Physics of an automobile, suspension, and weight transfer?

[wimms]

Originally posted by Mr. Robin Parsons
Although the COG shifts towards the rear when the car lifts, it is the torque application (wheels/axles to frame/chassis) that is lifting the car. Hence the manner of drive, axle, shaft, belt makes little (But not absolutley "NO") difference in the way it will cause lift.

The energy isn't really "lost", per say, but is not being used to drive the car forward as much as it is being used to lift the front end, counteracting gravity.

COG never shifts. Its a center of mass of the vehicle. Its weight that shifts, like every time you walk, your weight shifts from leg to leg. Your COG remains at same spot of your body.

How do you get the energy back when the front end drops back onto the ground? - meister
Originally posted by russ_watters
Where does it go if not toward driving the car? It must go somewhere, it can't just disappear.

There is almost no energy consumed on lifting the car as it almost doesn't lift. The energy that seems to go to lifting the car front against gravity in reality goes to pressing rear wheels against ground. Its a weight shift, purely due to inertia and COG being above ground. Front end of car almost becomes weightless as full weight of car goes to rear end.
Without the weight shift, rear wheels would simply spin too early.
There is no energy to be gained when front drops back to ground, as all that would be gained goes into front wheels contacting ground. The rest is simply weight shift back.

 

[Mr. Robin Parsons]

Originally posted by wimms
(SNIP) COG never shifts. Its a center of mass of the vehicle. Its weight that shifts, like every time you walk, your weight shifts from leg to leg. Your COG remains at same spot of your body. (SNoP)

So when I had mentioned the COG "shifting" I had meant 'relative to the ground' cause it does.

Originally posted by wimms
(SNIP) Its a weight shift, purely due to inertia and COG being above ground. Front end of car almost becomes weightless as full weight of car goes to rear end. (SNoP)

The lifting of the front end of the vehicule is due to the torque exerted by the driving wheels upon the frame/chassis of the car, not inertia.

 

[russ_watters]

Originally posted by wimms
COG never shifts. Its a center of mass of the vehicle. Its weight that shifts, like every time you walk, your weight shifts from leg to leg. Your COG remains at same spot of your body.

The way you put it is confusing and probably wrong - weight is a biproduct of mass, so center of mass and center of gravity (weight) are the same thing. MRP says it right:

The lifting of the front end of the vehicule is due to the torque exerted by the driving wheels upon the frame/chassis of the car, not inertia.

Yes. Its the torque that lifts the front wheels. However, this is wrong:

So when I had mentioned the COG "shifting" I had meant 'relative to the ground' cause it does.

The COG is simply a static physical property of the object.

The COG doesn't change, but where the COG is affects the torque required to lift the front end.

Now, this also makes something I said wrong (my statics teacher would be very upset).

A "funny car" with its bicycle front wheels and cog all the way back is set up at a standstill in very nearly this condition.

Not quite. If the COG were directly over the back wheel, the car would flip over backwards at the slightest applied torque. What is required is having the COG as far back as possible to maximize the weight on the back tires, while making the torque required to lift the front wheels as high as necessary to keep the car from flipping over backwards.

By making the front end long and skinny, you keep the COG far back while increasing the torque required to lift the front end.

Take a yardstick or a broomstick or something long and skinny and hold it horizontally in one hand at the center. The center is the COG. Notice you are not applying any torque to hold it level. Any small torque and you can spin it.

Now hold it towards one end, again keeping it level. The COG hasn't changed and the overall weight is the same, but now you need a pretty large torque to keep it level.

 

[Mr. Robin Parsons]

Originally posted by russ_watters

(SNIP) Take a yardstick or a broomstick or something long and skinny and hold it horizontally in one hand at the center. The center is the COG. Notice you are not applying any torque to hold it level. Any small torque and you can spin it.

Now hold it towards one end, again keeping it level. The COG hasn't changed and the overall weight is the same, but now you need a pretty large torque to keep it level.
(SNoP)

And by this example, (above) we can see the the center of gravity shifts relative to the ground, simply by lifting the far end of the long pole (that we hold by only one end) and noteing that the COG, relative to the ground, approaches the end of the stick that you are holding.

This is how the weight transfer occurs, towards the rear, when the car lifts. COG, (Perpendicular) relative to the ground, moves towards the rear.

 

[wimms]

Robin, COG NEVER moves. Max you can think of is movement of suspension and taking this as change in goemetry of vehicle.

Originally posted by russ_watters
weight is a biproduct of mass, so center of mass and center of gravity (weight) are the same thing.

While COG and COM are same, center of distribution of weight is not. Weight is meaningless without support, and it does depend on acceleration forces. If a-force were applied to a line parallel to ground and at height of COG, no front lift would occur.

Imagine this car in space. When you apply force to it, it either simply moves, or also starts rotating around its COG, depending on where the force is applied. In our case, front lift is not due to engine torque fighting gravity, but because car rotates around its COG. This rotation is stopped by rear wheels at ground, and this translates into weight shift, almost 100%. This is pure inertial behaviour. Weight of car never changes, nor is its COG lifted higher above ground. Thus, gravity is not beated. Its the only force we have to keep that car on the ground (besides aerodynamics).

MRP says it right: Yes. Its the torque that lifts the front wheels.

He seems to imply that its the angular momentum of rear wheels that causes front lift. Its not. Although angular momentum is conserved, its impact is way too small. Just compare angular momentum of wheel radius and their mass to that of chassis radius and its mass. Besides, angular momentum causes car to turn around its COG, not rear axle. This means that angular momentum forces rear axle to move towards ground again, being stopped by it.

Acceleration and deceleration are equal. Imagine this car moving backwards with high speed and blocked braking. Now wheels offers no angular momentum at all after they are stopped, but car continues to shift weight in direction of braking, to extent that now-rear wheels would get off the ground. This IS purely inertial effect.
The only thing that matters is location of COG and location of force vector relative to it.

The COG doesn't change, but where the COG is affects the torque required to lift the front end.

By torque, you mean accelerative force, not angular acceleration of rear wheel, right?

In regards to rear suspension travel, I'd even think it might have benefit. If COG is such that to get weight off front wheels requires rear axle to lower, then suspension allows that. Without it, full shift of weight wouldn't be achievable, and rear wheels would spin.
Other way to do it is to have COG higher and more at back, but this makes it harder to control acceleration force so that front wheels still touch ground for steering.

 

[Mr. Robin Parsons]

Originally posted by wimms
(SNIP) Robin, COG NEVER moves. Max you can think of is movement of suspension and taking this as change in goemetry of vehicle. (SNoP)

This is a dynamic event and requires that you examine the COG relative to something, the ground! and relative to the ground it moves, hence the weight transfer.

