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. [g)]

[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 displacment 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)
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

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.

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

[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!
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.
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

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

Take a board, lift one end only, the weight is transfered 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 seperate 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!
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. Lets say you lift the car so the angle between the wheels and the ground is 30 degrees. Lets 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, approachs 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 relitivity 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 occuring 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 explainations........
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 explainations 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 spacial 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 explainations 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 spacial 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%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.

Anyway, this isn't relevent 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 "[b]moment of[/i]intertia," not just "inertia." Big difference. http://dictionary.reference.com/search?q=moment%20of%20inertia

[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 occured 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 occured 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, thats 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 knowledgable, 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 lets assume its 0.5m. Now lets 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?
3rd 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
3rd 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.

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. 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 alot 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.

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. 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 transfered 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 transfered 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.

[wimms]

Originally posted by Mr. Robin Parsons
wimms, the energy transfered 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. Robin, I do assume that you know what you are talking about. Can't you realise that your position isn't easily acceptable? If you want others to understand, please try to explain it in more detail why it is that you are right. I posted this shocks function to express my point, not to teach you anything "new". And point is: energy that goes into springs already IS taken away from forward force. Springs compress only when vertical force is present. And shocks really transfer that vertical energy to the ground, they resist vertical compression. The only function of suspension is to doze vertical energy in controlled manner. The only energy loss is in the friction. Do you agree here? I'm not arguing that spring energy is recovered for forward motion. I'm only arguing that it already wasn't available for forward motion.

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. Chassis is not lifting from that. If you lock clutch, engine would stall. Engine is not lifting, it is twisting around gearbox and some of supports under the hood. At best, its jammed against the chassis. Its not same thing.

Again, please understand that locking wheels changes scenario completely. Then, energy has only one place to go - lifting chassis. When wheels are free, energy goes into accelerating chassis, not lifting it. Although the effect you support is probably there, it is so small that its irrelevant. At 5G accelerations, inertial effects dominate.

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. Fine. You got me here. I assumed we talk about tons of torque, like nm. But to compare things, we need to bring them to common metric. Lets approach from torque of engine. http://www.amatoracing.com/carspecs.shtml
It offers 6,250 lbft or 8500nm of torque. To give 50tons on rods, cam radius should be about 17cm. Sure, forces at pins are huge, maybe even larger (17cm sounds kinda excessive).

After some thought it seems that dragsters don't have any tranny. Rear diff ratio is 3.2, and wheel radius is 0.5m, basically adds ratio of 2:1. Notice that these ratios apply to increase torque at wheels. This translates into 8500x3.2x2=54000nm of torque at tire patches. a=F/m, and /975kg offers 5.7G linear acceleration.

To lift chassis, the only torque we have is 8500nm. If wheels are locked, then whole chassis has to accelerate at some rate about rear axle. We need to account for engine working rpm range, lets assume max torque is achieved at 2000rpm, 210 radians/sec. If engine can't make it, it'll either stall or fail. Something has to give. With free wheels, they'd spinup.
From standstill to max torque its angular acceleration of 210rad/sec2 at least. Now find how much torque is needed to give such angular acceleration to the whole chassis.

We've found that angular inertia of chassis is about 19 tons. To give such inertia angular acceleration of 210rad/ss, we'd need torque 3,990,000nm! http://hyperphysics.phy-astr.gsu.edu/hbase/n2r.html And thats without accounting for need to overcome gravity.

At 8500nm, angular acceleration of chassis about rear end is at best 0.45rad/ss, and thats when all energy goes there. No engine can work at such low rpm. It either stalls, or engine block blows up. When wheels are free, most of energy goes to linear acceleration, thus even less is avail to lift chassis. Even if you do not agree with 19tons of chassis angular inertia, by using plain 975kg and 1meter distance of COG from axle, needed torque is 204,750nm, which is much more than available.

