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From: Herman Trivilino on 19 Sep 2005 23:31 "Don1" <dcshead(a)charter.net> wrote ... >> And, as I said, you could improve on your formula, apart from fixing it >> so >> that each term has the same units, by adding a third order term. Would >> like >> me to do that for you? >> >> > Yeah. Then maybe I'll know what your talking about. When Motor Week tests a car, they give the time taken to travel the quarter mile (starting from rest) and the final speed. In this case it was 14.2 seconds with a final speed of 98 mi/h (or 144 ft/s). An additional figure given is the time it takes to accelerate from rest to a speed of 60 mi/h (or 88 ft/s). In this case, 5.7 seconds. So, to summarize, we have ... velocity v=0 at time t=0. v=88 ft/s at time t=5.7 s. v=144 ft/s at time t=14.2 s. In a previous post I ignored the middle data point and assumed the acceleration was constant. I calculated that acceleration to be 10.1 ft/s?, this is the so-called *average* acceleration. Using your formula, I got a displacement of only 1020 ft, quite a bit short of the 1320 ft length of a quarter mile track. Thus, it was revealed that your formula s = (v_o)t + (1/2)at^2 is flawed. It only works for the case where the acceleration is constant. In this case, apparently, the acceleration is not constant. You can verify this for yourself by plotting the above three data points on a graph. You can connect the first and last points with a straight line, and the slope of that line will indeed be 10.1 ft/s^2. But, plot that middle point, and you see it's well above that straight line! So, I told you that by adding a third order term to your equation, we could improve on it. Here it is ... s = (v_o)t + (1/2)(a_o)at^2 + (1/6)jt^3. Sparing you the details, we can use the above three data points to solve for the three unknowns and find that v_o = 0, a_o = 19 ft/s^2, j = -1.246 ft/s^3. Therefore, our final result is s = (9.5 ft/s^2)t^2 - (0.208 ft/s^3)t^3. You can check this for yourself by letting t=14.2 s and getting s = 1320 ft. It's a much better fit to assume the jerk j is constant than to assume the acceleration a is constant. (Actually, I'm surprised at well it fits!) In other words, as I told you, you could get a much improved model by adding that third order term. We can also check our work by taking the derivative of both sides. v = 19t - 0.624t?. and substituting, in turn, each of the three data points. If we let t=0, we get v=0. If we let t=5.7, we get v=88. If we let t =14.2, we get v=144. Now, wasn't that fun? ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
From: Don1 on 20 Sep 2005 11:23 Herman Trivilino wrote: > "Don1" <dcshead(a)charter.net> wrote ... > > >> And, as I said, you could improve on your formula, apart from fixing it > >> so > >> that each term has the same units, by adding a third order term. Would > >> like > >> me to do that for you? > >> > >> > > Yeah. Then maybe I'll know what your talking about. > > When Motor Week tests a car, they give the time taken to travel the quarter > mile (starting from rest) and the final speed. In this case it was 14.2 > seconds with a final speed of 98 mi/h (or 144 ft/s). An additional figure > given is the time it takes to accelerate from rest to a speed of 60 mi/h (or > 88 ft/s). In this case, 5.7 seconds. > > So, to summarize, we have ... > > velocity v=0 at time t=0. > v=88 ft/s at time t=5.7 s. > v=144 ft/s at time t=14.2 s. > > In a previous post I ignored the middle data point and assumed the > acceleration was constant. I calculated that acceleration to be 10.1 ft/s2, > this is the so-called *average* acceleration. Using your formula, I got a > displacement of only 1020 ft, quite a bit short of the 1320 ft length of a > quarter mile track. Thus, it was revealed that your formula > > s = (v_o)t + (1/2)at^2 > > is flawed. It only works for the case where the acceleration is constant. > In this case, apparently, the acceleration is not constant. > > You can verify this for yourself by plotting the above three data points on > a graph. You can connect the first and last points with a straight line, > and the slope of that line will indeed be 10.1 ft/s^2. But, plot that > middle point, and you see it's well above that straight line! So, I told > you that by adding a third order term to your equation, we could improve on > it. > > Here it is ... > > s = (v_o)t + (1/2)(a_o)at^2 + (1/6)jt^3. > > Sparing you the details, we can use the above three data points to solve for > the three unknowns and find that > > v_o = 0, > a_o = 19 ft/s^2, > j = -1.246 ft/s^3. > > Therefore, our final result is > > s = (9.5 ft/s^2)t^2 - (0.208 ft/s^3)t^3. > > You can check this for yourself by letting t=14.2 s and getting s = 1320 ft. > > It's a much better fit to assume the jerk j is constant than to assume the > acceleration a is constant. (Actually, I'm surprised at well it fits!) In > other words, as I told you, you could get a much improved model by adding > that third order term. > > We can also check our work by taking the derivative of both sides. > > v = 19t - 0.624t2. > > and substituting, in turn, each of the three data points. > > If we let t=0, we get v=0. > If we let t=5.7, we get v=88. > If we let t =14.2, we get v=144. > Clear as mud Herman. Don
