Speed of Rubber

I would think 100mph X ( radius of tire X 3.14 )

It makes no difference if the speed is 100 mph or 10 mph, if there is no slippage between the highway and the tire, the speed of the rubber and the highway at the point of contact are the same. Similar to two gears fully engaged.If there is a difference in speed, there has to be slippage.

Look at this problem in detail. When the tire is not touching the road (as in an airplane just before landing) the lower part of the tire is curved downward. When the tire is in contract with the road, the lower part of the tire is flat. The lower surface is then in compression, so it is shorter; thus when the center of the contact area is not moving with respect to the road, the ends of the contact area are slipping toward the center.
Of course, there is always slippage; on the driving wheel, the tire slips rearward with respect to the road surface, on the idling wheels the slippage is forward. When your wheels are spinning ("burning rubber") the tire is moving rearward with respect to the road, and at full braking, the tire is moving forward with respect to the road surface.

The answer is ZERO.

Explanation: The velocity of car and its body parts is described relative to road. So if the car is moving, their some relative velocity between road and chassis. But in case of wheel, we have some more physical quantities associated, mainly "angular velocity" which is an analogous to linear velocity. So when the car is in motion, wheel always does have angular velocity same throughout its particles but varies in linear velocity. The upper part of tire moves forward with the car having linear velocity while the part in contact with the road has zero velocity (as at the point of contact road and rubber have zero relative velocity) which does the job of pushing the car forward using friction. If its velocity would not be zero, then car would not move forward.

-Thank you

This is correct, although I didn't see the statement that the top of the tire is traveling at 200 mph, twice the car speed, while the center of the hub is travelling at car speed.

ZERO is correct! It might be useful to point out that the top of the tire is travelling slightly more than 200mph (more because compression makes the distance from the road to the axle less than the distance from the axle to the top of the tire). At any given instant, the tire as a whole is rotating around the point of contact with the road, while the car is travelling at roughly the average speed of the various parts of the tire.

This reminds me of the age old problem of how to make a helicopter fly in a straight line?!!

If the helicopter is flying forwards at an airspeed of 100 mph then if there were no variable pitch to adjust the blades one side of the helicopter's blades would be producing much more lift than the other side and the chopper wouldn't get far off the ground...

As the outer edge of the forward sweeping blade would be travelling at 300 mph (say) and the opposite blade would be travelling at 100 mph...

Hold on is that right?

Time for another coffee.... John :-)

a car going at 100 MPH has the tire on the road at 0 MPH and the tire at the top at 200 MPH

Outside of the car and the tire's linear speed of 100mph. You're saying the tire is going from 0mph to 200mph in a fraction of a second, or the time it takes to make half of a rotation?

I think that when the car is in motion, not at "an exact instance", both the top and the bottom are spinning at the same rate. Otherwise, the tire would not be in one piece. That's without taking in consideration of any strech, or the compression of the bottom of the tire.

At "an exact instance", the entire car and everything else in this world is moving 0mph altogether.

Somebody please correct me if I'm wrong...

You are wrong, and so is everybody that based their answer of ZERO on this notion that 'At "an exact instance", the entire car and everything else in this world is moving 0mph altogether.'

At an exact instance, you are not freezing time, that is impossible. You are merely marking a point in time and usually a location in space. A moving object still has velocity and even acceleration, if there are unbalanced forces acting on it.

Yes, the top and the bottom of the tire ARE spinning at the same rate, angular velocity and linear speed. However the velocity vectors are different for every point around the diamerter of the tire. To use the example another tried of the clock (forget AM or PM, a clockface is 12 hours!), from the right side of the car (which is important because the other side will be opposite in relation to the "clockface"), at the 12 o'clock position (12 OC) the outside diameter of the tire has a velocity vector of 100 mph (from rotational speed), plus 100 mph from vehicle speed. Since the direction of both is the same, towards the front of the car, the magnitude adds and the velocity is 200 mph. Opposite for the 6 OC position, since the rotation direction is now toward the rear and the vehicle direction is toward the front, these equally large vectors cancell each other, giving the velocity of ZERO. At 3 OC and 9 OC (and any other clock positions) vector math must be done based on the angle of the tangential vector of the tires rotation. For example, at 3 OC this vector is straight down. The resultant vector will be at 45 degrees below horizontal and the magnitude will be the hypotenuse of the right triangle formed by the two vectors as the legs. By Pythagoras, this gives a magnitude, V = (100² + 100²)^1/2 = 141.421... , that is, relative to the ground.