See? Key points I make:
- torque at tire patch is 3.2x2=6.4 times bigger than at ring gear
- angular acceleration needed accordingly 6.4 times higher at ring
- linear inertia of chassis is 19 times less than angular inertia about rear axle
- engine has not enough torque to flip chassis over at its working rpm range
- engine has enough torque to produce linear acceleration of over 5Gs.
- tires spin up and work as slipping clutch
- energy spent on acceleration is not available for angular lift of chassis about rear axle
- COG height and acceleration force is enough to account for front lift

[Mr. Robin Parsons]

Clearly wimms the problems here are, you don't read what I am writing, or, it is a comprehension problem, or, you are simply ignoring the facts and fishing me (ends HERE!)

I stated a car with an automatic tranny, you come back with popping the clutch stalls out the motor! (Please tell me how you pop any of the three clutches on an automatic transmission, cause I have never seen them popped, burnt yes, popped NO!)

A bit like the spring energy problem, lost in the translation.

So wimms, 1 (one) horsepower equals 33,000 lbs/ft/minute, or 550 ft/lbs per second. The rails engine has 6000 HP, so 3,300,000 ft/lbs per sec, or 1650 TONS/ft/sec, or 165 tons/ft/tenth of a second.

Is that enough to lift the car? (YES!!)

P.S. You seem to wish to use the "5 G" figure for the cars acceleration, it does not develop that 5 G immediately, first tenth of a second, but it does lift it's wheels

And please wimms you said; Originally posted by wimms
It offers 6,250 lbft or 8500nm of torque. To give 50tons on rods, cam radius should be about 17cm. Sure, forces at pins are huge, maybe even larger (17cm sounds kinda excessive).
Sooo, PLeeease Explain to everyone here how cam radius is related to wrist pin pressure!

(Don't know much about engines do you?)

[Mr. Robin Parsons]

P.S. wimms, I had mentioned that the pressure on the "boss/wrist" pins was 10 (ten) tons, and I had linked a site that told of 5 (five) tons in a race car.

Given F=ma, and the simplicity that a race cars pistons are as "shaved/trimmed" as much as is possible, it is very possible that the normal car engines pistons weigh in at twice the weight of the race cars pistons, hence the forces upon them, inertial forces would be twice that of the race cars engine, hence 10 (ten) tons.

So I am probably not off all that much. My source for that was a scientist, on television, to the best of my recollection.

[russ_watters]

Originally posted by krab
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. Yes. Again, torque. It magically converts vertical forces into horizontal ones.

For my next trick, I'll make something go up by pushing down on it!

[Mr. Robin Parsons]

Originally posted by Krab
Let go and the COG moves forward as the car levels. In fact, it Accelerates forward.
You mean because of inertial forces, right?

[wimms]

Originally posted by Mr. Robin Parsons
Clearly wimms the problems here are, you don't read what I am writing, or, it is a comprehension problem, or, you are simply ignoring the facts and fishing me Funny, I thought this about you. What I don't comprehend is why all disagreements in this site must develop into unpleasant confrontation.

I stated a car with an automatic tranny, you come back with popping the clutch stalls out the motor! chill. We talked about torque, not tranny. Effect you implied would have to manifest with any tranny. And of course I meant manual tranny when I said clutch.

So wimms, 1 (one) horsepower equals 33,000 lbs/ft/minute, or 550 ft/lbs per second. The rails engine has 6000 HP, so 3,300,000 ft/lbs per sec, or 1650 TONS/ft/sec, or 165 tons/ft/tenth of a second.Have it occured to you, that moving 1 lbs 3M feet in one second and moving 1650 tons 1 feet has a difference?
Engines generate torque. Horsepower is product of torque and rpm:
Horsepower = Torque x 2 pi x rpm / 33000 which simplifies to:
Horsepower = Torque x rpm / 5252
Torque = Horsepower x 5252 / rpm
for 6000hp @8200rpm this is 3842 lbs.ft of torque, meaning 3842 lbs of force at arm length of 1 ft, at tangential velocity of 858ft/sec, which is achieved only at the end of drag run with chassis speed of over 200 mph. Sure, if you had gearing ratio of 858:1, you'd have 858x3842 = 3M lbs x 1ft /sec, but your gearing ratio is 3.2:1 !
So gimme a break with this 1650 TONS of lifting force. This is not heavyweight lifter crane, its a vehicle. Max HP is developed only when vehicle is speeding. At low speed you have nearly no HP, and only torque.

