Interesting fact about a wheel.

[GSVR]

In one of my other threads I calculated that at 100 kph a 600mm diameter motorcycle tyre is doing approximately 884 rpm

When you put these numbers into the links below you will find the centrifugal force is huge.

http://www.msu.edu/~venkata1/gforce.htm

http://www.centrifuge.jp/cgi-bin/calc-e.cgi

Even with a reduced diameter to say the inside of the rim it would still be a large number.

[Badjelly]

Perhaps you really *should* be worried about your wheels!

[Steam]

For a radius of 300 mm and a speed of 885 rpm (100kph)
The G-force is 263.16 g units! Feck!

Apparently the relationship is non-linear, as when I put in 300kph or 2600 turns of the wheel per minute, I get a result of 2368g's. HOLY SPIT!
That's a lot of G's. I wonder at what point a wheel and tyre would come apart?
No wonder tyres have to be rated for certain speeds. The forces are huge.

[Finn]

In one of my other threads I calculated that at 100 kph a 600mm diameter motorcycle tyre is doing approximately 884 rpm

Hmmmm, maybe we should attach our engines directly to the wheel and get that sucker turning 14,000 rpm's.

[Patch]

Hmmmm, maybe we should attach our engines directly to the wheel and get that sucker turning 14,000 rpm's.
They already do - its called a propellor . . and it has wings :gob:

[GSVR]

This explains why a wheel is so hard to turn at higher speeds. But the faster you go the less you need to turn it so its also a good thing.

55rpm is the speed for 1g.

55 rpm times 1.884 metres (circumference) equals 103.62 metres per minute or 6217.2 metres per hour (6.2 kph)

Calculations

D = Tyre outside diameter (600mm for example)
Pi = 3.142 approx
Speed of motorcycle = 100kph

First lets get the distance travelled by the tyre in one revolution. Its the circumference which is Pi times the Diameter. All distance measurements will be in metres so 0.6 times 3.142 equals 1.8852 metres

With one revolution of the tyre the motorcycle travels 1.8852 metres.

If the motorcycle is travelling at 100kph what does this equal in metres per second?
100 kilometres = 100 times 1000 metres = 100000 metres
1 hour = 60 minutes = 3600 seconds
100000divided by 60 = 1666.666 mtres per minute (will use this to show RPM of tyre)
100000 divided by 3600 = 27.7777

At 100kph the motorcycle is doing 27.78 metres/sec.

If we divide the distance for one revolution of the tyre into the distance it covers in 1 minute we will get the RPM at 100kph.
1666.66 divided by 1.8852 = 884.08

Tyre is revolving at 884 RPM when the bike is travelling at 100kph

[Mikkel]

For a radius of 300 mm and a speed of 885 rpm (100kph)
The G-force is 263.16 g units! Feck!

Apparently the relationship is non-linear, as when I put in 300kph or 2600 turns of the wheel per minute, I get a result of 2368g's. HOLY SPIT!
That's a lot of G's. I wonder at what point a wheel and tyre would come apart?
No wonder tyres have to be rated for certain speeds. The forces are huge.

The acceleration, a, in a circular motion with a given speed, v, and radius, r, is given by:

a=v^2/r - so yes it's not linear, it's a potential relationship.

Don't underestimate the size of the wheel though - has a big impact!

As long as the wheel is balanced you shouldn't get any trouble with it coming apart. However, balancing something so that it's capable of doing e.g. 120,000 RPM requires very high precision. Take a turbine in a turbo - they typically operate at 100,000 - 200,000 RPM.

For car tyres there are speed ratings (http://en.wikipedia.org/wiki/Tire_code#Speed_rating_codes) which indicate how high a speed they are certified to work at. E.g. V up to 240 km/h and Z over 240 km/h.

[Mikkel]

They already do - its called a propellor . . and it has wings :gob:

The propellers on an airplane revolve at rather less than 14,000 RPM.

Propeller (http://en.wikipedia.org/wiki/Propeller#Aviation)

Turboprop (http://en.wikipedia.org/wiki/Turboprop)

[onearmedbandit]

For car tyres there are speed ratings (http://en.wikipedia.org/wiki/Tire_code#Speed_rating_codes) which indicate how high a speed they are certified to work at. E.g. V up to 240 km/h and Z over 240 km/h.

Bike tyres have speed ratings as well.

[yod]

this thread is far too relevant and informative

can someone PD it please?

[pritch]

Years ago I read that you should always ride with valve caps fitted as the centifugal forces generated at high speed can be enough to unseat the valve and deflate the tyre. Which event could create an "interesting" riding experience.

If the figures quoted here are correct, and I have no intention of checking them, it's is easy to see how such a thing might happen.

And I will be check the caps are tight before I head off next :scooter:

[Mikkel]

Bike tyres have speed ratings as well.

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


Years ago I read that you should always ride with valve caps fitted as the centifugal forces generated at high speed can be enough to unseat the valve and deflate the tyre. Which event could create an "interesting" riding experience.

If the figures quoted here are correct, and I have no intention of checking them, it's is easy to see how such a thing might happen.

And I will be check the caps are tight before I head off next :scooter:

I'd say you'd have to go fucking fast for that to happen - but indeed, why take a chance? Send Mythbusters an email - would be interesting to see. I guess they might take a motorcycle wheel and indeed rotate it at 14,000 RPM with the cap off to see if it deflates... ;)


Different issue effects maxium prop speed at the is the speed of sound. Exceed that at the tips and you have a problem.

Indeed - the aerodynamics when approaching the sound barrier becomes quite complicated.
However, it's not just the rotational speed of the propeller - it's the resultant speed, vR, of the planes airspeed, vP, and the tangential (rotational) speed, vT, that can't exceed the speed of sound.

vR^2 = vP^2 + vT^2 < (344 m/s)^2

[johan]

Any ideas how much the radius of the tyre would increase at 100kph due to the gforce pulling it out?

