Physics of an automobile, suspension, and weight transfer?

[Magnus]

The application is drag racing in a rear wheel drive vehicle.

The vehicle weighs approximately 3000lbs.
Traction is not an issue as both vehicles grab the ground perfectly.
The front and rear is suspended by a shock and spring in each corner.
All else is equal unless otherwise stated.

Car A launches and leaves the line completely horizontal with no suspension travel. This car has stiffer suspension setup.

Car B launches and leaves the line where the rear end squats and the front of the car raises a few inches. This car has a softer suspension setup.

Is energy diverted from the forward vector and lost in the suspension travel in car B? If so, can you please give me an explanation as to how this occurs?

I am fairly physics savvy but am having a hard time understanding how power is lost in minor suspension travel.

I understand that shocks are designed to absorbe energy (mainly from spring travel)

The way I see it, the engine turns the transmission, which turns the driveshaft, which turns the pinion, turns the differential/axles, and turns the tires. Tires just turn, and turn along the ground. They turn as fast as the engine will power them too regardless of what the suspension may do (within reason).

I understand that for the cars nose to lift, less rotation is seen in the tire, but is re-insert as the front comes down. I also understand that a car that may lift its nose will change its aerodynamics.

Please help me, I am confused. :)

This is my first post btw.

- Keith

 

[Integral]

Seems to me that any torque that contributes to lifting the front of the car is not available to turn the wheels, thus the measureable amount of energy would be used in lifting the front of the car would not be available for conversion to Kinetic Energy of motion. Thus lower speeds.

Just a gut feeling discription.

 

[meister]

Originally posted by Integral
Seems to me that any torque that contributes to lifting the front of the car is not available to turn the wheels, thus the measureable amount of energy would be used in lifting the front of the car would not be available for conversion to Kinetic Energy of motion. Thus lower speeds.

Yes I was thinking along the same lines. Putting energy toward raising the car clearly subtracts from the amount of energy used to spin the wheels.

Would it not also create a force in the opposite direction of his intended forward motion as the car jerked upwards?

 

[Sonty]

There's another factor that I'm thinking of: conservation of angular momentum. You know that second small blade mounted on the tail of helicopters? It's there to balance the force coming from the tendency of the helicopter body to rotate in the opposite direction to the main blade. It's the same with the car. When the wheels suddenly start to move fast the nose of the car tends to go upwards especially since its weight is mostly in the back. A part of the force needed to balance the car is covered by gravity. When you put in a very hard suspention the rest of the force needed is taken by the engine and the tension in the chasis (it can actually break). So you lose some energy there. When you put in a softer suspention the shock absorbers and the springs balance that force. So you win some energy.
So it's all a big compromise: you want more power but not more than the tires can put to the ground, you want to keep the nose down to stay aerodynamic and not to tip over or break the chasis, etc.

 

[Magnus]

I don't see it as torque lifting the front end, but the forward vector of the rear end of the car (axle) being greater than the down vector applied by gravity on the nose, thus being the reason the car comes up.

Sonty, are you saying you would lose more energy in stiffer suspension?

I still don't really understand how acceleration or force is lost when the rear end of the car squats for example.

f=ma... Mass is a constant.

The front of the car lifts when acceleration of the axle is greater than the down force applied by gravity of the front correct? If the suspension is stiffer, it will require a greater force for the front end to come up do to the suspension components weight being applied right away. With a softer suspension the body can come up and let the suspension hang and THEN pull up the suspensions weight.

I'm still confused.

 

[Sonty]

Originally posted by Magnus

Sonty, are you saying you would lose more energy in stiffer suspension?

I'm saying the optimum choice is somewhere in the middle towards the harder suspention. When you get enough speed the aerodynamic force pushing the nose down is enough to balance the car. Keeping the nose up too long you lose because of air drag.

[B}
f=ma... Mass is a constant.

The front of the car lifts when acceleration of the axle is greater than the down force applied by gravity of the front correct? If the suspension is stiffer, it will require a greater force for the front end to come up do to the suspension components weight being applied right away. With a softer suspension the body can come up and let the suspension hang and THEN pull up the suspensions weight.