Forgive me, but the rest of what you write seems rather off.

 

[wimms]

Originally posted by Mr. Robin Parsons
This is a dynamic event and requires that you examine the COG relative to something, the ground! and relative to the ground it moves, hence the weight transfer.

COG is defined as center of mass relative to geometric bounds of vehicle, not to the ground. And car moves forward, of course COG as point moves. But as long as mass doesn't move inside a car, COG does not move within it.

if COG moved away from ground, up, it would mean LOSS of contact with ground, its critical event, upto a flight.

Forgive me, but the rest of what you write seems rather off.

rather off, huh. Then show me how.

 

[Mr. Robin Parsons]

Originally posted by wimms
(SNIP) COG is defined as center of mass relative to geometric bounds of vehicle, not to the ground. And car moves forward, of course COG as point moves. But as long as mass doesn't move inside a car, COG does not move within it. (SNoP)

Yes, but in this instance we must adjudicate the effects upon the car as it is relative to the ground, otherwise we will not even know that it has lifted up.

What we are useing COG for, in this instance, is to find the balance point of the vehicule, hence we must measure it's relativity to the ground as to know how/why it is allowing the car to become unbalanced, hence front end lifting.

If you take a long plank, find the COG, measure that relative to the ground, lift one end of the plank and that measure shifts towards one of the ends. That is basically what is occurring with the 'rail' (car).

The rest, later, when I have a little more time, sorry, and thanks!

 

[Mr. Robin Parsons]

Originally posted by wimms
Imagine this car in space. When you apply force to it, it either simply moves, or also starts rotating around its COG, depending on where the force is applied. In our case, front lift is not due to engine torque fighting gravity, but because car rotates around its COG. This rotation is stopped by rear wheels at ground, and this translates into weight shift, almost 100%. This is pure inertial behaviour. Weight of car never changes, nor is its COG lifted higher above ground. Thus, gravity is not beated. Its the only force we have to keep that car on the ground (besides aerodynamics).

This statement makes some kinda strange non sense to me. It does have some sense to it, but it is NOT applicable to what the "original question asked" is about, as clearly we are NOT dealing with something floating in space we are dealing with something that is on the ground, and is affected by gravity, and is acting because of TORQUE that is being exerted upon the parts, not the angular momentum of the entire chassis which is what you seem to be implying in this quote...

Originally posted by wimms
He seems to imply that its the angular momentum of rear wheels that causes front lift. Its not. Although angular momentum is conserved, its impact is way too small. Just compare angular momentum of wheel radius and their mass to that of chassis radius and its mass. Besides, angular momentum causes car to turn around its COG, not rear axle. This means that angular momentum forces rear axle to move towards ground again, being stopped by it.

The angular momentum of the wheels, and the chassis, are relitivised by the exertion of torque through the pinion and the ring (King and Crown, old school) which is what allows for the overcoming of the forces as to cause the front end to lift.

The 'angular momentum' imparted to the wheels, does not cause the car to want to rotate around it's "Center of Gravity", it simply causes the wheels to rotate around the axle

 

[wimms]

Originally posted by Mr. Robin Parsons
Yes, but in this instance we must adjudicate the effects upon the car as it is relative to the ground, otherwise we will not even know that it has lifted up.

And why do you think that COG lifts? Are you saying that with sudden enough acceleration force, the car will jump off the ground with all 4 wheels??

You must have seen every single car with rear drive "squat" its rear end when accelerating, not just front end lifting. You must have seen every single car lowering its front when braking. You must have noticed cars tilting in turns upto inside wheels lifting off the ground. Every single effect with acceleration forces has to do with weight shifts, around COG.

Originally posted by Mr. Robin Parsons
This statement makes some kinda strange non sense to me. It does have some sense to it, but it is NOT applicable to what the "original question asked" is about, as clearly we are NOT dealing with something floating in space we are dealing with something that is on the ground, and is affected by gravity, and is acting because of TORQUE that is being exerted upon the parts, not the angular momentum of the entire chassis which is what you seem to be implying in this quote...

First, this wasn't polite. Second, are you claiming that inertial effects that occur in space do NOT occur in gravity?? Perhaps you've forgot that we have to do with acceleration not constant torque applied to stationary object. Perhaps you've forgot that parts that take the torque are located all way back behind COG. Perhaps you forgot that with 3G acceleration inertial effects become increasingly important.

Yes, I'm implying that angular momentum of whole chassis is important factor. I've never succeeded in lifting off with my chair by pulling it up while sitting on it, no matter what torque I exert on parts. The only thing that matters is point of tyre traction and location of COG. These two and gravity vector is all you need to predict front lift. Any torque inside chassis, engine, gears, etc has no effect unless viewed as opposite to acceleration vector. Tyre patch remains parallel to the ground, and so is acceleration vector. There is no way how it suddenly would cause antigravity lift.

The angular momentum of the wheels, and the chassis, are relitivised by the exertion of torque through the pinion and the ring (King and Crown, old school) which is what allows for the overcoming of the forces as to cause the front end to lift.

Seems you aren't relativizing enough. From your post it seems you imply that chassis lift occurs around rear axle. But you can't ignore inertia of car that is the only reaction force, concentrated to COG, which is located above rear axle and in front of it. Resultant acceleation force vector is parallel to the ground, any other idea would violate some conservation law. COG cannot depart from ground, as this would be liftoff of the whole mass of the car and reduction of traction. The only possibility that remains is force vector directed ahead between COG and ground, causing chassis rotation around its COG.

Dragsters are made so long for single reason - to increase rotational inertia of chassis around COG.

Sure, you can think of lift as result of torque between chassis and ground, but then you are stuck with idea that energy is spent on fighting gravity, while all energy is spent completely on fighting inertia, which is precisely the goal.

The 'angular momentum' imparted to the wheels, does not cause the car to want to rotate around it's "Center of Gravity", it simply causes the wheels to rotate around the axle

Although for our case its effect is small, your comment is wrong unless I've again misunderstood something. Car together with its wheels forms system. Angular momentum is conserved for the whole system.

 

[Mr. Robin Parsons]

Originally posted by wimms
(SNIP) Seems you aren't relativizing enough. From your post it seems you imply that chassis lift occurs around rear axle. But you can't ignore inertia of car that is the only reaction force, concentrated to COG, which is located above rear axle and in front of it. Resultant force vector is parallel to the ground, any other idea would violate some conservation law. COG cannot depart from ground, as this would be liftoff of the whole mass of the car and reduction of traction. The only possibility that remains is force vector directed ahead between COG and ground, causing chassis rotation around its COG. (SNoP)

The force vector of the COG (with respect to gravity) is perpendicular to the ground not parallel, and it shifts towards the rear, as the front end lifts, due to the torque that the pinion applies to the ring in the differential.