Check out the rear tires on these guys. The radius is increasing like 40 cm when the tires start to rotate.
http://www.youtube.com/watch?v=Ls_obl_fokc&feature=related

[Mikkel]

Any ideas how much the radius of the tyre would increase at 100kph due to the gforce pulling it out?

At the top of the wheel yes - not at the bottom where they contact the ground. Besides IIRC Top Fuel dragsters run a fairly low pressure on the rear tyres...

Doubt it would be even measurable on a motorcycle even at speeds exceeding 250 km/h.

[Mr Merde]

At the top of the wheel yes - not at the bottom where they contact the ground. Besides IIRC Top Fuel dragsters run a fairly low pressure on the rear tyres...

Doubt it would be even measurable on a motorcycle even at speeds exceeding 250 km/h.

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.

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.

[onearmedbandit]

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





Here (http://www.webbikeworld.com/Motorcycle-tires/tire-data.htm) 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 (http://en.wikipedia.org/wiki/Ideal_gas#Classical_thermodynamic_ideal_gas).



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/physics/phys01/velocity/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 (http://en.wikipedia.org/wiki/Frame_of_reference).

[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/physics/phys01/velocity/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.

[Mikkel]

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.

:eek: :crazy:

Are we talking about momentum or angular momentum?

Bottom halves, top halves, that's the part that doesn't make sense. It's a wheel for christ's sake... Not two hemi wheels or something else, it's a wheel and should be treated as such.

Unless of course you're not interested in the mechanics behind the motion...

[GSVR]

:eek: :crazy:

Are we talking about momentum or angular momentum?

Bottom halves, top halves, that's the part that doesn't make sense. It's a wheel for christ's sake... Not two hemi wheels or something else, it's a wheel and should be treated as such.

Unless of course you're not interested in the mechanics behind the motion...

I am ignoring any vertical displacement as these are equal in this case.
So out of any angular value your only taking the horizontal component. The same direction as the bike in negative or positive values but all positve after combined with the forward displacement.

[Mikkel]

I am ignoring any vertical displacement as these are equal in this case.
So out of any angular value your only taking the horizontal component. The same direction as the bike in negative or positive values but all positve after combined with the forward displacement.

Que?

Just tell us what you're trying to achieve and stop trying to make sense, because it doesn't work.

[GSVR]

Que?

Just tell us what you're trying to achieve and stop trying to make sense, because it doesn't work.

Experiment:

Lets set up a 17 inch wheel in a cradle that can pivot the wheel horizontally about and axis that goes through the centreline of the axle from front to back of the wheel maybe ever so slightly above center.
Place this in a train moving at 100 kph and spin the wheel at 884 rpm. Now move the entire cradle from side to side in relation to the forward movement of the train . What happens?

[Mikkel]

If you only move the cradle from side to side - the wheel will sit rock solid in the cradle not tilting or changing it's direction of rotation.

[Finn]

Think we should all delete this as its a bit off topic!

Dude, your bike is off topic.

[marty]

so the 250-odd g's, causing substrantail centrifugal stability, would react in what fashion when i input a force to the axle 90 degrees before i want the reaction to happen?

[Patch]

:stupid: Since when did KB ever start being on topic with anything?? :stoogie:


stop being a farkin whinger otherwise we'll shrinkwrap ya, and send ya down to Dunedin with all the rest of 'em

[Steam]

...send ya down to Dunedin with all the rest of 'em

Eh wot?? Bastidge!

[Steam]

EDIT: OOps, posted in wrong thread.

Look, GSVR, the whole wheel has the same momentum, it just looks like the top half has more momentum when you do the calculations as you would with the ground as your frame of reference. This cannot be done.
The calculation must be done with the axle as the frame of reference, then the thing becomes clear - the whole wheel has the momentum, not just one part of it.

[Steam]

What are you on about? Are you sure your posting in the right thread!

OOps!:2thumbsup
You are right, wrong thread.

[Mikkel]

Can you delete all of it then as otherwise this will end up in senseless drivel

Why bother with deleting it. I doubt anyone is going to find this thread "prune"-worthy.

Either it dies from old age or gets relocated to PD.

[GSVR]

If you only move the cradle from side to side - the wheel will sit rock solid in the cradle not tilting or changing it's direction of rotation.

Made a drawing. The sideways movements will be reasonably forcefull over say 1 metre.

[Badjelly]

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

Yes, all the time. But it might be worth remembering that the "top half" and "bottom half" of the wheel that we're talking about at one instant will not be the top half and bottom half a short time later. Then again it might not be worth remembering: it depends (again) on where you're going to go with this argument.

[Mikkel]

Made a drawing. The sideways movements will be reasonably forcefull over say 1 metre.


If you only move the cradle from side to side - the wheel will sit rock solid in the cradle not tilting or changing it's direction of rotation.

There's your answer.

[Badjelly]

Made a drawing.

Why is the curved blue line with arrows on the end in the right-hand panel curved? This suggests the wheel is tilting as it moves from side to side, which is quite an important detail.

[GSVR]

Why is the curved blue line with arrows on the end in the right-hand panel curved? This suggests the wheel is tilting as it moves from side to side, which is quite an important detail.

Its showing that it can rotate about that axis. Could do the full revolution if it wanted.

[GSVR]

There's your answer.
And if I started it spinning in the cradle as well as the wheel spinning?

[Mikkel]

And if I started it spinning in the cradle as well as the wheel spinning?

That's quite different. Now you're talking about applying a torque that'll try to change the direction of the axis...

Around which axis do you want to spin the cradle? The vertical?

If so you'll find that the cradle won't spin - but the wheel will tilt.