[/B]

f=ma right, m=constant right, the engine gives the same force right, but where do you spend that force makes the difference.
Am talking about the back suspention, which one are you talking about?

 

[Magnus]

Aerodynamics isn't a concern at all. My real concern is just the launch... and how a car should launch for ideal energy savings.

I'm talking about front AND back.

The Force of the engine just rotates the tires in the end. The tires just rotate along the ground moving the vehicle forward and because the vehicle accelerates so fast in a direction, momentum kicks in and the suspended chassis wants to go backwards. Because the center of gravity is higher than the forward force vector, when the forward force vector exceeds the downard vector of the nose of the car, the front end lifts.

Regardless if the suspension is stiff or soft, if the nose comes up, or it doesnt, is the forward force vector the same?

I don't really know.. my entire post is basically a question.

 

[Mr. Robin Parsons]

Originally posted by Magnus
(SNIP) Regardless if the suspension is stiff or soft, if the nose comes up, or it doesnt, is the forward force vector the same? (SNoP)

The forward force vector changes based upon the changes in the suspension, and the nose lifting.

So when a dragster takes off, the torque generated by the engine, driving the wheels, lifts the front end, resulting in a minor addition to the "co-efficients of friction" at the rear end, and a slight loss of the power, due to suspension compression.

It acts a little like a lever, hence you get a minor transfer of center of gravity, towards the rear, as the machine attempts to balance the energy lifting the front, and the traction (and slip) at the rear driving the thing forward.

Softer suspension absorbs power, but lowers the rear end resulting in a minor addition to traction due to the lever/balancing of the center of gravity.

It's all pretty well vector physics, Newtonian.

Really neat to watch the 'slo-mo' of the rear tires during the "Bleaching" process (WATER poured out onto the track, and the tires spun to smoking to heat, and clean) as you can see the displacement of the sidewalls of the tires as they are gripping at a 'tractionable surface' and then lossing that traction and 'rounding out' due to centrifugal force.

 

[OldHubcap]

Auto suspensions are very interesting. In the case of drag racers, taken to an extreme, some racers have the power to lift the front wheels completely off the ground. This does cost them time and to counter that, they may have "wheelie casters", small wheels that stickout from the the rear of the car that prevent the front from leaving the ground.

Also, not all suspension types behave the same when the engines full power is applied to them. Leaf spring rear suspensions have a tendency to be twisted by the rear axle under torgue into an "S" shape. Actually causing the rear of the car to lift up on some models. This is controlled with traction bars, bars that attatch to the spring under the axle and extend forward with a rubber snubber that hits the frame and minimizs spring twist. (some designs). Independent rear suspensions have a tendency to squat under full power launches.

 

[Magnus]

Originally posted by Mr. Robin Parsons
So when a dragster takes off, the torque generated by the engine, driving the wheels, lifts the front end, resulting in a minor addition to the "co-efficients of friction" at the rear end, and a slight loss of the power, due to suspension compression.

friction is increased at the rear and decreased at the front.

I don't understand how power is lost due to suspension compression though.

The front end will lift depending on how much power you have. With a softer supsnsion your chassis will lift first then the suspension will hang and carry up the wheel assemblies with it (if the front end lifts that far)..

So say you have a soft suspension and the car only lifts up 2". Its not enough to carry the wheels. Wouln't it be easier for the car to lift like this, the 2", then it would be if the suspension was stiff and thus the added weight of the wheel assemblies was too much for the car to lift at all?

 

[Mr. Robin Parsons]

Originally posted by Magnus
friction is increased at the rear and decreased at the front.
If it leaves the ground, front end friction is gone/eliminated (temporarily, until it touches back down)[/color]
I don't understand how power is lost due to suspension compression though. The loss is due to the amount of energy that is required to cause the compression in the first place, as that is as a result of the cars weight tranfers caused by it's acceleration forward[/color]

The front end will lift depending on how much power you have. With a softer supsnsion your chassis will lift first then the suspension will hang and carry up the wheel assemblies with it (if the front end lifts that far).. With a soft suspension the drive wheel torque effects will cause suspension compression, prior to any lifting.[/color]

So say you have a soft suspension and the car only lifts up 2". Its not enough to carry the wheels. Wouln't it be easier for the car to lift like this, the 2", then it would be if the suspension was stiff and thus the added weight of the wheel assemblies was too much for the car to lift at all?[/color]

Sorry don't understand what you are driving at here. ("Pardon the pun" AKA PTP)[/color]

 

[Magnus]

The car will accelerate forward, that's a given.

