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".
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.
Believe me, I have a bit of trouble getting it, even visualizing it, but there's really no other way to see it. The difference in rotational speed has to be taken up in the wrinkling in the tread and all the sidewall flexing, which is why a low tire is a very bad thing, since all that action creates a lot of heat.
| | Perhaps now, but not when I was in the business. | In any case, the discussion is about semi tires | which have a split rim that can do grievious | bodily harm if they pop out and hit you. I've | known of 2 people seriously injured that way, | and nobody injured by flying rubber.
If it was a long time ago, perhaps there was just bias ply, and now everything is steel belted, and they do deflate rather explosively. The rubber of course carries lots of steel wire with it, and for all those who've had wire brushes lose strands into your skin you get the picture. I'll agree that the risk on passenger tires is a lot lower, but there's lawyers everywhere, spoiling all the really cool accidents we can laugh about later.
| > 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.
Thank you for making it clearer than I did. I was so wrapped up in the big picture I missed the simple explanation!
On Fri, 19 Aug 2005 04:54:53 GMT, "carl mciver" wrote: *snip*
Or, perhaps this was a direct object lesson from a Higher Power, telling you to get smarter about driving defensively, and, leaving a COMFORTABLE stopping distance between you and the vehicle in front of you, and, a parable for the rest of us about it. Glad you survived that one...and I hope you (and everyone that reads this) will take a second to think about it, and, leave a bit more space the next time. Drive more defensively folks...keep the insurance companies from posting record profits in the coming years! Don't use the excuse that if you leave room someone will pull in front of you. Trust me...they are not making you take ANY significant amount of time more to get to your destination. Even if you are going 2 MPH slower than surrounding traffic, you are STILL moving along at a good clip. Which gives me an opportunity to post an interesting question that came to me. If I am driving through a congested area (say, downtown Atlanta, or, a construction zone) at the posted speed limit and YOU are blasting through at 10 MPH or 20 MPH over that...which is better - For ME to speed up to an illegal (and likely more dangerous) speed, or for YOU to slow down a little to approach the LEGAL speed for the area? Regards Dave Mundt
The distance from axle to ground is less, but calling that distance a radius (rolling or otherwise) is misleading. In the first place, that distance does not equal half of the deformed diameter, so it's not a radius in any conventional sense; secondly, and more importantly, the shape of the under inflated tire is not circular, so even the semi-diameter of the deformed shape does not have a 2 pi relationship to the tire's circumference.
but the overall circumference, particularly on a steel
Yes.
No. You're assuming that the relationship between the circumference and the axle-to-ground distance (what you call the "rolling radius") remains constant; it does NOT.
under inflated
I doubt there is much slippage at all. The car will tend to pull in the direction of the under inflated tire because there is more friction, not less, as would be the case if the tire were slipping. (Of course, the lean of the car also contributes to the pull.)
Nor does it mean that the radius has changed.
Neither the linear velocity of the rubber nor the angular velocity of the hub has to vary. What "gives" is the mathematical relationship between the two, due to the departure from a circular form.
The sidewall probably
That is true, but the scrubbing occurs during turns, not during straight travel. Because the under inflated tire has a longer distance between the foremost and aftmost points of contact with the road, more scrubbing will be involved when turning, compared to a properly inflated tire. Hence, more wear.
It also isn't analogous because the rigid rim retains a circular shape.
No. The only meaningful definition of "effective radius" in this context is c/2*pi. If the circumference doesn't change, the effective radius doesn't change.
The question is, how much does the circumference change as the pressure changes? Will it change enough for the sensors to classify the associated change in rotation speed as significant? Anyone have a compressor and a tape measure handy to gather a little empirical data?
And from scrubbing during turns.
It could be that you're not giving enough weight to Ray's penchant for hyperbole. Maybe by "a heck of a lot" he really means "a little bit"...
No. See discussion above about effective radius. As for the "effective circumference", that's just the circumference, since each point will contact the road during each revolution. The circumference will change a little due to the change in pressure, and possibly a little due to compressive effects within the contact zone, but not because the axle is closer to the ground.
