Car physics- D'Alembert forces

Imagine a car that's reached its top speed due to aerodynamic drag acting exactly at the center of mass of the car, hence no torques exist on the body from this force. The car's center of mass is 1 foot above the ground, and the longitudinal force at the wheels is 1000 lb. The acceleration of the car is 0 ft/sec^2. Is there rearwards weight transfer or not? It seems that D'Alembert forces indicate that there is no torque on the chassis, but that conflicts with the standard Newtonian approach... Is there 1000 lb*ft torque, or is there 0? (My vote is for 1000 ;-)) It seems that RCVD indicates in a few places that the answer is actually 0, but maybe I'm misunderstanding this. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com My little car sim screenshots: http://performancesimulations.com/scnshot4.htm

In cases like this it is best to throw the terminology out the window and draw a free-body diagram. You have a tractive force acting on the contact patch, and a drag force acting on the center of mass. From a force balance standpoint they cancel out, but they are not collinear, therefore they form a moment. Assuming that the car is not rotating in pitch, there must be some other moment opposing it. This other moment is the change in the normal force distribution on the tires, aka weight transfer. "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > Imagine a car that's reached its top speed due to aerodynamic dragacting > exactly at the center of mass of the car, hence no torques exist onthe body > from this force. The car's center of mass is 1 foot above the ground,and the > longitudinal force at the wheels is 1000 lb. The acceleration of thecar is 0 > ft/sec^2. > > Is there rearwards weight transfer or not? It seems that D'Alembertforces > indicate that there is no torque on the chassis, but that conflictswith the > standard Newtonian approach... Is there 1000 lb*ft torque, or isthere 0? (My > vote is for 1000 ;-)) It seems that RCVD indicates in a few placesthat the

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> answer is actually 0, but maybe I'm misunderstanding this. > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com > > My little car sim screenshots: > http://performancesimulations.com/scnshot4.htm >

I would like to thank the two of you for making me feel even more stupid than I thought I was. "Haqsau" <[email hidden]> wrote in message news:ar9ltb$g92dn$[email hidden]...

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> In cases like this it is best to throw the terminology out the window > and draw a free-body diagram. You have a tractive force acting on the > contact patch, and a drag force acting on the center of mass. From a > force balance standpoint they cancel out, but they are not collinear, > therefore they form a moment. Assuming that the car is not rotating in > pitch, there must be some other moment opposing it. This other moment > is the change in the normal force distribution on the tires, aka weight > transfer. > > > "J. Todd Wasson" <[email hidden]> wrote in message > news:[email hidden]... > > Imagine a car that's reached its top speed due to aerodynamic drag > acting > > exactly at the center of mass of the car, hence no torques exist on > the body > > from this force. The car's center of mass is 1 foot above the ground, > and the > > longitudinal force at the wheels is 1000 lb. The acceleration of the > car is 0 > > ft/sec^2. > > > > Is there rearwards weight transfer or not? It seems that D'Alembert > forces > > indicate that there is no torque on the chassis, but that conflicts > with the > > standard Newtonian approach... Is there 1000 lb*ft torque, or is > there 0? (My > > vote is for 1000 ;-)) It seems that RCVD indicates in a few places > that the > > answer is actually 0, but maybe I'm misunderstanding this. > > > > > > Todd Wasson > > --- > > Performance Simulations > > Drag Racing and Top Speed Prediction > > Software > > http://PerformanceSimulations.Com > > > > My little car sim screenshots: > > http://performancesimulations.com/scnshot4.htm > > >

"Doug Hook" <[email hidden]> wrote in message news:ELZB9.36102$%[email hidden]... > I would like to thank the two of you for making me feel even more stupid > than I thought I was. Doesn't it make your head hurt just LOOKING at that stuff?? dave henrie

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> > >In cases like this it is best to throw the terminology out the window >and draw a free-body diagram. You have a tractive force acting on the >contact patch, and a drag force acting on the center of mass. From a >force balance standpoint they cancel out, but they are not collinear, >therefore they form a moment. Assuming that the car is not rotating in >pitch, there must be some other moment opposing it. This other moment >is the change in the normal force distribution on the tires, aka weight >transfer. >

Ok, thanks for verifying that. That's how my stuff has been working so far, but reading some mentions of D'Alembert inertia forces lately got me thinking maybe I was missing something. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com My little car sim screenshots: http://performancesimulations.com/scnshot4.htm

On 18 Nov 2002, J. Todd Wasson wrote:

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> >In cases like this it is best to throw the terminology out the window > >and draw a free-body diagram. You have a tractive force acting on the > >contact patch, and a drag force acting on the center of mass. From a > >force balance standpoint they cancel out, but they are not collinear, > >therefore they form a moment. Assuming that the car is not rotating in > >pitch, there must be some other moment opposing it. This other moment > >is the change in the normal force distribution on the tires, aka weight > >transfer.

