and inadvertently proved them 100% correct. I'll repeat, I'm not asking you to fall on your sword, I just think you should now accord my equations the respect you denied previously, and admit that the answer is "IT DEPENDS!"" Lawwdog, I don't think trading phone numbers would be a very good idea, do you?? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 578 Registered: 2-2003 |
Why not admit that I've re-learned my physics, and derived an equation that explains far more of this difficult concept than a simple F=MA?? I'm no longer contending that EVERY car accelerates maximally at peak Power, or even that most cars do. I've proposed an equation that proves with particularity that "it depends!" You said you'd disprove my equation; in fact, you proved it correct. Is there NO meeting of the minds with you? Will you now continue to argue that I'm completely wrong, even after you inadvertently proved me correct? I'm not asking you to fall on your sword, this is a discussion, not a duel at high noon with holstered slide rules!! Examine the proofs you've laid bare. Can you not now admit that the answer to the question, "at what rpm does a motor accelerate the vehicle at max A" is: "IT DEPENDS!" | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 577 Registered: 2-2003 |
Alsup, PLEASE STOP!! You know EXACTLY what I mean when I say you've algebraically done away with variables; that is the ESSENCE of "simplifying" an equation! My equations can be derived FROM F=MA, proving their truth, that doesn't mean they are the SAME as F=MA!!, only that their truth is implicit, given that they can be DERIVED from that fundamental truth!!! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 576 Registered: 2-2003 |
Alsup, you've posted equations that prove, beyond doubt that: Power = Delta Kinetic Energy over time You've proven that equation can be DERIVED from F=MA, thus proving that it is true. If Power equals the rate of change of kinetic energy, and therefore acceleration, how can you continue to argue any further? You've also posted graphs proving that in at least one vehicle, acceleration is maximal past peak torque. How can you argue that "it depends" isn't the correct answer to this conundrum? My equations, (which you've proven obey Newtonian Law), explain a host of things you haven't been able to explain (or admit, at least). ALSUP!! You've proven my equations to be true beyond any doubt, how can you now say they aren't true? | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 561 Registered: 4-2002 |
"However, that doesn't mean that simplification yields the same result," Each means arrives at the same conclusion: max acceleration occurs at max TQ in any particular gear. You have tried the "I'm a good guy Mitch is a bad guy--therefore, I get to win the argument" argument. Sorry, but the math, the physics, and the web litterature are on my side. "because you've algebraically done away with variables implicit in the understanding of this dynamic situation," Lets see, we still use F, m, a, E, t, and P. What fundamental stuff has been left out? " e.g., a motor creating the ability to change velocity at an increasing or decreasing RATE of change." We solved this through F=M*a, We solved this through E=1/2*m*v**2 and now We have solved this trough W=F*d "In proving that my equations are resolvable into F=MA, you've proven their implicit truth." That you are wrong, and that I am right! "That doesn't mean they don't shed more light on the subject than that simple truth, F=MA!! " Since any of these can be converted into their other forms, the form a complete system. From 'E' I can get to 'a', and from 'a' I can get to 'E'. From 'P' I can get to 'a', and from 'W' I can get to 'a'. You, however, don't seem to be able to get it! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 560 Registered: 4-2002 |
For the record I am claiming that peak TQ occurs at the following points:  While max HP occur higher up in the RPM band. "What can, must and will be disproved is that you can arrive at an understanding of acceleration over time and distance using F=MA." Just exactly what is wrong with integrating acceleration over time to get velocity and then integrating velocity over time and getting distance? Seems pretty straightforward to me! But then I actually understand calculus. Lets see: we have established that F=M*a. This establishes the connection between accleration and force. Velocity is the integral of accleration over time. Distance is the ingeral of velocity over time. The only way you can avoid arriving at the conclusion that Force can be used to determine distance is by a complete lack of understanding of calculus! "It necessarily follows that F=MA and W=Delta KE are TRUE, that doesn't mean they describe the same thing!!" I did not say ther were the same 'thing', I said you could derive F=m*a from W=F*d. Since I have shown this derivation, it follows that F=M*a and W=F*d are simply differnt manifestations of the same underlying physics. And since they are simply different ways of computing the same physical relationship, it also follows, that if you can solve the F=M*a and get max acceleration occura at max TQ, then you will reach the same conclusion through W=F*d. | ||
| Mario B (Lawwdog)
New member Username: Lawwdog Post Number: 18 Registered: 12-2002 |
Something (more) to think about: You could both spend less time hashing this out over a telephone call. How about trading numbers? Then again, this may change the odds on the office pool. : ) | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 575 Registered: 2-2003 |
