Re: Tyre stuff

["Stephen Bigelow" <sbigelow@sprint.ca>]

>sbigelow@sprint.ca writes:
>>A 2800 lbs car with say, 60% of its weight on the front axle, has 1680 lbs
>>on that axle, or 840 lbs per tire.
>>@ 32 lbs per square inch in the tire, how many square inches of patch does
>>it have?840/32 = 26.25 square inches.
>>Wanna change its size? Change vehicle weight, or tire pressure.
>
>Or change wheel width.  There's a point at which 840lbs on a wide tire on a
>narrow rim, isn't enough to give you 26.25 sq in.  Why?  Because you have
>effectively done the same thing as increasing pressure, you changed the
>diameter of a given tyre.  Not by pressure, by rim width. Statically, we
can
>change wheel width to change contact patch @ the same tire pressure.
>Dynamically, a small change in wheel width, has a great affect on contact
>patch.
>
>Look at it this way, put 40psi into the tire you had 32 in above.  What
>happens?  The rolling resistance decreases due to an effective larger
>diameter tyre which gives a smaller contact patch.  Take the same tyre, put
>it on a narrow rim. Rolling resistance decreases due to an effective larger
>diameter tyre and a smaller contact patch.  We can change overall diameter
>with tyre pressure OR rim width. Both effectively do the same thing:
>Decrease rolling reistance due to a reduced contact patch


840/40=21 square inches.

By going to a narrower rim you are changing the _shape_ of the patch, of
course, not it's size.

There is still the weight of the car, working against a given air pressure.
That equation hasn't changed.