Agreed. > Ok, thanks for verifying that. That's how my stuff has been working so far, > but reading some mentions of D'Alembert inertia forces lately got me thinking > maybe I was missing something. Some of the stuff in RCVD is heavily tied to the axis system definitions, for example, the aero axis system has origin at ground level. Maybe this is what was confusing you? BTW, we keep correcting errors in RCVD, so if you think you have found an error, please let us know. -- Doug Milliken www.millikenresearch.com/olleyfl.html <-- gift for the hardcore<grin>

On Mon, 18 Nov 2002, Dave Henrie wrote:

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> "Doug Hook" <[email hidden]> wrote in message > news:ELZB9.36102$%[email hidden]... > > I would like to thank the two of you for making me feel even more stupid > > than I thought I was. > > Doesn't it make your head hurt just LOOKING at that stuff?? >

That's nothing, you should hear the calls we get from the users of our analysis software, or even our dinner table conversation... Once you know that you are ignorant there isn't any more excuse<grin>. Time to sign up for a physics course and learn how the world works.

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>Some of the stuff in RCVD is heavily tied to the axis system definitions, >for example, the aero axis system has origin at ground level. Maybe this >is what was confusing you? BTW, we keep correcting errors in RCVD, so >if you think you have found an error, please let us know. > >-- Doug Milliken > www.millikenresearch.com/olleyfl.html <-- gift for the hardcore<grin> > >

I'm not quite sure what it was exactly. I read your book quite often (comes to the bathroom with me 70% of the time too ;-)) and have occasionally run across reference to D'Alembert's inertia force there. For some reason, I was left with the impression that even a free body diagram (like I always have used) might not be entirely correct. Needless to say, that would throw a wrench into my new dynamics code, so I wanted to be sure that the good old fashioned way was still correct. I'm relieved to hear that it is :-) My brain did a little stutter there when I read that. If this is how it worked, there would suddenly be a lot more serious problems with high speed oversteer. I wouldn't jump to the conclusion that this was an error in RCVD though. This was just an impression I got off a few sentences, most likely without reading enough, and my imagination ran off with the idea while I blankly continued "reading". I'll check it again and drop you a note if I think I found a problem. Thanks Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com My little car sim screenshots: http://performancesimulations.com/scnshot4.htm

Funny, that's exactly how I think of physics - not as science, but simply as how things work. I guess that might be one of the reasons why it makes sense to some people and not to others. "Doug Milliken" <[email hidden]> wrote in message news:[email hidden]... > > Time to sign up for a physics course and learn how the world works. >

On 18 Nov 2002, J. Todd Wasson wrote:

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> >Some of the stuff in RCVD is heavily tied to the axis system definitions, > >for example, the aero axis system has origin at ground level. Maybe this > >is what was confusing you? BTW, we keep correcting errors in RCVD, so > >if you think you have found an error, please let us know. > > > >-- Doug Milliken > > www.millikenresearch.com/olleyfl.html <-- gift for the hardcore<grin> > I'm not quite sure what it was exactly. I read your book quite often (comes > to the bathroom with me 70% of the time too ;-)) and have occasionally run > across reference to D'Alembert's inertia force there. For some reason, I was

.... From <http://www.britannica.com/eb/alpha?search=D>: "...alternative form of Newton's second law of motion, stated by the 18th-century French polymath Jean le Rond d'Alembert. In effect, the principle reduces a problem in dynamics to a problem in statics. The second law states that the force F acting on a body is equal to the product of the mass m and acceleration a of the body, or F = ma; in d'Alembert's form, the force F plus the negative of the mass m times acceleration a of the body is equal to zero: F - ma = 0. In other words, the body is in equilibrium under the action of the real force F and the fictitious force -ma. The fictitious force is also called an inertial force and a reversed effective force. "

I like the term inertial force better, so much easier to remember. ;o) "Doug Milliken" <[email hidden]> wrote in message news:[email hidden]...