See this video:
http://www.youtube.com/watch?v=dCcfKBfmyP4

[Badjelly]

The rotation is important. If you get a spinning wheel (doesn't matter how fast) and push it around without changing the direction of the axis of rotation then it will behave just like a body of the same mass that's not spinning. As soon as you try to change the direction of the axis of rotation (or the rotation rate) then you bring angular momentum effects into play. These can be very strong and they work in surprising directions, as anyone who has played with a gyroscope will know.

[GSVR]

The rotation is important. If you get a spinning wheel (doesn't matter how fast) and push it around without changing the direction of the axis of rotation then it will behave just like a body of the same mass that's not spinning. As soon as you try to change the direction of the axis of rotation (or the rotation rate) then you bring angular momentum effects into play. These can be very strong and they work in surprising directions, as anyone who was played with a gyroscope will know.

Yes the axis it can rotate in train example is the same as steering a bike except horizontal instead of vertical. And no rake or trail.

[Mikkel]

Honesty I get tired of people that post youtube video as I can't see it on dailup.

Patience is a virtue!

Honestly, I think the mods get tired of people quoting embedded media as they have to correct such mistakes.


Yes the axis it can rotate in train example is the same as steering a bike except horizontal instead of vertical. And no rake or trail.

Honestly, I get tired of trying to explain how classical mechanics work as it seems people either already know or don't want to listen anyway.

What Badjelly said is correct.

If you know vector calculus (http://en.wikipedia.org/wiki/Vector_calculus) wikipedia will be able to help you:

Classical Mechanics (http://en.wikipedia.org/wiki/Classical_mechanics)

Angular momentum (http://en.wikipedia.org/wiki/Angular_momentum)

Frame of reference (http://en.wikipedia.org/wiki/Frame_of_reference)

Cheers
Mikkel

Case in point:

I also don't need to see static gyroscopes if the thing was flying trough space at several thousand kph that would be interesting.

See FoR...

[GSVR]

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?

I say that a static spinning wheel behaves differently to one that is moving. ie the center of rotation is moving. In my case its moving at 100 kph same as the surface speed of the tyre.

[Badjelly]

OK, train moving at 100 km/h on tracks, cradle on train with wheels that let it move sideways (ie perpendicular to train tracks), wheel on crade mounted so it can tilt (still not 100% what directions it can tilt in). What now?

[GSVR]

OK, train moving at 100 km/h on tracks, cradle on train with wheels that let it move sideways (ie perpendicular to train tracks), wheel on crade mounted so it can tilt (still not 100% what directions it can tilt in). What now?

It can revolve so the bottom becomes the top and the top the bottom. Which effectively reverses the direction of rotation. But the axle is alway perpendicular to the direction of travel of the train carraige.

[Badjelly]

I say that a static spinning wheel behaves differently to one that is moving. ie the center of rotation is moving. In my case its moving at 100 kph same as the surface speed of the tyre.

I think that's probably wrong, but before I can say for sure, I need to know what you mean by "behaves differently".

Einstein used to explain some of his arguments as "thought experiments". Here's a thought experiment: you're in a box that is either stationary or moving at constant velocity, you don't know which. (Think of it as an aeroplane, if you wish.) Can you, by doing experiments with (say) a spinning bicycle wheel, tell whether you're moving and how fast? The answer is no.

[Badjelly]

It can revolve so the bottom becomes the top and the top the bottom. Which effectively reverses the direction of rotation. But the axle is alway perpendicular to the direction of travel of the train carraige.

OK, I think I've got that.

[Mikkel]

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?

Sorry, I forgot that one...

Yes true - the upper part has a forward momentum of X and the lower half has backward momentum of X. They cancel out and there is no momentum due to the rotation of the mass.
If the centre of mass moves - then there's a momentum equal to the product of the speed and the mass.

What you have for a stationary rotating wheel is an angular momentum equal to the product of the angular frequency and the rotational inertia.

My point is - if you want to consider a wheel from a mechanical perspective it makes no sense to treat the upper and the lower part of the wheel as two seperate entities. It's ONE wheel and should be treated as such.

[GSVR]

OK, I think I've got that.
I'll turn this around a bit and say which direction would the wheel be the most stable. Turning like it would going down the road or in reverse.

Or maybe it doesnt matter. ( By saying this I'm going against what I believe to be true)

[Mikkel]

I'll turn this around a bit and say which direction would the wheel be the most stable. Turning like it would going down the road or in reverse.

Or maybe it doesnt matter.

The fact that the train is moving doesn't matter as long as it's a steady motion(constant velocity).

The direction the wheel turns does not impact on it's stability - only the direction of the angular momentum.

Have you read anything posted in this thread or followed any of the links that has been provided?

[Badjelly]

My point is - if you want to consider a wheel from a mechanical perspective it makes no sense to treat the upper and the lower part of the wheel as two seperate entities. It's ONE wheel and should be treated as such.

Agreed. I only said you can calculate the momentum vectors of the top and bottom halves separately. I didn't say you that doing so was a useful step towards understanding the mechanics. The mechanics of a spinning wheel are best understood in terms of the concepts you have been suggesting: mass, centre of mass, linear momentum, moment of inertia, angular momentum, etc.

If GSVR is wrong about something, let's at least be clear about where that is. That way we can all learn.

[GSVR]

The direction the wheel turns does not impact on it's stability - only the magnitude of the angular momentum.

I think I'm making real progress. What impact does it have on the angular momentum?

[Badjelly]

I'll turn this around a bit and say which direction would the wheel be the most stable. Turning like it would going down the road or in reverse. Or maybe it doesnt matter. ( By saying this I'm going against what I believe to be true)

My answer: it doesn't matter. (Bearing in mind that I'm not 100% sure what you mean by "stable".)

Like I said 1 or 2 posts ago, put walls around your train carriage so you can't see the scenery going past and there's no way you can tell from the behaviour of the spinning wheel (or any other mechanical object inside the carriage) which way the train is moving.

