On Sat, 20 Aug 2005 01:47:46 -0400, Jeff Wisnia
wrote:

Ned Simmons wrote:
In article ,
says...

carl mciver wrote:

"Jeff Wisnia" wrote in message
...
| Ned Simmons wrote:
|
| In article ,
| says...
|
| On Thu, 18 Aug 2005 13:09:28 -0400, Ned Simmons
| wrote:
|
|
| In article ,
| says...
|
| The "NPR "Car Talk" show's "Puzzler" a couple of weeks ago gave an
| answer stating that some car's computer "knew" a front tire was low on
| air because the ABS system noted that wheel was rotating "a heck of a
| lot faster" than the other wheels when the car was driven.
|
| I didn't buy that one.
|
| Sure, the rolling radius of a low tire is less than that of a fully
| inflated one, but the overall circumference, particularly on a steel
| belted tire, remains the same. Barring slippage, that circumference
must
| lay its whole length on the road once per revolution, just like the
| circumference of a full tire does.
|
| But if the circumference remains constant as the rolling
| radius decreases there has to be slippage. Underinflated
| tires run hot, and some of that heat surely comes from
| excess flexing of the tire, but I imagine a large
| proportion is a result of the rubber scrubbing against the
| pavement.
|
| "a heck of a lot faster" may be exaggeration, unless the
| tire is seriously under inflated, but I'm sure the effect
| is measurable under controlled conditions even with small
| changes in pressure. I guess the question is how sensitive
| can the system really be without causing nuisance alarms?
|
| Ned Simmons
|
| Picture a spoked wheel with string instead of spokes, and the strings
| 1/2" too long. Just because the axle is closer to the road doesn't
| mean the tire is slipping,
|
|
| I don't think it's the fact that the axle is closer to the
| road that's causing the tire to slip relative to the
| pavement. When the tire deforms the radial distance from
| the axle to the ground across the length of the contact
| patch is not constant. So either the linear velocity or the
| angular velocity of the rubber on the road has to vary - in
| other words, something's got to give. The sidewall probably
| absorbs most of the difference when the tire is properly
| inflated, but can only do so much. Keep in mind that
| underinflated tires wear more rapidly, which implies at
| least some scrubbing.
|
| Your example of a loosely strung wheel with a rigid (I
| assume) rim really isn't analogous since the rim only
| contacts the road at a point.
|
|
| or that the tire's radius has actually
| changed.
|
|
| If the axle is closer to the ground, hasn't the effective
| radius of the wheel been reduced?
|
|
| The heat is probably almost exclusively from the flexing,
| primarily in the sidewall.
|
|
| I'm skeptical, especially in a seriously underinflated
| tire.
|
| Ned Simmons
|
|
| I didn't prased my OP post clearly. I know that that part of the ABS and
| couputer sytem will report a difference in the revolutions of the wheels
| after integrating the revolutions over some time period long enough to
| let you make a few consecutive turns in the same direction without
| trigering a warning.
|
| What I was incredulous about was the part of the puzzle's answer saying
| the tire with low air pressure would be rotating "a heck of a lot faster".
|
| The specific wording of the answer, by Ray, of Bob and Ray's "Car Talk"
| show was:
|
| ***************
|
| RAY: But when a tire loses air pressure and its diameter gets smaller,
| when the car is going down the road, in order for that tire to keep up
| with all the others and not get left behind, it has to turn faster. And
| your car does have something that is constantly monitoring the speed of
| all the wheels and comparing them to one another.
|
| What most modern cars have is ABS-- antilock brakes. And there's a
| sensor at every wheel that's reading how fast each of the wheels is
| turning. So, if it notes that the right front wheel is going a heck of a
| lot faster than the other wheels, it can either assume that you're
| making a lot of left hand turns or driving around a circle...or that
| your right front tire is going flat.
|
| **************
|
| It sounded to me like Ray somehow tricked himself into thinking that the
| increase in rotations per unit distance would be in direct proportion to
| the decreased rolling radius, and I don't believe that could be the
| case, for the reasons I already stated.
|
| Jeff
|
| --
| Jeffry Wisnia
|
| (W1BSV + Brass Rat '57 EE)
|
| "Truth exists; only falsehood has to be invented."


If the tire is low, the axle is therefore lower to the ground. That
means the effective radius is shorter. Since the radius is shorter, the
effective circumference must be smaller. Following the progression of basic
geometry, more revolutions are required to move the same distance.

Now Carl, that explanation is what I had trouble with in the first place.

Imagine if you would that the tire had side to side notches on the tread
like an inside out timing belt and the pavement had mating pitch grooves
on it. (Sort of like the ones which make a warning sound if you start to
wander off the side of the road?)

That would create a "rack and pinion" configuration.

Would you still say that the number of revolutions per mile that tire
makes would vary with the air pressure in it, or as you put it "the
effective radius".


I think Carl has explained the paradox. Imagine that as your inside out
timing belt engages with the rack it develops a bubble in the center of
the engagement such that there are x+1 pitches of belt between x teeth
on the rack. The overall length of the belt hasn't changed, but its
effective length has been reduced by one tooth.


That's where my skepticism to the "Car Talk" answer stemmed from. I
don't doubt that second order effects come into play to make the
rotations per unit distance increase somewhat with lower tire pressure,
but I'm willing to bet that the effect is nowhere near as large as being
fully inversely proportional to the rolling radius, at least not until
the tire jumps right off the rim.


It seems to me no matter what the shape the tire is forced into, the
distance traveled will be 2*pi*r per rev, where r is the distance from
the ground to the axle. Think about it from the standpoint of torque. If
the tire is driving the auto, the torque at the axle is clearly equal to
F/r, where F is the force required to move the car. If the car travelled
more than 2*pi*r per rev you'd have the basis for a perpetual motion
machine.

Ned Simmons

Well, you sure changed my thinking with that one Ned. I guess the
circumference of a tire which is low on air must develop a large enough
moving "ripple" in it to accept the decreased radiusin order to get its
entire circumfrential length moved around the axle once per revolution.

Excuse me while I go off and fall on my sword. G


Hi Jeff,

This has been beaten to death, but maybe a few calculations
or examples would help.

Say the radius from the ground to the axle is 10 inches
properly inflated. The seen/working circumference would be:

2*3.1416*10 = 62.8320 inches

This would work out to 5280*12/62.8320 = 1008.40 revolutions
per mile.

Now soften the tire so that the radius drops a mere 1/2 inch
and you get this:

2*3.1416*9.5 = 59.6904 inches

This would work out to 5280*12/59.6904 = 1061.48 revolutions
per mile.

Try to visualize the distance from the axle to the ground.
This seems to be the key part. The rest of the tire seems to
be just a distraction.

I did this kind of quick and didn't recheck my calculations
all that well, but the theory should be okay...

--
Leon Fisk
Grand Rapids MI/Zone 5b
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