"Mike Marlow" <mike6963REMOVE@alltel.net> wrote in message news:3a6mevF65pcaqU1@individual.net...>
"Martin X. Moleski, SJ" <moleski@canisius.edu> wrote in message> news:086s31ppp2gs5mn638qjc0n4joeav82ot5@4ax.com...> > On Sun, 20 Mar 2005 15:02:20 CST, Chuck_Steak@nospam.com (Chuck Steak)> wrote:> >
The bottom of the tire, is a point.> > >And that point moves, if the vehicle is moving.> > >If the vehicle is not moving, and the tires are> > >spinning, the point still moves.> > >Just around an axis.> >
Sure. But the axis is in motion with respect to the earth.> > The point has its own speed plus the speed of the axis> > at the top of the tire and its own speed minus that> > of the axis at the bottom of the tire. If the tires are> > not spinning, that point is still for a moment with> > respect to the racing surface down at the bottom of> > its circuit.> >
Remove the ground for a moment. Please explain this with no ground> reference. Just a spinning tire. Please explain the forces that cause an> object in motion to not be in motion momentarily at some point within it's> travel about an axis. Please explain how the ground patch becomes the> arbitrary point at which motion stops. Why not at 17 degrees, or at 181> degrees?>
I'm not arguing the physics that John and a couple of others have
attempted> to interject in this, but I am disputing its presentation, or perhaps even> its applicability to the discussion.> -- >
-Mike-> mike6963REMOVE@alltel.net
Ok, if you want to talk about a rolling wheel now, lets do that.
I'm going to talk about vectors here -- for those who don't know (probably nobody in the audience -- skip down if you like) a vector is a line drawn from a point whose direction indicates the direction of movement and length indicates the velocity of movement. When you know the movement in two planes, you can draw the vector component of appropriate length in the direction of the first plane, then from the end of it, draw the vector in the other plane. The complete vector now extends from the origin to the tip of the second component vector, describing some angle -- the actual direction of the overall movement, with it's length indicating the overall velocity of movement. By solving this right angle triangle you can now get the actual velocity of this motion which is the length of the vector, the two component vectors being the other two sides of the triangle.
So...
If you assume an ideal wheel -- ie one that does not deform in any way and is perfectly round, you can watch the vector of motion as it goes around. Also my car will be in constant, even motion, not cornering or changing speed in any way, on a perfectly flat and level track.
Alright....
As a given point on the exact perimeter of the wheel it approaches the top of it's rotation, a vector drawn from it will be extending towards the front, approaching it's maximum horizontal value, as the vector comes down from vertical to point exactly horizontally (270 degrees) at the very top of the rotation. In other words, at the very top of the rotation, it's horizontal velocity will hit twice the car's ground speed, and it's vertical velocity will go to zero. From there, the vector of motion will reduce in size and swing towards pointing down. When your point is at the same height as the axle of the wheel, the vector will be pointing on a 45 degree angle downwards and will be reduced to somewhere between 2x and 1x ground speed -- solve the right angle triangle with two equal sides to get it, it's the (square root of 2) x ground speed actually. In other words, that point on the perimeter will be moving forward at the ground speed of the vehicle but also moving downwards at that same speed, so it is moving at a 45 degree angle downwards. As the point approaches contact with the ground, the vector will continue to reduce in size and swing towards pointing straight down. At the instant just before the point touches the ground, the vector points almost straight down and is very small. In the next instant it is zero -- not pointing in any direction, there is no movement. Only in this very instant is it perfectly still. Just as a (straight) bouncing ball is perfectly still for an instant at the top of it's arc and at the bottom of it's bounce, this point of the tire is at rest with respect to the earth. In the very next instant the vector will sprout upwards and begin to fall forward from straight vertical, as the point begins to move up from the ground and be carried forward in rotation. At halfway up it's forward component will equal it's upward component and it will point 45 degrees up and to the front at (sqrt 2) x the vehicle's ground speed. The vector will now continue to swing down towards the horizontal as it stretches out to indicate twice the car's ground speed in exactly the forward direction once again, at the top of the rotation where it's vertical velocity hits zero again.
Similarly, each point on the outside perimeter of this ideal wheel will be at some instant perfectly at rest with respect to the earth at the bottom of it's rotation, and then at another opposite point at the top of the rotation it will be moving at exactly twice the ground speed of the vehicle with respect to the earth.