Just as if I fire a gun straight down from a very fast moving airplane, the speed of which is the same as the muzzle velocity. Neglecting the slight air friction, assume we are low enough that terminal velocity does not come into play, the bullet will make a hole in the ground at an angle, not straight down!

"You're saying the tire is going from 0mph to 200mph in a fraction of a second, or the time it takes to make half of a rotation?" Well, yes. This is the velocity vector, relative to the ground. The tire still only has a rotational velocity or speed of 100 mph, relative to the axle of the car, so it does not "tear itself apart". On the other hand, that is precisely why the tire needs polyester or steel belts because the resulting acceleration (change in direction, not speed) causes a force which tends to expand the diameter of the tire unless constricted by material strength. This why we have speed ratings for tires, because at higher speeds these belts are under much greater tension and need to be stronger. Also, there is heat buildup inside the tire, because no tire is perfectly round, due to the weight of the vehicle, giving the resulting "footprint" of a tire we see in so many tire commercials. Since the tire actually changes its shape as it revolves, work is being down on the belts in varying the tension and friction between the layers. This results in heat buildup, which if not properly dissipated, would quickly elevate the temperature of the material to a point where it would disintegrate.

I disagree with your statement that at an exact moment everything is moving at 0 mph. If that was the case, then everything would be torn apart by forces of acceleration.

Velocity is distance divided by time. Reducing the time frame toward zero, the velocity calculation becomes more and more accurate. After all velocity is defined as dX/dt where t approaches zero.

A bigger puzzle is this. Take a typical tire, 28 inches in diameter, or almost 88 inches in circumference. However setting on the ground, the tire compresses such that the center of the axle is perhaps only 13 inches from the ground. This means that for every revolution, at that distance from the ground, only about 82 inches should be traveled. What happened to 6 inches of tread.

Howetwo, you said, "I disagree with your statement that at an exact moment everything is moving at 0 mph." I never said that. All I said was the velocity of a point on the outside diameter of a tire (assuming it is perfectly round of course) is zero at the lowest point of its revolution. Remember high school physics? Velocity is a vector, speed is a scalar. While the point on the tire may have a linear speed of 100 mph WRT its own axis (rotational velocity), it has a velocity of zero WRT the ground at that point.

As to your "bigger puzzle", that is just too easy. Your calculation of 82 inches is based on the radius of the 28 in. dia. tire becoming 13 inches, all the way around! It is not. In fact, it is only 13 inches directly below the center point (or center line in 3 dimensions) of the tire. Directly above the center and for most of the rest of the tire, the radius is still 14 inches. "What happened to 6 inches of tread" you ask? Nothing at all. The tire perimeter is no longer a circle and so your calculations are totally inaccurate.

The tire is basically deformed, with the flat spot forming a chord from one point on the circular tire to another. The small amount of perimeter that may be "lost" (less than one inch I am sure) is due to the compressibility of air and rubber under a load.

I think Howetwo was responding to my post, lol...

I didn't say anything has stopped, just frozen in that "instance". At that instance, nothing is currently moving.

A turning tire at 100mph never has an instant when nothing is currently moving. Did you ever see anything do that? Except maybe a ufo!

I think the cold has gotten to ArticZone's brain!

Could somebody let me know how long an instance is for me to put in my calcualations.

Thanks for the reply. I was thinking of it as if I taken a picture of a car driving down the road at 100 mph. Being that a picture captures objects, even moving ones, at an instance of time.

I'm not a master mathematician but, when any point on the tire returns to it's origin, it will always have a velocity of 0, right? Any point of the circle could potentially have a velocity of 0. When the car was stopped you could mark the bottom as it's starting point. Once the car is up to speed, and the mark is at the bottom, it's velocity has become 0 relative to the axle. Still stating without consideration to compression and stretch.

And I do understand that we cannot freeze time, but an instance does not have a realistic value of time. I even looked it up to make sure.

1. An almost imperceptible space of time. See synonyms at moment.
2. A particular or precise time: at the instant of combustion.
3. (Abbr. inst.) The current month: your letter of the 15th instant.
4. A food or beverage designed for quick preparation.

If time doesn't "move forward", neither can the distance of the car. Putting it in a 'frozen' state, having a speed of 0 along with everything else.

Speed = c / 0 = undefined

ArticZone,

Your are not making any sense!

You are floundering!

Watch out! There be icebergs ahead!

There are two ways to freeze time. 1. Use a high speed camera. 2. Use a strobe light.

Maybe there are other ways to do this? Perhaps a mental image? Or a head on crash?