If you want launch forces, talk about torque. I specially searched for you and found a spec with 6,250 lbft of torque, which is more than is produced at 8200rpm, normal for any engine.
I spent time to find the calculations needed. You obviously didn't read anything. You see 6000HP -> boom - 1650 tons of lifting force.

P.S. You seem to wish to use the "5 G" figure for the cars acceleration, it does not develop that 5 G immediately, first tenth of a second, but it does lift it's wheelsInteresting. What is the supporting reaction force? Surely not inertia of car to acceleration, right? Some magical 'sticking' to the ground. Wheels spinning, producing 50000nm of traction torque, and for the first 10th of a second vehicle instead of accelerating, opts to spend some fun time on lifting its fronts instead.
In first 1/10 seconds, 5Gs produces 25cm displacement. Calculate. 5Gs all the way! I can agree that 5Gs aren't achieved immediately, because of engine reving up and tires warming up, but I can't see how lifting can be independant from acceleration.

You're arguing wrong thing. So far angular moment of inertia of rear wheels haven't been discussed, but spinning up rear wheels takes considerable amount of energy. That might be your first 10th of a second you are seeking for.

Sooo, PLeeease Explain to everyone here how cam radius is related to wrist pin pressure!
(Don't know much about engines do you?)English is not my language. I meant crank. If you'd not mean disrespect you'd not try to make fun of it. Or, is it that I'm having personality comprehension problem here?

[VeeP]

hey yall


just to clear something up (and help some of you understand this particular idea).

Softening the rear suspension (to a limit) on a RWD car in a drag launch situation is there to help provide the rear subframe recieve as much of the cars weight as possible (giving the nose up tendency).
As the nose lifts, it removes wieght from the front wheels (which serve no purpose other than to steer the vehicle) to the rear wheels. More wieght over the rear wheels means that the tire contact patch grows (and u still have the same amount of pressure per square unit) - which means that your tire can generate more traction, which means that you can apply more force to the ground without the tire lighting up - which means you can accelerate harder. . . .

However - this does NOT apply to certain types of cars (specifically dedicated drag cars) - particularly rail cars - why?
Becuase in rail cars, most of the weight is already over the rear tires. Same goes for funny cars, alot of the weight already sits over the back wheels.
Oh, rail cars dont even have suspension actually. . . its just the wheels directly bolted to the chassis (obviously through very low friction means) - the tire provides most of the suspension. . . .

FWD cars are a completely different story, maybe if someone specifically requests it i will explain.


oh. . . btw, i hope i can positively contribute to these forums and learn much from yall too :D

[Mr. Robin Parsons]

Originally posted by MRP
Clearly wimms the problems here are, you don't read what I am writing, or, it is a comprehension problem, or, you are simply ignoring the facts and fishing me (ends HERE!)
wimms, did you read this?? (above and below)

Originally posted by wimms
So gimme a break with this 1650 TONS of lifting force. This is not heavyweight lifter crane, its a vehicle. Max HP is developed only when vehicle is speeding. At low speed you have nearly no HP, and only torque.
Where did I call it lifting force, I simply stated force, aside from that, listen to the car's engine at the start line, revvvvvving up, well past 2000 rpm, as it is engine speed that counts, NOT the cars speed.

Originally posted by wimms
English is not my language. I meant crank. If you'd not mean disrespect you'd not try to make fun of it. Or, is it that I'm having personality comprehension problem here?
Should have told me it wasn't your first language from the start, it would have helped. No. I am not making fun of you, simply pointing out, from your posting the appearance YOU created of NOT knowing what you are talking about. Sorry if it caused you any negative emotive, it was NOT my intention.

Once again, Originally posted by MRP(ends HERE!)