hahaha, Sudhir means G.M.I. dropout in english. ;)

On Mon, Apr 19, 2010 at 18:33, Mike Miller <mwmiller@cwnet.com> wrote:
> 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
>
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