Which direction of rotation do you think would make the wheel more stable and why?

[Badjelly]

The direction the wheel turns does not impact on it's stability - only the magnitude of the angular momentum.

A slip of the keyboard? The direction the wheel turns does not impact on the magnitude of the angular momentum, only on the direction of the angular momentum vector.

[Mikkel]

I think I'm making real progress. What impact does it have on the angular momentum?


A slip of the keyboard? The direction the wheel turns does not impact on the magnitude of the angular momentum, only on the direction of the angular momentum vector.

Yes, of course. Silly me - the magnitude is affected by the rate of rotation, not the direction.

And regarding the direction - change the direction of rotation and you "flip" the vector representing the angular momentum.

[GSVR]

My answer: it doesn't matter. (Bearing in mind that I'm not 100% sure what you mean by "stable".)

Like I said 1 or 2 posts ago, put walls around your train carriage so you can't see the scenery going past and there's no way you can tell from the behaviour of the spinning wheel (or any other mechanical object inside the carriage) which way the train is moving.

Which direction of rotation do you think would make the wheel more stable and why?

I would have thought that if the wheel was rotating in the same direction as the train is going. ie same way as the train wheels the top would be more resistant to direction changes than the bottom half as its got more momentum.

The effect would be the same no matter which way the wheel rotated but it would be different to if it wasn't spinning at all.

[Mikkel]

I would have thought that if the wheel was rotating in the same direction as the train is going. ie same way as the train wheels the top would be more resistant to direction changes than the bottom half as its got more momentum.

Mate, seriously - stop asking more questions and read the Wikipedia page on Frame of Reference it should help to clear out that misconception that the fact the train is moving has any impact at all upon how the wheel behaves.


The effect would be the same no matter which way the wheel rotated but it would be different to if it wasn't spinning at all.

Which effect are you thinking about here.

The resistance to direction changes are derived from the angular momentum of the wheel - not that it is moving one way or the other, but that it's rotating. This is a gyroscopic effect. All gyroscopic forces are proportional to the angular frequency - as such there are no gyroscopic forces if the wheel is not turning.

[GSVR]

The top half has its spinning momentum(positive) plus the trains forward velocity. The bottom half has the spinning momentum(negative) plus the trains forward velocity.

For a wheel spinning same way as the train wheels.

[Mikkel]

The top half has its spinning momentum(positive) plus the trains forward velocity. The bottom half has the spinning momentum(negative) plus the trains forward velocity.

For a wheel spinning same way as the train wheels.

Admit it - you're just trolling. Nothing else!

[GSVR]

Admit it - you're just trolling. Nothing else!


Mikkel I know your a smart guy. But I really thought you might be able to explain yourself without refering me to youtube or Wickypedia who require an expert to fix their page.

"This article or section is in need of attention from an expert on the subject." Well they aren't getting me lol.

[Badjelly]

The top half has its spinning momentum(positive) plus the trains forward velocity. The bottom half has the spinning momentum(negative) plus the trains forward velocity.

For a wheel spinning same way as the train wheels.

I'm (finally) with Mikkel on this. To understand the mechanics of rotating objects you can work it out for yourself or you can try to understand what other people have found out in the last 300 years. Your line of reasoning is not going anywhere useful as far as I can see. The fact that the equations of motion are the same in any inertial frame of reference (ie one moving at a constant speed, eg our train carriage) is pretty basic and if you have reasoning that leads to a different conclusion, then your reasoning is probably wrong. (Sometimes it can be useful to work out why. Sometimes not.) Take it from someone who has made plenty of mistakes himself.

[Badjelly]

Mikkel I know your a smart guy. But I really thought you might be able to explain yourself without refering me to youtube or Wikipedia who require an expert to fix their page.

Wikipedia is a marvellous resource, not perfect of course, and if something is explained there, why reinvent the wheel. (Yes, that's a pun.)

[Badjelly]

Admit it - you're just trolling. Nothing else!

Most trolls don't put in quite as much work as GSVR has. He may be a bit stubborn, but I don't think you can call him a troll.

[GSVR]

Wikipedia is a marvellous resource, not perfect of course, and if something is explained there, why reinvent the wheel. (Yes, that's a pun.)

When I was at intermediate school many years ago a teacher was explaining electricity to the class. His theory was that cables had many strands to carry more power as electricity only moved over the surface of the wire.

Go figure

I'm actually not concerned about any gyroscopic coupling here its something that doesn't come into this equation. All its about is measuring momentum in one direction.

Hey got to go thanks to Mikkel and Badjelly its be good reading your posts and made me think. The decoy thread seemed to work well too.

[Mikkel]

Mikkel I know your a smart guy. But I really thought you might be able to explain yourself without refering me to youtube or Wickypedia who require an expert to fix their page.

Being a smart guy I have no intentions of writing a book on a subject that is already well covered. I am sorry if you feel that youtube and wikipedia are not good enough to be used as demonstration tool and textbook reference respectively. However, my own mechanics book is in danish and scanning and posting stuff from that won't help you. I can't show you the demonstrations online since I haven't got a webcamera. I'm truly sorry if you feel I can't help you with the resources that are available.

If I was sitting next to you making sketches and explaining the maths would be easy. Seeing as I actually did teach classes of approximate 25 engineering students at my old university for two years while studying to supplement my scholarship I have a reasonable confidence in my ability to communicate this subject. What I may not be confident about is whether your background is sufficient to deal with the subject at the level I am prepared to teach.

On the other hand if you're too lazy or too impatient to peruse the resources I have provided you with - then I can only say that it's extremely arrogant wanting to understand without wanting to learn. And learning requires study.


"This article or section is in need of attention from an expert on the subject." Well they aren't getting me lol.

It's classical mechanics. It's well understood. If there was anything directly wrong in the page it would have been corrected a long time ago. Classical mechanics is not a subject which is under contention.

