Thank you! Many people whom I was sure knew what they were talking about
said that bigger wheels mean not as good stopping but I couldn't understand
why, because the added weight of the wheels/tires is negligible compared to
the weight of the vehicle.

This explains it so even I can understand it.

Hooray.

Anyway I knew my syncro/countryhomes/westy stopped WAY better with the G60's
up front and Audi 5000 turbo rear disks and calipers. Now I know why.

Thanks again, Sudhir [Does that mean "smart" in some language other than
English?]

Mike

On 4/19/10 2:39 PM, "Sudhir Desai" <sudhir.desai@GMAIL.COM> wrote:

> Sorry about the simplicity, I figured I would gloss it over HARDCORE.
>
> the 0 represents the hub, the ) the brake rotor/drum, the | the
> outside of the tire.
>
>
> STOCK
> distance of force from hub:
> 0--5"----//---12.5"
>
> direction of force:
> 0--up-------down
>
> illustration:
> 0--)-----//----|
>
> lets say the car is rolling along, and the torque needed to stop the
> wheels abruptly (ignoring/subtracting any sort of rotational inertia,
> and friction, etc) is 5000in-lb.
>
> that means, we would have to exert a force of 5000in-lb/12.5in = 400lb
> (at 12.5" out from the hub) to stop the wheels abruptly.
>
> now, we go in more to where we can stop the wheels, the brakes. at 5"
> out from the hub, that 400lb force needed is now (400lb*12.5in) =
> (xlb*5in), or x = ((400lb*12.5")/5"), so
>
> x = (5000in-lb)/5" = 1000lb
>
>
> SO... to stop the wheel abruptly, we need 1000lb of force at 5" out
> from the hub, and that's the torque our stock brakes can put out.
>
>
>
> when we increase the tire size to a 28" tire:
>
> OVERSIZE
> distance of force from hub:
> 0--5"----//---14"
>
> direction of force:
> 0--up-------down
>
> illustration:
> 0--)-----//----|
>
>
>
> SO NOW WE GO BACKWARDS!!!! lol~
> due to our original assumption of 1000in-lb,
> we have just increased the diameter of our tires to 28".
>
> this means that our 1000 lb force (5000in-lb) that was enough to stop
> the 25" tire will not be enough to stop a 28 inch tire by the
> following equation.
>
> 5000in-lb/12.5in = xin-lb/14in, or xin-lb = (5000in-lb*14in)/12.5in
>
> x=(5000*14)/12.5 = 5600in-lb is our new stopping torque needed.
>
> our stock brakes can only exert 1000in-lb of torque, so we won't be
> able to stop the wheel abruptly anymore.
>
> hopefully y'all are following me still...
> we just increased the amount of force needed to stop the wheels, and
> we see that our stock brakes are not enough to do so.
>
> NOW, we need to calculate the size of brakes we need to stop the tire
> abruptly (ignoring all other factors).
>
> we take the known stopping torque, and our original brakes (10" brakes
> LOL), and see how that compares with our new 5600in-lb
> x-in/5600in-lb = 5in/5000in-lb
>
> x = (5*5600)/5000 = 5.6in
>
>
>
> so we'd need 11.2" diameter brakes to put out the 5600in-lb torque
> needed to stop the wheels abruptly.
>
> which is another inch diameter of brake needed.
>
>
> I hope my simple explanation (with calculations to show i wasn't just
> pulling numbers out of the air) was satisfactory. :)
>
> Sudhir