EQUALIZED REAR TIRE LOADINGS FOR DRAG RACING

BillyShope · Sep 5, 2003

I have been searching, since my days with the Ramchargers in the early sixties, for means by which rear tire loading could be equalized in a RWD, suspended car without IRS. I've summarized much of my work in an article at:

http://racingarticles.com

I did not include, in that article, a discussion of pinion shaft angle. It would appear that initial efforts, by drag racers, to adjust the pinion shaft angle, were merely to compensate for suspension deflection during launch. But, I strongly suspect that some have realized a beneficial effect when the pinion shaft is angled down more than necessary. With a "beam" axle, it is very apparent that a horizontal pinion shaft will result in an unloading of the right rear tire due to driveshaft torque. If the pinion shaft is visualized as pointing down (vertical), it becomes apparent that the right tire unloading is gone, but, instead, driveshaft torque would tend to push the right tire forward, wedging it underneath the instant center and resulting in increased vertical loading. So, with a load reversal between horizontal and vertical configurations, it is also quite apparent that there exists a pinion shaft angle, somewhere between horizontal and vertical, at which rear tire loading would be equal. The following defines that condition:

sin(alpha)tan(beta) = cos(alpha) - V

where alpha is the angle, positive down from the horizontal, of the pinion shaft, beta is the angle of a line through the tire patch and the instant center, measured positive up from the horizontal, and "V" is the ratio of rear roll stiffness to total roll stiffness.

Insertion of reasonable values quickly indicates that the alpha angle would normally be so large as to be impractical. The interesting point is that the drag racers have realized the value of even a partial achievement of equalized rear tire loading.

GregLocock · Sep 6, 2003

Interesting observation.

I don't understand your point about roll stiffness in the article-if the car has completely rigid suspension at the rear it will still see an increase of load on the rear wheel, won't it? I'm 99% sure of this, just by conservation of angular momentum.

Cheers

Greg Locock

BillyShope · Sep 6, 2003

Greg, since motion is not a consideration, angular momentum is not a factor. In other words, you could anchor the car by means of a horizontal chain at the center of gravity height and the tension in the chain would be equivalent to the inertial force as the car launches.

Reaction to the driveshaft torque is carried into the chassis through the engine and transmission mounts. The chassis then responds in the same manner as it would if the car was cornering (except that, since this is a pure couple, roll center heights are not considered). So, if all the roll stiffness is at the rear, driveshaft torque is completely cancelled and rear tire loading is equalized.

Thank you for your comment.

GregLocock · Sep 6, 2003

Now I'm puzzled.

Sit in a car with the engine running in neutral. Blip the throttle. Are you telling me that the resulting body roll does not increase the loading on the tyres on one side?

Cheers

Greg Locock

BillyShope · Sep 6, 2003

No, Greg, I'm not. But, if all the roll resistance is at the rear, front tire loading is unaffected. Obviously, this is not easy to achieve. Try to picture the front crossmember connected to the rest of the chassis by a frictionless bearing at the center. Not a practical design, but it will give you the idea. As you approach this extreme, you also approach complete cancellation of the driveshaft torque, as far as rear tire loading is concerned, for all of the reaction torque, which is equal in magnitude to the driveshaft torque, but opposite in sense, is fed back to the rear axle housing by the springs.

PTwizz · Sep 8, 2003

Billy, I beleive that the point Greg is trying to make is that fundamentally, the engine produces a nett torque which must be transmitted from the car to the remainder of the universe. Since the primary contact is at the rear contact patches, this is where the torque will be reacted, resulting in uneven loading.
With regard to your original point concerning pinion shaft angle, you appear to be ignoring the effect the axis of torque tranmitted to the pinion shaft via the U.J. from the propshaft. I assume your engine/ gearbox is not situated below ground level on the pinion shaft axis (although that would be a nice solution for a drag racer!). If (for example) the propshaft runs horizontal and is coupled to an angled pinion shaft via a U.J., then the nett torque transmitted to the axle still acts about the propshaft axis.

Pete.

patprimmer · Sep 8, 2003

As our engine and gearbox are as low and as far back as we can get them without moveing the firewall (moveing firewall is against class rules) and as the tall 33" OD tyres raise the pinion to something like 13 or 14 " above the pavement, out tailshaft slopes noticably up.

We are right hand drive in OZ, so we move the driver back and keep him as close to the right hand side and as far back as possible.

As we are about 1100 HP and probably about 700 foot lbs of torque, we need a very robust axle assembly. A broken CV joint on the low to top gear change would not be a good thing I'm sure, and appart from pinion reaction on the crown gear, the torque reaction of the motor must find it's way to the outside world somehow, or at least that's how I understand Isaac Newtons explanation.

Regards
pat

BillyShope · Sep 8, 2003

Greg wrote of blipping the throttle while the car was in neutral and I let that "get by me." The observed torque reaction, under such conditions, is a transient. My analysis is concerned with steady state forces and torques. Both the beam axle and IRS car will react in the same way to Greg's blip.

PT, the torques acting at each end of the driveshaft are equal in magnitude and opposite in sense. The one torque acts on the rear axle housing and the other on the chassis. The torque on the chassis is distributed, front-to-rear, in proportion to roll stiffness. If all the roll stiffness is at the rear (a situation which can only be approached in practice), all of the chassis torque is carried back into the rear axle by the suspension, thus completely cancelling the tendency to lift the right rear.

