NEED SIMULATION INFO

[BillyShope]

I'm retired and am requesting a favor from one of you guys who has access to chassis simulation software. I'd like to know the ratio of right front spring rate to left front spring rate, for a beam axle RWD car, that will provide cancellation of the driveshaft torque. Obviously, many parameters are involved, but I'd like a ballpark figure for an "average" car. Yes, I realize that asymmetric rear suspension links would provide a much better solution, but, if the ratio is reasonable, this would be a quick and dirty solution for the dragracer who either has adjustable coilovers or...as in the Chrysler products...has a torsion bar adjustment. Please keep the sum of the spring rates constant. Thanks in advance.

 

[85chief]

I would have to disaree about no weight tranfer, after 35 years of racing cars, there is very significant weight transfer, do to roll center location in vertical and lateral location, front to rear,relative to the center of gravity.
With shocks your are controlling how fast it happens and where it is going first.
At this point in time shock packages are the most crucial adjustments you can make for transitional responses.
And in our form of racing, the getting into and getting off the corner is where you are gaining the most.

 

[BillyShope]

Okay, let's try to "common sense" this problem through to a solution.

We have a RWD, beam axle car with symmetrical suspension linkage (i.e., mirrored in the XZ plane). RR and LR spring rates are equal. And, we'll start with RF and LF spring rates also equal. Although it doesn't affect the final outcome, we'll assume 100% anti-squat, just to make it a bit easier to visualize. We'll assume that 2/3 of the total roll stiffness is at the front.

On launch, driveshaft torque tends to unload the RR tire (and, of course, load the LR). The reaction to that driveshaft torque tries to prevent this unloading, but, with 2/3 of the roll stiffness at the front, only 1/3 of the necessary reaction torque is fed back to the rear axle assembly through the rear suspension springs.

With the right rear spring being compressed, there is necessarily a lifting of the left front of the car relative to the right front. If full cancellation of the driveshaft torque is to be realized, the sum of the jounce of the RR and the rebound of the LR must be tripled, meaning that the same deflection multiplier must be used at the front. Since the anti-squat removes the opportunity to use the rear springs to achieve cancellation, efforts must be directed to the front.

But, here's where it gets really interesting! Since the rear wheel loadings are to be equal, that means the front loadings will also remain equal. Since they remain equal, the front springs cannot contribute to the front roll stiffness and that 2/3 figure originally assumed becomes meaningless. Yet, somehow, that right rear of the car has to come down to load that right rear tire, through the suspension spring, and cancel the driveshaft torque.

The task is left, then, to the front swaybar. It must provide all of the resistance to the driveshaft torque WITHOUT affecting front wheel loadings. This can be done with the proper relationship between front spring rates and front swaybar rate. I ended up with the following relationship:

(KR - KL)/(KR*KL) = 2*L*R/(H*X*KB)

Where "KR" is the RF spring rate, "KL" the LF spring rate, "L" the wheelbase, "R" the effective rear tire radius, "H" the CG height, "X" the axle ratio, and "KB" the sway bar rate (using the deflections at the tire).

As would be expected, the difference between LF and RF spring rates becomes infinite as the sway bar rate goes to zero.

 

[BillyShope]

Arrgh! How can I be so stupid? Don't answer that! I lost my "common sense" half way through that analysis. I turn 70 in eleven days and my only excuse is my obvious senility.

Okay, the right rear of the chassis has to come down. This will happen with a symmetrical setup, but the roll resistance at the front takes up some of the reaction torque so the driveshaft torque isn't completely cancelled. The front roll stiffness is from the suspension springs and the sway bar. What is needed is enough chassis deflection to completely cancel the driveshaft torque, but, at the same time, the front loadings must remain equal. So, at the rear, the rear roll rate and the driveshaft torque determine the necessary chassis angle. All that's necessary, then, is to achieve that angle without, again, upsetting the front.

The solution is simply to get rid of that pesky front sway bar entirely and then adjust the spring rates, at the front, so that the front tire loads are equal with that necessary deflection. (The problem is really very simple, but it's taken a long time for this tired old brain to "see" it.)

The solution is similar...in appearance...to that in my last post:

(KR -KL)/(KR*KL) = 2*L*R*T/(H*X*Kr)

Where "Kr" is the rear roll rate and "T" is the front track.

Now, I'll go take my medicine and return to bed.

 

[mloew]

85Chief,

Please re-read my comment. I am distinguishing between weight (mass * acceleration due to gravity) and load. Assuming g is constant, weight is directly proportional to the mass. Mass is NOT being transferred (unless there is movable ballast). Load is being transferred by the lateral and longitudinal accelerations of the sprung mass (and the torque reactions, as well). Anti-dive, anti-lift, etc. all reduce the load transfer from suspension geometry.

Regarding the damping and transients, I thought the discussion was all about steady state (static) load transfer. I did not recall a discussion about transients.
Please correct me if I am wrong on this part.


Best regards,

Matthew Ian Loew


Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[BillyShope]

Sorry. Those have to be WHEEL rates, not spring rates.

(I had started with the assumption that I'd end up with a simple ratio, in which case it wouldn't have mattered.)

 

[bhart]

Billy,

Your last expression appears very similar to my original expression which I stated I only had partial confidence in (same terms anyway)... have you tried to solve for the ratio of RF to LF wheel rate to see if it comes out the same, that would be encouraging if it we independantly arrived at the same result? The only thing that looks appreciably different to me is that you have a 2*T factor instead of a T/2 factor, which may come out in the wash. One of the finer points might be chassis flex, which would probably mean that you need an even greater difference between the RF and LF wheel rate to get the effect you're looking for... which is where the rig (or track) testing would be handy.

 

[BillyShope]

No, bhart, can't get a simple ratio. I'm quite confident in my result, though an algebraic manipulation error is always possible. As for a practical application, it would appear that a hefty rear sway bar would be needed to keep the difference between front spring rates reasonable. Handling would be terrible, of course, but we're considering only dragstrip conditions here. By the way, that rear roll rate relates to radians, not degrees.

 

[BillyShope]

Bhart, here's the equation set with which I started. I hope we agree to this point:

Driveshaft torque = N*L*R/(H*X)

where "N" is that which is called the "weight transfer" by the dragracers.

Driveshaft torque = Kr*A

where "A" is the chassis roll angle.

A = (DL - DR)/T

Where "DL" is the left front chassis rise (measured at the wheel) and "DR" is the same for the right.

N = KL*DL + KR*DR

KL*DL = KR*DR

The remaining variables are defined in earlier posts.

 

[85chief]

Billy,
I am sorry, I thought it was a discussion in what really happens, not an exercise for the mind.
You know nothing is static at the hit of the throttle.

Have a nice day