Anti-dive

[Flieger]

Hi, There has been a thread going about anti-dive suspension dynamics over in the Turbo forum but I think it would be interesting to see if anyone here can make some measurements or calculations regarding the amount of anti-dive effect felt when switching from a 911 to a 930 front suspension. Is it significant enough for a "good" driver to notice? Will it help handling all that much or just change its style?

Any input is appreciated.


P.S. "Burgermeister", this one would be right up your alley, I think.

[Superman]

The great and powerful Tyson Schmidt was at my house this afternoon changing passenger side glass for the ride home (he drove six hours for our world-famous autocross event), and it so happened he was waxing poetic on a variety of subjects including this one. He even showed me, on a bare tub, the difference. The rear torsion mount on a 930 tub is raised. That angle the torsion bar down slightly, toward the front. This is why the 930's tend to dive less in hard braking.

The tradeoff is, basically, what happens after the dive is lessened. The short answer to your question, I think, is that many other factors and variables are more significant than this one. Tires and shocks, for example.

[Flieger]

Thanks but I was more hoping for a solution to the equation in post#33

(k × Θ) ÷ (b × cosine(Θ)) = ((W/g) × H ÷ L) - (Fbraking × (h + ½d) ÷ ℓ)

solve for Θ to find vertical suspension deflection y using y = b × sine(Θ)

[euro911sc]

With just the front change I doubt anyone could really tell...

Turbo:


Regular:


Not a very big difference. Pics courtesy of Equality72521.

-Michael

[Superman]

The control arm is not the only difference. On a Turbo tub, the rear mounts are raised. Slightly. The fronts are lowered, as shown, and the rears are raised.

[Craig_D]

Is this anti-dive geometry added to the Turbo because the brakes are significantly stouter/stronger than the 911? ...followup question: if so, should this modification be done when upgrading the brakes on a standard 911 to Turbo brakes?

Also, what are the 'tradeoffs'?

Just curious..


- CraigD

[Steve@Rennsport]

Craig,

It was done for homologation purposes for the 934, the race car based on the 930.

Tweaking the brake bias toward the rear (930 brakes are heavily front-biased) will really help keep the nose from diving under hard braking. We do this in a multitude of ways including the use of dual MC's, stepped MC's, different pad compounds front-to-rear, etc.

[Flieger]

Would there be any interest in making some custom A-arm bearings that provide anti-dive geometry? They would be Teflon-lined steel sleeves and Aluminum mounting blocks. No rubber, no lubrication. The idea would be as a bolt-on for a 911 crossmember to give anti-dive similiar to, maybe more than the 930.

If there were enough interest, there is a small chance I could get some made. Do not hold me to any deadlines, though.



Not To Scale

[equality72521]

Originally Posted by euro911sc

With just the front change I doubt anyone could really tell...

Turbo:


Regular:


Not a very big difference. Pics courtesy of Equality72521.

-Michael

I think I have a new career as a hand model.

[equality72521]

Originally Posted by Flieger

Thanks but I was more hoping for a solution to the equation in post#33

(k × Θ) ÷ (b × cosine(Θ)) = ((W/g) × H ÷ L) - (Fbraking × (h + ½d) ÷ ℓ)

solve for Θ to find vertical suspension deflection y using y = b × sine(Θ)

I'm a child of Florida public schools, I have no idea what this is.

[Flieger]

I explained the symbols in the linked thread. I can re-post here.

[equality72521]

Originally Posted by Flieger

I explained the symbols in the linked thread. I can re-post here.

That's alright, I'll just check out the other one.

[Flieger]

#32


r = vertical distance between torsion bar center and top strut mount
h = SVIC height

ℓ = r ÷ (tangent Θcaster + tangent Θarm)

h = r - ℓ × tangent Θcaster

h = r – ((r ×tangent Θcaster) ÷ (tangent Θcaster + tangent Θarm))


Fanti-dive = Fbraking × (h + ½d) ÷ ℓ
d = tire rolling diameter


#33


Red is the inertial reaction forces and the green is the actual force from the wheel, divided into the separate components in black.

At the instant of brake application, the spring force is zero since the displacment is zero (we are talking dynamics here, not steady-state).

The Fanti-dive is the instantaneous front load transfer (added to static front weight).

Steady state load transfer (once the springs are fully deflected and the pitch has stablized) is (W/g)×H/L where H is center of gravity height, L is length from tire contact patch to the center of gravity, W is weight and, g is gravity= 32.2 ft/s^2 for pounds.

vertical Fspring = ky = ((W/g) × H ÷ L) - (Fbraking × (h + ½d) ÷ ℓ)

Spring deflection = y = b × sine(Θ)
b is the length of from the torsion bar to ball joint
Θ is the angle of the A-arm viewed from the front.

vertical Ftorsion = (k × Θ) ÷ (b × cosine(Θ))

(k × Θ) ÷ (b × cosine(Θ)) = ((W/g) × H ÷ L) - (Fbraking × (h + ½d) ÷ ℓ)

solve for Θ to find vertical suspension deflection y using y = b × sine(Θ)

Add the spring deflection to the tire deflection to get the full pitch angle. This is assuming a race car with a rigid chassis and monoball bearings with negligable compliance.

The anti-dive force is reacted through the tire "spring" because it is a part of the SVSA. The anti-pitch reaction will therefore be delayed by the amount of time it takes to squish the tires, at which point the main springs will come into action to stop the roll.


#37
I made a clarification to the above post regarding spring deflection.

Both the anti-dive force and the braking force are transmitted through the SVSA, which includes the tire "spring". The tire will be deformed by these horizontal and vertical forces in a somewhat elastic/linear way assuming no sliding. Therefore, the braking force is delayed by the F=k1x; x = ∫ax dt lag time. The anti-dive is linked to braking force so once the anti-dive starts, there will be further vertical deformation of the tire to generate the vertical component. Fanti-dive = (k1∫ax dt) × ((h + ½d) ÷ ℓ) so vertical deformation y = (k1 × (h + ½d))/(k2 × ℓ) × ∫ax dt

Since the acceleration is not consant, one uses integrals. Since the tires are very hard to quantify, being a "black art" of mysterious non-linearity, the k1 and k2 spring rates in the horizontal and vertical directions. are not known, nor are they likely the same.

Anyway, that math right there is just to show that anti-dive and the anti-pitch reactions will be affected by tire pressures and stuff like that.