Back again for another one. If we had two tires that were identical, except
one was wider than the other, what generally might be expected to happen to the
lateral force vs. slip angle curve? It seems to me it would widen/flatten out
and peak at a higher slip angle (but not necessarily more force because the
contact patch size stays the same at a given pressure) since each tread element
doesn't get distorted as much on it's way through the shorter contact patch,
which should indicate a stiffer sidewall must be used to keep the peak in the
same spot.. Same would go for the longitudinal force vs. slip ratio curve...
Right?
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm
Hi Todd,
that's a good question, and I believe the answer can go both ways. If one
would look at the tyre stiffness (the force vs. slip angle slope) one
could also argue that, since the tyre is wider, the deformations at the
given lateral force would be smaller since the structure can better resist
the twisting (the sidewalls have a larger arm from the center of the
tread, not to mention that the internal structure itself resists
deformation better the wider it is) thus resulting at a smaller slip angle
per given force. This means that the stiffness is higher and that the peak
should occur sooner.
However, this is just me playing the devil's advocate, as I seriously
have no clue whether the shortening of the contact patch that you
mention overcomes the effect of a broader and hence more stiff
structure. Doug :)?
-Gregor
On 4 Apr 2002, J. Todd Wasson wrote:
Quoted text (23 lines)
> Back again for another one. If we had two tires that were identical, except > one was wider than the other, what generally might be expected to happen to the > lateral force vs. slip angle curve? It seems to me it would widen/flatten out > and peak at a higher slip angle (but not necessarily more force because the > contact patch size stays the same at a given pressure) since each tread element > doesn't get distorted as much on it's way through the shorter contact patch, > which should indicate a stiffer sidewall must be used to keep the peak in the > same spot.. Same would go for the longitudinal force vs. slip ratio curve... > Right? > > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com > > My little car sim screenshots: > http://performancesimulations.com/scnshot4.htm > >
J. Todd Wasson <[email hidden]> wrote in message news:[email hidden]...
> Back again for another one. If we had two tires that were identical, except
> one was wider than the other, what generally might be expected to happen to the
> lateral force vs. slip angle curve? It seems to me it would widen/flatten out
I *think* that the lateral force peaks at a lower slip angle, and
more sharply because the shorter (though wider) contact patch allows
less room for the tread to break away progressively towards the rear
of the contact patch.
I'm open to being corrected on that, however.
Jonny
The premise "If we had two tires that were identical" is the problem.
Tires in the same "family" (say, all "Eagle GT" at your local Goodyear
store) come in different sizes and look similar externally. That is as far
as the "family relationship" goes. Internally, they have wide variations
in construction, even though they all act like "high performance" tires.
Likewise, tire designers can produce tires of the same general shape and
size with a wide variety of characteristics. Just for one simple
example, one tire may include 20 different kinds of rubber, and this
says nothing about the cord reinforcement.
Sorry, you are going to have to fake this on your own. Better give the
users some idea of _how_ you faked it...or they will hate you forever<grin>.
-- Doug
On Thu, 4 Apr 2002, Gregor Veble wrote:
Quoted text (47 lines)
> Hi Todd, > > that's a good question, and I believe the answer can go both ways. If one > would look at the tyre stiffness (the force vs. slip angle slope) one > could also argue that, since the tyre is wider, the deformations at the > given lateral force would be smaller since the structure can better resist > the twisting (the sidewalls have a larger arm from the center of the > tread, not to mention that the internal structure itself resists > deformation better the wider it is) thus resulting at a smaller slip angle > per given force. This means that the stiffness is higher and that the peak > should occur sooner. > > However, this is just me playing the devil's advocate, as I seriously > have no clue whether the shortening of the contact patch that you > mention overcomes the effect of a broader and hence more stiff > structure. Doug :)? > > -Gregor > > On 4 Apr 2002, J. Todd Wasson wrote: > > > Back again for another one. If we had two tires that were identical, except > > one was wider than the other, what generally might be expected to happen to the > > lateral force vs. slip angle curve? It seems to me it would widen/flatten out > > and peak at a higher slip angle (but not necessarily more force because the > > contact patch size stays the same at a given pressure) since each tread element > > doesn't get distorted as much on it's way through the shorter contact patch, > > which should indicate a stiffer sidewall must be used to keep the peak in the > > same spot.. Same would go for the longitudinal force vs. slip ratio curve... > > Right? > > > > > > > > Todd Wasson > > --- > > Performance Simulations > > Drag Racing and Top Speed Prediction > > Software > > http://PerformanceSimulations.Com > > > > My little car sim screenshots: > > http://performancesimulations.com/scnshot4.htm > > > > > > >
Thanks a lot Jonny, Doug, and Gregor, for your usual excellent responses :-)
>Sorry, you are going to have to fake this on your own. Better give the
>users some idea of _how_ you faked it...or they will hate you forever<grin>.
