## Wheel frequencies VS track surface

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Rustem 1988
0
Joined: Tue Sep 05, 2017 10:38 am

### Re: Wheel frequencies VS track surface

It is believed that stiff springs should have more stiff shock absorbers. To what extent does this theory work in practice?

DaveW
242
Joined: Tue Apr 14, 2009 11:27 am

### Re: Wheel frequencies VS track surface

The theory is good so model results should match the vehicle. Otherwise question the model.

p.s. Some of the publications you have been looking at use inadequate models (hub masses, tyre stiffness, insulation stiffness, etc.)…

Rustem 1988
0
Joined: Tue Sep 05, 2017 10:38 am

### Re: Wheel frequencies VS track surface

If the car travels on a flat road, then in this case, the shock absorbers do not bring any benefit? Is there any benefit from shock absorbers when accelerating and braking the car?

gambler
1
Joined: Sat Dec 12, 2009 6:29 pm
Location: Virginia USA

### Re: Wheel frequencies VS track surface

I have test driven a few track cars with way too stiff springs on the front , (presumably to plant the nose on heavish caveman braking.) They follow the tracks in the road that trucks mash in, or even the irregular pavment at the track almost like there are bias-ply tires on the car, ...very darty. I can see the need for "tuned" shocks that take up that slack when needed, but still be able use a softer spring....as you can imagine it is extremely complicated before you even whip out the pocket calculator.

Rustem 1988
0
Joined: Tue Sep 05, 2017 10:38 am

### Re: Wheel frequencies VS track surface

DaveW wrote:
Mon Mar 05, 2018 3:31 pm
Rustem 1988 wrote:
Mon Mar 05, 2018 12:20 pm
...the load on the wheel W=Wst+ks(Xs-Xu)+Cs*(dXs/dt-dXu/dt)+kt(Xu-Xr)+Ct*(dXu/dt-dXr/dt)
I believe you have forgotten the inertia force acting on the wheel. If I simply rewrite your equation:
Load on the wheel = Wst+(Ks+ Cs*D)*(Xs-Xu) + (Kt+Ct*D)*(Xu-Xr), where D is the operator d/dt.

Then dyamically, Mu*D*D(Xu) = (Ks+Cs*D)*(Xs-Xu) + (Kt+Ct*D)*(Xu-Xr). Here D*D(Xu) is the acceleration of the wheel, and Mu is the wheel (unsprung) mass. Wst disappears because it is balanced by static offsets of the springs. A similar equation can be written for the acceleration of the sprung mass, and the two can be solved together to compute the various responses of the vehicle per unit road input.

It is a little more than that, because the sprung mass connects the front & rear suspensions (so the sprung mass has "heave" & "pitch" inertias).

I hope this helps...
Now I think that the load on the wheel is determined by the force with which the tire presses on the road. Changing the load at a given time: (Kt+Ct*D)*(Xu-Xr). And we must find the maximum value of this expression.
(Kt+Ct*D)*(Xu-Xr)=Mu*D*D (Xu) - (Ks+Cs*D)*(Xs-Xu) .

Rustem 1988
0
Joined: Tue Sep 05, 2017 10:38 am

### Re: Wheel frequencies VS track surface

It turns out we need to either reduce the acceleration, or increase the force in the spring, to reduce the change in the load.

gambler
1
Joined: Sat Dec 12, 2009 6:29 pm
Location: Virginia USA

### Re: Wheel frequencies VS track surface

The last I heard was that they are Dyno testing some of the shocks together with the springs? Talk about a massive machine....