Originally posted by wimms
(SNIP) Second, are you claiming that inertial effects that occur in space do NOT occur in gravity?? (SNoP)

NO!, but it is notable that forces acting in the two different "spaces", will offer different intertial results.

Thats it!

 

[wimms]

Originally posted by Mr. Robin Parsons
The force vector of the COG (with respect to gravity) is perpendicular to the ground not parallel, and it shifts towards the rear, as the front end lifts, due to the torque that the pinion applies to the ring in the differential.

Its unfortunate that I placed this 'force vector' in the middle of talks about COG, but I meant acceleration force vector. COG vector doesn't shift anywhere, unless you imagine spacetime curvature changes or car geometry changes. COG is center of MASS in gravity.

I repeat here as you didn't comment that: All torque that exists at ring of diff is result of reaction from tire patch. Tire patch never becomes nonparallel to the ground, thus you can't get nonparallel acceleration vector (3rd law), and your idea that rear diff is the reason for lift of the front of car is unwarranted. It has no structural strength to do that.

Do you at all understand what I'm saying? If you disagree, please explain my error. And perhaps we can learn something.

check this out http://www.miata.net/sport/Physics/01-Weight-Transfer.html

NO!, but it is notable that forces acting in the two different "spaces", will offer different intertial results.

Whats that? Do you agree or do you disagree? Are you saying now that in gravity, the results would be COMPLETELY different, and therefore example of inertial effect in free space is, as you put it, non sense?

 

[Mr. Robin Parsons]

Originally posted by wimms
Its unfortunate that I placed this 'force vector' in the middle of talks about COG, but I meant acceleration force vector. COG vector doesn't shift anywhere, unless you imagine spacetime curvature changes or car geometry changes. COG is center of MASS in gravity.

Sorry, but that simply isn't true.

Take a plank of wood/board, find it's COG, draw a force vector down to the ground, (perpendicular) lift one end, and NOTICE that the force vector has now changed (Rotated) POSITION, RELATIVE to the GROUND.

From your linked site; http://www.miata.net/sport/Physics/01-Weight-Transfer.html these explanations...

from; The Physics of Racing, Part 1: Weight Transfer, Brian Beckman, physicist and member of No Bucks Racing Club, P.O. Box 662
Burbank, CA 91503, ©Copyright 1991

The braking forces create a rotating tendency, or torque, about the CG.
AND
It is a fact of Nature, only fully explained by Albert Einstein, that gravitational forces act through the CG of an object, just like inertia. This fact can be explained at deeper levels, but such an explanation would take us too far off the subject of weight transfer.
And from part 2; Keeping Your Tires Stuck to the Ground
That is, we explained why braking shifts weight to the front of the car, accelerating shifts weight to the rear

So your link deals mostly with "braking forces" and does NOT offer explanations of why/how accelerative forces work to produce similar, but NOT identical results. He took the 'simplicity route' to explain weight tranfers, Braking effects. He avoided explaining how the accelerative effects cause front end lift, other then inertial mass, as inertial mass alone will not succeed in lifting the front end off the ground as the force vector does not go UP, it goes parallel to the ground, in accelerative force.

It is the force vector of the torque on the ring/pinion set, that causes the lift to arise, pun intended!

Does that help?

 

[Mr. Robin Parsons]

PS from that link you can actually calculate just how many G's it would take to actually succeed in getting the front end off of the ground on inertial means alone, but I suspect it is a substantial number, similar in nature to how much braking force is required to get enough weight transfer to get the rear end to lift, off of the ground, on braking alone, again substantial.

That is (partially) why we know that the lift of the front end is accomplished by the force vector of the torque at the ring/pinion set.

If you were to lock the ring in place, you would see the pinion "walking" itself around the ring, pulling whatever it was attached to, with it. (given all forces being what are needed to do that!)

 

[Mr. Robin Parsons]

post-post-script-script, If you would like to envision the absence of torque on an accelerating vehicule, the rocket sled used to test astronauts abilities to withstand G is an excellant example.

Although it is attached to the tracks (if I remember that properly) that is to prevent take off at speed (as that might cause lift) but when the force vector forward, is maintained parallel to the ground, there is no lift, even though there is a tremendous G force being applied. Inertia alone doesn't tend to lift the front end of the sled, not in/with the same force/manner that the rail lifts, as it is absent of what the rail has, torque exerted upon the axle to the frame/chassis.

Simply put it is a combination of the two, but I would respectfully suggest that the torque, from the pinion/ring combo, has more to do with the lift the perhaps you had realized.

As for spatial examples, (in absence of gravity) if your rail was in free space, and you started the engine, the torque from that alone, would initiate lateral rotation around the COG. Engaging the drive-train would probably initiate rotation of the frame/chassis around the axle, same as a helicopter without it's counterballancing rear rotor blades.

Is that clearer? better? do we now agree?

EDIT P.P.P.S.S.S. Nice link! BTW

 

[wimms]

Originally posted by Mr. Robin Parsons
Sorry, but that simply isn't true.

Take a plank of wood/board, find it's COG, draw a force vector down to the ground, (perpendicular) lift one end, and NOTICE that the force vector has now changed (Rotated) POSITION, RELATIVE to the GROUND.

I don't understand you here. COG is point. Gravity Vector from it is directed always perpendicular down to the ground. No matter in what position object you hold is, no matter what other forces act on it. Are you talking about rotated direction of G vector relative to object geometry, or are you talking about location of COG within object, as COG shift would imply?
Force vector doesn't change, or I'd build antigravity device.

So your link deals mostly with "braking forces" and does NOT offer explanations of why/how accelerative forces work to produce similar, but NOT identical results.

Not at all. The very start of explanation goes with words:

Let us continue analyzing braking. Weight transfer during accelerating and cornering are mere variations on the theme.
...
These equations can be used to calculate weight transfer during acceleration by treating acceleration force as negative braking force. If you have acceleration figures in gees, say from a G-analyst or other device, just multiply them by the weight of the car to get acceleration forces (Newton's second law!).

So, although his approach is indeed simplified, he makes it very clear that acceleration and braking are equal here.

It is the force vector of the torque on the ring/pinion set, that causes the lift to arise, pun intended!

Funny thing about this is that in the end its true, as rear diff is the only source of torque for acceleration too, but no, I believe you too much overestimate contribution of this effect.

To go further, we'd need to include some sample numbers. Perhaps we should. I'd only ask you to think about this: to lift chassis of full weight from lever arm with length L towards ground (ring gear ->tires patch), compared to length of chassis 50-300 x L - what kind of gear would withstand this?
COG, or Center of Mass, does NOT lift higher from ground, as it would mean antigravity effect.