Any other point on the wheel, slightly inboard of the perimeter, will never be at rest. It's vector will never reach zero, though it may get close if it is at the outer edge. If you think about the angle of the vector from the perimeter, it swings over 180 degrees in total --- swinging between 180 through 270 and up to 360, then back down through 270. It's vertical speed varies from ground speed (halfway up or down) to zero (top or bottom), and it's horizontal speed varies from zero (bottom) to twice ground speed (top). As you move in towards the center of the wheel, each point swings over a smaller sweep of angles, and a smaller range of speeds. Halfway up the wheel, you would expect the horizontal vector to stretch between some amount less than ground speed and some equal amount more than ground speed, and the vertical to range between some amount less than half of ground speed and some equal amount more than half of ground speed. This results in an angle sweep of 90 degress, swept between 315 and 225 degrees. The center of the axel has no vertical component, sweeps zero degrees, and is fixed at exactly ground speed. It's vector never changes. The point just outside the axel has a horizontal vector that nudges up and down slightly, growing and contracting just a bit as it does.
To properly describe points in between the outside, halfway, and the middle, you need to use calculus, and I'm way to far out of school to do that. The calculus function should intersect my examples as pointed out, but the rest would be a pretty ugly problem unless you're still enrolled in university calculus.
I used an ideal wheel for my explanation. In reality, our rubber tires have flex, and because of that, a portion of the outside of the tire will rest in contact with the track for more than an instantaneous amount of time -- from the time it first touches, all through the contact patch, until it is lifted off at the back to begin it's rotation. Wheel pressure would affect how long this period of rest is. As you move inward from the outer part of the tire, those portions of the tire are never at rests with respect to the ground.
It was proposed to discuss the rotation of the tires without considering the ground. All motion must have a frame of reference. If there is no ground, what is your frame of reference? The car? Ok, we can certainly do that. Same basic deal, only now the outer rim of the tire is always in motion, while the center of the axel is at rest relative to the car, although it has rotational motion. But it's vector with respect to the car is zero. Motion is all about frame of reference. Even on the ground while parked, the entire car is in motion with respect to the moon as the earth spins. If the earth were to stop spinning it would be in motion with respect to the sun as we orbit it. If our orbit were to cease it would be in motion with the other components of our galaxy etc. etc. No motion is absolute. No state of stillness is absolute. You must, must, must have some frame of reference to discuss motion.
Back to the original discussion then... yes, the exact outside edge of a rolling tire is at rest with the ground it rolls across for a brief period as it contacts the ground. You could drive across a hundred lug nuts at an even 30MPH and not hurt anyone. They would be distrubed by the flex of the tire appying a small force to them and then releasing it, perhaps by "squirting" out the side as it went by, or by sticking to the rubber compound and then releasing at some non-zero velocity point of the rotation, but they will not spit out like deadly projectiles the same way they would from a spinning wheel peeling out from a pit stop.
"Somebody" <somebody@nospam.russdoucet.com> wrote in message news:cDt%d.94995$vO1.587379@nnrp1.uunet.ca...>
Ok, if you want to talk about a rolling wheel now, lets do that.
Actually, I was trying to get past the rolling wheel discussion to the burning out wheel which was where this started. The only problem is John went and introduced something very intriguing and off we go in another direction...
< major snippage of a lot of really good effort that you stayed up way too late to work on Russ>
It was proposed to discuss the rotation of the tires without considering
ground. All motion must have a frame of reference. If there is no
ground,> what is your frame of reference? The car? Ok, we can certainly do that.> Same basic deal, only now the outer rim of the tire is always in motion,> while the center of the axel is at rest relative to the car, although it
rotational motion. But it's vector with respect to the car is zero.
Motion> is all about frame of reference. Even on the ground while parked, the> entire car is in motion with respect to the moon as the earth spins. If
earth were to stop spinning it would be in motion with respect to the sun
we orbit it. If our orbit were to cease it would be in motion with the> other components of our galaxy etc. etc. No motion is absolute. No state> of stillness is absolute. You must, must, must have some frame of
reference> to discuss motion.
I do understand that... the reason I asked about the reference point, or absense of one was to get to the discussion of rotation versus direction. The explanation you provided explains velocity in a direction, but it does not address rotational speed - correct? The rotational speed remains constant (assuming the perfect wheel), it is only the vector of forward direction that changes, correct? So - take out the direction component and examine this from strictly a rotational persepective and you really have something that looks a lot like a car tire durring a burn out, right? I would think that the reference should be the axle. Rotation about it. The theoretical point on the perimeter of the wheel would approach the ground at exactly the same velocity that it would depart from it at, correct? As well, it would approach the very top of the circumfrance at exactly the same velocity, correct? Since the car we're talking about is doing a burnout in the pits, isn't this more about rotational speed than forward direction?
