Contour Enthusiasts Group Archives Forums General Autocross and Open Track Adjustable Anti-Roll Bars - One or Both End?

I've done three Evolution Schools and they are worth every penny. Jean has also setup a discussion group that is very informative. Its at: http://www.egroups.com/group/evolution-discussions. Sign up (if you haven't already). They just finished a thread about FWD vehicles and Ackerman steering.



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John Coffey
johnc@betamotorsports.com

John,

let me get this straight. You are talking about sway bar which is sort of U shaped:



Right?

You can then adjust lengths of each end. By varying lengths of this ends you vary torque applied on the center piece. Your statement is that if you change length of only one end (say, 4mm) you changed twisting torque as much as if you changed lengts of two ends by half the value (say 2mm)

This is true. It's true however as far as twisting torque is concerned. But that's not all what's in the game here. To twist the bar you need to do some work. May be work is different?...

Suppose you've shortened left end. Now hold the right wheel and push the left wheel up. Amount of energy stored in the sway bar (amount of work done) is proportional to the angle between ends.

Twisting angle (theta) is proportional to vertical displacement of the wheel. In other words you need to produce more displacement for a shorter end to twist the bar by the same amount. Right? Well, not quite...

The point is you twist both ends, and it doesn't really matter which one YOU pull, there has to be as much torque applied to the other end or the bar will rotate around it's central section (much like when you are going over a bump and you hear that squeak ) Soo... you are saying that even energy is the same.

Yes, but that is not all yet. Torque equals F x r where r is arm's length, and F is force. You need to apply different force on the rignt and on the left to twist the bar the same amount. This is assymetric part. Forces (one force is from the ground pushing on the tire and the other one is from the spring to oppose the first one) are different and that's what matters in the end.

In summary, you need to apply different forces in order to produce the same torque, therefore you need to load tires on the right and on the left differently.

I hope I confused you enough

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Tour I: 96 GL(sort of) V6 ATX -- mild enhancements
Tour II: 98 SE V6 MTX Sport -- getting better

I already said that. Just kidding....



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Chris
2000 SVT

Alright! We've got geometry and everything. I like it when the discussion goes this way...

> Suppose you've shortened left end.
> Now hold the right wheel and push
> the left wheel up.

Unfortunately, in a real world situation the unsprung weight (the car's chassis) is reacting to this jounce (pushing up) of the left wheel (you're in right hand turn) and the right wheel is now in droop (pushing down). The right wheel is now reacting to the torque at the center of the anti-roll bar.

Again, it goes back to the false premise that each end of the anti-roll bar is working in isolation.

> You need to apply different force
> on the rignt and on the left to
> twist the bar the same amount.

But that force is compensated for because the opposite side of the bar reacts accordingly. In a perfect laboratory scenario, where you have a rigid, stationary mass that the anti-roll bar is attached to in the center, your example works.

On a vehicle going around a race track, the unspruing rate rolls. And the unsprung weight of the vehicle is what the anti-roll is affecting, not the suspension. So, the unsprung weight could care less about 100 lbs of jounce on the left and 50 lbs of droop on the right, it just sees 150 lbs of resistance (weight transfer) as transmitted to it via the center of the anti-roll bar.

Easy huh?

P.S: Oops! When I said "unsprung wieght" in the above, I actually menat "sprung weight."


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John Coffey
johnc@betamotorsports.com

[This message has been edited by johnc (edited November 13, 2000).]

John,

here's the picture.


Compare pictures two and three. Force which compresses both springs remains the same, but torque changes. Assuming that energy stored in the sway bar didn't change we come to a conclusion that torques had to be redistributed. The only way to compensate for this change is to change forces acting on the tires... Torques would redistribute differently if we shorten opposite end...

That's as much as I can tell. We would need to talk to a suspension engeneer to figure out how to calculate things, because frankly I'm not sure how the sway bar moves/works under load.

There is no lateral displacement unless you use SCA bar , so the bar IS attached in the center. There is some fore/aft movement which is affected by the bushes and there is twisting effort which we are talking about.

gotta go,
--alex

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Tour I: 96 GL(sort of) V6 ATX -- mild enhancements
Tour II: 98 SE V6 MTX Sport -- getting better

Now I'm confused...

> Assuming that energy stored in the sway
> bar didn't change we come to a conclusion
> that torques had to be redistributed.
> The only way to compensate for this change
> is to change forces acting on the tires...
> Torques would redistribute differently if
> we shorten opposite end...

Most people assume the loads coming into an anti-roll bar come in from one of the ends. Actually, loads come into the anti-roll bar from the center and both ends react to it (at the exact same time), thus creating the twisting load (torque) on the anti-roll bar. This load is introducd via the sprung weight rolling in reaction to centripedal force from a corner. This sprung weight cannot differentiate between the left and right sides of the vehicle. Its one solid mass that's leaning one way or another.

Look at it another way:

An anti-roll bar is a torsion spring with a specificed rate based on degrees of twist. The degree of twist of measured at the center of the bar. So, let's say the bar twisted 2 degrees and generates 500 lbs of force. That 500 lbs is exactly the same if the right side of the bar moved up .5 degree and the left side moved down 1.5 degrees or vice versa.

Where the confusion starts is the assumption that this 500 lbs is divided up between the left and right sides of the vehicle. Its not. Its just 500 lbs of weight transfer that's not happening.

How's that?

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John Coffey
johnc@betamotorsports.com

what I'm trying to say is that the bar is not an isolated torsion spring. to truly understand what's going on you need to write down two eqations:
Sum of all forces = 0
Sum of all torques = 0

and solve the system. I assure you, the result will be different. notice that two forces remain the same (see picture)

so much for general comments, I need a book on suspension to continue. unfortunately MSU is a big university and its cold outside, so I'm not going to engeneering building to get an appropriate book any time soon.

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Tour I: 96 GL(sort of) V6 ATX -- mild enhancements
Tour II: 98 SE V6 MTX Sport -- getting better

That's why I dropped out of college. All the libraries were too far away and it was damn cold outside. Wait a minute, the bars were even farther away and I still made it there.



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Chris
2000 SVT

oh, but it's natural. it's cold on the way to AND from library, but it's never cold on the way from a bar.

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Tour I: 96 GL(sort of) V6 ATX -- mild enhancements
Tour II: 98 SE V6 MTX Sport -- getting better

Its probably not worth going out into the cold, but this book is the bible on vehicle suspension development:

Race Car Vehicle Dynamics
by William F. Milliken, Douglas L. Milliken

I have it and I've read as much as I can but you really need to be a mechanical engineer to get the most out of it. I've searched it to find an answer to this very topic and couldn't find any obviouse relevant information. That doesn't mean its not there, it just means that I'm not smart enough to find it.




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John Coffey
johnc@betamotorsports.com