I'm not sure if I buy the notion that the area of the contact patch would remain unchanged, but in any case, the *length* of the patch will be longer, resulting in a greater deviation from a circular shape, and thus a greater deviation from the relation c = 2*pi*r, especially if you're using the axle-to-ground distance as r.
That's a static condition. I don't know if that bubble would be maintained under rolling conditions, since that would require each point on the circumference to travel up and over the bubble (kind of a standing wave thing). My hunch is that the bubble would be diminished when rolling, but I could be wrong...
The guys on Car Talk have a style of exaggeration that should be regarded as for listening enjoyment, not quantitative accuracy.
What quantity is "a heck of a lot faster" anyway?
I'm guessing that being 10 PSI low (out of say 30 or 35 PSI) is only a few percent of difference in rotation speed. It's certainly not
33% difference in rotation speed! But it'll be visibly low, bulging out to anyone who cares to look, you don't need ABS or pressure sensors to see that much.
To an ABS system with wheel revolution counters that's easily detectable (if not "a heck of a lot"!)
And anyway the indirect system won't tell you if all your wheels are equally deflated...
r is not constant because the tire has a flat patch. C = pi *2 r comes from integrating dC = r(theta) d(theta) thru 2 pi radians with r constant. If r(theta) is not constant, then the formula for circumference of a circle (C = pi *2 r or C = pi * D) is no longer valid.
Even a properly inflated tire has a flat patch. An underinflated tire just has a bigger flat patch. Circumference can remain unchanged, so revs/rolled_distance also can remain unchanged.
I don't know when you were in the business. Tubeless truck tires have been around for 30 years or more.
Fact: Very few new big trucks have tube type tires (split ring/lock ring). Almost everything except roadable cranes have tubeless tires. Reason for placing the tubeless tire/wheel in a cage is to restrain the big pieces if one lets go.
More and more truck tires have steel body plies; not just belts. When one is run under inflated or flat, there is a tremendous amount of flexing in the sidewall. This flexing is just like bending a piece of wire until it breaks. Except in this case the broken or weakened wire is concealed by rubber. When the tire is inflated, as pressure builds the sidewall lets go in what is known as a zipper failure. A ragged rip in the sidewall that is parallel to the tread. It may be a foot long. That sudden release of air can send a tire and wheel flying.
Next month will mark 40 years in the business. Twelve with Goodyear and 28 for myself.
Then where is the length of tread that compensates for the length that is lost to the flat spot? It seems to me that it either must be in an inward wave in the middle of the contact patch, or causing an outward bulge just outside the contact patch, or possibly both.
Note that we all seem to be accepting the fact that the tread length is fixed. Do we really know this to be the case? I'm sure the belts are pretty effective at limiting the length of the tread in tension, but how do they really behave in compression? If the tread can compress slightly as it rotates into contact with the road that would resolve the entire controversy.
In any case, I just can't accept that the car travels anything other than 2*pi*r per rev (as before, r is the distance between the axle and the road), regardless of what the tread does. To carry the torque and work argument further, consider that the horizontal reaction of the driving tire on the road is equal in magnitude to the horizontal force at the axle pushing the car forward, call it F. Work is equal to F * d, d being the distance the car travels. Work is also equal to Torque * angular displacement in radians, T * theta. The torque at the axle is F
r. So... F * d = T * theta = F * r * theta
But if d per rev is greater than 2*pi*r, the work moving the car forward is greater than the work input to the system by the torque turning the axle.
If the radius is less at the flat spot, then it must be greater elsewhere. No length of tread is lost, it just isn't all at the same distance from the axel.
This is true if you use the r that the tire had when it was circular in shape.
These formulae were derived for circular geometry. The tire covers once circumference per rev regardless of its shape, so work output = work input (minus losses that go to heat the tire). Torque is exerted on all partsof the periphery, not just the part that touches the road. The parts of the tire not touching the road still have torque due to "pulling" the tread around with circumferential tension.
The total torque is the sum of the various moments (at various radii) around the axel.