Alsup, to reiterate, I never said my equations were at odds with F=MA, to be correct, by definition, they must be resolvable into that fundamental law. However, that doesn't mean that simplification yields the same result, because you've algebraically done away with variables implicit in the understanding of this dynamic situation, e.g., a motor creating the ability to change velocity at an increasing or decreasing RATE of change. In proving that my equations are resolvable into F=MA, you've proven their implicit truth. That doesn't mean they don't shed more light on the subject than that simple truth, F=MA!! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 574 Registered: 2-2003 |
Alsup, on initial analysis, you've done the same thing you always do. Apply accurate equations to reach inaccurate findings. No one can dispute that F=MA, nor can anyone dispute that W= delta KE, or that Power is the ability to accomplish that set amount of work in a set amount of time. What can, must and will be disproved is that you can arrive at an understanding of acceleration over time and distance using F=MA. It necessarily follows that F=MA and W=Delta KE are TRUE, that doesn't mean they describe the same thing!! Go home, have a cold drink, and I will post my reply when I'm ready. And, while you're at it, explain how the Beetle accelerated harder "as power ramps up" than it did at torque peak. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 558 Registered: 4-2002 |
Derviation was too easy to wait a day: Given: W=F*d P=W/t Then: P=F*d/t v=d/t // by definition therefore: P=F*v F=m*a // by definition So: P=m*v*a // by derivation But: E=1/2*m*v**2 dE/dt=m*v*dv/dt dv/dt=a dE/dt=m*v*a P=m*v*a So: P = dE/dt = F*v // so it all fits together nicely. So your equations are not at odds with F=m*a or at odds with E=1/2*m*v**2. I submit that since these equations can be converted into the original equations (F=M*a and E=1/2*m*v**2), that these equations (W=F*D and dKE=W/t) will lead to the same conclusion. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 572 Registered: 2-2003 |
Alsup, once again, please stop, you're behaving like a child. I'm referring, of course to F*D, where D equals zero. When D in that equation equals Zero, there is NO WORK. Please stop trying to take things out of context. Similarly, as I said, this is NOT my life, and with a wonderful wife and family at home, parents I take care of, my cars, and my profession (where I'm held in high esteem, oddly enough), I'm not lonely because you disagree with me! | ||
| DES (Sickspeed)
Advanced Member Username: Sickspeed Post Number: 3568 Registered: 8-2002 |
Sorry, EF, but i needed to pick out the spelling error to compensate for my total lack of anything remotely knowledgeable on physics... i'll comb through this thread when it dies and pick out ALL the spelling errors... | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 557 Registered: 4-2002 |
"Remember, my construct is that W=Delta KE, and that Power equals W/T. Simple. Please don't tell me you can refute that? " No, what I will show is that using these equations, physics will arrive at the same conclusion that it arrived at through F=M*a and E=1/2*m*v**2 | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 571 Registered: 2-2003 |
DES, you're right, damn you!! Perhaps you'd better go through and collect Alsup's many misspellings too. Can't have an unfair advantage!! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 556 Registered: 4-2002 |
"there is no need to "solve" equations. When D is not equal to zero, there is Work," Misleading at best: Consider the case where no force is being applied to a vehicle already moving. Here Distance accululated, but since not force is applied, no work is performed. So, just how much work is done in the first picosecond? Show us that your math skills are up to the task. "I said your statements such as "KE=1/2mvsquared implies" are fraudulent" But you are unable to show a nonfradulent set of equations. All you have done is argue without portfolio. You have presented no data, no statistics, and no derivations. You have been presented with equations, graphs and demonstrationg that the whole net belives like I do. Does it feel lonely out there, all alone, believing in things that are not true? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 570 Registered: 2-2003 |
Oh, Alsup, does that mean that I have to meet you at high noon tomorrow? Count on something, I will post when I'm ready and not before. Remember, my construct is that W=Delta KE, and that Power equals W/T. Simple. Please don't tell me you can refute that? Please stop calling me out, I find it ridiculous and offensive. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 569 Registered: 2-2003 |
Alsup, I'll have some stuff by Friday or Monday; unlike you I need to make a living. You are welcome to post whatever you want tomorrow or whenever, and to claim "three strikes" or whatever will make you feel most manly (or childish). You've taken the concept of a discussion and turned it into an exceedingly vicious argument, without regard for courtesy or respect. I apologized (after reading Peter Sedlak's post) when I descended to what I now see is your natural level, you remain there casting invective and spraying spittle. Understand that there is no equivocation about the torque peak of the Turbo S Beetle, it is maximum at 1,950rpm, and your graph shows that acceleration does not peak there, but at 3,000rpm "as Power ramps up." No amount of assertions on your part that it peaks elsewhere avail. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 554 Registered: 4-2002 |