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> From <http://www.britannica.com/eb/alpha?search=D>: > > "...alternative form of Newton's second law of motion, stated by the > 18th-century French polymath Jean le Rond d'Alembert. In effect, the > principle reduces a problem in dynamics to a problem in statics. The > second law states that the force F acting on a body is equal to the > product of the mass m and acceleration a of the body, or F = ma; in > d'Alembert's form, the force F plus the negative of the mass m times > acceleration a of the body is equal to zero: F - ma = 0. In other

words, > the body is in equilibrium under the action of the real force F andthe > fictitious force -ma. The fictitious force is also called an inertial > force and a reversed effective force. " > >

Don't be fooled, we just make all this stuff up as we go. ;o) "Doug Hook" <[email hidden]> wrote in message news:ELZB9.36102$%[email hidden]... > I would like to thank the two of you for making me feel even morestupid > than I thought I was. >

Remember also about rolling resistance, which is a force acting at the contact point of the tyres with the road, and so *will* create a torque opposite to the logitudinal force at the wheels, unless your longitudinal force of 1000lb is a resultant force of forward drive and rolling resistance added together. "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > Imagine a car that's reached its top speed due to aerodynamic drag acting > exactly at the center of mass of the car, hence no torques exist on thebody > from this force. The car's center of mass is 1 foot above the ground, andthe > longitudinal force at the wheels is 1000 lb. The acceleration of the caris 0 > ft/sec^2. > > Is there rearwards weight transfer or not? It seems that D'Alembertforces > indicate that there is no torque on the chassis, but that conflicts withthe > standard Newtonian approach... Is there 1000 lb*ft torque, or is there 0?(My > vote is for 1000 ;-)) It seems that RCVD indicates in a few places thatthe

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> answer is actually 0, but maybe I'm misunderstanding this. > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com > > My little car sim screenshots: > http://performancesimulations.com/scnshot4.htm >

On Tue, 19 Nov 2002 04:08:32 -0500, Dastardly Dan <[email hidden]> wrote: >On Mon, 18 Nov 2002 21:32:30 -0500, "Haqsau" <[email hidden]> >wrote: > >>Don't be fooled, we just make all this stuff up as we go. ;o) > >Yea really. Who's the dufus that decided 1+1=2 anyway? That was before 1+1 became 10. I guess we're stuck with 2 though, even though I really think the rest of the world should learn to think in hexadecimal. ;-) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

On Sun, 17 Nov 2002 22:17:34 -0500, "Haqsau" <[email hidden]> wrote:

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>In cases like this it is best to throw the terminology out the window >and draw a free-body diagram. You have a tractive force acting on the >contact patch, and a drag force acting on the center of mass. From a >force balance standpoint they cancel out, but they are not collinear, >therefore they form a moment. Assuming that the car is not rotating in >pitch, there must be some other moment opposing it. This other moment >is the change in the normal force distribution on the tires, aka weight >transfer.

Wasn't d'Alembert the acceleration described as a force? That'd mean it's just there to get a system with F_sum=0, using the net acceleration of a body actually as a force, cancelling the rest out. Need more pencils... :) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

On Tue, 19 Nov 2002 13:19:07 GMT, [email hidden] (Ruud van Gaal) wrote:

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>On Tue, 19 Nov 2002 04:08:32 -0500, Dastardly Dan ><[email hidden]> wrote: > >>On Mon, 18 Nov 2002 21:32:30 -0500, "Haqsau" <[email hidden]> >>wrote: >> >>>Don't be fooled, we just make all this stuff up as we go. ;o) >> >>Yea really. Who's the dufus that decided 1+1=2 anyway? > >That was before 1+1 became 10. >I guess we're stuck with 2 though, even though I really think the rest >of the world should learn to think in hexadecimal. ;-)

well, there's 10 sort of peoples, those who know binary numbers and those who dont :) Sebastien TIXIER - Game Developer Dynamics and Car Physics http://www.eden-studios.fr GPLRank Normal:-44.24 Monster:-124.44

On Tue, 19 Nov 2002 13:19:07 GMT, [email hidden] (Ruud van Gaal) wrote: >That was before 1+1 became 10. >I guess we're stuck with 2 though, even though I really think the rest >of the world should learn to think in hexadecimal. ;-) F=c*k you Ruud. =) Jason

On Tue, 19 Nov 2002 15:41:26 GMT, [email hidden] (Jason Moyer) wrote:

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>On Tue, 19 Nov 2002 13:19:07 GMT, [email hidden] (Ruud >van Gaal) wrote: > >>That was before 1+1 became 10. >>I guess we're stuck with 2 though, even though I really think the rest >>of the world should learn to think in hexadecimal. ;-) > >F=c*k you Ruud. =)

LOL! That 'k' is not a hex digit, Jason, you still have some way to go! ;-) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