If you had a platform to rest on, holding a strobe light to point at your tire at 100 mph, you could then see the frozen instant of that point in time, and see no rotation of the tire or mabe even a frozen moment of the road. Perhaps this discussion can be recorded as frozen?

If the rubber is moving at 100 mph and the road at zero, there sure will be a lot of smoke!

The question is ambiguous since it doesn't state a frame of reference. Relative to the center of the earth, the rubber is moving somewhere in the vicinity 1000 miles per hour depending on which direction the car is going and at what altitude and where it is on the planet.

Hmmm.

An instance of time has no time change (velocity must be 0 or undefined) so the molecules on the road that get kissed by the tire for that instant would have a zero velocity. I guess, if we asked for how long will the molecules get kissed by the tire, than we have to do some math associating the rotational velocity of the tire. Or when will the molecules 100 miles away get to kiss the tire than we can say 1 hour and the tire is going 100 miles an hour.

Just my $0.02

the rubber is going zero.... the road is going 100mph?

It is not going, it is revolving as a part of the car.

Reply to Hmmm

Your comment is worth more than "my two cents worth"!

"My two bits worth" is $0.25.

Well, since the outer surface of the tyre (contact patch) is in contact with the asphalt, it cannot truely make but line contact, but I digress for the topic of conversation here. Since "how fast" is part of the question, then speed must be part of the answer.

Tyres wear, so the contact surface is losing material, no? Therefore, for the car to maintain a speed of 100 miles/hour the surface of the tyre must move very slightly faster than 100 miles/hour. So perhaps neither 0 miles/hour nor 100 miles/hour is the answer? And since molecules were mentioned; they (and their atoms and their components) are moving a bit as well. but again I digress.

Forgive me for being a realist, or perhaps an ornery gadfly.

Ing. Robert Forbus

Rubber tires in straight line, low acceleration services will last hundreds of thousands of miles. Montreal subway tires have very long lives.

Once you turn corners you get the main cause of wear, differential rubbing as the inner and outer radius of the tire both travel at the same speed, but cover a larger/lesser amount of road = rubs off the tread. Acceleration and braking also wear tires more. The extreme burning rubber wears a lot. Very hard cornering also adds wear du to load transfer

Have you heard of centrifugal force on a tire at 100 mph? I think perhaps the tire slows down relative to your speedometer speed as you increase in velocity. What did you say about tire wear. One hundred mph is still 100 mph except the rpm of the tire is greater with with a worn out tire.

Group C believes that "at the exact instant" speed does not exist, because there is no time to change position!!

Neil

I like the way you think.

Molecules slapping molecules. How fast does your hand move when slapping another person's face, when the molecules of your hand come into contact with the molecules of the other person's skin? Zero MPH! That's the sissy answer.

Of course if you want to get technical about tires..........

It depends on the alignment of the tire, the balance, the shape of the tire (as far as how far from being a perfect circle it is...no tire is a perfect circle), if it is on the front or rear axles, if the vehicle is in a turn and how tight the radius of the turn is at the time of contact, the inflation of the tire, whether the vehicle is light enough to allow air under the body (small aircraft take off between 50 and 80 mph and dragsters have wings to keep the wheels on the ground) and lastly, the condition of the asphalt upon which it is riding or jumping.

Before you smirk at the seeming small differences of the latter, think of the practical value of this question: Why does the wear on our tires come from? OOOPs (that's another thread!) Sorry. Speed, the discussion which prompted the question was speed. 100 mph IS speeding in any state in the USA and most of the world, except for places like Germany, where the Autobahn restricts the speed only through metropolitan areas and other places deemed worthy of a speed limit!

the important thing to know is the diameter of the tyre and RPM of the tyre atthat speed.

then we can calculate the surface speed of the tyre surface contcating the road.

regs

aarshi

aarshipathan@hotmail.com

The correct answer is the tire is not moving relative to the road at the point of contact (6PM position). At the 3 AM and 3PM position it is going at the car velocity and at 12 PM it is going at twice the car velocity.

Tire wear in straight line travel is a function of pressure. The higher the pressure the lower the wear. WHy? The higher the pressure the smaller the contact patch. A veyr hard(diamond) tire on a diamond road would have the smallest possible contact patch( and very little maneuverability either).

Tire wear in turns is a function of pressure and tire width. Most wear in low speed travel comes from tire scrub on turns. You can hear it if you open the windows and turn a corner with engine off(coasting) you will hear the scrub abs the tire breaks contact and elsstically jumps to a new position.