If the suspension is soft enough to allow weight transfer though, where and how is drive energy being lost? That's what I don't understand.

Another example is this. You have trains of equal power on parallel tracks. Each carries a flat bed which a vehicle on it died down by the wheels. Vehicle A has solid suspension (no give) where as vehicle B has soft suspension.

The mass of the trains are equal and so is the power output of the locomotive.

If forward energy is lost due to weight transfer, then the train with the solid suspended vehicle on it would accelerate faster correct?

 

[Mr. Robin Parsons]

Originally posted by Magnus
The car will accelerate forward, that's a given. OK![/color]
If the suspension is soft enough to allow weight transfer though, where and how is drive energy being lost? That's what I don't understand. Reguardless of the suspensions compressability, weight transfers will take place, so we see that a softer suspension will absorb more of the energy because it will likely travel more.[/color]
Another example is this. You have trains of equal power on parallel tracks. Each carries a flat bed which a vehicle on it died down by the wheels. Vehicle A has solid suspension (no give) where as vehicle B has soft suspension.

The mass of the trains are equal and so is the power output of the locomotive.

If forward energy is lost due to weight transfer, then the train with the solid suspended vehicle on it would accelerate faster correct? No[/color]

Because in your train example the will be no 'leveraging' action. (Transfer of weight)

Take a board, lift one end only, the weight is transferred towards the end that is still on the ground, A soft suspension will abosrb more of the weight transfer's energy (quicker and longer motion/distance travelled) then will a stiffer suspension.

 

[OldHubcap]

Magnus, the engine's energy used to lift the front end of the car is lost to you for purposes of accelerating. Gravity brings the front end down again and you don't regain the energy. Therefore if you can keep the car front end from lifting, you keep more of the engine's energy available for acceleration. Fairly straight forward.

Folks who drag race production automobiles try to stiffen up the suspension as much as possible. If one looks at purpose built drag cars like top fuel dragsters, you see that they have no suspensions and very long wheelbases to control front end lift.

 

[Magnus]

In the train example though, if the trans where able to accelerate as fast as the cars did at the track, and only the rear wheels where tied down.. wouldn't there be leveraging action due to momentum? The center of gravity of the car is higher than the point of forward force thus it goes upward? And if this is correct, then would the trains still accelerate evenly?

OldHubcap, I must have a bad mental image of the equation then. I just picture the wheels moving forward along the track. I picture them moving so fast that the front end comes up because of the rate of acceleration. I don't think of it as the engines power cranking up the front end as if the wheels where not moving.

Example: You have an RC car. You grab its rear wheels and give it a quick JERK forward.. front end comes up. Because the front end comes up it takes more energy for you to jerk the car forward than it would if the car where to remain parallell?

 

[Mr. Robin Parsons]

Originally posted by OldHubcap
Magnus, the engine's energy used to lift the front end of the car is lost to you for purposes of accelerating. Gravity brings the front end down again and you don't regain the energy. Therefore if you can keep the car front end from lifting, you keep more of the engine's energy available for acceleration. Fairly straight forward.
Folks who drag race production automobiles try to stiffen up the suspension as much as possible. If one looks at purpose built drag cars like top fuel dragsters, you see that they have no suspensions and very long wheelbases to control front end lift.

Nice, but on 'minor details' (which is what you need to finesse in competative drag racing) lifting of the front end, and the resultant weight transfers, increases the co-efficients of friction on the rear end of the car, giving it greater traction for forward propulsion.

Balancing it out well, gets the best result. That is also why Wheely bars let it lift the front, somewhat.

Better weight transfer means better rear end traction = greater accelerative possiblity.

 

[OldHubcap]

Very well put Mr. Parsons. You do need some degree of weight transfer to increase traction. Drag racing, like all motorsports, is about finding the proper balance.