On Tue, 19 Nov 2002 14:36:37 +0100, Sebastien Tixier <stixier@wanadoo[NO_SPAM].fr> wrote:

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>On Tue, 19 Nov 2002 13:19:07 GMT, [email hidden] (Ruud >van Gaal) wrote: > >>On Tue, 19 Nov 2002 04:08:32 -0500, Dastardly Dan >><[email hidden]> wrote: >> >>>On Mon, 18 Nov 2002 21:32:30 -0500, "Haqsau" <[email hidden]> >>>wrote: >>> >>>>Don't be fooled, we just make all this stuff up as we go. ;o) >>> >>>Yea really. Who's the dufus that decided 1+1=2 anyway? >> >>That was before 1+1 became 10. >>I guess we're stuck with 2 though, even though I really think the rest >>of the world should learn to think in hexadecimal. ;-) > >well, there's 10 sort of peoples, those who know binary numbers and >those who dont :)

I think as for VRally driving I fit into the 0 side of those 10. :) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

"Ruud van Gaal" <[email hidden]> wrote in message news:[email hidden]... > That was before 1+1 became 10. > I guess we're stuck with 2 though, even though I really think the rest > of the world should learn to think in hexadecimal. ;-) I thought 1+1= a window (try writing the equals on top of the '1+1' bit - bad primary school joke) Also I prefer to think in binary and convert to hexadecimal - I have 2 fingers but unfortunately not 16...

On Tue, 19 Nov 2002 17:57:12 -0000, "Nick" <[email hidden]> wrote:

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>"Ruud van Gaal" <[email hidden]> wrote in message >news:[email hidden]... >> That was before 1+1 became 10. >> I guess we're stuck with 2 though, even though I really think the rest >> of the world should learn to think in hexadecimal. ;-) > >I thought 1+1= a window (try writing the equals on top of the '1+1' bit - >bad primary school joke) > >Also I prefer to think in binary and convert to hexadecimal - I have 2 >fingers but unfortunately not 16...

My mind is actually analog it seems but that makes it hard to add up really large numbers. :) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

On Tue, 19 Nov 2002 12:39:12 -0500, Dastardly Dan <[email hidden]> wrote:

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>On Tue, 19 Nov 2002 13:19:07 GMT, [email hidden] (Ruud >van Gaal) wrote: > > >>That was before 1+1 became 10. > >In my way of thinking 1+1=11.

Wear a condom next time! :) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

On Tue, 19 Nov 2002 12:39:12 -0500, Dastardly Dan >>That was before 1+1 became 10. >In my way of thinking 1+1=11. Wear a *** next time! :) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

> >well, there's 10 sort of peoples, those who know binary numbers and > >those who dont :) > > I think as for VRally driving I fit into the 0 side of those 10. :) Why can't a computer programmer tell the difference between halloween and christmas day? Because halloween is OCT31 and christmas day is DEC25 Paul

On Thu, 21 Nov 2002 04:12:28 -0000, "Paul Laidlaw" <[email hidden]> wrote:

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>> >well, there's 10 sort of peoples, those who know binary numbers and >> >those who dont :) >> >> I think as for VRally driving I fit into the 0 side of those 10. :) > >Why can't a computer programmer tell the difference between halloween >and christmas day? > >Because halloween is OCT31 and christmas day is DEC25

Wow, that was a deep one. :) Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

>Haqsau" <[email hidden]> wrote in message >news:arc7kp$h5aqa$[email hidden]... > > Don't be fooled, we just make all this stuff up as we go. ;o) Just making it up?? Let me give it a try..... "The coeffecient reactive principal is not conversly attracted to the Dusenbery hypothesis therefore can not be applied to the friction equals mass times acceleration theory. However if you fart under the covers your wife will get pissed" You are right. It's not that hard once you get the terminology down. :-) "> "Doug Hook" <[email hidden]> wrote in message > news:ELZB9.36102$%[email hidden]... > > I would like to thank the two of you for making me feel even more > stupid > > than I thought I was. > > >

Hmm, not bad, but you misspelled coefficient. I give it an 8.5, pretty good for a first try. ;o) "Doug Hook" <[email hidden]> wrote in message news:WM5E9.120755$NH2.8319@sccrnsc01... > > Just making it up?? Let me give it a try..... > "The coeffecient reactive principal is not conversly attracted to the > Dusenbery hypothesis therefore can not be applied to the frictionequals > mass times acceleration theory. However if you fart under the coversyour > wife will get pissed" > You are right. It's not that hard once you get the terminology down.:-) > >