How do I know this? I worked for Dunlop, and we tested tires this way( the techs did).

How do you ruin a new tire the fastest? Run it at 120 MPH with 4 times the load and half pressure. Most would blow up on our enclosed tire dynamometer in 10-20 minutes or less. (with a bang, I might add and a stink too). Side wall flexure and the quad load kills them fast

You forgot to mention that the tire has a 100 mph continuous impact, therfore the steel belted radial tire. There is a big difference between a 1/4 inch impact and a 3 inch. I'm surprized the tires lasted even that long underinflated. You should have been there to defend the tire companies; the judges would have thrown the lawyers out of court.
Judging from the answers, some people sure don't have a clue about specific details, or perhaps they spoke too soon.

Where on a 12 hr clock is 3AM? Then I would ask where 3PM is... If your referring to a 24 hour clock, I believe your statement would not be accurate.

It is 360 degrees; this is why you have continuous road noise. Remember the wheel is turning????

Maybe I just don't get it.

A clock face has 12 numbers on it. If you start at 12 and end on 12, you've completed the perimeter of the circle and 360 degrees.

So wouldn't 6 o'clock be 180 degrees and 9 o'clock represent 270?

It depends on your viewpoint, maybe 0 degrees has slightly shifted due to road impact and has become 1 degree relative to the original location?

Correct! you got it spot on.

The tire surface is traveling at the same speed af the road which is a static situation at zero. When the tire is stopped and the car continues the tire as a whole is moving at 100 mph. However if the tire is turning and the car is moving at 100 mph, the axel of the car is traveling at 100 mph. How fast is the top of the tire traveling, 100, 200 or zero mph.

The top of the tire is going 100 MPH same as the rest of the car. The top of the tire is always the top of the tire, the bottom of the tire is always the bottom of the tire and it too is travelling at 100 MPH! This of course is relative to a certain point in location, not in time! If you don't believe me drive your car 10 miles and see where the bottom of your tires are. They are right there with the rest of the tires!

Some scientist may ask how fast a certain molecule is going at a certain point in time like a strobe light shining on a timing mark. Sure, it is not moving at that instant!

Because the road is stationery, it acts more like a rack and pinion gear. So we have 2 different types of motion: Linear and circular.

The question is a tricky one about point of contact in combined motion.

Position or juxtaposition= tire in contact with road = exact position data by definition is not motion specific, lacks motion, defies motion.

Speed = distance/time. Distance - time = No speed. Time - distance = No speed.

Therefore: The speed of anything at a certain point in time is 0. The speed of anything at an exact location is 0. The term exact location would ultimately be a universal exact location, because our universe is in a contant state of flux, however planetary location would suffice for a local definition of exact location for most of us earthbound mortals.

So rubber at 100 MPH spinning around and around will travel at the rate of 100 miles in one hour. Pavement travels at zero (except during mudslides, earthquakes etc.) Since tires travel in circular motion, a perfectly round 15 inch (lower radius) tire will cycle 67227 times in one hour at 100 MPH (1120 RPM). This means that roughly, every 94 inches, the first spot on the tire to touch the pavement will touch it again and do so a little more than every 20th of a second (0.053549912981391405238966486679459) it will pass by the same spot on the wheel well as many times and as often.

If you rotated the wheel to where the bottom of the tire had the valve stem in the 6 o'clock position, every 94 inches the tire would look the same if the vehicle were to stop exactly at that point. If you were riding parallel to the vehicle at 100 MPH and was taking pictures and your camera was ready every 20th of a second and you snapped a picture at that instant, you would have 67227 pictures of the wheel in the same position, appearing not to move.

You could almost take that to court! :)

Ah, but the pavement has speed as well. Due to Earth's rotation it could be travelling if it is in an east-west direction as much as a postive 1,000+ miles/hour or a negative 1,000+ miles/hour. Near the Equator, of course, 90 degrees to the axis of rotation (yes it wobbles a bit) and as far from the axis as possible whilst remaining on usable land.

Now for north-south, the speed of pavement movement would be much slower, as full positive (or negative speed) could not be attained until the Earth's axis went from full positive to full negative tilt. It is a lot of distance but it takes a long time to move.

Forever the gadfly,

Ing. Robert Forbus

If the top of the tire could hold something and throw it forward, that forward flying object would be at 200 mph at the instant of thrust.

Ah yes! On the molecular level then: Does the molecule that comes into contact with the asphalt remain on the tire and continue it's circular voyage? Or does it rub off and become a part of the scenery?