From: Don1 on 20 Sep 2005 11:43 Herman Trivilino wrote: > "Don1" <dcshead(a)charter.net> wrote ... > > > In the beginning when grampa starts pushing grandson on a sled the > > acceleration increases > > Right. Ok. So, the acceleration is not constant, but it's still equal to > the ratio of the net force to the mass. The net force is not constant, is > it, Grandpa? > The acceleration increases, and the force is as constant as I can keep it, but lessens as the sled gains speed. This is what happens with any vehicle and propellent - why a rocket can't go faster than the propellent is expelled. Don > The formula for displacement that gave us, where the displacement is the sum > of two terms, (v_o)t and (1/2)at2, won't give us the sled's displacement. > It's a formula that gives results that don't match what's observed. It may > fit your notion of how things ought to be, but it won't make for good, or > even acceptable, physics. > > > > ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- > http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups > ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
From: Randy Poe on 20 Sep 2005 12:06 Don1 wrote: > This is what happens with any > vehicle and propellent - why a rocket can't go faster than the > propellent is expelled. Incorrect. A rocket is not limited to its propellant speed. - Randy
From: TomGee on 20 Sep 2005 12:12
Randy Poe wrote: > TomGee wrote: > > Sam Wormley wrote: > > > Newton is much more precise > > > http://scienceworld.wolfram.com/physics/NewtonsLaws.html > > > > > > 1. (Law of inertia): A body at rest remains at rest and a > > > body in motion continues to move at a constant velocity > > > unless acted upon by an external force. > > > > > > > > Well, well, Worms, you're starting to support your beliefs with actual > > excerpts from your websites! Very good, keep it up. > > No matter what well-known fact you choose, somewhere on the > internet is somebody who disbelieves in it and writes volumes > on his (occasionally her) disbelief. > > > According to your buddy PD, though, the 1st law includes a claim that > > the body has no internal force to keep it in motion. Did you leave > > that out, or don'tcha agree with PD? > > That's correct. There is no force that keeps bodies in motion. > Forces only act to change motion. > > So you agree with PD and Worms. What else is new? They only believe that because they were taught to believe that. Anyone with a real brain would be able to support such a wild statement, but I see you offered none, so that's just your opinion. > > > > > 2. A force F acting on a body gives it an acceleration a > > > which is in the direction of the force and has magnitude > > > inversely proportional to the mass m of the body: F = ma . > > > > > > Here F is the applied force, m is the mass of the particle, > > > and a = dv/dt is the particle's acceleration, with v being > > > the particle's velocity. This equation, together with the > > > principle that bodies act symmetrically on one another > > > (Newton's third law--so that the force particle A feels > > > from particle B is equal to the force B feels from A--is > > > the basis for understanding particle dynamics". > > > > > > > > PD claims such accelerations do not amount to a change in kinetic > > energy. Do you agree or disagree? > > You are misreading something. A change in speed constitutes > a change in kinetic energy. > > No, no misreading. Read it for yourself word for word in this same thread. > > > Perhaps you garbled a description of circular motion, in > which the force acts perpendicular to the velocity. In that > case, the acceleration results in a change of direction, > but not of speed, and the kinetic energy is constant. > > No, no garbling by me. It was all PD's garbling, as I said. > > > > > 3. Whenever a body exerts a force on another body, the latter > > > exerts a force of equal magnitude and opposite direction > > > on the former. This is known as the weak law of action and > > > reaction. > > > > > > "Newton's [second] law completely describes all the phenomena > > > of classical mechanics...." > > > > > So the force in 2. above is different from the force of bodies striking > > each other? > > When bodies strike each other, the forces obey Newton's laws, > including the third law. > > Then PD must think those laws apply only to bodies in classical physics and not to quantum particles. > > > > In 2., the force makes a body move in the same direction > > of the force, > > The acceleration is in the direction of the force. That may or > may not cause motion toward the force. Generally not. If > something is moving past you and you give it a kick as it > goes by, it won't change direction and start going straight > out from the direction of your kick. It will still have a > component due to the original motion. > > Yes, I know, but PD does not know that. > > > > but in 3., the force of one body makes the other move > > away from it. > > No. That is really garbled. > > No, it's not, read it up above, it's exactly what PD said. If it proves anything it's that his brain is garbled. > > > If A pulls on B, then A gives an > acceleration of B toward A. > > Yes, but PD did not say, "IF A _pulls_ on B,...." He makes no distinction of pulling or pushing. > > > That's law number 2. At the same > time, A is pulled toward B. That's law number 3. The fact that > A is pulled toward B does not contradict the fact that B is > pulled toward A. > > No, that's not law number 3. Now you're getting confused just like PD. > > > If you hit a piece of wood with your fist, you might dent > the wood (law #2). That does not prevent the wood from denting your > hand at the same time (law #3). > > No, not so. Both relate to the 3rd law of action/reaction., |