[Patar]

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.

You're point of view is from a stationary position on the ground which is wrong, at least if you are considering the effect it has on the bike. Imagine the bike to be floating in mid air and the wheels are spinning at an rpm equal to 100kph, it is easy to see that both the top and bottom of the wheel carry equal momentum as relates to the bike.

If you get hit by the wheel of a bike then yes the top of the wheel would be carrying more momentum than the bottom, but i fail to see how that would be relevant as there would be bigger things to worry about.

[Mikkel]

When I was at intermediate school many years ago a teacher was explaining electricity to the class. His theory was that cables had many strands to carry more power as electricity only moved over the surface of the wire.

I'll make no excuse for teachers who don't understand the rudiments of their subjects!


I'm actually not concerned about any gyroscopic coupling here its something that doesn't come into this equation. All its about is measuring momentum in one direction.

If you want to know the momentum the rotation is irrelevant. You want the momentum of the wheel in the setup on the train?
Easy - it's just the product of the mass and the velocity of the centre of mass (in this case it's 100 km/h when seen from an inertial FoR).
So if the wheel weighs 10 kgs - the momentum is 277.8 kg*m/s or 277.8 N*s.

[Badjelly]

I'm actually not concerned about any gyroscopic coupling here its something that doesn't come into this equation.

Maybe not, but gyroscopic effects do come into play when you start changing the direction of the wheel's axis of rotation, or when you talk about the stability of the wheel to such changes.

I think that if you followed down the route of considering the difference in linear momentum between the top and bottom of the wheel, and its effect on the wheel's dynamics, you could eventually reinvent the concept of angular momentum. Better to accept that this concept was invented 200-odd years ago and has been developed and tested since then.

[GSVR]

I do actually agree with pretty much all that you guys have said.

There must some flaw in what I've said and calculated. Saying you can't look at two halves in isolation doesn't actually work for me becuase I already have.

Yes its so easy just to get the wheels mass and multiply it by the velocity. Breaking it down further may be pointless but it does demonstrate that although something appears to make perfect sense on paper in reality it could be totally wrong.

Liked the Einstein example Badjelly but when your in a train don't you find it slightly easier to sway from side to side if your looking out the window as opposed to down the corridor? To accelerate from side to side might be a better term to use

[GSVR]

Ok heres a final train carriage example.

All the figures are the same as the last one just the pivot point has changed. The pivot passes through where the contact patch would be on the road if the wheel was on the motorcycle. I'd better also say the the red pivot points are bolted solid to the carriage deck.

Two experiments:

One with the wheel revolving the direction indicated in the drawing.
The other with the wheel revolving anticlockwise.

Does one require more work. (moving the mass though the same angular movement in the same amount of time) than the other.

[Badjelly]

Two experiments:

One with the wheel revolving the direction indicated in the drawing.
The other with the wheel revolving anticlockwise.

Does one require more work. (moving the mass though the same angular movement in the same amount of time) than the other.

Actually, neither requires any work at all, using "work" in its technical sense. There are 2 equivalent ways of looking at work. One is that work done on a system changes its energy. If I take a spinning wheel and change the direction of the axle, the wheel does not spin faster or slower, so no work has been done. The other way of looking at work is that it is equal to the force applied multiplied by the distance moved in the direction of the force. If you take a spinning wheel, grab the axle and change the orientation of the axle, there will be forces generated due to the rotation of the wheel, but they will always be at right angles to the movemement. So again, no work. (We had to get the same answer using both ways of defining work, because they are equivalent.)

Try this:

http://en.wikipedia.org/wiki/Mechanical_work
http://en.wikipedia.org/wiki/Gyroscope

Furthermore, whatever question you are asking about the behaviour of your hypothetical wheel in your hypothetical train carriage moving at constant velocity, the answer doesn't depend on the speed or direction of movement of the carriage. Really.

[GSVR]

Maybe I choose the wrong word there. Effort?

Lets just say that the angular side to side travel is limited to 30 degrees against solid stops. The assemby is resting on one stop and a constant force applied till it hits the other stop. The time taken for this to happen is accurately measured.

Then the direction of the wheel is reversed and the same experiment done.

Will these times be exactly the same or will they be different?

[Badjelly]

Maybe I choose the wrong word there. Effort?

Lets just say that the angular side to side travel is limited to 30 degrees against solid stops. The assemby is resting on one stop and a constant force applied till it hits the other stop. The time taken for this to happen is accurately measured.

Then the direction of the wheel is reversed and the same experiment done.

Will these times be exactly the same or will they be different?

If I understand you correctly, exactly the same, as the only thing you change between the 2 experiments is the direction of rotation and there's nothing in the experiment that would be affected by that.

You might like to read this again:


Furthermore, whatever question you are asking about the behaviour of your hypothetical wheel in your hypothetical train carriage moving at constant velocity, the answer doesn't depend on the speed or direction of movement of the carriage. Really.


Really.

[Badjelly]

Furthermore, whatever question you are asking about the behaviour of your hypothetical wheel in your hypothetical train carriage moving at constant velocity, the answer doesn't depend on the speed or direction of movement of the carriage. Really.

To back this up, the Earth is moving through space at some tremendous speed. What effect does this have on your wheel in your train carriage, or on the handling of your motorcycle? The Earth is also rotating (1 rpd) and this does have effects on winds and ocean currents, that I spend most of my working life calculating. But the linear movement has no effect whatsoever.

[GSVR]

Thats great so now we can comfortably say that anyone who reads this thread should get it as its been totally flogged to death by me.

Lets move on to something else now the two threads have been merged.