Or was it a silly question to begin with, knowing we were going to putz around with the obvious technical answer?

At least the Dunlop tire tester guy knows what I meant. In parking lots, tight turns and limited slip differentials can cause tires to leave marks. Not the sign of Zero MPH, but not 100 MPH either as the test question indicates...

Sorry guys, the answer is, that neither answer is correct and both answers are correct. (Or, more succinctly: the question has no definite answer.) The question states that the car is travelling at 100MPH (relative to the road). Now imagine that the tire tread (its OD) is (continuously unwrapped to form) a continuous linear strip of rubber (of equal length as the strip of road being envisioned) that is moving just above the roadway at a speed of 100MPH (relative to the road; or the road relative to it). Imagine further that, at a point on the road (or the rubber), the speed of the rubber (or the road) is being measured--this is the analogy to the point of contact between road and tire tread--as it passes over (or under) the road (or rubber). In a broad sense (as inferred from the syntax of the riddle being posed), the speed measuring sensor must measure a speed of 100MPH (same as the car's speedometer) so long as the measurement is taken over a finite duration of time. So the answer in a broad sense is 100 MPH. However, if the measurement is attempted to be made instantaneous--if the time duration for the MPH measurement is reduced to zero--if the rubber passing over road (or vice versa) is reduced to zero dimension (the analogous point of tire-pavement "contact" reduced to zero) then the speed of the rubber with respect to the road (or road with respect to the rubber) must be zero (or null). 100 mi/hr x 0 hr resolves to 0 miles during an undefined (i.e., non-existant, i.e., zero) duration of elapsed time.

Now (if I correctly construed the original post) we can talk about...propylaxis?

What the...!?! ....propylaxis? ....propyl-axis? How does the direction of the hydrogen bond of the hydrocarbon radicals, CH₃CH₂CH₂ or CH₃CHCH3 (1-propyl and 2-propyl) , which derive from propane, have anything to do with the Speed of Rubber?

MAD-MAD-MAD You are Mad right.

#35,

Methinks #24 won't fall for that one, about the axis and such. You're just trying to get him to mention direction, speed, and traction--in context--and be out-moderated, aren't you?

If the tyre is not skidding then the speed is zero.

It's along this logic the bottom edge of the flange of a train wheel is going the opposite way.

Whilst on trains, and this logic, you can stop a moving train with a golf ball. Stand on the track and throw the ball at the approaching train. The instant the golf ball bounces back is when at velocity is zero. That instant was when in contact with the train thus the train speed was also zero. Youv'e stopped the train.

It's a theory I prefer not to test myself.

like the

It depends on the size of the tire!!

Yes, It does. If the tire diameter is too small to reach the road the matter becomes aerodynamic, esp if the same condition applies to all the other tires!

Explain yourself further???? Tire diameter too small????Explain what happens, when a tire is underinflated???

If both your legs are too short to reach the ground, how can you walk?

You do not, you must fly

underinflation = excess flexure = internal tire heat = early failure, if at high speed, catastrophic failure

You are then stating that the tire has no contact with the road. Very interesting! You are saying that you fly, like maybe a hovercraft effect, to have underinflated tires??

Yes, a car can fly, with enough thrust and something to vector it. Once in the air, no need for tires, as you can see with many planes.

They have no choice but to stay up if they have flat tires, and many are up there to this day, 'tis sad, yet 'tis true.

BTW, if you agree with the foregoing, do you want to buy some Florida land?

If you could see the tire, I think that the impact is more advanced and not straight down.

No way-it depends on the make of the car!

Ing. Robert Forbus

Yes, the finned cars from the 60's were best for this

#35 might say: It depends on the size (and oscillatory velocity, wrapping, and lubing) of the axis.

ROFLOL

And you call yourselves a bunch of Engineers!

"If a car is travelling at 100 miles per hour, how fast is the part of the tire that contacts the road travelling at the exact instant of contact with the road. Group A believe the answer is zero miles per hour. Group B believe the answer is 100 mph."

Figures, with only two possible answers, almost half would get it right and almost half would get it wrong, with some of you just on another planet I guess.

Does nobody remember solving problems with Free Body Diagrams?

