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Calculating tire pressure's effect on spring rate?
While we're talking about obscure handling effects, I've wondered about
something off and on for a long time, but especially lately.
On my first blast through Highway 9 in my '83 Coupe back in June, I
noticed an unhappy wobble in the right rear. Felt like a shock not
quite right. I decided to check out the suspension and see what I could
come up with.
Well, wisdom costing nothing in the application and everything in the
acquisition, I checked my tire pressures after I got home. Sure enough,
the tires were pretty low, 26 psi in all except the right rear, which
was 21 or so.
A quick stint with the air compressor brought them all back up to 38
PSI, where I was MUCH happier with the car's handling, steering,
transient response, stability in corners -- you name it. Yeah, they
thumped a bit more, but it was MORE than well worth it.
And it got me thinking. In the right rear, I nearly doubled the air
pressure in the tire, and one purpose for raising the air pressure in a
modern street radial is to raise the effective spring rate.
But... what *is* the effect? It's more complex than I can figure out by
sitting here and thinking about the bits and pieces: take the
square-cube drag law and apply it to the ratio of vertical wheel travel
(that's the travel of the rim relative to the ground, not the wheel
relative to the chassis) over the cumulative surface area of the inside
of the tire... it may not *quite* be Newton's three-body problem, but
it's beyond my limited engineering know-how.
So. Anybody got a formula that says "raise the tire pressure X% and
it's like raising the spring rates Y%"? (I'm guessing it's, oh, on the
order of X = 10Y, but that's pure seat of the pants...)
(And in case there's any question about what I'm REALLY up to, I'm
trying to convince myself that yeah, I really WOULD enjoy the car more
with stiffer springs, even if they WERE a little more rumbly... :-)