Interesting fact about a wheel.

[onearmedbandit]

I'd expect as much - I'm just not familiar with the ratings (are they the same).

Here you go.

[Badjelly]

At the top of the wheel yes - not at the bottom where they contact the ground.

Are you suggesting that the centrifugal force(*) that is tending to increase the radius of the tyre acts only at the top of the wheel, not the bottom? Or just that the weight on the wheel would prevent the radius of the tyre from increasing?

If the former, then, sorry, you're wrong.

If the latter, then I think this may also be wrong. However the forces acting on the tyre at the contact patch are pretty complicated so I don't want to stick my neck out on this one.

Not that I'm suggesting that tyres do grow much due to centrifugal force. Those steel belts are pretty strong.

(*) Yes, I know centrifugal force is a virtual force, i.e. it does not exist in an inertial frame of reference. However it's a handy concept in a rotating frame of reference.

[Badjelly]

How does the mass of the wheel feature into all these calculations?

Into calculations on the centrifugal force acting on the tyre? Not at all.

Into calculations on the gyroscopic effect of the wheels? Yes.

[Mikkel]

From my understanding these cars run at something like 8psi when staging but due to the burnout and the run the pressure goes up substantially. Something to do with the heat generated and the expansion of the air due to the heat.

Indeed - a gas will expand when heated. If the gas cannot expand then the pressure will increase. Likewise, if you expand a gas the temperature will drop - if you've been diving you will notice that the expansion valve (the 1st stage) gets rather cold after 30 mins of depressurising 200 atm down to 50 (or whatever the 1st stage does)...

A good approximation is the Ideal Gas Equation.

As to this thread.

A question.

How does the mass of the wheel feature into all these calculations?

From personal experiencve of using non factory wheels on one of my bikes I found that the acceleration, deceleration and handling of the bike dramatically increased due to a reduction of the inertial mass of the wheels.

The mass of the wheel doesn't feature unless you start talking about changing the motion. Mass is inertia - mass and it's distribution around an axis is rotational inertia. The higher the inertia - the more effort it takes to change the motion... For the steady state that is irrelevant.

Acceleration (negative or positive) and handling (turning) changes the magnitude and the direction respectively of the motion - as such the rotational inertia is important. (Less is more )

[GSVR]

The path any point on a wheel except its center travels when the bike is moving is a cycloid.

The dissplacement forward for a point on the wheel in the bottom half of the wheels revolution is half the circumference minus the diameter. And for the top half it is half the circumference plus the diameter.

Here are the figures for the periphery of a 0.6 metre diameter wheel (17 inch motorcycle wheel)

3.142 times 0.6 equals 1.8852 divided by 2 equals 0.9426 meters (half the circumference)

0.9426 minus 0.6 equals 0.3426 metres ( forward displacement of that point for 1/2 revolution in bottom half)

0.9426 plus 0.6 equals 1.5426 metres ( forward displacement of that point for 1/2 revolution in top half)


The rate of change in displacement with respect to time is velocity

Since the point takes the same time for each displacement the average velocity for each half is very different.

eg. If it took 0.068 seconds for the wheel to do 1 revolution. ie. That is 0.034 for half a revolution then:
0.068 seconds is the approximate time for 1 revolution at 100 kph.

1 second divided by 0.034 equals 29.41176 multiplied by 1.543 metres equals 45.3823 metres per second average velocity
and
1 second divided by 0.034 equals 29.41176 multiplied by 0.343 metres equals 10.08825 metres per second average velocity


Momentum equals mass times velocity.
Introduce a figure for mass and its clear that the top half of the wheel has much more forward momentum.
What are the implications of this? Well the greater the momentum the more the mass wants to stay on its current path.

Speed and velocity are qiute different things so heres and explaination:
http://regentsprep.org/Regents/physi...ty/default.htm

[Mikkel]

Are you suggesting that the centrifugal force(*) that is tending to increase the radius of the tyre acts only at the top of the wheel, not the bottom? Or just that the weight on the wheel would prevent the radius of the tyre from increasing?

I didn't imply the former. The latter is very likely to be the case - I'm not saying it doesn't increase, just that the magnitude of the increase would be much less than at the top of the wheel.

But yes, it would be extremely complicated to make any quantitative observations on this matter. Indeed it would require a proper numerical simulation since any analytical treatment, if at all possible, would be based on numerous assumptions.

(*) Yes, I know centrifugal force is a virtual force, i.e. it does not exist in an inertial frame of reference. However it's a handy concept in a rotating frame of reference.