"Alsup, I'm sorry to see that you're still calling me names. I am neither an imbecile nor an idiot, but rather a National Merit Scholar and Ben Franklin Scholar at U of P" With credentials like these, its a wonder they let you graduate without an education in math and physics. "who admitted to needing to bone up on physics that was 20+ years old," Point of reference: I have not opened up a math or physics textbook, nor have I interrogated math or physics on the web durring this discussion. I am 50 years old, and still remember the math and physics I got when in school (Hint: I use this stuff all the time). "and admitted when my early posts were "silly." " "I have, however, evolved and posed some equations that you cannot refute," at best, 'have not yet refuted' but come tomorrow...... "well-founded in simple Newtonian physics," You have not refuted the F=M*a approach, nor have you refuted the E=1/2*m*v**2 approach. "while you have posted some clunkers that you refuse to admit were silly. " If I recall correctly, you are talking about the unit of time used in differential calculus. I stand my be statements. You have called those equations wrong, but not been able to solve the system to derive a solution different than mine. Is you NMS and BFS gone to waste? "Please stop with the invective, you're starting to foam at the mouth." Then come up with the math required to get from W=F*D and show that it leads to a different conclusion that F=M*a and E=1/2*m*v**2. You have--today. | ||
| DES (Sickspeed)
Advanced Member Username: Sickspeed Post Number: 3566 Registered: 8-2002 |
If i had half a brain, it would throb in pain from trying to comprehend this thread... What i can contribute to it, though, is that perjorative is really spelled pejorative... | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 568 Registered: 2-2003 |
Mitch, there is no need to "solve" equations. When D is not equal to zero, there is Work, plain and simple, whatever fraction of a nanosecond you choose to select. I never said your Newtonian equations were a fraud, I said your statements such as "KE=1/2mvsquared implies" are fraudulent, that expression implies nothing other than what is on its face. You have taken similar liberties with other statements and equations. Again, I'd ask you to stop with the perjorative terms. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 553 Registered: 4-2002 |
"You know, Mitch, you're just plain wrong. " Oh, really, care to use math to show the error in my derivations? Care to show why all the web sites support my position? "You haven't shown anything, really, you haven't even explained your own post showing acceleration maximum far past torque peak." Care to site the post number that shows this? "Similarly, the whole "in Mitches [sic] favor and in Efwun's favor thing is exceedingly childish, why stoop to that?" Because it summarizes the situation susinctly! "I've derived equations that show why Power is important, and those equations also explain the variations even your own posts show. You've merely attacked those equations" After attacking your 'use' of those equations, I went in and understood how to use them, then derived a solution to those equations that is in line with other means to solve the same problem. Both solutions arrive at the same conclusion. So if I am wrong, get in there, do the math, and show the error. Or do you want me to go in and use the W=F*D equation and derive a solution in accordance with the other two solutions? Would that satisfy you? "You are simply ignoring things you've posted that make no sense," Would anyone other than EFWUN like to point out wher I have ignored my own errors? "and also that actually prove you wrong." I have shown/demonstraited/proven thatn F=M*a and E=1/2*m*v**2 lead to the same conclusion: max acceleration occurs at max TQ. You have not shown anything, demonstrated anything, nor proven anything. I challenge you to get out your math book and derive a solution based on W=F*D that shows where maximum acceleration occurs. Tomorrow, I will show my derivation--if yours is not ready, then its three strikes against your postulate: "peak acceleration occurs somehwere other than at peak TQ." | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 567 Registered: 2-2003 |
Alsup, I'm sorry to see that you're still calling me names. I am neither an imbecile nor an idiot, but rather a National Merit Scholar and Ben Franklin Scholar at U of P who admitted to needing to bone up on physics that was 20+ years old, and admitted when my early posts were "silly." I have, however, evolved and posed some equations that you cannot refute, well-founded in simple Newtonian physics, while you have posted some clunkers that you refuse to admit were silly. No wear at 7,000rpm, slices of time so small that despite the fact that F*D is not equal to zero, no Work is done? I won't give the exhaustive list here. You've posted and adopted a graph showing peak acceleration well past peak torque; you can't explain that, nor can you explain your other rather far out postulates. Please stop with the invective, you're starting to foam at the mouth. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 552 Registered: 4-2002 |