It is true that the *average* radius is always r, which is the (constant) radius of the tire when it is circular in shape. If you use that r in your assertion, then your assertion is correct. If r varies with theta, then the average radius is (1/(2*pi)) * integral ( r(theta) d theta) integrated over 2pi radians. If r is constant, as in a circle, this comes out to r, fancy that!
With my nearly last breath I just rescinded my DNR document and asked to be kept on life support until this one gets settled.
Don, I definitely would have picked you to be on the topside end of 100 feet of manilla line back around 1960 when we thought it great fun to SCUBA dive under the ice around here.
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That's me on the left. All that's left from those halcion days are some great memories and that very same length of manilla line, which I used just yesterday while felling a tree in the backyard.
As I explained in my previous post (and as others have subsequently explained), the equation c=2*pi*r applies only if (a) you're dealing with a circle and (b) r is measured from the center of the circle. Neither of those are true in this case. Why would you expect to be able to use that equation when neither of the basic premises are met?
Sorry, but an analysis of the physics isn't going to do you much good unless you first get the geometry right.
You *are* dealing with a circle; r is constant regardless of the shape the tire assumes. Would you argue that you can't use the distance from the driven axle of a tracked vehicle to the road surface to calculate the speed of the vehicle as a function of axle RPM simply because the track isn't circular in shape?
Then where is the propelling force F acting, if it's not at the interface between the tire and road, i.e., perpendicular to r and distance r from the axle?
You must be using the distance from axel to pavement as r. That distance is constant, but the distance from axel to periphery of the tire is not constant.
The problem with this analogy is that the wheel on a tracked vehicle is a rigid circle with constant radius. A tire is an elastic wheel with varying radius. Not all points on the periphery of the tire are moving at the same angular velocity wrt the axel. The (scalar) surface speed stays pretty constant if the tread doesn't stretch circumferentially, but some stretching and compression obviously goes on in the sidewalls. If surface speed remains constant while radius varies, then angular velocity must vary over the course of a rev. It all averages out over a rev because the tread makes no net progress relative to the periphery of the rigid steel wheel.
Force acts on the road at the point(s) of contact, but torque is applied all round the tire as tangential force transmitted from the wheel to the tread thru the sidewalls. Net torque is the sum of the moments, i.e. the definite integral of the differential moments.
First you say that r (defined by you as the axle-to-ground distance) varies with pressure, and now you say it's constant. You can't have it both ways.
If you define r = c/2*pi, your statement above true. But that's not what *you* are calling r.
Not really an analogous example because the track is not affixed to the drive wheel, but yes, I would argue that you can't compute speed based solely on RPM and the distance from the axle to the ground, and furthermore, that the distance from axle to ground is irrelevant. You
*can* compute the vehicle speed as a function of axle RPM *if* you know the radius of the drive wheel. Since the wheel is circular, the speed is (rotational rate) * radius. But this calculation is valid
*only* because the drive wheel is circular and *only* if the radius of the drive wheel (measured from center to edge) is used, NOT the distance from the axle to the ground. It doesn't matter where the axle is relative to the ground; for a given wheel size and RPM the vehicle speed is the same.
With your definition of radius as the distance from the axle to the ground, the tracked vehicle would go faster or slower (at the same axle RPM) depending on whether the drive wheel is at the bottom of the track, inside the top of the track, or above the track.
You're asking the wrong question. The problem with your analysis isn't where the force is acting, it's in your attempt to use equations derived for a circle when you're not dealing with a circle. Once the tire is deformed, it is NOT a circle, and you can no longer use equations derived for a circle.
A point on the periphery of an arbitrarily-shaped must be at the same angular position after each revolution. If it were not, the periphery would make net angular progress relative to the axel and "wind up".
If there is no slippage between road and wheel periphery, then the vehicle must progress one peripheral distance per revolution, whatever the various radii might be.
If the wheel were a circle this peripheral distance would be circumference. Call it whatever you like for a non-round wheel.
A tank-tread analogy does not apply because the tread is driven only where the tread is in contact with the rigid round wheel, where all points on a tire's periphery are mechanically (albeit elastically) coupled to the axel by the sidewalls.
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