The long answer (:-P) is due to where these longitudinal forces come from.
This is a quick write up, so bear with me if it's a little shoddy :-D
Following is a simplified example, but the same principle applies to real
tires.
Imagine a contact patch 1 foot long. When you have a "slip ratio," this simply
means that the wheel is rotating at a different speed than it would be if it
was rolling freely. First, let's paint a little dot on the tire tread:
If the tire was rolling freely at a speed at which you could watch the little
dot move the through contact patch at 1 foot per second (our camera is moving
along with the wheel,) for the purpose of this post you could say the tire is
moving at 1 foot per second.
Now, let's apply a driving torque to the wheel so it's now running at a 0.1
slip ratio. As you know, this means that the wheel is rotating 10% faster than
it would be if it was rolling freely. Meanwhile, we keep the tire moving along
at 1 foot per second.
The little dot will enter the contact patch and stick to the ground. It'll
stay stuck to the ground of course because there's a vertical force on the
tire. Since it's stuck to the ground, the little dot will move at 1 foot per
second through the contact patch, matching the tire's free rolling speed (the
axle speed.)
However, the tire is rotating 10% more quickly than the free rolling velocity.
The dot will not stay in the same position as it would if it was NOT stuck to
the ground. In essence, the little dot *should* be moving through the contact
patch at 1.1 foot / second instead of 1 foot per second. As a result, the tire
of course flexes and some stress (force) is produced.
When the dot first enters the front of the contact patch, it's exactly where it
"should" be. The tire isn't distorted yet. However, as the dot moves through
the patch, it is progressively pulled further and further away from where it
should be. By the time it leaves the contact patch, it has been stretched 0.1
feet out of shape.
The contact patch acts like a spring. If you stretch the "dot" out of place by
0.1 feet, you could basically multiply that by the tire's "longitudinal spring
rate," or the stiffness of the tire, to find out how much force was produced.
Of course, in a real tire, the rear of the contact patch slips, but this still
serves the point.
Now, what happens if the tire is moving 10 feet per second and also has a 0.1
slip ratio? The "slip velocity" is quite different, but the force remains
basically the same. You're question is "why?"
Let's follow the dot...
The dot "should be" moving at 11 feet per second. It's stuck to the ground
though, so it moves at the same speed as the axle: 10 feet per second. The
contact patch is 1 foot long, so it will enter the front and leave the rear in
only 1/10 = 0.1 seconds.
How far will it be stretched out of the way at the end of the contact patch?
The speed difference in the contact patch betwen actual and free-rolling is 1
foot/second. The dot is stretched out of place by 1 * 0.1 = 0.1 feet.
It's the same amount of stretching, regardless of the velocity. The amount of
stretching is proportional to the amount of force produced.
So... The tire velocity doesn't matter in this example. The ratio causes the
force, not the slip velocity.
After all, if the contact patch is 1 foot long and try to make the dot go 10%
"too far," it will be stretched out of place by .1 feet, regardless of the
tire's speed.
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm
