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| Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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Don Foreman wrote:
On Sun, 21 Aug 2005 02:10:50 -0400, Ned Simmons wrote: In article , says... On Fri, 19 Aug 2005 20:27:08 -0400, Ned Simmons wrote: 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 ^^^^^F*r more than 2*pi*r per rev you'd have the basis for a perpetual motion machine. 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 If r(theta) is not constant, then the formula for circumference of a circle (C = pi *2 r or C = pi * D) is no longer valid.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. 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. 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. 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. This is true if you use the r that the tire had when it was circular in shape. 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. 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. http://home.comcast.net/~jwisnia18/temp/ice2.jpg 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. Jeff -- Jeffry Wisnia (W1BSV + Brass Rat '57 EE) "Truth exists; only falsehood has to be invented." |
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Ned Simmons wrote:
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. 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? Think about it from the standpoint of torque. Sorry, but an analysis of the physics isn't going to do you much good unless you first get the geometry right. Bert |
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On Sun, 21 Aug 2005 14:40:42 -0400, Ned Simmons
wrote: In article , says... Ned Simmons wrote: 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. 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? You *are* dealing with a circle; r is constant regardless of the shape the tire assumes. 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. 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? 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. Think about it from the standpoint of torque. Sorry, but an analysis of the physics isn't going to do you much good unless you first get the geometry right. 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? 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. |
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On Sun, 21 Aug 2005 13:31:12 -0400, Jeff Wisnia
wrote: 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. http://home.comcast.net/~jwisnia18/temp/ice2.jpg Topside is exactly where I'd rather be, Jeff. Brrrrrr! |
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Ned Simmons wrote:
In article , says... Ned Simmons wrote: 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. 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? You *are* dealing with a circle; Nonsense. r is constant regardless of the shape the tire assumes. 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. 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? 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. Think about it from the standpoint of torque. Sorry, but an analysis of the physics isn't going to do you much good unless you first get the geometry right. 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'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. Bert |
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On Sun, 21 Aug 2005 14:13:11 -0500, Don Foreman
wrote: We're making this far too complicated. 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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In article ,
Ned Simmons wrote: In article , says... On Fri, 19 Aug 2005 20:27:08 -0400, Ned Simmons wrote: 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 ^^^^^F*r more than 2*pi*r per rev you'd have the basis for a perpetual motion machine. 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. 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. An inward wave doesn't seem likely on a tire that's anywhere near to normal inflation: that would mean the tire tread lifting off the road; but the air pressure inside is pressing the tire tread down onto the road. Even if the tire is only at 15 psi, with a contact patch of several tens of square inches that's still a lot of force. For the tire to lift off the road in the middle is plausible for a completely flat tire, which is where someone who posted to this thread observed it happening; but it's different if the tire is just low, not flat. 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. The nitpicker in me says that the tread will certainly compress, even if the belt doesn't, since the tread is outside the belt and is normally of a larger radius, yet as the two go over the contact patch, they are flattened out to the same length, or rather, to be more precise, to the same (infinite) radius. But this happens even with a tire at normal pressure, and happens about the same amount, since the initial radius is almost the same, and so is the final (infinite) radius. To address your real question, the steel belt is prestressed by the tire's air pressure. For it to compress much, it'd have to lose all that prestress. Then it'd have to buckle -- which the individual wires in the belt are restrained from doing by the rubber surrounding them and binding them together. Either the rubber would have to disintegrate, or the tread as a whole would have to buckle. It can't buckle downward into the road, because the road is hard, so it would have to buckle up off the road, which gets us back to the question of whether the tire lifts up off the road in the middle of the contact patch; as mentioned above, I don't think this is likely for the sorts of tire pressures which you'd like to catch and report to the driver via an automated system of the type that started this thread. 