Look at the FBD of the tire. It has a linear velocity, at its center of 100 MPH. We know that the revolution of the wheel has some angular velocity that is measured by the odometer (revolution counter) and translated into a rotational speed for a given tire diameter (assuming it is the correct diameter). Therefore the tire has a linear speed of 100 mph at its outside diameter. This means that anywhere on the surface of the tire (including where it meets the road) the rubber molecules will have a velocity VECTOR of 100 MPH tangential to the diameter and in the direction the wheel is travelling. If we are looking at the right side of the car, the wheels are spinning clockwise and vice versa for the left side of the car. In both cases (left or right), the velocity vector is toward the rear of the car, 100 MPH. However, since this point is also attached to the tire (remember FBD!) it also has a velocity vector of 100 MPH toward the front of the car. Being opposite in direction, but equal in magnitude (speed), the two velocities cancel each other and the answer is: (ta-dah) ZERO mph.

Einstein would have fun with this one. The theory of relativity. How about this added confusion. The tire is rolling forward, backwards, up and down; including every other angle all relative to the axle. Even a little wobble if the rim is bent, plus a little extra bounce if the wheels are out of balance. How about a little out of alighnment..and to top it all off; how about different tire pressures in each wheel to really anti the relativity? Now put your second thinking cap on!

On the molecular level though, the answer could be discribed as 'near zero', but the interesting answer on that level is that at the front end of the contact patch, the rubber starts to take a bit of a short-cut in the diameter of the tire (sweeping through a lesser radius compared to that of the unloaded state).

I figure that 'nearly' instentaneous change in direction must cause a lot of force to be applied to the rubber in that region, and makes anyone wonder how much of a rebound reaction occurs (like a rubber ball bouncing) compared to what force the rigidity of the rubber and steel belts have on that region keeping it in place.

Assuming there isn't a big rebound, and that the contact patch is not so complex to describe, then for the brief moment the tire is in contact with the road, there is little to no appreciable relative velocity, if I understand the question correctly.

One thing to note is, for anyone who has done any reading on tires, is that tire grip comes at some small amount of slippage in a rolling state. If you consider that it takes at least 15hp to cruise on the freeway at legal speeds, and that it would take considerably more power to sustain 100mph in your average vehicle on the road, then there would be a small amount of slippage as a result of the forward motive force.

Just because no one has included any slick pictures yet...

(from http://mathworld.wolfram.com/Cycloid.html)

The vertical velocity of the point on the tire clearly goes through a zero point because it changes direction. And I seem to remember having a dynamics professor that asked this question one time. The speed of the tire horizontally has to equal zero when it contacts the road (relative to the road). In the FBD of when this point of the tire contacts the road, the road reacts equally against the force of the tire through friction and so the car goes forward. The tire does pass through a zero point unless there is sliding.

Just adding my effort to the confusion.

No, I think you illustrated the point perfectly, at least for the vertical component of the velocity vector. How about doing a graphical illustration for the horizontal component, or for the velocity vector itself (speed and direction angle)?

No matter where the red dot is, that is where the zero point is. Go ahead and put a red dot with a pointer on your tire. Turn it! The red dot continuously changes direction in that circular movement for 360 degrees.

The point zero can only be established by an opposite 180 degree which is relative to the red dot, and then the fine line is a constant vertical unless you are going up or downhill, hit a bump or a rut.

i was wondering...

why did u need 2 discuss that??

and by the way, which group were u in?

I can't see what all the fuss is about...

I don't know Elika, I'm sure I wouldn't want to be in that group!

It seems to me that everyone is chatting about speed and velocity as if they were interchangeable.

Isn't it all a matter of relativity antway?

John

so very true!!

and yet it is 1 of d most viewed, commented, etc. threads and yet so non-sense!

but the comments r really funny and seem 2 me more imp than d original question itself!!! yeah?

Here is another discussion about speed:

If a car is travelling at 100 miles per hour, and a bug is flying at 20 miles per hour in the opposite direction, towards the car, what is the last thing that goes through the bug's brain as he hits the windshield of the car?

Group A believe the last thing to go through his brain is "Wow, how fast was I going?".

Group B believes the last thing to go through his brain is his anus.

Wow! This is a lot worse than hitting the fan.

Wouldn't the anus swallow the brain?

And by his own negligence, the bug becomes a butthead.

Ing. Robert Forbus

I've known some people who have their heads...

I've a simple mind (after 30 years of Coast Guard service as an electronics technician. I believe the correct answer is somewhere close to 75 mph.

