Elephant sway bar, how you adjust yours?

[Burint]

Hi guys,
I have a set of Elephant racing street track one on my 1970, 3.2, All front/rear are yellow Bilstien.
I always had my sway bar set softest in front and medium in rear.
On track, I have never been able to catch up with BMW E30 and older Alfa 1750.

I met with an old friend, a German engineer who used to worked for Mercedes in Stuttgart.
Now he lives in Bangkok. He made an observation and comment on my set up.

"as every 911 is butt heavy it naturally tends to oversteer
accelerating with engine power increases oversteer
which is desirable to a certain degree for racing, but not to much
so you need a stiff sway bar in front and a soft one in the rear to oppose oversteering
because a stiff sway bar causes the end where it's installed to drift off sooner.

So the front is always stiffer except for the first years which are known to have badly oversteered.
BTW stiffness of a sway bar increases with the 3rd power of the diameter of a non adjustable sway bar
i.e. double diameter means 8 times stiffer
I would start with the softest setting rear and the hardest in front.
This will probably result in a neutral or understeering result"

Would you please advice how I should adjust my sway bar.
Thanks
Burint

[callard]

I have Elephant front and rear w/Bilsteins on a 72E. Full soft on the rear and middle of adjustment on the front gives me the best times. And E-30s are no problem.
BOL

[dad911]

What torsion bars are you using?

Looks like the other 911 in the pic has flares, wider tires?

Do you have stock flares with the 3.2?

What other mods have been done to the suspension & car?

[Burint]

Dad911
I don't remember what size torsion bar I have but it was in the set up of the "street track 1" It could be 21/27.
The blue in front is the RSR spec with 3.3 bore to 3.4 and a set of 50mm. PMO.
My viper green has the RS for the rear flare, I have a set of Fuchs 6/15 front and 8/15 rear. My 3.2 is the U.S. spec, ported and polish, 964 cam, standard ECU.

[Flieger]

It depends on how much power you have. If you have a reasonable amount of tire on the back compared to the power, then accelerating when turning will cause understeer because of the weight transfer.

I think the general rule is that in 911s with limited slip differentials accelerating in a turn causes understeer.

911s have turn-in understeer since there is no weight on the front, it is all hung off the back of the rear axle trying to lever the nose into the air.

911s oversteer in a somewhat slow, steady manner.

If you have an open differential you can spin the inside wheel and cause oversteer.

By the way, that is a 2002 in the picture, not an E30. You will have a hard time keeping up with him in the turns if it is well set up.

Also, torsion bar stiffness increases with the 4th power of the diameter, not 3rd.

Aerodynamic drag increases with the 2nd power of speed and power required to maintain a speed increases with the 3rd power of speed.

[gtc]

Some good reading in here:
Shocking Truth about Racing

[burgermeister]

I set mine so the car still understeers under neutral & slight trailing throttle. That way, you can adjust to oversteer / understeer with the throttle as the situation warrants.

I have 19/25 tbars (similar balance to 21/27), 23mm ARBs, 180mm front arm length, 150mm rear arm length (so the rear bar is set stiffer than the front).

Stiffness of the ARB is inversely proportional to the length of the ARB arm ^2. So an 80mm arm is 4* stiffer than a 160mm arm, when attached to the same bar.

[Flieger]

How are you arriving at that lever arm relationship?

[burgermeister]

I feel like I'm being baited (Flieger, I don't buy that you don't know the answer to your question), but I'll bite & see.

For a given suspension deflection, the twist of the stab bar (using small deflection assumptions - angle <10 deg) is:

twist (in radians)= susp displacement / lever arm

The force resulting from the twist is (torsional stiffness) * twist / lever arm

Substituting the first eq into the 2nd,

Force = (torsional stiffness) * (susp displacement) / (lever arm) ^ 2


So if the lever goes from 1 to 0.5, the force exerted by the stab bar per unit suspension deflection goes up by a factor of 4 (1/0.25).

[Flieger]

OK, I was just looking at the amount of force at the end of the lever for a given angular deflection. That is linear wrt lever arm length.

The suspension deflection goes down with the lever arm length so force on the wheel is constant. I guess rearranging that equation works out to

(suspension displacement) = Force * (lever arm)^2 / torsional stiffness

Which is more clearly correct to me.

I think where I was confused is that it is linear wrt force, and 2nd order wrt displacement.

[burgermeister]

Phew - I lucked out!!