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. Much as I like conservation-of-energy type arguments, they are made inapplicable here by the energy loss that heats up the tire. The actual torque at the axle is going to be greater than the number yielded by your above formula, because the torque at the axle must also provide the energy that goes into heating the tire. The basic situation seems to be this: -- The tire can't slip on the rim; one revolution of the tire always means one revolution of the rim. -- At the middle of the tire contact patch, the tire wants to move the wheel at the angular speed omega given by the formula omega = v/r, where r is the distance from the axle to the road, and v is the car's speed. -- At the very front of the tire contact patch, the tire wants to move the wheel at approximately omega = v/r0, where r0 is the distance from the axle to the front of the tire contact patch, which is about the normal radius of the wheel. The two parts of the tire are thus fighting each other: each wants to move the wheel at a different angular velocity. The resulting angular velocity will be between the two figures, and there'll be scrubbing of the tire against the road, with the front of the contact patch trying to go faster over the ground, but slipping backwards, and the middle of the contact patch trying to go slower over the ground, but slipping forwards. (Some of the people in this thread have been arguing for omega=v/r; others have been arguing for omega=v/r0; well, you're not the only ones; the tire is arguing with itself too.) Sharp eyes will have noticed that there's an "approximately" in my above argument for omega=v/r0. That's largely because the tire tread is hitting the ground at an angle; that's really a formula for the velocity of a point on the tread just before it hits the ground, not just after. The correction factor for this is roughly cos(theta), where theta is the angle between vertical and the line between the axle and the front of the contact point: really the formula should be more like omega = v*cos(theta) / r0, except even that isn't correct, since it's based on the approximation that the tread makes a perfect circle interrupted only where it touches the ground, where it's perfectly flat. Still, it's not too far off; and for small theta, cos(theta) is approximately equal to one, yielding the approximation-to-an-approximation formula omega=v/r0. -- Norman Yarvin http://yarchive.net |
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On Sun, 21 Aug 2005 22:45:49 -0400, Ned Simmons
wrote: In article , says... Ned Simmons wrote: In article , says... Ned Simmons wrote: 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. 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? You *are* dealing with a circle; Nonsense. From the frame of reference of the axle, r sweeps out a circle. The shape the tire assumes where it's not contacting the road has no effect on the velocity of the vehicle - it's irrelevant. R sweeps a circle of circumference 2*pi*r per revolution of the axle. The peripheral length of the tread is greater than 2*pi*r, but the same spot on the tread is in contact with the pavement each revolution. Must be some wrinkles in the tread? (snip) In the case of the tracked vehicle r is the distance from the axle to the surface of the track that contacts the road. In other words, assuming rigid track and wheel, r is the pitch radius of the drive wheel plus the distance from the pitch line of the track to the driving surface. No. The part of the tread in contact with the ground doesn't move relative to the ground (if it isn't mud), so the r there is the pitch radius of the (circular) drive wheel. Imagine a tread all laid out rather than wrapped around the wheel. The drive wheel would be riding on stationary tread that is laying on the ground surface. Making the tread thicker wouldn't make the vehicle go faster. Think about it from the standpoint of torque. Sorry, but an analysis of the physics isn't going to do you much good unless you first get the geometry right. 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'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. I don't know how many more ways to say it. The car isn't bouncing up and down as the tire rotates - r is constant; from the axle's frame of reference r sweeps out a circle; the only propelling force is applied perpendicular to r at surface of the road; the formulae for circular motion are applicable. Plug some numbers into this calculator, it will agree with my model. http://www.club80- 90syncro.co.uk/Syncro_website/TechnicalPages/TRC% 20calculator.htm Don, if your following, I'll hopefully get a chance to respond to you tomorrow night. Ned Simmons |
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On Thu, 18 Aug 2005 11:52:04 -0400, Jeff Wisnia
wrote: 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. From my TSD rallying days I remember that low tire pressures made some slight differences in odometer measurements, but these were in the second decimal place, hardly "a heck of a lot". Am I missing something here? What do the great minds on rcm think about this one? Jeff This is about as plain as it can be. Jeff, would you hand out those DNRs for those who can't understand this? http://www.desser.com/tech/centrifugal.html Want more? Google for tire traction wave. |
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Andy Asberry wrote:
On Thu, 18 Aug 2005 11:52:04 -0400, Jeff Wisnia wrote: 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. From my TSD rallying days I remember that low tire pressures made some slight differences in odometer measurements, but these were in the second decimal place, hardly "a heck of a lot". Am I missing something here? What do the great minds on rcm think about this one? Jeff This is about as plain as it can be. Jeff, would you hand out those DNRs for those who can't understand this? http://www.desser.com/tech/centrifugal.html Want more? Google for tire traction wave. That I believe I can understand, Andy, and it was a quite interesting page you pointed us to. The forces involved were impressive. I'll try not to think about them too much the next time Im boarding a flight. But, I can't tell if you were presenting it as evidence for or against my position when I started this thread. If I read the diagram correctly, the traction wave takes place *behind* the tire's contact patch, so I don't understand waht effect, if any, that could have on the wheels rotational speed vs inflation pressure? ******************* I was discharged from intensive care yesterday and driven home from the hospital by a masked avenger who initialed the release register "DF". I'm back home, believing again that the tire's change in rotational speed is NOT inversely proportional to the "axle height", but is a LOT less than that. Jeff -- Jeffry Wisnia (W1BSV + Brass Rat '57 EE) "Truth exists; only falsehood has to be invented." |
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In article ,
Norman Yarvin wrote: :The basic situation seems to be this: : : -- The tire can't slip on the rim; one revolution of the tire : always means one revolution of the rim. : : -- At the middle of the tire contact patch, the tire wants to : move the wheel at the angular speed omega given by the formula : omega = v/r, : where r is the distance from the axle to the road, and v is : the car's speed. : : -- At the very front of the tire contact patch, the tire wants to : move the wheel at approximately : omega = v/r0, : where r0 is the distance from the axle to the front of the tire : contact patch, which is about the normal radius of the wheel. I'm going to pick up on the words "wants to" in the above. We are dealing with an elastic system (the tire sidewalls), so there is no requirement that the speed of a point on the periphery of the tire is always given by (omega * r), where omega is the angular speed of the hub and r is a variable that changes as the point moves in its non-circular path. Indeed, I postulate that omega at the periphery of the tire is not a constant, but in the region of contact with the road is higher than the angular speed of the hub. The only constraint is that, integrated around one complete revolution, the average value of omega at the periphery equals omega of the hub (i.e., there is no slippage of the tire on the rim). Conveniently, this invalidates everyone else's caclulations to date, so I can smugly wave my hand and ride off into the sunset, at least until such time as someone comes up with evidence that the angular position of a point on the periphery of an underinflated tire is indeed fixed relative to the hub. -- Bob Nichols AT comcast.net I am "rnichols42" |
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In article ,
Ned Simmons wrote: In article , says... The basic situation seems to be this: -- The tire can't slip on the rim; one revolution of the tire always means one revolution of the rim. -- At the middle of the tire contact patch, the tire wants to move the wheel at the angular speed omega given by the formula omega = v/r, where r is the distance from the axle to the road, and v is the car's speed. -- At the very front of the tire contact patch, the tire wants to move the wheel at approximately omega = v/r0, where r0 is the distance from the axle to the front of the tire contact patch, which is about the normal radius of the wheel. The two parts of the tire are thus fighting each other: each wants to move the wheel at a different angular velocity. The resulting angular velocity will be between the two figures, and there'll be scrubbing of the tire against the road, with the front of the contact patch trying to go faster over the ground, but slipping backwards, and the middle of the contact patch trying to go slower over the ground, but slipping forwards. (Some of the people in this thread have been arguing for omega=v/r; others have been arguing for omega=v/r0; well, you're not the only ones; the tire is arguing with itself too.) Sharp eyes will have noticed that there's an "approximately" in my above argument for omega=v/r0. That's largely because the tire tread is hitting the ground at an angle; that's really a formula for the velocity of a point on the tread just before it hits the ground, not just after. The correction factor for this is roughly cos(theta), where theta is the angle between vertical and the line between the axle and the front of the contact point: really the formula should be more like omega = v*cos(theta) / r0, except even that isn't correct, since it's based on the approximation that the tread makes a perfect circle interrupted only where it touches the ground, where it's perfectly flat. Still, it's not too far off; and for small theta, cos(theta) is approximately equal to one, yielding the approximation-to-an-approximation formula omega=v/r0. Do the trig and I think you'll find that the horizontal component of omega * r0 is exactly equal to omega * r . Hmm, I was imagining that factor wouldn't cancel, but you're right: it does. Even if there's a bulge in the tire, making r0 greater than the original radius of the tire, it still does. At any point on the ground plane (assuming the ground to be flat), the horizontal component of the rotational velocity vector is the same. So my above-quoted explanation is wrong. There's still the problem of the tread having to make one complete revolution for every complete revolution of the rim, though. If the angular velocity of every piece of the tread were constant, then what would have to happen would be that just as the tread hit the ground, it would compress in length by a