Given that rotational inertia of chassis with its full length is many hundreds of times larger than rotational inertia of wheels, what would happen sooner - chassis lift or wheel spin/acceleration? Wheels DO spin in every single drag run. They on purpose do that to store rotational energy for later boost.

If you were to lock the ring in place, you would see the pinion "walking" itself around the ring, pulling whatever it was attached to, with it. (given all forces being what are needed to do that!)

Yeah, but if you'd try that with dragster, it'd simply blow up, without walking anywhere.

Although it is attached to the tracks (if I remember that properly) that is to prevent take off at speed (as that might cause lift) but when the force vector forward, is maintained parallel to the ground, there is no lift, even though there is a tremendous G force being applied. Inertia alone doesn't tend to lift the front end of the sled, not in/with the same force/manner that the rail lifts, as it is absent of what the rail has, torque exerted upon the axle to the frame/chassis.

But you are completely ignoring the fact that rocket exhaust is directed away from box at height of COG, not at ground height! Infact, it is more probable that acceleration vector is applied slightly ABOVE COG and slightly at angle towards ground to provide downforce. Same with that car that broke sound barrier on salt lake, its thrust was applied at or above COG.

As for spatial examples, (in absence of gravity) if your rail was in free space, and you started the engine, the torque from that alone, would initiate lateral rotation around the COG. Engaging the drive-train would probably initiate rotation of the frame/chassis around the axle, same as a helicopter without it's counterballancing rear rotor blades.

Perhaps you should recall what rotational inertia is.
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi
Check out samples about moments of inertia. Notice that its proportional to Square of arm length. Helicopter main blades have helluva moment of inertia, especially at working rpms.
Now apply moment of inertia to full chassis length, and meanwhile ask yourself, why did they build dragsters so long, not forgetting that COG is placed as far back as possible. Also think what this would mean to ring gear stress if full 6000 hp were about to lift it all. Compare this to final gearing of wheels and linear inertia of chassis.

http://www21.brinkster.com/jimsideagarage/dragster/dragstersci.htm
(I find Acceleration Graph interesting, I think the boost at sec 3 is from stored rotational inertia in spinning wheels)

Is that clearer? better? do we now agree?

Now we almost agree. I'd only like to clear up what effect dominates, inertial rotation around COG as I understand, or ring/pinion torque to lift full length of chassis, as you say.

 

[Mr. Robin Parsons]

Originally posted by wimms
I don't understand you here. COG is point. Gravity Vector from it is directed always perpendicular down to the ground. No matter in what position object you hold is, no matter what other forces act on it. Are you talking about rotated direction of G vector relative to object geometry, or are you talking about location of COG within object, as COG shift would imply?
Force vector doesn't change, or I'd build antigravity device.

Cheese I had thought you would get that one, cause it is simple.
Take a board, find the COG, tie a string to it, let the string hang down by gravitational activity, and let the string represent effective center of gravity (means where the COG acts).
Lift one end of the board, and note that, the string remains perpendicular to the ground, hence effective COG has shifted towards the lower end of your board.
(the point on the ground that COG acts upon)

Originally posted by wimms
So, although his approach is indeed simplified, he makes it very clear that acceleration and braking are equal here.

Yes he does, but he is not dealing with forces strong enough to lift one end of the car from the ground, as in with rails the accelrative force available exceeds the braking forces that the car can generate.

Originally posted by wimms
I believe you too much overestimate contribution of this effect.

Just that that is the contibuting factor in actually getting the front end off of the ground, forward acceleration alone isn't enough.

Originally posted by wimms
Yeah, but if you'd try that with dragster, it'd simply blow up, without walking anywhere.

If you bolted the tires to the ground, the machine would attempt to do an "a$$ over tea-kettle" flip, using the rear axle as the point of rotation.

AHH, went to the second link, note the distance in the data points (a) (b) and (c) because for inertia to be the effective lifter of the car, the car MUST travel forward, if you have ever watched drag racing you will have observed that dragsters lift their front ends RIGHT FROM THE START LINE, (not enough forward motion for inertia to have worked with the kind of force required) torque from the axle/chassis combination (matches, same as, the ring/pinion example)

Nice link though.....

 

[russ_watters]

Originally posted by Mr. Robin Parsons
Cheese I had thought you would get that one, cause it is simple.
Take a board, find the COG, tie a string to it, let the string hang down by gravitational activity, and let the string represent effective center of gravity (means where the COG acts).
Lift one end of the board, and note that, the string remains perpendicular to the ground, hence effective COG has shifted towards the lower end of your board.
(the point on the ground that COG acts upon)

MRP, you have all the pieces, but you aren't putting them together. COG is the place on the board where the string is tied, the string itself is the force vector. The string is NOT the COG. Clearly the place the string is tied to remains in the same place no matter how you orient the board (as implied by your post).

Example: a yardstick:

Hold it horizontal. COG: 18 in.
Rotate it vertical. COG: 18 in.

MRP, this is a simple definition issue. There really isn't any room to argue: The force vector is not the COG. The force vector acts FROM the cog. http://dictionary.reference.com/search?q=center of gravity

I understand what you mean, but you are saying it wrong. What you mean to say is that the angle of the force vector from the COG with respect to the car changes as it rotates when the front wheels lift .

The point on the ground does move, but it doesn't have a name. I'm thinking you decided to call it "effective cog" because you now realize you were wrong but don't want to admit it. This really isn't important enough for that. Let it go.

Anyway, this isn't relevant since the cases I proposed had the front end just barely lifting off the ground. And the case originally proposed only had the car shifting on its shocks. So there is very little change in the angle of the force vector.

Another definition issue:

not enough forward motion for inertia to have worked

Besides being wrong on its own, wimms said "moment of[/i]intertia," not just "inertia." Big difference. http://dictionary.reference.com/search?q=moment of inertia

 

[russ_watters]

Originally posted by wimms
Imagine this car in space...

Picky, picky. Yeah, I know COG and COM aren't exactly the same thing, but I'm not driving my car in space. On the surface of the Earth in a car, they are at the same point.

By torque, you mean accelerative force, not angular acceleration of rear wheel, right?

By torque, you mean accelerative force, not angular acceleration of rear wheel, right?

Yes. What lifts the car is best modeled as the force between the ground and the two back wheels. The actual component of the torque that goes to lifting the car depends on the mass and moment of inertia of the car: some of the force pushes the car forward and some of it lifts the front end. I think we're on the same page about this.

