Physics of an automobile, suspension, and weight transfer?

[wimms]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Is that clearer? better? do we now agree?

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

 

[Mr. Robin Parsons]

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

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

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

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

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

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

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

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

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

Nice link though.....

 

[russ_watters]

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

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

Example: a yardstick:

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

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

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

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

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

Another definition issue:

not enough forward motion for inertia to have worked

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

 

[russ_watters]

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

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

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

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

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

 

[Mr. Robin Parsons]

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

Ummm, russ...

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

Butt out!

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

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

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

 

[Mr. Robin Parsons]

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

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

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

 

[russ_watters]

Originally posted by Mr. Robin Parsons
Butt out!

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

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

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

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

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

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

 

[Mr. Robin Parsons]

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

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

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

 

[russ_watters]

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

Thank you. Finally.

But...

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

Huh? Where exactly does it go?

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

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

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

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

 

[Mr. Robin Parsons]

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

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

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

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

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

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

 

[Mr. Robin Parsons]

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

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

Not too much different from inertia itself;

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

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

Now here we have your statement from PG3 seventh post down

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

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

And you agree with this here...[/color]
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.[/color] 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?[/color]
So the answer to the initial question is no: suspension issues have no effect on the total acceleration of a car.
One little catch though: In a drag race, its not the final speed that matters, its the avearge speed. The two cars would have identical 0-60 times, but the car with the stiffer suspension will have traveled further in that time (and thus win a drag race).

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

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

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

 

[krab]

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

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

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

 

[wimms]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[Mr. Robin Parsons]

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

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

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

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

Does that answer your question? (tee hee)

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

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

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

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

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

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

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

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

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

 

[wimms]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[Mr. Robin Parsons]

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

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

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

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

 

[russ_watters]

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

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

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

 

[Mr. Robin Parsons]

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

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

 

[Mr. Robin Parsons]

Originally posted by wimms

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

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

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

Robin, I do assume that you know what you are talking about. Can't you realize 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. Let's 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, let's 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 that's without accounting for need to overcome gravity.

At 8500nm, angular acceleration of chassis about rear end is at best 0.45rad/ss, and that's 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 occurred 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 wheels

Interesting. 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]

first post :D

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 receive 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, a lot of the weight already sits over the back wheels.
Oh, rail cars don't 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!)