factor of cos(theta), and stay compressed by that factor until it lifted off the ground. But really the tread length is rather incompressible, due to the steel belt. So the angular velocity can't be constant: instead of the radial reinforcing cords staying purely radial at all times, they must deviate from being radial as they go over the contact patch: they must become, in angular terms, more divergent from each other, although no farther apart in absolute terms. The sidewall then has to flex to accomodate this. Since the motion of the tire isn't a pure rotation, the formula v=omega*r doesn't apply in the first place. This is about where I give up trying to juggle the variables in my head, and think in terms of writing a computer program. It's not necessary for anybody to have analyzed this in detail: even the guys developing the system might just have noticed that the wheel ran at a different rate when the tire pressure was low, and that the value was significant enough to be seen over the noise. But the original poster to this thread seems to have been right that the change can't have been huge. -- Norman Yarvin http://yarchive.net |
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"Robert Nichols" wrote in
message ... SNIP| | Conveniently, this invalidates everyone else's caclulations to date, so | I can smugly wave my hand and ride off into the sunset, at least until | such time as someone comes up with evidence that the angular position of | a point on the periphery of an underinflated tire is indeed fixed | relative to the hub. | | -- | Bob Nichols AT comcast.net I am "rnichols42" Damn! Nuthin' like a couple physics majors arguing with each other! |
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On Mon, 22 Aug 2005 12:38:41 -0400, Jeff Wisnia
wrote: Andy Asberry wrote: On Thu, 18 Aug 2005 11:52:04 -0400, Jeff Wisnia This is about as plain as it can be. Jeff, would you hand out those DNRs for those who can't understand this? http://www.desser.com/tech/centrifugal.html Want more? Google for tire traction wave. That I believe I can understand, Andy, and it was a quite interesting page you pointed us to. The forces involved were impressive. I'll try not to think about them too much the next time Im boarding a flight. But, I can't tell if you were presenting it as evidence for or against my position when I started this thread. If I read the diagram correctly, the traction wave takes place *behind* the tire's contact patch, so I don't understand waht effect, if any, that could have on the wheels rotational speed vs inflation pressure? ******************* I was discharged from intensive care yesterday and driven home from the hospital by a masked avenger who initialed the release register "DF". I'm back home, believing again that the tire's change in rotational speed is NOT inversely proportional to the "axle height", but is a LOT less than that. Jeff Yes, I caught that "behind" the tire thing also. I was positive I remembered the wave being in front of the tire. Dug into my notes; couldn't find the photo but found the notes. It was the right rear wheel of a 1962 Pontiac on a dyno. Perhaps the drive/free-rolling wheels are opposite. One other possibility is because the tire wasn't belted. |
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In article ,
says... On Sun, 21 Aug 2005 14:13:11 -0500, Don Foreman wrote: Sorry I haven't come back to this sooner, as I'm sure everyone is on tenterhooks anticipating a resolution g, but I've worked about 30 hours in the last couple days, and got an afternoon's sailing in to boot. We're making this far too complicated. Or perhaps not complicated enough. Rather than continuing with the speculative arguments I poked around a bit and found info that was either overly simplistic, or complicated by analyses of traction, drag, stability, and handling. This quote, from the description of Matlab function... http://www.mathworks.com/access/help...p/toolbox/phys mod/drive/tire.html ....is the most concise explanation I found. ************************************************** ******** The Tire block models the tire as a rigid-wheel, flexible- body combination in contact with the road. The model includes only longitudinal motion and no camber, turning, or lateral motion. At full speed, the tire acts like a damper, and the longitudinal force Fx is determined mainly by the slip. At low speeds, when the tire is starting up from or slowing down to a stop, the tire behaves more like a deformable, circular spring. The effective rolling radius re is normally slightly less than the nominal tire radius because the tire deforms under its vertical load." ************************************************** ******** I haven't found anything that's definitive on the relationship between what I've been calling r (the height of the axle above the ground) and what is usually referred to as re, the effective rolling radius. As best as I can tell r and re are close to equal at rest, and re increases with speed. r apparently increases with speed as well; whether it increases at the same rate as re is not clear to me. Mark's Handbook has a table that lists the free diameter of, and the number of revs/mile made, by several sizes of common auto tires. It indicates that the effective radius of the tires listed is from around 1/4" to 1/2" smaller than the free radius of an unloaded tire. These figures are at highway speed and would presumably be greater if measured at low speeds. This calculator agrees closely with the table in Mark's... http://www.club80- 90syncro.co.uk/Syncro_website/TechnicalPages/TRC% 20calculator.htm This paper details several tire models... http://www.trombi.com/docs_trombi/do...175-report.pdf The model that is detailed starting on p.27 has some pertinent info. Ned Simmons |
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