--
-Mike- mike6963REMOVE@alltel.net>
Back to the original discussion then... yes, the exact outside edge of a> rolling tire is at rest with the ground it rolls across for a brief period> as it contacts the ground. You could drive across a hundred lug nuts at
even 30MPH and not hurt anyone. They would be distrubed by the flex of
tire appying a small force to them and then releasing it, perhaps by> "squirting" out the side as it went by, or by sticking to the rubber> compound and then releasing at some non-zero velocity point of the
rotation,> but they will not spit out like deadly projectiles the same way they would> from a spinning wheel peeling out from a pit stop.>
"Mike Marlow" <mike6963REMOVE@alltel.net> wrote in news:3a7t50F6963rpU1@individual.net:
I would think that the> reference should be the axle. Rotation about it.
Well, we started out talking about lugnuts, which makes it convenient to use the ground as the frame of reference for that discussion, since that's where the lugnuts are resting.
We can go off into all sorts of interesting sidelights if we want to look at the axle as the reference (things like why big brake disks are bad...which leads me to the thought that most everyone in Cup uses Honda main bearings in their engines).
"Somebody" <somebody@nospam.russdoucet.com> wrote in news:ACz%d.95001$vO1.587710@nnrp1.uunet.ca:
on my street car, if I am stationary (say, front bumper against a> wall) and I am in second gear and redlining it, the bottom of my tires> are moving at 85kph with respect to the ground, and a mighty cloud of> smoke is raised.
Couple of thoughts here - the first being that apparently no-one ever told Jeff Gordon about doing burnouts in second gear, which was why his of a couple of years back tended to be lame.
Apropos of the lugnuts, tho, the key point is that the car isn't stationary, it's heading out of the pits at some velocity. Even tho the rear tires are spinning and have some net velocity with respect to the ground (and any lugnut which happens to be in the path), that velocity is not as great as if the car were stationary, it's reduced by the speed of forward motion of the car.
"John McCoy" <igopogo@ix.netcom.com> wrote in message news:Xns9620C124EFC51pogosupernews@216.168.3.30...>
Apropos of the lugnuts, tho, the key point is that the car isn't> stationary, it's heading out of the pits at some velocity. Even> tho the rear tires are spinning and have some net velocity with> respect to the ground (and any lugnut which happens to be in the> path), that velocity is not as great as if the car were stationary,> it's reduced by the speed of forward motion of the car.>
Agreed. Thus the argument that the lug nuts would be expelled at approximately that delta between the wheel speed and the forward ground speed.
Apropos of the lugnuts, tho, the key point is that the car isn't>> stationary, it's heading out of the pits at some velocity. Even>> tho the rear tires are spinning and have some net velocity with>> respect to the ground (and any lugnut which happens to be in the>> path), that velocity is not as great as if the car were stationary,>> it's reduced by the speed of forward motion of the car.>>
Agreed. Thus the argument that the lug nuts would be expelled at> approximately that delta between the wheel speed and the forward ground> speed.>
And, if said nut connected with ankle, shin, thigh or other nut, would HURT significantly!
"John McCoy" <igopogo@ix.netcom.com> wrote in message news:Xns9620C124EFC51pogosupernews@216.168.3.30...> "Somebody" <somebody@nospam.russdoucet.com> wrote in> news:ACz%d.95001$vO1.587710@nnrp1.uunet.ca:>
on my street car, if I am stationary (say, front bumper against a> > wall) and I am in second gear and redlining it, the bottom of my tires> > are moving at 85kph with respect to the ground, and a mighty cloud of> > smoke is raised.>
Couple of thoughts here - the first being that apparently no-one> ever told Jeff Gordon about doing burnouts in second gear, which> was why his of a couple of years back tended to be lame.>
Apropos of the lugnuts, tho, the key point is that the car isn't> stationary, it's heading out of the pits at some velocity. Even> tho the rear tires are spinning and have some net velocity with> respect to the ground (and any lugnut which happens to be in the> path), that velocity is not as great as if the car were stationary,> it's reduced by the speed of forward motion of the car.
It starts stationary, but there is a very real possiblity of running over one of the rear lugnuts after having moved only a foot or so. Those are the scary ones. Running over the fronts will happen at a reduced "net contact patch ground velocity" as you pointed out.
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