Edward,

thanks for your valuable input. It is great to have this forum to get your questions answered.

It has been a great resource for the thesis for a first year team.

Somebody suggested in a former thread, when I was still determining what to write about in the thesis, to

write a sort of "handbook" for a team at the old university to follow along and to get good inputs. I followed that route

with a concentration on the suspension.

It is indeed the final chapter I had to finish. All the others have been finished before!

For the case of pure braking:

If we take 1,5 G and assume a mu of 0,9 in the longitudinal direction (which I recon is a little bit on the low side for a FS tire) the front axle

would experience a vertical loading of 239,4 kg = 2348,5 N. With a coefficent of 0,9

that would give me a braking force of 2113,6 N on the front axle. Making it 1056,7 N at one wheel.

As for pure cornering at 1,5 G:

Taking Wl=W/2+(W*Ay*h)/t

Wl= 148,84 kg/2 + (148,8 kg * 1,5 G * 0,3) = 141,38 kg = 1386,94 N vertical load. Higher than

just braking!

In the lateral direction we have to look at the centrifugal force. The centrifugal force acts on

the CG towards the outside and both tires have to counteract that force.

The outer tire takes more load because of the weight transfer. The higher the CG, the higher the weight transfer.

1,5 G * 148,8 kg * 9,81 m/sē = 2189,59 N

Lateral force is not linearly increasing with increased wheel load. Fy= Fz*mu is not entirely

correct for tires, but I will use it here:

Right tire:

1291,29 N*0,9 = 1162,16 N lateral force

Left tire:

168,43 N*0,9 = 151,587 N lateral force

combining these two give 1313,747 N lateral force that the axle can produce. Countering the 2189,59 N under the 1,5G lat. acc

the tire with that mu won't even go there and sliding before.

As I have said the mu in reality is probably higher than that.

I've seen diagrams with 2,47 mu lateral friction coefficient at 1000 N Fz.

So this would become more a question of not tipping the car

So concluding the findings out of this single investigation, my value I have now found for Fy are now higher

than in the last post. The vertical force is higher in the combined case.

What I really need, as you have said Edward, are some realistic numbers for the coefficient of friction for the tire.

But soon this will clear up