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Tire contact calculation was Extra load tires

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Hi folks,
The tire contact calculation post reproduced below is so elegant. Mathematics by keeping track of units is an amazingly simple way to get things right.

But this physics exercise has curious implications that I am having trouble putting my finger on (despite a month of occasional reflection)! For those on digest I'll repeat the salient part.

Doug wrote,
“also here's calculation to help explain weight 2700 lbs
+ 2 passengers...say 350 lbs + cargo say 50 lbs..= 3100 lbs.
if tires ARE at 50 psi (assume equal weight distribution for now)
then the average required area in contact to road is = 3100/50 = 62 / 4 = 15.5 per tire..approx. equal to a 3 x 5 spot on
the ground..conclusion ..lower pressure requires more surface area in
contact..and 175 tires don't deform uniformally to accomodate a
larger surface contact area..thus cupped wear pattern. and why at 35
psi michelins need to be 185's.”

<me again>
Could this mean that
1. On the moon or on Jupiter the tire contact area is the same (independent of gravity).

2. Two narrow tires instead of one wide tire, would give half the contact area.

But why oh why?

Puzzled in Pennsylvania
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1. Vehicle weight (including two passengers) is MUCH different on the moon and on Jupiter. It figures prominently in the equation for average required contact area.

2. Do you mean two wheels side by side (like the back wheels of semi's)? In any case, the total contact area would tend to stay the same, so each tire in the pair would need to provide about half the contact area of one wide tire at the same spot.

Any less puzzled?

I think what has you puzzled is that length of the contact
of the tire changes as it deforms to equlize the pressure.

If it had no load, it would be a perfect circle and contact the
road with an infinitesimally small length. Seen edge on:


As you load it (or lower the pressure under load), the bottom
edge gets flat - it's no longer a perfect circle. The more weight,
the flatter the bottom. Once the total area is enough to equalize
the pressure, it stops deforming. So if the tire is twice as wide,
the length of the contact area is half as much (remembering
that the length can theoretically go to zero).

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Thanks guys,
Upon reflection it makes perfect sense.
Except for run-flat tires.
And it nicely explains why you can reliably fill the spare unloaded in the trunk.
Also a nice way to experimentally determine asymmetric weight distribution.
cheers, nathan
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