 

[Mr. Robin Parsons]

Originally posted by russ_watters
MRP, you have all the pieces, but you aren't putting them together. COG is the place on the board where the string is tied, the string itself is the force vector. The string is NOT the COG. Clearly the place the string is tied to remains in the same place no matter how you orient the board (as implied by your post).
MRP, this is a simple definition issue. There really isn't any room to argue: The force vector is not the COG. The force vector acts FROM the cog. http://dictionary.reference.com/sea...%20of%20gravity
I understand what you mean, but you are saying it wrong. What you mean to say is that the angle of the force vector from the COG with respect to the car changes as it rotates when the front wheels lift .
The point on the ground does move, but it doesn't have a name. I'm thinking you decided to call it "effective cog" because you now realize you were wrong but don't want to admit it. This really isn't important enough for that. Let it go.

Ummm, russ...

Originally posted by wimms
(SNIP) COG vector doesn't shift anywhere (SNoP)

Butt out!

Cause apparently you missed this one from page three...

Originally posted by Moi
(SNIP) So when I had mentioned the COG "shifting" I had meant 'relative to the ground' cause it does. (SNoP)

cause "relative to the ground" the COG does shift, towards the rear, the force vector motion/movement, towards the rear, PROVES that.

 

[Mr. Robin Parsons]

Oh yes, BTW russ, I learned aout COG's in High school physics classes that occurred when I was in grade 10. That was waaaaaaay back when I was ~15/16 years old, or ~1971/72, OOOOOooops that's right, you were not even born yet!

Please, forgive my overt rudeness, (or not, your choice) and beware OLDER MEN (and Women!) as they have had more time to learn more "things".

PS Russ, from the start line, to the finish line, the real, and actual, 'COG' moves towards the rear of the car(!), without question!

 

[russ_watters]

Originally posted by Mr. Robin Parsons
Butt out!

Cause apparently you missed this one from page three...

I'm here correcting misconceptions and misunderstandings. The two quotes you provided are one of each.

cause "relative to the ground" the COG does shift, towards the rear, the force vector motion/movement, towards the rear, PROVES that.

PS Russ, from the start line, to the finish line, the real, and actual, 'COG' moves towards the rear of the car(!), without question!

The COG is a point, not a vector.
The COG is a point, not a vector.
The COG is a point, not a vector.
The COG is a point, not a vector.
(is there an echo in here?)
Thats part of the definition. YOU MUST ACCEPT THAT. But hey - I posted the definition, so its up to you to read it.

But anyway, I'll bite: When a yardstick is horizontal, the COG is at 18" on the yardstick. So tell me then: when you lift one end 30 degrees above horizontal: where EXACTLY does the cog shift to? Express the answer as inches from either end (specify) of the yardstick. Providing the equation for figuring out the COG would be helpful as well.

 

[Mr. Robin Parsons]

Originally posted by russ_watters
(SNIP) But anyway, I'll bite: When a yardstick is horizontal, the COG is at 18" on the yardstick. So tell me then: when you lift one end 30 degrees above horizontal: where EXACTLY does the cog shift to? Express the answer as inches from either end (specify) of the yardstick. Providing the equation for figuring out the COG would be helpful as well. (SNoP)

@ 45 degrees (much simpler) it, the COG, remains exactly where it was when the stick was level, BUT in the interaction of the COG with gravity, and the vector from the COG to the ground, that shifts towards the end that remains on the ground, ~9" from it.
That, BTW, is how you can measure the actual weight tranfer as opposed to percieved weight transfer.

But I'll still tell you, when a rail, or a funny car, (for that matter) goes from the start line, to the finish line, the real, and actual 'COG' (THE POINT, Not the vector) shifts towards the rear of the car. No question! No doubt!

 

[russ_watters]

Originally posted by Mr. Robin Parsons
the COG, remains exactly where it was when the stick was level

Thank you. Finally.

But...

But I'll still tell you, when a rail, or a funny car, (for that matter) goes from the start line, to the finish line, the real, and actual 'COG' (THE POINT, Not the vector) shifts towards the rear of the car. No question! No doubt!

Huh? Where exactly does it go?

I still think you're confusing different concepts here (static and dynamic forces). Again, the COG (if you don't believe me, read the definition) is a point on a body defined only by its geometry and gravitational interaction. The acceleration of the car has nothing to do with that.

COG is a STATIC (ie, best measured when the body isn't moving) property of a body and is independent of any dynamic forces (imbalanced forces which cause motion) on the body.

Oh yes, BTW russ, I learned aout COG's in High school physics classes that occurred when I was in grade 10. That was waaaaaaay back when I was ~15/16 years old, or ~1971/72, OOOOOooops that's right, you were not even born yet!

Yeah, I probably should have let it go, but this kind of thing gets on my nerves. I hate breaking out resumes because no-one no matter how smart knows everything. COG isn't the only thing you got wrong, MRP. Moment of inertia is another one and its not a high school level concept. Its not a big deal, but this thread is sophomore Engineering Statics and Engineering Dynamics. It looks to me from wimm's posts that he has taken these courses.

 

[Mr. Robin Parsons]

Originally posted by russ_watters
Huh? Where exactly does it go?
I still think you're confusing different concepts here (static and dynamic forces). Again, the COG (if you don't believe me, read the definition) is a point on a body defined only by its geometry and gravitational interaction. The acceleration of the car has nothing to do with that.

Sorry russ, but no confusion, no error, no mistake, no question, no doubt, the COG (the point, not the vector through it} shifts towards the rear of the car/rail between the start line, and the finish line.

See russ HUGE dirfference 'tween persons who can repeat what they have learned in school, and people who can actually think.

It is the difference 'tween a mechanic who is simply a part changer, and a mechanic who can diagnose the problem. One repeats what they have learned (from others) the other continues to learn.

As for 'moment of inertia', where do you see me "getting it wrong"?

PS given the acceleration of the rail, the timing, motion forward, and the speed at which it lifts it's front end, it must be adjudicated that it is the torque 'tween the axle/chassis that is the major component of force at work there.
(As for resumes, Gauranteed! mine's Loooooooonger...)

 

[Mr. Robin Parsons]

So russ, from your link for 'moment of interia'

A measure of a body's resistance to angular acceleration, equal to:The product of the mass of a particle and the square of its distance from a reference.

Not too much different from inertia itself;

Same link
Physics. The tendency of a body to resist acceleration; the tendency of a body at rest to remain at rest or of a body in straight line motion to stay in motion in a straight line unless acted on by an outside force.

Canadian curriculums might be slightly different then US ones, none the less, I also took physics in University, soooo...

Now here we have your statement from PG3 seventh post down

Originally posted by russ_watters
The way you put it is confusing and probably wrong - weight is a biproduct of mass, so center of mass and center of gravity (weight) are the same thing. MRP says it right:

Originally posted by MRP
The lifting of the front end of the vehicule is due to the torque exerted by the driving wheels upon the frame/chassis of the car, not inertia.