 

[Magnus]

I understand the effects of weight transfer on traction...

did you guys read my last reply? That's where I'm getting confused.

 

[OldHubcap]

With respect to front end lift on a drag car.
The lift occurs when the car initially accelerates and still has little forward speed, thus allowing us to discount significant lift caused by airflow. This means that the energy to lift the front end ultimately comes from the engine. If you break it down into vectors, if the car had no lift, the only vector is pointing forward.

<<<<<<<<<< Imagine this is the vector for a perfectly rigid suspension. All 10 vector arrows go forward


If the front end lifts, the vector picture looks like this

^
^
^<<<<<<< In this case 3 vector arrows used to lift front of car and only 7 vector arrows accelerate the car.

Its a rather crude illustration, but I hope it sheds light on what is happening.

 

[Magnus]

I can understand that.

If your forward vector is so great though that the front end cannot come down due to the fact that the down force applied by gravity isn't enough to overcome the acceleration, what are you supposed to do to conserve energy?

In that scenario, isn't there no enrgy lost?

 

[Mr. Robin Parsons]

If your question is "why is the front end lifting?", it is as a result of torque that the drive wheels apply to the car frame.

If you look at the wheels, as they begin to spin, you can figure out that there MUST be an equal, and opposite, torque force being applied to stop the frame of the car from being the item that turns instead of the wheels turning.

Imagine you hold the car in your hands (God that you would be) and hold it by the back wheels (only) so they cannot turn, apply the force of the motor to the drive wheeels and you will realize that, then, the frame of the car will be the thing that turns. (in large circles)

Does that help?

 

[Magnus]

I understand that. I separate the force that gravity applies though.

Say you throw a baseball... you have an X and Y vector. The X vector is depandant on your arm and air resistance. The y factor is dependant on gravity.

You throw the ball hard enough and it will never hit the Earth (escape velocity exceeded)... however, if you don't, gravity affects the Y vector completely independant of what X is doing.

Thanks for being patient, I'm just looking for understanding. :)

 

[wimms]

Time is what makes the difference. Why are dragsters so long with center of gravity (cog) so far back? They need lightweight car, and they need to keep nose down. When traction kicks car forward, car starts turning around its cog, driving nose up. To slow this down, nose is made so long, to work as a lever of rotational inertia. This forces rear wheels to push against ground harder. Weight transfer is used to gain more traction at wheels. Lowering cog isn't good idea at all.

The very instant of start is what makes the most difference. Any subsecond lost there translates into lost run. With hard suspension, weight transfer happens faster, softer suspension wastes time on compression before max weight transfer occurs, and lowers cog. So, perhaps istead of wondering about energy, wonder about when it gets applied.

When wheels spin, some power is lost for acceleration, and stored in wheel rotational inertia. This rotational energy is regained, but later. By that time, competitive car has gained some advantage. So it seems that storing rotational energy makes sense only when max traction possible is achieved for given vehicle geometry and cog position. That means pushing rear wheels to the ground as hard as possible, as fast as possible, using inertial levers around cog.

 

[Magnus]

I think I got it now...

When you initially long the drag car, the roation of the pinion on the ring gear is what lifts the car (that and the torque arm pushing up on the body)... NOT the rapid acceleration of the tires. It's easier for the car to go up than it is forward under the instant shock.

So its not really a matter of your forward vector being greater than the down vectors applied by gravity, but instead the rotational vector applied by the ring/pinion/torque arm being greater than the forces applied by gravity.

Correct?

 

[OldHubcap]

Magnus, you got it!

Not to cloud things, but the driveshaft also exerts torgue on the axle, causing the axle to rotate on its pinion and move one wheel up and the other wheel down. Just wanted to point out that auto suspensions are subject to interesting stresses.

 

[Magnus]

See, I just thought though that the front end jerked up because the forward acceleration was so great, and the forward vector was lower than the COG thus making the front lift.

 

[Mr. Robin Parsons]

Originally posted by OldHubcap
Magnus, you got it! AGREED![/color]
Not to cloud things, but the driveshaft also exerts torgue on the axle, causing the axle to rotate on its pinion and move one wheel up and the other wheel down. Just wanted to point out that auto suspensions are subject to interesting stresses.