Going from the calculator in the very first post:

55rpm is the revolutions required for 1g about a 0.6m wheel (0.30m radius)

55 rpm times 1.884 metres (circumference) equals 103.62 metres per minute or 6217.2 metres per hour (6.2 kph)

So at around 6.2 kph the part of the tyre at the top of the tyre is in equalibrium with the 1g of gravity and is infact weightless.
And also the part of the tyre at the bottom is experiencing 2 g of force (gravity plus the centrifugal force)

Comparing this to the 200 and something g at 100kph its seems insignificant.

[scumdog]

I'd say you'd have to go fucking fast for that to happen - but indeed, why take a chance? Send Mythbusters an email - would be interesting to see. I guess they might take a motorcycle wheel and indeed rotate it at 14,000 RPM with the cap off to see if it deflates... ;)

It's true, a Kiwi drag racing team back in the day had problems with the front tyres on their rail deflating with each run, they found out the valve spring pressure was being overcome by the centrifugal forces.

Simply putting on valve caps fixed the problem.':yes:

[GSVR]

I think most people have rolled a hoop or a tyre down the road and at a certain speed it will stay upright by itself.

For a hoop of 0.6m diameter would this be when the hoops speed exceeds 6.2kph or would the actual speed be somewhat lower! I'm pretty sure anything above 6.2 kph maxium effect has taken place and the forces holding it upright will be will the same. But I'm also sure others will say the faster it goes the greater the gyroscopic forces are and these also hold the wheel upright!

Read my last post if you need more info.

[Hitcher]

The Bogan's Guide To Physical Laws. Please make it stop.

[NighthawkNZ]

Interesting fact about a wheel. (http://www.kiwibiker.co.nz/forums/showthread.php?p=1434552#post1434552)
Wheels go wound and wound weelie weelie fast... :baby:

[madandy]

Click on the Yellow text!!!! wound and wound, wound and wound.
Who let the geeks out?

[GSVR]

The Bogan's Guide To Physical Laws. Please make it stop.

I'm dissapointed in you Hitcher. You couldn't help yourself could you. What are you going to do next move it to PD or merge it with another thread?

[Hitcher]

I'm dissapointed in you Hitcher. You couldn't help yourself could you. What are you going to do next move it to PD or merge it with another thread?

A deliciously tempting prospect. I was thinking "Jokes & humour"...

[GSVR]

A deliciously tempting prospect. I was thinking "Jokes & humour"...

Maybe you should quote the bits you find funny and say why? I could do with some more laughs.

[NighthawkNZ]

Click on the Yellow text!!!! wound and wound, wound and wound.
Who let the geeks out?

Some people have no sense of humour... :lol: he gave me a juicy red rep for it... :lol: kewl :woohoo:

[Hitcher]

Maybe you should quote the bits you find funny and say why? I could do with some more laughs.

No. I want this Living Hell to cease.

[Badjelly]

No. I want this Living Hell to cease.

Then why poke the inert corpse?

[avgas]

Hmmmm, maybe we should attach our engines directly to the wheel and get that sucker turning 14,000 rpm's.
haha then a bmw would spoke itself

[avgas]

Click on the Yellow text!!!! wound and wound, wound and wound.
Who let the geeks out?
The troll obviously
quick get the hall monitor to round em up

[GSVR]

Grabbed this out of a real geeky forum but it explains whats going on in my head visualizing the wheel in rotation. American tire = tyre

If we are lucky someone will show up who knows a good online or other textbook chapter on rotation/ang.mom.

But until that happens...

The picture of the tire is an instantaneous one. Pretend you had a camera that could take a picture of the tire just as it rolled past you and that it would show, painted on at each point, a little arrow that is the instantaneous velocity of that point-----at the instant the shutter clicked.

The snapshot of the tire would be all covered with little arrows.

The point touching the road would have a zerolength arrow.

The point at the top of the tire would have a horizontal forwards pointing arrow twice the length of the one at the center of the hubcap

Up the center of the picture would be layer after layer of arrows getting longer the higher off the road, till reaching the top of the tire. Their length would be proportional to height.

Just like on a regular stationary turning wheel the speed of a point is proportional to how far out from the center.

None of the arrows in the snapshot are correct except for that one instant (well, the center of hubcap arrow stays the same but the rest dont)

The very next instant you would need to erase them all and draw the picture over, because a new point would be touching the ground. But the picture would look the same! It will always look,
for just that one instant, like rotation around the point touching the ground.

One can ask, well OK but what is the big deal? why should a physics teacher want me to realize this. But as long as one doesnt get off onto that and just looks at the tire it is not such a hard picture to get. Actually I like it---it helps solve something in an elegant way, doesnt it? Did this discussion help at all?

[avgas]

Did this discussion help at all?
Good point!
Well kind of....but no - as long as tyres dont fly to bits....no one really "pays-for-research".....unfortunately for us the wheels and tyres havent disintegrated for these reasons in a while (80's?).
However i would be interested to know the difference between sitting the wheel weights in the centre circumference in comparison to towards the edge.
Or the effect of carbon defects on carbon wheels at these speeds.

[GSVR]

And heres a little bit more...

To undertsand tyre, move with it. Then in your system of reference the tyre center is not moving (0), the tyre bottom is moving back (i.e. with -v versus you) and the tyre top - forward (with +v). Now return back to non-moving system (=add v to all), and you'll get 0 at bottom, v at center and 2v at top.

(Called addition of velocities due to coordinate transformation from moving to non-moving system).

In non-moving system tyre is seen as rotating around its bottom ( instant axis of rotation constantly moving forward with tyre).

[Badjelly]

Then why poke the inert corpse?

I take it back. I saw the corpse move!

Seriously Hitcher, if you don't like the thread, don't read it. Are you secretly worried there might be something on it that you would find interesting, so you have to keep checking it? If so, don't worry, I assure you there's no chance of that happening.

[Badjelly]

It will always look,
for just that one instant, like rotation around the point touching the ground.
Amazing, eh? (No I am not taking the piss.)