And you agree with this here...
Yes. Its the torque that lifts the front wheels. However, this is wrong:

then you come back later and tell us...

Originally posted by russ_watters
Yes. What lifts the car is best modeled as the force between the ground and the two back wheels. The actual component of the torque that goes to lifting the car depends on the mass and moment of inertia of the car: some of the force pushes the car forward and some of it lifts the front end. I think we're on the same page about this.

Authoritatively adding in the "moment of inetia" stuff, which seems to be something you do, perhaps, without realization, cause you did a similar thing here...

Originally posted by russ_watters
I didn't read the whole thread, but let me clarify a misconception in the first few posts.
There is no loss of engine power whatsoever to the springs in the suspension. They are springs. You get energy back as the car accelerates. The only way you lose any of it is the damping from the shocks, which is insignificant.
Similarly if the front end lifts off the ground, some of the energy that would have accelerated it goes into lifting the front end - but you get that back as well when the front end drops back to the ground.
So according to you when the front end drops back down this somehow translates back into forward motion? HUH?
So the answer to the initial question is no: suspension issues have no effect on the total acceleration of a car.
One little catch though: In a drag race, its not the final speed that matters, its the avearge speed. The two cars would have identical 0-60 times, but the car with the stiffer suspension will have traveled further in that time (and thus win a drag race).

Those two emboldened statements of yours, at the bottom, are complete contradictions of each other.

As for the emboldened in green, that's simply wrong, the energy that the springs get (and retain till 'release') is energy that was intended to accelerate the car forward, would translate into better traction if there were NO springs at all, and adds NOTHING to the forward motion of the car when it is re-released, because the springs PUSH UP, NOT FORWARD!

A fine example of book learning, awaiting practical, experiantial knowledge, as to gain the missing understanding of the operation of the integrated system. (respectfully suggested as it is really clear to me, russ, that both you, and wimms, are smart, and knowledgeable, people!)

 

[krab]

Let's break this problem down into what happens right off the line, and what happens in the long run, say after a quarter mile.

We assume both cars have the same power, and so after say a tenth of a second, car A is at a certain velocity v. Car B, however, has a distribution of velocities. Since the suspension is compressing, the top of the car is going a tiny bit slower than the bottom. As well, the front is lifting and the back is sinking. Energy is being stored in the suspension springs. The total of the total kinetic energy found by integrating over the whole car, plus the potential energy in the springs, is the same as that of car A.

As you, know, because of the way air resistance scales as v^2, and force to the road scales as 1/v, the car's acceleration tails off drastically and is much smaller at the end of the quarter mile than at the beginning. This allows the springs to give back the energy to the car; both cars arriving at the end, more or less level. So whatever was stored in the suspension is given back, and to first order I would expect the E.T.'s to be the same.

 

[wimms]

Robin, either you're too smart for me, or you are pretty confused. In either case you'd need to elaborate more what you mean, as it sounds quite unconventional.

Originally posted by Mr. Robin Parsons
@ 45 degrees (much simpler) it, the COG, remains exactly where it was when the stick was level, BUT in the interaction of the COG with gravity, and the vector from the COG to the ground, that shifts towards the end that remains on the ground, ~9" from it.

Ah, now I see. You take top-down projection of chassis dimensions onto the ground. Basically, you say that with sun in zenith, COG shifts relative to rear edge of its shadow. Is this correct?

That, BTW, is how you can measure the actual weight tranfer as opposed to percieved weight transfer.

uh, what is the difference between the two? Concept of weight afaik implies scales. Collisions and accelerations afaik impliy inertial forces.

But I'll still tell you, when a rail, or a funny car, (for that matter) goes from the start line, to the finish line, the real, and actual 'COG' (THE POINT, Not the vector) shifts towards the rear of the car. No question! No doubt!

Front wheels of dragster lift no more than few inches. http://bmeltd.com/ If they lift more, there is something terribly wrong with their chassis.
Given wheelbase of 300in, find angle. How much would COG shift back there?

A measure of a body's resistance to angular acceleration, equal to:The product of the mass of a particle and the square of its distance from a reference.
- Not too much different from inertia itself;

http://www.amatoracing.com/carspecs.shtml
Lets try some numbers. Moment of inertia around rear axle is equal to that of rod abound end: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi
Given weight 975kg, length 7.62m, moment of inertia around rear axle is 18.8tons (I=ML²/3). Thats 19 times linear inertia! Think about it. To lift such moment of inertia off ground you'd need to accelerate it at over 1G up to overcome gravity. You DON'T have such torque at ring at all. Linear inertia is 19 times lower resistence for energy to go. And total weight of car makes it impossible for rear wheels to stick to the ground without spin.

Then, to lift COG away from ground, you NEED to defy gravity. For that you need to accelerate COG upwards at over 1G. You say that ring gear is enough to give support for chassis to lift. Do you respect 3rd law? Action-reaction. To lift something up, something has to go down. What? Wheels rotate, transferring all of the torque to rotation, and through patch to the ground as strictly parallel to the ground force vector. Based on you, if you put engine in middle of chassis and use front-drive, then ring torque would instead of lifting front end, produce downforce?

When we bolt wheels to the ground, its completely different case - then linear inertia is infinite and only angular inertia remains. Then, only one thing can happen - either car flips over its back in very short time, dictated by gearing and engine's rpm range with most tourque, which translates into insane angular acceleration of 18 tons, or, more likely, something mechanical will give, engine stalls or blows up. Please, don't forget that this is real metal and loong chassis, this isn't some plastic toycar.

http://www.mytsn.com/publ/publ.asp?pid=6811 This is best I could find that show where COG of top-fuel dragster is. It talks about additional twisting torque from aero drag, but in 2nd half there are pics with COG shown.
There we see that roughly, COG is such that 75% of weight is on rear axle, and 25% on front. I can't say its height exactly, but let's assume its 0.5m. Now let's calculate load on fronts with launch Gs that for dragsters are typically 5G
Lf = dM - Fh/w (h=height of COG, w=wheelbase, M=car weight, d=static weight distribution as a fraction of weight in the front, and F=accelerative G-force)
F=975kg x 5G = 4875
Lf = 0.25 x 975kg - 4875 x 0.5m / 7.62m = 243 - 319 = -75kg. Its a front lift, 319kg weight shift to back, or 33%.
So I'd say weight shift due to COG height is enough to lift fronts.

Funny cars, they have much shorter chassis, wheelbase, higher COG, and that translates into much larger weight shift, and much lower angular inertia! Thats why they do wheelies. not because they have such torque at ring. Every single bicycle can do wheelies. And that simple huge wheelbase of dragsters is to increase moment of inertia of chassis, to _avoid_ wheelies.