Again, AGREED, some of the usful additions are things like differential locks, positraction. (Lose due to weight and slippage) Tire 'circumfrence' size is absolutely critical as differtiation in tire sizing can cause the rear end to steer the car on takeoff, as would differentiated wheel slip. (hence the diff lock)

OldHubcap is right about the interesting stresses, but if you would want a real challenge, try to figure out how to stop the "pitching" that a wheel loader gets protection from (now-a-days) by the 'ride control system' in them, complex.

 

[Magnus]

Let me as you this then... ingore wind resistance.

If the force on the axle wasn't applied by the engine/tranny, but perhaps by a strap that was directly even with it on a horizontal level..

If the car accelerates fast enough to the point where the front end DOES lift due to its higher COG, there is no energy lost correct?

IE: Say you accelerated fast enough to where the front was just 1 foot off the ground as opposed to a car of equal weight that has a COG on the same axis as your forward vector (no lift).

 

[Mr. Robin Parsons]

Originally posted by Magnus
Let me as you this then... ingore wind resistance.
If the force on the axle wasn't applied by the engine/tranny, but perhaps by a strap that was directly even with it on a horizontal level..
If the car accelerates fast enough to the point where the front end DOES lift due to its higher COG, there is no energy lost correct?
IE: Say you accelerated fast enough to where the front was just 1 foot off the ground as opposed to a car of equal weight that has a COG on the same axis as your forward vector (no lift).

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.

If it could be kept/preserved to drive the car forward, it would be better, but at moments like that, in a drag racers/rails actions, the amounts of energy at play already has losses, due to tire spin, by way of a loss of traction, so that balancing acts actually can help to conserve some of that energy, that would, otherwise, be completely lost(?).

 

[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 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).

 

[meister]

How do you get the energy back when the front end drops back onto the ground? You use engine power to lift it up, then whatever potential energy you gained while in the air doesn't all go completely towards accelerating the car.

 

[Mr. Robin Parsons]

Originally posted by meister
How do you get the energy back when the front end drops back onto the ground? You use engine power to lift it up, then whatever potential energy you gained while in the air doesn't all go completely towards accelerating the car.

You do not really get the energy back, BUT as the car accelerates down the track, the wheel slipage diminishes to the point where all of the engines energy is effectively used driving the car forward, only!
No longer enough torque on the frame/chassis to lift it, so it transfers to the wheels to drive forward

 

[meister]

Originally posted by Mr. Robin Parsons
You do not really get the energy back, BUT as the car accelerates down the track, the wheel slipage diminishes to the point where all of the engines energy is effectively used driving the car forward, only!
No longer enough torque on the frame/chassis to lift it, so it transfers to the wheels to drive forward

How is this different from a drag car that keeps all four wheels on the ground at all times?

 

[russ_watters]

Originally posted by meister
How do you get the energy back when the front end drops back onto the ground? You use engine power to lift it up, then whatever potential energy you gained while in the air doesn't all go completely towards accelerating the car.

Where does it go if not toward driving the car? It must go somewhere, it can't just disappear.

Thinking about the car in motion makes it seem more confusing than it really is. Attach a jack to the front of the car and lock the wheels. Wheat happens when you lift the front? It rolls back on the back tires. Now drop the front, what happens? It rolls forward to its original position.

And its easy enough to calculate how much. Let's say you lift the car so the angle between the wheels and the ground is 30 degrees. Let's say the tires are 36" in diameter.

30/360 * pi * 36 = 9.4"

How is this different from a drag car that keeps all four wheels on the ground at all times?

There is a slight dip in the acceleration curve, but the overall average acceleration remains the same.

The perfect scenario in fact would be one where the torque was exactly balanced between lifting the front of the car and staying right on the edge of spinning the tires. Lifting the front of the car increases the weight on the rear tires, increasing the traction, and increasing the maximum appliable torque (and thus maximizing acceleration). 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.

 

[Magnus]

Ok, I was right in line with russ_watters ideas, however then I was swayed the other way.

Now I'm just completely confused.

 

[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[/color], 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[/color]
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[/color]
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