[Hitcher]

Seriously Hitcher, if you don't like the thread, don't read it. Are you secretly worried there might be something on it that you would find interesting, so you have to keep checking it? If so, don't worry, I assure you there's no chance of that happening.

It's like a train wreck. Morbid fascination, if you will. I'm waiting for the Flat Earth Society to arrive and erect a stall.

[GSVR]

Amazing, eh? (No I am not taking the piss.)

Interesting not Amazing.

[GSVR]

It's like a train wreck. Morbid fascination, if you will. I'm waiting for the Flat Earth Society to arrive and erect a stall.

I thought they already had when you showed up!

[Mikkel]

Don't worry - Hitcher's just here to check that we don't fuck up our spelling and grammar too much.
Dunno if he's on the ball as far as mechnics and vectors goes - but hey, he might be checking in on that as well.

Besides, he's at least in the top-5 as far as constructive input in this thread goes.


It's true, a Kiwi drag racing team back in the day had problems with the front tyres on their rail deflating with each run, they found out the valve spring pressure was being overcome by the centrifugal forces.

Simply putting on valve caps fixed the problem.':yes:

Awesome. Cool to hear that it can actually happen. :yes:

Good thing that they have moved on to maglev trains - imagine what could happen on the TGV.

(And yes, that's a joke - sarcasm if you'd like...)

[Bonez]

My wheels went round and round today. They also went sideways, up, down and then they stopped. Wheels are very usefull things.

[avgas]

Good thing that they have moved on to maglev trains
Tell that to the account in Shanghai. Only a few billion US over the proposed budget.....and it never got finished due the fact that the EMF was enough to stop it being build closer than 100m from any buildings.
Shanghai is not known for big back yards either.
It did make me wonder - at what speed and weight do you consider the maglev a threat to derailment though (back to you G-force idea)

[Mikkel]

Tell that to the account in Shanghai. Only a few billion US over the proposed budget.....and it never got finished due the fact that the EMF was enough to stop it being build closer than 100m from any buildings.
Shanghai is not known for big back yards either.

Funny that, I actually have friends who recently went on that maglev train.

And it has cut down the travel time from the Shanghai centre to the airport significantly.

Magnetic fields can be shielded quite well. Besides - if you want to make a maglev train efficient you'd aim to make a strongly curved and very dense magnetic field which wouldn't have much of an outreach effect.


It did make me wonder - at what speed and weight do you consider the maglev a threat to derailment though (back to you G-force idea)

Huh? :crazy:

[avgas]

And it has cut down the travel time from the Shanghai centre to the airport significantly.
Errr people square/garden would be where i put city centre.....and its about 3/4 hour drive (or Subway) from there. Mind due you have to keep into comparison that Shanghai is like 2-3 hours wide (driving/subway) and is infact broken up between Pudong and Pushi (spelt incorrectly so it sounds right) and a few other less minor cities. Anything they can do to improve the infrastructure there has to be good. But unfortunately most Shanghainese never use the Maglev.
Still the new Siemens subway set works good for most people.
And there still is the People's Busses.
Both of which branch out from that old race track they call the Peoples Park

[Hitcher]

Still the new Siemens subway set works good for most people.

One hopes that's not a euphemism for uterus?

And one also suspects that Captain Pugwash would be interested in Siemens trains...

[GSVR]

If the wheel exploded in an instant into lots of little pieces all the same size what directions and speeds would the pieces initially head in?

Lets say the red dots are 1 dollar coins attached at 45 degree intervals around a 0.6 metre PCD. The wheel is stationary and spun up to 27.78 metres/second (100 kph 884 rpm). As one of the red dots gets to exactly the bottom they are all released.

Because the wheel is stationary they all fly off at a tangent to the radius they are travelling around. After they have travelled 150 mm or about 0.0050 seconds a high speed photo is taken. The blue spots show where they will be. They will also be all travelling at near enough to 27.78 metres/sec. Because the wheel is sitting stationary apart from the rotation theres no other vectors to add.

To image the purple dots you have to be travelling alongside the wheel so that it appears to be stationary but it actually isnt as its rolling down the roadway at 100kph.
Again at the instant a red dot gets to the bottom all the red dots are released and after 0.0005 seconds a high speed photo is taken. The purple dots show where they would appear on the photo. These positions are the resultant vectors after adding the 27.78 metres/sec direction of travel vector to the angular vector of the tangent they release at.

[GSVR]

You're point of view is from a stationary position on the ground which is wrong, at least if you are considering the effect it has on the bike. Imagine the bike to be floating in mid air and the wheels are spinning at an rpm equal to 100kph, it is easy to see that both the top and bottom of the wheel carry equal momentum as relates to the bike.

If you get hit by the wheel of a bike then yes the top of the wheel would be carrying more momentum than the bottom, but i fail to see how that would be relevant as there would be bigger things to worry about.

Quote:
"In non-moving system wheel is seen as rotating around its bottom ( instant axis of rotation constantly moving forward with the wheel)."

The relevance is that it actually is carrying more momentum at the top when the bike is in motion so the top requires more force to alter its path.

Another way would be to sit a bike on a rolling road then the only parts that carry forward momentum are the chain. some engine components and the wheels. Of course they also carry the same amount of backawards momentum as well as up and downwards.

[Sanx]

Any ideas how much the radius of the tyre would increase at 100kph due to the gforce pulling it out?

Check out the rear tires on these guys. The radius is increasing like 40 cm when the tires start to rotate.
http://www.youtube.com/watch?v=Ls_obl_fokc&feature=related

To answer a much earlier question, the increase in radius on a bike tyre is measurable. At speeds of us to about 270, the rear 190/60 Dunlop I had on the Blade increased by about 1cm in radius. Enough to wear a hole through the rear hugger, anyway.

[Badjelly]

The relevance is that it actually is carrying more momentum at the top when the bike is in motion so the top requires more force to alter its path.