AHH, went to the second link, note the distance in the data points (a) (b) and (c) because for inertia to be the effective lifter of the car, the car MUST travel forward, if you have ever watched drag racing you will have observed that dragsters lift their front ends RIGHT FROM THE START LINE, (not enough forward motion for inertia to have worked with the kind of force required) torque from the axle/chassis combination (matches, same as, the ring/pinion example)

Acceleration in above data is average, integrated value. Can you quickly say how much distance would car travel from v=0 in 0.1sec at constant 10G? 0.5 meters! In next 0.1secs, it would travel 1m... So at 5G launch, in 0.1 sec you can notice, it doesn't even move more than 0.25m, or 10 inches.

Its acceleration Gs that matter. COG tends to stay static, wheels accelerate, of course lift due to rotation around COG starts immediately. Its not only front lifting, its also rear squatting, and trying to roll under the COG. Its 4 times lower inertia to rotate around COG, and 19 times lower inertia to boost forward than to lift around rear axle.

Inertia of car is the only reaction force that allows for torque to even develop at ring gear. F=ma. You can't have lifting torque without acceleration either. And reaction you get from linear inertia is maximum you can get at ring gear, its 19 times less than needed to lift the chassis about rear axle!
Although the effect maybe there, its too small to be dominating.

 

[Mr. Robin Parsons]

Originally posted by wimms
Ah, now I see. You take top-down projection of chassis dimensions onto the ground. Basically, you say that with sun in zenith, COG shifts relative to rear edge of its shadow. Is this correct?

YES, I see the COG as, simply, the center of the compass that tells of the direction of the forces at work.

Originally posted by wimms
Front wheels of dragster lift no more than few inches. If they lift more, there is something terribly wrong with their chassis.
Given wheelbase of 300in, find angle. How much would COG shift back there?

BTW russ, when it goes from "start to finish" it uses most of it's fuel, hence the COG moves slightly backwards in the process as the 'total mass' and distribution of 'total mass' changes.

Does that answer your question? (tee hee)

Originally posted by wimms
Then, to lift COG away from ground, you NEED to defy gravity. For that you need to accelerate COG upwards at over 1G. You say that ring gear is enough to give support for chassis to lift. Do you respect 3rd law? Action-reaction. To lift something up, something has to go down. What? Wheels rotate, transferring all of the torque to rotation, and through patch to the ground as strictly parallel to the ground force vector. Based on you, if you put engine in middle of chassis and use front-drive, then ring torque would instead of lifting front end, produce downforce?

3^rd law, action/reaction, hummmmmm, hows about equal and opposite action, as in, torque exerted by wheels is balanced out by equal and opposite force upon frame and chassis.

Lets see, all we are dealing with here are levers and fulcrums. There is an internal fulcrum that is being called a COG, and there are external fulrcrums called axles, front and rear.

We deal with the rear one, in looking for the leverage action, in accordance with the "internal fulcrum's" placement, and the levers length/weight distribution.

As for the springs, anyone who has actually ridden a mountain bike with front shocks/springs, knows that the energy used to compress the springs/shocks is lost, nearly completely, in re-translation, as it is not imparting forward motion, but upwards motion.
(Then try riding one with no springs/shocks, notice the difference, cause it is really obvious)

It is/was energy that was meant to be "forwards motion" that got 'caught'/translated by/into the shocks, and translated into downwards pressure, re-translated/re-emitted as upwards pressure, or lift, on decompression.

The shocks/springs softness translates into energy loss!/Lost!

As for tons of force, do you mean like the "boss pin" in a normal car engine that can be subjected to (10) ten tons of pressure??

Lets see, if it were inertia lifting alone, then jets (High G accelerational force) could take of semi vertically, BUT, they cannot, no torque exerted upon them to create initial lift.

 

[wimms]

Originally posted by Mr. Robin Parsons
3^rd law, action/reaction, hummmmmm, hows about equal and opposite action, as in, torque exerted by wheels is balanced out by equal and opposite force upon frame and chassis.

Yep. Point is, torque at wheel patch causes action parallel to the ground. Reaction causes forward force of chassis. Look, rear diff has input shaft and rear axle at right angle, are at same plane, and both are contained in rigid body. Torque between the two because of gearing, is not able to lift anything. The only reaction force that could lift, is traction force at wheel patches. And they are unable to offer up force, unless they'd be doing something we call jumping.

Let's see, all we are dealing with here are levers and fulcrums. There is an internal fulcrum that is being called a COG, and there are external fulrcrums called axles, front and rear.

We deal with the rear one, in looking for the leverage action, in accordance with the "internal fulcrum's" placement, and the levers length/weight distribution.

So, rear axle is fulcrum, and lever arm that goes to ground is directed straight down, forming right angle between chassis and point of contact with ground. Question you must answer is, can lever arm being at right angle to reaction support (ground) generate any other kind of force vector but that which is at right angle to lever arm.

http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

As for the springs, anyone who has actually ridden a mountain bike with front shocks/springs, knows that the energy used to compress the springs/shocks is lost, nearly completely, in re-translation, as it is not imparting forward motion, but upwards motion.

It is/was energy that was meant to be "forwards motion" that got 'caught'/translated by/into the shocks, and translated into downwards pressure, re-translated/re-emitted as upwards pressure, or lift, on decompression.

The shocks/springs softness translates into energy loss!/Lost!

Although you are in a sense right here, you must understand why springs and shocks are used in first place. Their function is damping of wheel movements so that at contact patch there would be always optimal amount of downforce without excessive compresion of tires. Purpose of suspension is to avoid bumping off ground and high-Q resonanace that would cause up/down jerking of chassis, to increase amount of time best possible traction contact is maintained. They are force limiters.

What they do, is translate sharp and short 'collision' force into smooth and spread over time force. This allows to avoid overloading of wheels and jumping up off ground when downforce is removed, but introduces delay. The force applied to springs is delivered to the ground, but with delay. The only energy lost in them and shocks, is that which goes into heat from friction.

Springs and shocks in cars, do not consume forward energy, because they are compressed only when applied with vertical force vector, that already is excluded from forward motion.

The decision of using suspension and its softness depends on a lot of factors, and can deal with them very reasonably. The only idea I have about why would suspension be unwanted in dragsters, is its added delay (spring force is not linear) in delivering weight shift to the tire patches. This would mean less Gs of traction for the short duration of spring compression. Cars that are not balanced well enough, would use some suspension. Perhaps also cars that have very low COG, to create initial full weight shift with limited torque. Perhaps also to compensate for centrifugal rear tire expansion to void rear jumping off ground.

As for tons of force, do you mean like the "boss pin" in a normal car engine that can be subjected to (10) ten tons of pressure??