Paul B. Searles, MCPO, retired

Good grief! I wish this thread would die. The question is logically flawed. There is no stated reference frame. Any answer that does not clearly indicate to what the motion you are describing referenced gets a "C" even if it would be right otherwise. Any number of answers may be correct depending on the frame of reference. I pick my frame of reference to be that of the molecules in contact with the pavement and they have no motion at all excepting the normal vibration from heat and quantum effects. The entire remainder of the universe is moving in relation to said molecules. The world is moving around the molecule as is the tire. Physics and the lack of reference in the question permit this. This is a correct description though you may not like my choice of reference frame. Please ask logical questions. Please state your reference frame in the answer so that it is not ambiguous and most of all please let it stop now.

THANK YOU!

Although it has been more entertaining than most threads.

This reminds me of a question another engineer would always ask (after a few beers): "Is it quicker to Chicago than by bus or should I take a lunch"?

"Nuff said".

Speed of Rubber thread - R.I.P.

I agree in part But, since the the questioner had the intent to find the one "correct" answer--that the question itself was rhetorically flawed--then the question itself could not be said to have been flawed per se but, rather, to have been shortsighted in not contemplating that many if not most respondents would not sense (even those that saw) the correct answer when it was posted on a few occasions. But it appears the questioner did try to anticipate this, since the initial post also invited "humorous" (i.e., off-topic) responses from viewers. Unfortunately--again--it was not anticipated that only a few would understand that it was another kind of rubber(s) which the questioner perceived would inspire humorous (perhaps even entertaining) asides and retorts. Failing on both counts, the questioner must certainly share the angst that you now bespeak: that, speaking in a biblical sense, these threads have become a vexation of the spirit.

Ok viewing from a set location on the road the surface touching is zero mph the top of the tire is 200 and the x axis which is the foward and back of the tire 100 mph. what is the actual speed of the tire one moment before and after contacting the surface at zero mph. 100 mph, greater than 200 mph, less than 200 mph greater than zero and less than 50 mph or slightly faster than 100 in front and slightly lesser than 100 behind.

There is no such "set location" on the road.

I think the pre-impact-after velocity of the tire is 100 mph because the tire is turning at 100 mph.

RCAPPER said, "Good grief! I wish this thread would die."

Then why are you breathing new life into it? Your post is rhetorical CPR!

"The question is logically flawed." No, it's not. It is a simple question which you chose to overanalyze.

"There is no stated reference frame." You are correct, there is no stated referece fram. The reference plane is implied to be the one from which the car would be traveling 100 mph, i.e. some fixed point on the road. Don't make this more than it is.

"I pick my frame of reference to be that of the molecules in contact with the pavement and they have no motion at all..." Your point is moot, of course an object has no motion releative to itself.

I believe the question was a good one, because it points out the difference between speed, a scalar, and velocity, a vector, as well as angular velocity versus linear velocity. The answer also requires the understanding of vector math. It also shows how a lack of understanding those principles leads to faulty conclusions about the world around us.

Thanks for your comments. We all have days when we let little peeves bug us. I can't say why this annoyed me but I guess it did.

Let me try to illustrate the matter as I see it this way.Lets use a toy car, on a toy conveyor.The conveyor belt is powered by the toy car tires that are in contact with it.

At ANY speed of the tires, the conveyor speed and the tire speed are the same, if there is no slippage, and I presume no slippage for this experiment.

Now, take your little toy conveyor and car, and put them in the back of a pickup truck and have someone drive you down the highway.Now the tires are moving at the same speed of the truck, RELATIVE to a stationary object beside the road, but still zero relative to the conveyor.

The question did not specify a certain diameter of tire, so rotational speed should not figure into the answer.

Does this clear it up, or muddy the water?

"The question did not specify a certain diameter of tire, so rotational speed should not figure into the answer.

Does this clear it up, or muddy the water?"

I am afraid you mucked it up a bit. I don't understand what your "toy car on a conveyor belt" has to do with the original question. And regarding the diameter of the tire, that is certainly irrelevant, only that it is fixed and equal all around the tire. For the purposes of the question, the flattened area at the bottom is discounted. It would have only a very small effect anyway, if the tire were at correct pressure.

By rotational speed, we mean the linear velocity tangential to the perimeter of the tire, NOT angular velocity (RPM).

Consider the highway to be analogous to the conveyor belt.If there is no slippage, the tire is moving at the same speed as the highway under it.The same could be said of 2 gears in mesh.Their surface speeds, regardless of rpm, are the same.Pull along side this hypothetetical car, and put a tachometer on the highway.Then put the tach on the tire. Speeds should match.Therefore,where the rubber meets the road, the relative speed is zero, relative to the point of contact.Anything else shows slippage.

Put a mark on the tire. Roll forward 1 revolution.Now roll back 1 revolution.Do the marks line up?if so,then no slippage. If no slippage, then no speed differential between tire and road.