You think you can sneak that one in again without anyone noticing?

I'll paraphrase what I said about this earlier:


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.

The thing is, when you say that a body "requires more force to alter its path" as its speed increases, you're correct if you're making a statement about the rate of change of the direction of motion, but you're wrong if you're making a statement about the rate of change of velocity (AKA acceleration).

[GSVR]

I'll make it as simple as I possibly can.

1/ A mass is 10 grams is travelling at 100 meters per second and a force of 1 newton is applied at 90 degrees to the direction of travel.

2/ A mass of 10 grams is travelling at 200 meters per second and a force of 1 newton is applied at 90 degrees to the direction of travel.

After one second the the mass in the second example (2/) has travelled twice as far forward but the same amount 90 degrees to the direction of travel sideways as the first example (1/).

So to get it to the to move the same amount sideways in the same distance four times as much sideways force would have to be used. ie 4 newtons.

Now apply this logic to the 1 dollar coins in my last wheel example.

[Badjelly]

I'll make it as simple as I possibly can.

A mass is 10 newtons is travelling at 100 meters per second and a force of 1 newton is applied at 90 degrees to the direction of travel.

A mass of 10 newtons is travelling at 200 meters per second and a force of 1 newton is applied at 90 degess to the direction of travel.

After one second the the mass in the second example has travelled twice as far forward but the same amount 90 degrees to the direction of travel sideways.

So to get it to the to move the same amount sideways in the same distance twice as much sideways force would have to be used. ie 2 newtons.

There are a couple of things wrong with this: the units of mass are kilogrammes; and you'd actually need 4 times as much force to move the the second body the same distance sideways in the same distance of forward travel (because you have half the time to do it and s = (1/2)*a*t^2, well let's not go into that).

But, basically, yes. A faster-moving body requires more force to deflect it the same distance sideways in the same distance of forward travel.

[GSVR]

There are a couple of things wrong with this: the units of mass are kilogrammes; and you'd actually need 4 times as much force to move the the second body the same distance sideways in the same distance of forward travel (because you have half the time to do it and s = (1/2)*a*t^2, well let's not go into that).

But, basically, yes. A faster-moving body requires more force to deflect it the same distance sideways in the same distance of forward travel.

Kilograms are weight due to acceleration due to gravity
Mass is always newtons

Thanks for pointing out the error so example instead of 2 newtons I should have said 4 yes I get that now and have corrected it.

[Badjelly]

Kilograms are weight due to acceleration due to gravity
Mass is always newtons

Other way round, basically. The kilogram is a unit of mass.

http://en.wikipedia.org/wiki/Kilogram

The newton is a unit of force

http://en.wikipedia.org/wiki/Newton

One newton is the amount of force that is required to accelerate one kilogram of mass at a rate of one meter per second squared.

To confuse things, we usually estimate the mass of objects by weighing them, ie by measuring the force exerted on them by gravity. But the kilogram is definitely a unit of mass, so Dani Pedrosa would still have a mass of 51 kg if he were on the moon, even though his weight would be 1/6 that.

[Mikkel]

To answer a much earlier question, the increase in radius on a bike tyre is measurable. At speeds of us to about 270, the rear 190/60 Dunlop I had on the Blade increased by about 1cm in radius. Enough to wear a hole through the rear hugger, anyway.

Cool! What pressure were you running?

I've seen bikes where the front wheel guard has been ground away slightly due to wind pressure at high speed deflecting it down to the tyre (but I guess it could be a combination...).


And GSVR - Badjelly has got it right! Perhaps look up SI units on wikipedia or somewhere and that should also confirm it.

[GSVR]

And GSVR - Badjelly has got it right! Perhaps look up SI units on wikipedia or somewhere and that should also confirm it.

Yes I should have just said the dots are 1 dollar coins.

1 KG times 9.8 is the force in newtons you would feel if you you were to hold 1 KG against gravitys pull.

[Mikkel]

Yes I should have just said the dots are 1 dollar coins.

1 KG times 9.8 is the force in newtons you would feel if you you were to hold 1 KG against gravitys pull.

You're getting close...

1 kg * 9.81 m/s^2 = 9.81 N

N = kg*m/s^2

[GSVR]

You're getting close...

1 kg * 9.81 m/s^2 = 9.81 N

N = kg*m/s^2

Yes ok I got weight and mass confused.

But it actually doesnt matter if the mass is 1 gram or 1000 kg in the example the vectors will still be the same.

[Mikkel]

Yes ok I got weight and mass confused.

But it actually doesnt matter if the mass is 1 gram or 1000 kg in the example the vectors will still be the same.

Only if you assume the wind resistance doesn't have an impact... ;)

[GSVR]

Only if you assume the wind resistance doesn't have an impact... ;)

Well if you would like to give me the same example and add it your quite welcome to. Personally I think the effects of wind and gravity over the course of half a hundredth of a secound to be unimportant.

[Mikkel]

Well if you would like to give me the same example and add it your quite welcome to.

Nope I wouldn't like to do that...


Personally I think the effects of wind and gravity over the course of half a hundredth of a secound to be unimportant.

...and that is one of the reasons why.

[GSVR]

Nope I wouldn't like to do that...



...and that is one of the reasons why.


Well just great as I can only assume that there aren't any major errors on the Explodingwheel.jpg or you would have most certainly pointed them out by now.

[Mikkel]

Well just great as I can only assume that there aren't any major errors on the Explodingwheel.jpg or you would have most certainly pointed them out by now.

...that would be a safe assumption, assuming that I had looked at it.

[GSVR]

...that would be a safe assumption, assuming that I had looked at it.

Say no more a nod is as good as a wink to a blind man.

[Hitcher]

Say no more a winks as good as a nod to a blind man.

What's a *winks*?

[007XX]

What's a *winks*?

And excitable Pygme's fascial tick.