NEED SIMULATION INFO
NEED SIMULATION INFO
(OP)
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.





RE: NEED SIMULATION INFO
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: NEED SIMULATION INFO
I realize that the reaction torque is distributed, front-to-rear, in proportion to the roll stiffness, so that which I fear is that the required rate ratio would be infinite, since only that would mean zero front roll stiffness. But, the problem is sufficiently "cloudy" that I hope I'm wrong and that a reasonable ratio will be found. Certainly, if the problem is simplified to the point that the fronts are assumed to rise equally (no twisting of the chassis), the ratio can be easily found and is quite reasonable. Intuitively, the idea seems to be a "slam dunk." The sum of the right side loads must remain equal, so, with a high rate spring at the right front, this would mean the majority of the weight transfer would be going to the right rear. But, my intuition has gotten me into trouble so many times in the past that I no longer trust it.
RE: NEED SIMULATION INFO
That's from a 3d model, I'll look at the free body diagram again, just to make sure.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: NEED SIMULATION INFO
I should have a solution to post later today.
RE: NEED SIMULATION INFO
Certainly, the ease with which the effectiveness of such a scheme can be tested in the shop...using d'Alembert's Principle...should be considered. Simply slipping wheel scales under the front tires and observing the load changes as a horizontal chain is tensioned out the rear is a worthwhile test.
RE: NEED SIMULATION INFO
I recently went through the painstaking process of writing out long hand expression for all of the dynamic load transfer components on my type of racecar. I have a live axle @ the rear w/ a torque tube, and a solid axle up front. I think most drag cars probably have independent front, but adapting my expressions is just as simple as making front spring rates = front wheel rates, and front spring width = front track width. When I do that, in order to get the cancellation that you speak of, I get a somewhat different expression than you:
K_R/L = ratio of RF to LF wheel rate
= N/D
N = %FRS + %RRS*(t_f/2)*(h_cg/L)*(eta/R_L)
D = %RRS*(t_f/2)*(h_cg/L)*(eta/R_L) - %FRS
%FRS = your "Z" value
%RRS = fraction of total roll stiffnes @ the rear
t_f = front track width between contact patch centers
h_cg = CG height from grd.
L = wheel base
eta = torque multiplier between where the driveshaft pivots
on the vehicle and the rear wheels (rear end ratio
usually)
R_L = rear tires loaded radius (assumed = side to side
which is actually true if the tires are loaded
equally)
The %FRS terms come from the engine torque compenent of load transfer, and all that other junk has to do with the fact that asymmetric spring rates @ the front will cause the car to roll whenever it is required to pitch, which will transfer load across through the rear roll stiffness.
My confidence in these expressions is probably 70% as it is easy to make a mistake when doing that much algebra by hand, but it's up for discussion now anyway.
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
You might try a little trick we use on roundy round cars and use rebound adjustable front shocks and set the LF with MORE rebound than RF. The weight will transfer to the LR more quickly than the RR and you can tune it with the twist of a knob.
Wayne Malmstrom
RE: NEED SIMULATION INFO
Best regards,
Matthew Ian Loew
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: NEED SIMULATION INFO
Norm
RE: NEED SIMULATION INFO
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.
RE: NEED SIMULATION INFO
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.
RE: NEED SIMULATION INFO
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.
RE: NEED SIMULATION INFO
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.
RE: NEED SIMULATION INFO
(I had started with the assumption that I'd end up with a simple ratio, in which case it wouldn't have mattered.)
RE: NEED SIMULATION INFO
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.
RE: NEED SIMULATION INFO
RE: NEED SIMULATION INFO
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.
RE: NEED SIMULATION INFO
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