No, point of ring gearing inside rear diff, as would your explanation imply. Engine is decoupled by tranny and sees much less torque, although I have no idea what gearing ratios are used in dragsters. In any case, normal cars accelerate linearily instead of excercise angular acceleration, thus they don't face anything even remotely close to tons.

Let's see, if it were inertia lifting alone, then jets (High G accelerational force) could take of semi vertically, BUT, they cannot, no torque exerted upon them to create initial lift.

Geez, Robin, can't you understand that Jet engine thrust vector is on same line as COG of chassis. In Cars, thrust vector is located ALWAYS at the ground, thus being BELOW height of COG. The difference is immense. And most modern fighter Jets DO use chassis rotation around COG by redirecting their thrust relative to COG.
And I repeat, forces relative to COG do not result whole chassis lift, they result only rotation around COG. Front wheels lift only because rear squats, at least for dragsters.
When COG is stupidly high and chassis is short, vector addition would yield considerable up vector for COG, and indeed car can flip over, if forward force is large enough.

 

[Mr. Robin Parsons]

wimms, the energy transferred to the shocks is lost, why, thereafter, you seem to decide that I somehow need to have the reasons for the "use of shocks" explained to me, well you've lost me as to why you are doing this. (not entirely, I can/could 'speculate', easily)
Heck wimms, I had training from people in the automotive industry, batteries, shocks, tires, exhaust systems, braking systems, ignition systems, transmissions, electrical systems, etc. etc.

If you are still missing the point, try practicallity, goes like this, get a rear wheel drive car, automatic tranny, open the hood so you can see the motor from inside the car, get into the car, start the car, place your foot solidly on the brake-and hold down firmly, shift car into drive, place your other foot on the throttle, press gently to slightly accelerate the engine, note the lifting of he engine by the torque that is generated.

This is also how a car can be tested for having broken motor mounts as the exerted torque will lift the entire engine block, quite high, if one, or both, of the mounts are not solidly attached to the frame.

BTW the "Boss pin" is the pin that connects the connecting rod to the piston, and it is subjected to tons of force inside a very normal car engine. Just because you seemed to think that such forces would be unbearable by the components of cars, they aren't they are, kinda normal.

 

[russ_watters]

Originally posted by Mr. Robin Parsons
wimms, the energy transferred to the shocks is lost

The first law of thermodynamics disagrees with you. A spring is a near perfect 1st law system.

Lets try looking at it another way: when you brake heavily in a car it does a nose dive. When the car finally comes to a complete stop, you feel the car lurch (or maybe just settle depending on the car) backwards. Thats the feeling of the energy being recovered from the shocks.

 

[Mr. Robin Parsons]

Originally posted by russ_watters
The first law of thermodynamics disagrees with you. A spring is a near perfect 1st law system.
Lets try looking at it another way: when you brake heavily in a car it does a nose dive. When the car finally comes to a complete stop, you feel the car lurch (or maybe just settle depending on the car) backwards. Thats the feeling of the energy being recovered from the shocks.

Yes, the shocks tranfer the energy back, in a vertical direction, not in a horizontal direction which, in the case of a rail is where that energy originated from, horizontal acceleration.
The shocks cushion the tranfer of weight, they do nothing to the acceleration/decceleration of the car. (other then to take some of that energy, out)
The sensation of the "lurch" is due to the re-balancing of the cars inertial mass, little, if any, input, into forward, or backwards, motion.
Try a mountain bike, with shocks, and without, (stand on it to peddle) you will feel the difference immediately!

 

[Mr. Robin Parsons]

Originally posted by wimms

Originally quoted by wimms, written by MRP
As for tons of force, do you mean like the "boss pin" in a normal car engine that can be subjected to (10) ten tons of pressure??

No, point of ring gearing inside rear diff, as would your explanation imply. Engine is decoupled by tranny and sees much less torque, although #1)I have no idea what gearing ratios are used in dragsters. In any case, normal cars accelerate linearily instead of excercise angular acceleration, #2)thus they don't face anything even remotely close to tons.

#1)So apparently you don't read what you have recomended, from the site YOU linked, http:**bmeltd.com, rear end Chrisman, 12-in. 3.20:1 ratio
#2)and again, same site, 'nuther page, http://bmeltd.com/newpages/Wristpin.htm this info;

Originally published at Bill Miller Engineering Ltd
Inside one cylinder of a 700 horsepower, NASCAR Winston Cup engine running at 9000-9500 rpm, a crushing force of five tons hammers the wrist (I said/called it was a "Boss pin", same thing as a 'wrist pin') pin about 77 times each second and this punishing, cyclical loading lasts for up to 600 miles in some races. Needless to say, wrist pins are subjected to unbelievably high levels of both bending and radial stress and they must sustain those stresses for a considerable period of time.

Remember that force on the piston top we talked about earlier? In a supercharged, nitromethane-burning engine in a Top Fuel Dragster or a Nitro Funny Car, that force is even more extreme, perhaps as much 50 tons. BME also manufacturers a line of Wrist Pins for alcohol-burning and nitromethane-burning supercharged drag race engines. There are very few raw materials with the incredible strength required by wrist pins in a blown-fuel, drag race motor. BME blown-fuel Pins are made of VascoMax C-300, an exotic, very expensive, nickel-cobalt-titanium steel "superalloy" with very high ultimate tensile strength (294,000 psi) and an extreme fatigue endurance limit (one billion cycles at 125,000 psi).

So when I had said ten, I have either errored, (sorry) or I remembered/suplanted what a diesel is subjected to.
None the less, my fault, sorta!
Plus these three lovely FACTS, 6000 horsepower at 8200 rpm, clutch AFT 10 in. dia., 5-disc, bell housing Trick Titanium
Hummmmmmm 6000 Horse power, nuff to lift the front end right over backwards if it were NOT for those "wheely bars".

 

[krab]

Originally posted by Mr. Robin Parsons
Yes, the shocks tranfer the energy back, in a vertical direction, not in a horizontal direction which, in the case of a rail is where that energy originated from, horizontal acceleration.
The shocks cushion the tranfer of weight, they do nothing to the acceleration/decceleration of the car. (other then to take some of that energy, out)

That isn't true. Here's a thought experiment which should make things clearer. Let's say the car with the soft suspension (that the original poster brought up remember? we weren't talking about dragsters) has its rear tires glued to the road. With a system of rods or some such, imagine I have a mounting point for a cable at the exact COG of the car. Pull back on the cable. What happens? The car begins to tilt. Pull the cable about 1 inch back and the rear springs compress maybe about 4 inches, the front springs extend a similar amount. Let go and the COG moves forward as the car levels. In fact, it Accelerates forward.