"F=MA works only in the construct of the application of a force as in an inelastic collision" F=M*a works in all sorts of situations 1) elastic colisions, 2) inelastic colisions 3) plain acceleration 4) cornering 5) gravitational acceleration 6) explosions If ever there were a universal 'law' this is it. "because if that force is more than the instantaneous overcoming of inertia, there would be an increasing rate of acceleration," No, F=M*a, so if the force is constant, the acceleration is constant. A constatn is neither increasing nor decreasing (thats why its called a constant). If the force increases wit time, the acceleration increases with time. If the force increases with distance, the acceleration increases with distance. All a force is is tthe relationship between acceleration and mass. "and distance and time become factors." If you were paying attention, I have been very careful with respect to distance and time. "Remember, F*D =Work." Remember, you also posited E=1/2*m*v**2 as your solution. In that case, I showed that it supports my position. "Therefore, one physical construct that works for acceleration over distance and time is the change in kinetic energy of the vehicle, or W=delta KE." As I have shown, this leads to the conclusion that max acceleration occurs at max TQ. If you want to demonstrait otherwise, get out your math book and derive a different solution that shows where max acceleration lies. But I bet you can't. Your math and physics skills are too thin to allow you to solve your way out of a paper bag. "Let me finish by stating that my final understanding of this conundrum is that a motor that can hold within percentage points of its torque peak as revolutions ascend will accelerate hardest at the point where ascending revolutions optimally intersect with descending torque." My claim is simpler at peak TQ. it is backed by math and physics. "Those vehicles accelerate hardest well above torque peak, perhaps even at �power� peak. Vehicles with steeply descending torque curves reach that optimal combination of torque and revolutions much earlier, and therefore accelerate hardest closer to, or at, torque peak. Remember, torque is the measured resistance on a dyno, while power is that output calculated as the ability to accomplish a set amount of work in a discrete amount of time." There is a principle in science that applies here. It is called Occams razor. Occams razor states that when there are competing theories, the simpler one is usually the better one. In this case, the simpler theory is that peak acceleration occurs at peak TQ in any gear; with the caveat that CVT equipt cars accelerate hardest at peak HP. EFWUN postulates that some cars do this and some cars do that, and that in the final analysis, "it all depends". Which IS the simpler theory? Which theory expalins the available data? Which theory is in accordance with available litterature (web sites)? Which theory is based on math and physics? and Which theory is based on inuendo? "TimN, I'm not "spewing" the same nonsense as Alsup, I'm saying that his equations are a fraud," F=M*a is a fraud--I'm sure Newton would be surprised! E=1/2*m*v**2 is a fraud--its EFWUN's equation(#1) a=dv/dt is a fraud--its the friggen definition of acceleration! v=dd/dt is a fraud--its the friggen definition of velocity! Perhaps you meant my derivation--and if it is to fradulent, why haven't anyone found a hole in the math? "(just as he ignored several of my defined terms previously to reach his conclusions)" I use commonly understood definitions for words, I use commonly defined calculus methods for transforming equations. The conclusions speak for themselves. "and that, while I've publicly admitted when I've posted "silly" responses, his misunderstandings run so deep that he has told us, unrepentantly, that no wear occurs in a 355 at 7,000rpm," I am still willing to stand by this statement. However, the real statement "From the point where the engine is fully lubricated until stresses start to overcome basic friction, wear is basically linear and basically zero*. Modern oils and engine materials and surface treatments provide very hard, low friction, long wearing surfaces. This range extends upwards to about 80% Red line or around 7000 RPMs for an F355/F360 engine. " With the footnote: *engine wear is basically zero, however ancillary wear continues unabated." "and that manufacturers set redlines where stuff begins to break for "know-nothing" customers, and that Distance can be so small that F*D, while not equal to zero, is nevertheless no W!!" At best this is a mischaracterization of what I said or implied. Differential calculus deals with units of time that are infinitessimal. Take a force of 350 lb-ft and apply it to a 3200 lb car for 1 nanosecond. How far has the car moved in the first nanosecond? In all practical terms the car has not moved. This is true because in computing distance from acceleration ther are two integrals over time, and time is very small, leading to an amount of movement that is not measurble by commmonly available measuring devices. Apply this same force over a full second, and anyone can measure the distance traveled with a ruler. Furthermore, W is F*D and since D ~=0 so is W, but since we have only been pushing on the car for 1 nanosecond, we would expect that W would be tiny (it is not zero BTW just close). "Finally, he posted a graph showing a car accelerating at maximum well past torque peak, and first disavowed it because "it was Motor Trend's graph", then simply lied about where the Turbo's torque peak occurs. " I still submit that peak TQ on that car is 3000. However, notice the EFWUN reamin unwilling to use the other two examples--they support my position. "I am not dismissing calculus, I am dismissing the ridiculous notion that D can be so tiny that F*D results in no W. Don't be so quick to pile on, I thought you were a deeper thinker than that." OK, einstein, just how much work was acomplished in that first nanosecond, in the first picosecond, in the first femto second. The equations are all up above, can you solve them? These are real units of time considerably larger than dt. |