I touched upon this subject in another thread a couple of hours ago. Indeed - the centrifugal force is only present in a rotational FoR.

[Mikkel]

The path any point on a wheel except its center travels when the bike is moving is a cycloid.

...

Momentum equals mass times velocity.
Introduce a figure for mass and its clear that the top half of the wheel has much more forward momentum.
What are the implications of this? Well the greater the momentum the more the mass wants to stay on its current path.

I fail to see your point with this - and I fail to see why you didn't post this in the thread that you started and is still going strong...

Speed and velocity are qiute different things so heres and explaination:
http://regentsprep.org/Regents/physi...ty/default.htm

As I said... Velocity is the vector, speed is the magnitude.

Anyway, worthwhile mentioning in relation to that link is that by going to infinitesimal difference you get the current speed and velocity instead of the average over the time (delta t).

[GSVR]

I fail to see your point with this - and I fail to see why you didn't post this in the thread that you started and is still going strong...

So you agree with what I'm saying here? That the top half of the wheel carrys almost 4.5 times the momentum of the bottom half at any speed.

[Badjelly]

Introduce a figure for mass and its clear that the top half of the wheel has much more forward momentum.
What are the implications of this? Well the greater the momentum the more the mass wants to stay on its current path.

Your last statement might mean something, but only if you make it more precise.

A body of mass m subject to a force F will undergo an acceleration of a = F/m, irrespective of how fast it's already going.

If a body is moving in a straight line at speed s and you then subject it to a sideways force F, then the rate at which the direction of motion changes will be ... well I can't be bothered doing the maths, but it will become smaller as s become larger. So, yes, you might say the mass will "want to stay on its current path" more. However the word "want" does not really belong in discussions about mechanics.

As I've said before, I'm curious about where you're taking this.

[Mikkel]

So you agree with what I'm saying here? That the top half of the wheel carrys almost 4.5 times the momentum of the bottom half at any speed.

Not at all... I'm saying that I can't see the relevance of starting a new thread to speculate on something like that.

Besides momentum is measured at the centre of mass... As such the momentum of a wheel travelling along a flat surface is given by:

P = v*M, where the speed v = 2*r*Pi*f and M is the mass of the wheel.

Additionally it'll have a rotational momentum of:

L = f*I - where I is the rotational inertia given by M*r^2 times a fraction that expresses the distribution of the mass around the axis of rotation.

So, no the top half of the wheel doesn't carry more momentum than the lower half - it's a rigid system.

[GSVR]

So, no the top half of the wheel doesn't carry more momentum than the lower half - it's a rigid system.

Can you show me where I have gone wrong? I know what your saying but I'm not looking at the wheel a a whole I'm looking at just the top then just the bottom. Assume in each case the other half doesn't exist.

Only measuring positive and negative dissplacement relative to the direction of the the wheel/bike

Another thing the other thread is all about rotational forces and this one is ignoring them.

[Badjelly]

So, no the top half of the wheel doesn't carry more momentum than the lower half - it's a rigid system.

I can't agree with you here, Mikkel. I think its undeniable that, in a stationary frame of reference, the top half of a rotating wheel has more forwards momentum than the bottom half. To prove this, just integrate the product of mass and velocity over each half.

As you say, it's a rigid system, so pretty quickly the top half will become the bottom half (for what that's worth). And I'm curious, as you are, about what implications GSVR thinks this difference in momentum has.

But, does the top half have more forward momentum than the bottom at any instant? Yes. Can we move on?

[Mikkel]

Can you show me where I have gone wrong? I know what your saying but I'm not looking at the wheel a a whole I'm looking at just the top then just the bottom. Assume in each case the other half doesn't exist.

Only measuring positive and negative dissplacement relative to the direction of the the wheel/bike

Another thing the other thread is all about rotational forces and this one is ignoring them.

That's part of the problem. Quite simply you can't...

It's a rigid body - as such you can not decouple it.

And what you hope to achieve by doing so, I fail to see.

[GSVR]

But, does the top half have more forward momentum than the bottom at any instant? Yes. Can we move on?

Cool you just said it. Now which instant are you talking about becuase every point in time is an instant in time. So you are actually saying all the time.

[GSVR]

That's part of the problem. Quite simply you can't...

It's a rigid body - as such you can not decouple it.

And what you hope to achieve by doing so, I fail to see.

I would say if you had another identical wheel rotating at the same rpm in the opposite direction then the momentum of the top and bottom halves would be the equal.