Case closed.Call your next witness.

Change RCAPPER to Re-Thread?

I like to take an odd perspective now and then to stimulate thought. It has the tendency to evoke unexpected viewpoints and I'm also used to getting some flack. I try to stay focused on the topic. If I have nothing intelegent or supportive to offer I try not make uneccessary comments. I don't see a need for name calling. We aren't on the playground anymore, are we?

You ask "Does ontogeny recapitulate phylogeny? "

The answer: NO

Good answer.

Knowing that a tire has a flat spot against the road couldn't I look at this problem as if it were the tracks on a dozer? You can visually see that the track relative to the ground is moving at zero mph while the dozer moves forward at a constant speed. So, the tire relative to the road would have a smaller, but similar flat spot. Correct? Is This logical?

Obviously we have intelligent design here, or does someone suggest that cars evolved from wheeled ancestral forms?

In that case we are lucky they did not evolved from kangaroos

Yes, you can look at it that way, but you are missing the point of the question. For a logical explanation see my posts #26, 29, and 44 above. For a good graphical illustration of the vertical position vectors and why the vertical velocity vector must be zero at the top and bottom points, see Nate's post (#45).

This reminds me of my high school Physics class. The class had sophomores, juniors, and seniors in it. We sophomores had not yet studied Trigonometry in math and the Physics teacher was trying to teach us about distance, velocity, and acceleration as vectors. Without Trig, he had an extrememly difficult time trying to explain it, let alone work problems. He ended up taking a week to Teach us sophomores "Trig in a Nutshell" while giving the Juniors and Seniors who knew Trig a one week pass to the library.

If this discussion continues, we will have a pyramid of discussions. Will this create a problem?

Not a probelm as long as we place the heavier comments at the bottom, and maintain a low center of gravity.So if we keep it light from now on, it's a cinch.

sounds like a hoosier answer if ever we heard one.

When a car is parked outside in -40 below weather, the tires have a frozen impression of where they are as they support the weight of the car, which you can see. When you start driving, you can hear this continuous womp, womp, womp, as the tires rotate past the flat spots. You can also feel the shock absorbers working as the car is womped up and down. With time and more rotations of the tires the flat spots and the womps vanish. The next morning, you have similar flat spots, in a diffferent location. These flat spots are a problem too; especialy if you are parked on ice. When this happens, you wish you had tracks like a dozer instead of tires, because the back wheels womp out trying to get over the flat spot of the front tires creating a zero ground speed. The facts are real, only the circumstances are changed. Correct?

I'm sure your probably tired of reponses, but it's an interesting question. Here's how I see it.

In order for a car to travel 100 miles per hour, the tires must be rotating at a specific rpm. Given a tire of radius r, the distance travled due to one rotation of the tire is defined as 2πr. So Car Velocity/2πr should give the number of rotations per time.

So lets say the tire radius is 1 ft. Then one tire rotation = 6.18 ft ~ 6 ft. 100 mph is equal to 528000 ftph. So 528000 ftph / 6 ft = 88,000 rph. So a tire has to rotate 88,000 times per hour in order for a car to travel 100 mph, or 24.4 rotations per second. 1 rotation is 360 degress. So the tire rotates 8784 degrees per second (48.8π/s=153.2 1/sec), this is the angular velocity of the rubber molecules in the tire. The linear velocity (tangential velocity) of the rubber molecules "where the rubber meets the road" are equal to;

radius of the tire x the angular velocity of the tire =

153.2 1/sec * (1 ft.) = 153.2 ft/sec = 104.45 mph (this should be 100 mph, but I cut corners and approximated)

The rubber molecules "where the rubber meets the road" are traveling 100 mph relative to the road, in the opposite direction as the car due to the rotational motion of the tire and 100 mph in the direction the car is traveling due to linear motion of the tire resulting in a total speed of 0 (after you add the vectors up).

It's important to note that this is only true "where the rubber meets the road" since in all the other parts of the tire, the vectors don't cancel out.

Didn't I say basically the same thing in my earlier posts, #26 and #29, without the unnecessary angular velocity calculations?

Yes, you were right, just like the guy who said that it depends on your reference frame was correct. If you're in the car, the rubber molecule has speed of -100 mph with respect to you. I like doing the calculation because it gives me insight into the whole situation. For instance, at the top of the tire, it's speed is 200 mph, which means the rubber molecule on the surface of the tire is accellerated from 0 mph to 200 mph to 0 mph every rotation. I find that pretty amazing.