U anti roll bar

[silente]

Hi!

Let's say you have a formula car with a U anti roll bar fitted at the rear and connected to the suspension rocker with a certain Mr (=wheel movement/bar drop)

Let's say you have to calculate the wheel rate of the bar.

L = lenght of the bar

B = Force arm

To calculate the effect of the bar at the wheel do i have to calculate the torsion stiffness of the bar considering 1 mm of arm drop (and the angle that is subsequent considering B) and the total lenght L of the bar or just L/2 (beacuse i will have a torque dued to a force F on one arm and another equal Force F on the other arm but with opposite direction)?

Thx

[boyracer]

Force at the wheel will be different to force at the bar, unless the bar is connected to the control arm co-incident with the wheel centre line.
There are formulas on the web that will help you work this out.
So B is the length of the arm on the bar, then assume the other end is rigid, If it's not rigid it's a dynamic system and good luck trying to analyse that. So you're looking for angle of twist in the bar, length L, equating to 1mm at the control arm. Then work back to the force required to cause that twist, giving you N/mm.
That's how I'd attack it. Then use it as a baseline only because you'll get some pretty serious binding in the mounting bushes (unless they're bearings) that will probably increase your effective spring rate due to the reduced length.

[silente]

Hi Boyracer,

this is the way i would attack it too.

the problem is this: the other end of the bar (the one we suppose to be rigid) has a blade as well as the end we suppose to move to perform our calculation. This is because one end is connected to one wheel (via pushrod and rocker with a certain motion ratio) as well as the other is connected to the other wheel of the rear axle.

In my opinion, to calculate the effect of the bar on the car roll stiffness it is correct to use the stiffness in N/mm calculated like we said. But somebody says that, because during a corner one end of the bar move in a direction and the opposite end move in the opposite direction, you should use a stifness value equal to two times the stiffness you have calculated in our way, because it is something like you are working with a bar with a lenght equal to L/2.

What do you think about it?

THX

[boyracer]

I think this is where Einstein steps in, as it's all relative. As I said before, analysing dynamic devices is complicated so forget about it. So we have one end compressing, the other extending, now reach out and stop the compressing end from moving, the extending side now has to do all the work so it moves as far as it would have PLUS the amount the compressing end would have compressed. So in a simplistic world, you've got twice the amount of twist, or if you wanted to play with numbers, you could say you've got the same twist over half the bar length (L/2). At least I think this is what they're doing. It's probably not as simple as that in real life as I doubt that the outisde wheel will compress by as much as the inside wheel extends.

As I said in my original post, play with the calcs until you're confident with it, then build it and go to the track and test it.