The whole idea of a U-joint is to redirect torque. Free body diagrams of the yokes and cross indicate that the torque entering the rear axle assembly is determined by the pinion angle. If it were as you suggest, what would happen if you had a dozen shafts coupled with U-joints, with the initial shaft at a right angle to the final?

PTwizz · Sep 22, 2003

Billy,
Taking your second point first, if you couple shafts and UJ's as you describe, they will try to 'wind up' when loaded. To transfer torque from one plane to another, you would need to support the intermediate shafts of the arrangement in bearings, which in turn need to be supported to 'ground' to react the loads created within the system. In the case of your beam axle, it would require an outrigger bearing on the pinion shaft, anchored to the chassis, to 're-direct' torque to the pinion shaft axis. In the free body diagram of the propshaft/UJ/pinion shaft, note that the cross changes it's angle relative to each yoke as it rotates, generating loads axial to the pinion shaft, which resolve to a secondary couple force at 90° to its axis.
With regard to your second point, you are entirely correct in that torque is equal within a rigid chassis in the steady state. My point (and I beleive Greg's) was relating to the accelerating mass of the rotating parts of the engine and transmission. This may be regarded as transient, but it is just as transient as the acceleration of the car as you power out of a corner, and is not to be ignored. Anyone who has ever ridden a BMW or MotoGuzzi motorcycle will be well aware of the effect.
One possible advantage of pinion shaft angle would be moving the axis of the propshaft away from the rear roll centre, so that more of the load is taken through the links, and less through the springs.
Pete.

jloudon · Sep 22, 2003

Billy,

Im very interested in your findings on this subject. As I describe on another thread (Mallock TAM and Mumford links) I am interested in improving my live axled sprint cars handling. This includes standing start take off with maximum traction and without impairing overall handling. Someone has once told me that inclining the diff nose one way or another would help, but wasn't sure which way. From your reasoning, it would seem that its with the nose down, but by how much is the problem?

I am following the thread with great interest.

John

BillyShope · Sep 23, 2003

John, I've evidently failed to make it clear that the thoughts on pinion angle, while interesting, are not of any practical use. The angle for total driveshaft torque cancelation is far too excessive and, while an acceptable angular step in that direction would have some value, it is too small to be considered.

I would, instead, recommend the asymmetric trailing link suspension, as used by Jaguar.

Pete, as the title of this thread indicates, I'm interested in addressing the tire load problems facing the dragracer. While all the loads can certainly be considered transient (what else can be said about an event that lasts only a handful of seconds), transients involving drive train inertias are of such duration that they can be safely ignored. (For performance calculations, drivetrain inertias are usually converted to equivalent masses and added to the mass of the total car.) For those cases which you cite, where the car (or bike) is performing at the periphery of the friction circle, even small inertia loads can cause problems. But, with a drag car, that extra load is simply included in the thrust required of the tires.

I said that the purpose of a U-joint is to redirect torque. That is a simplification, of course. Instantaneously, the pinion torque vector is always conjoined with the pinion shaft axis, but it is not constant in magnitude. With a cross-type U-joint, the torque received by the pinion shaft is equal to the driveshaft torque when the pinion arms are perpendiculat to the plane of U-joint angularity and equal to the driveshaft torque divided by the cosine of the angularity when the pinion arms are in the plane of angularity. Since power out cannot exceed power in, the pinion shaft speed must change accordingly. The preceding is assuming constant driveshaft torque and speed. As is evident, large U-joint angles are to be avoided, which gets us back to my comments to John.

PTwizz · Sep 26, 2003

Billy,
Sorry to dwell on this point, but I think we must be at crossed purposes. The effect I have been trying to explain is easily observed in a T-handle spark plug spanner equipped with a universal joint. If you try to use this tool on a normal nut, with about 45° angle on the UJ, it will tip off the nut when torque is applied. The only reason it works OK with the spark plug is because the plug itself maintains the alignment of the spanner portion.
The function of a UJ is to allow the redirection of torque. When the torque changes axis, the difference between the input and output vectors must be reacted to the outside world. In the case of a live axle, this happens through the pinion shaft bearings.
I have discussed this subject with a couple of my colleagues. We are of the opinion that there is scope for changing the dynamics of the live axle suspension by alteration of the pinion shaft angle, but the observed effect is the result of offset between the propshaft axis and the rear roll centre.
Incidentally, there is a secondary torque effect from the gearbox shaft angle, but the length of the propshaft makes its effect small at the axle.

Pete.

BillyShope · Sep 26, 2003

Pete, I do understand what you're describing. The torque (which tips the spark plug wrench off the nut) is equal to, as a minimum, twice the input torque times the tangent of the half angle. And, its effect on the axle housing would be to require an even larger pinion angle than that described by the relationship in my first post of this thread. That relationship assumed no driveshaft angularity, including the absurd extreme of the driveline pointing straight down in the analysis leading to its derivation. So, if one was to attempt to achieve equal rear tire loading through pinion angle and, further, the driveline was not to be concentric with the pinion shaft, then I agree that the torque of which you speak would probably destroy any chances of ever reaching the goal, even if efficiency losses and speed variations were to magically disappear.

In short, except for a unique situation, which comes to mind, where the driveline was angled steeply up from the axle and cancelation was to have been achieved through an asymmetric trailing link suspension, I consider this a discussion without any value.

PTwizz · Sep 28, 2003

Billy,

In short, I agree. Thanks for indulging me.

Pete.

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EQUALIZED REAR TIRE LOADINGS FOR DRAG RACING

BillyShope

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GregLocock

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BillyShope

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GregLocock

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BillyShope

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PTwizz

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patprimmer

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BillyShope

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