>
>-- Doug
Hehe ;-) In time, in time... If it's good enough...
Ok, guys, I wrote up two big things earlier and didn't post either one. The
first one was too full of ramblings, and half way through the second one I
decided to look up and study stress, strain, and modulus of elasticity a bit
for clues. I've come to the quick conclusion that modulus of elasticity,
stress, and strain won't help much in my simplistic view right now (unless I
look at bending sidewalls, something that I'm not prepared for), because I'm
not talking about actually stretching a section of the tire to be a different
width (where strain could develop along that width), but merely displacing the
entire section sideways somewhat against the sidewall's resistance (and
whatever else).
Gregor, you wrote that the wider tire would be structurally stiffer. Meaning
that if you put it on a test stand, grabbed hold with 200 hands around the
perimeter of the tire and pulled as hard as you could, the tire wouldn't flex
as much as it would with a skinny tire with the same sidewall stiffness and air
pressure... Meaning the (I think this is the right term) "foundation stiffness"
should be greater in the wider tire. Right? If so, I wonder if it would
double when the width is doubled, to a point...?
I've tried things a couple ways here. First, if I assume that the lateral
force at one little sliver of the contact patch (extending through the width of
the patch) has a "spring rate" that does not change with tire width (foundation
stiffness is fixed), a skinny tire has a much steeper lateral force curve
(higher cornering stiffness) than a wide tire does. This happens because the
longer the contact patch is, the more lateral distortion there is at the
rearward area of the patch at any given slip angle. For instance, if the slip
angle was 45 degrees and the patch could not slip at all (not realistic, I
know), a skinny tire's patch at 0.2 ft long would distort 0.2 ft at the rear,
while a wide tire's shorter 0.1 ft patch would only distort 0.1 ft at the rear,
resulting in far less lateral force at any given sip angle in the linear
region.
Clearly, if this assumption is made, then a wider tire has very much less
cornering stiffness. The same happens in the slip ratio area too, but I don't
think this is right, hence the question :0)
If on the other hand, as Gregor said, the wider tire's "lateral spring rate" or
foundation stiffness would increase, then I wind up with almost exactly the
same force curve regardless of tire width; the shortened contact patch at a
given load is exactly cancelled out by the increased lateral strength if I
assume doubling the width causes the stiffness to double... Same slope, peak,
everything (well, maybe very slightly different.)
Intuitively, if I imagine getting a 1 inch wide tire and sticking it to a stand
of some kind and trying to pull the tire toward me, I'd think that it would
bend more easily than an 18 inch wide tire would, even if the construction was
pretty much identical. Lowering the profile I bet would increase this
stiffness even more. So it looks like this is more of a "bending sidewall"
problem, complicated by air pressure.
To me, this would mean that this article about why a wide tire usually has more
grip than a narrow one would seem to be rather right:
http://www.autospeed.com/A_0996/page1.html
If the compounds and everything else were identical, and the lateral/foundation
stiffness (spring, not the cornering stiffness slope on force vs. slip angle
graph) is really based on the width, then it wouldn't much matter how wide a
tire was at all. The force curves would be very close to the same in all
respects. The only reason why it might be steeper and peak higher is because
it allows you to use a higher coefficient of friction compound (and related
stuff) that isn't as heat resistant, which would let the cornering stiffness
and peak force both increase...
Any thoughts?
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
My little car sim screenshots:
http://performancesimulations.com/scnshot4.htm
Since my newsfeed has been bad lately (I often see replies but not the
original...), I've quoted your whole post below, in case someone else
didn't get the original.
Going off topic:
I had a few spare moments and read this page you found:
http://www.autospeed.com/A_0996/page1.html
and came across the term "Armonton's Law" which I'd never heard of. Of
course I know the equation, [ Force = mu x load ] but never heard of that
person.
Near as I can tell, Amontons (note different spelling, appears as both
Amonton and Amontons) studied friction in France in the 1600's. So he
pre-dates Coulomb (and the common term in mechanics, "Coulomb friction) by
roughly 100 years. And in some fields (earthquake geology for one) they
give credit to both of these Frenchmen for the concept of friction
independent of velocity and proportional to load.
But the bottom line is that tire-road friction is _not_ "independent of
velocity" or "proportional to load".., so the author of:
http://www.autospeed.com/A_0996/page1.html
appears to be blowing a rather large amount of smoke... And Google finds
several other tire articles on the www that mimic this usage and spelling--
beware of authors that don't do their homework and/or plagarize.
End of today's history lesson!
-- Doug Milliken
http://www.millikenresearch.com
On 5 Apr 2002, J. Todd Wasson wrote:
