Wheel upgrade - a question

neeb

Yup, no doubt if you are doing lots of short accelerations as in racing in the mountains, light wheels are going to spin up faster, same as they do on the flat.

Maybe that's all that's needed to explain the subjective advantage of light wheels for climbing vs. the theory that they shouldn't be any faster. Even if you are not racing, if you can accelerate quickly that's going to feel faster, and because you are traveling slower in any case when you are climbing, the impression could be greater than on the flat. You might not get up the hill any quicker by the clock, but being able to accelerate quickly out of the hairpins could make it feel faster, and if you are racing this will be a real advantage.

Davey C

Wirral Paul has gone mental!

Slo Mo Jones

neeb wrote:

P_Tucker wrote:

So no evidence then. We'll file this one right next to the valve weight concern, shall we?

Before you find evidence you need to know where to look for it. That's how the scientific method works - the first thing you do is come up with a hypothesis, then you look for hard data to support it, then you try to falsify it. Without speculation you never get anywhere, it's imagination that separates doing science from being a technician.

I would suggest that a place such as Bike Radar is more suited to the speculative side of the science surrounding cycling than it is to rigorous presentation and analysis of data.

Incidentally, you still haven't come up with any figures to support your dismissal of rim weight asymmetry as being necessarily insignificant. We're talking up to 15g or so (for a valve). At about 30mph with a 33cm radius I calculate the centrifugal force of that 15g to be about 8 Newtons. That's 8 N squashing the tyre once every wheel revolution,

That makes no sense. The centripetal force is generated within the tension of the spokes. It is not 8 newtons pushing down on the tyre.

The difference 15g in your valve makes in terms of force pushing down into the tyre is 0.015kg*9.8m/s^2=0.1N

Bar Shaker

Er, Neeb is right. That isn't the formula for centripedal acceleration, it is a=v2/r or Force = mass times v2/r

At 30 mph, v=13.333m/s r=.33m so a=537... times 15g mass = 8N.

smidsy

Riding up hills is essentially constant acceleration as you need to combat gravity.

With this in mind I would prefer a lighter rim to make those accelerations that much easier.

Then again the heavier rim deccelerates slower - oh no, now I am in a tiz.

Edit: I have not read the previous 5 pages so apologies if I am going over old ground

drlodge

smidsy wrote:

Riding up hills is essentially constant acceleration as you need to combat gravity.

Er, no it isn't. Going uphill at a constant pace means zero acceleration. There is, however, a constant force (gravity) that is trying to pull you down the hill and you have to work against that. Which I guess is a bit like constantly accelerating but its not the same thing

smidsy wrote:

With this in mind I would prefer a lighter rim to make those accelerations that much easier.

Yes I agree here, since most people will not ride constantly up the hill but be putting in tiny accelerations at each pedal stroke. All a heavier wheel will do will make the acceleration less (since there is more mass) and make accelerating the bike up to a given speed harder.

smidsy wrote:

Then again the heavier rim deccelerates slower - oh no, now I am in a tiz.

But its gravity that is providing the deceleration, the rider is providing the acceleration

smidsy wrote:

Edit: I have not read the previous 5 pages so apologies if I am going over old ground

I haven't ready all the pages either! All things being equal lighter wheels are obviously an advantage when hill climbing for 2 reasons:
- less weight to lug up the hill (same applies to all aspects of the rider and bike)
- easier to accelerate the bike on each pedal stroke and/or if trying to accelerate the bike uphill due to less rotating mass.

The relative advantages of the above can be debated, depending on the amount of weight at issue. And all things are not always equal e.g. lighter wheel might mean weaker wheel so there is more energy being lost as the wheel distorts more.

Slo Mo Jones

Bar Shaker wrote:

Er, Neeb is right. That isn't the formula for centripedal acceleration, it is a=v2/r or Force = mass times v2/r

At 30 mph, v=13.333m/s r=.33m so a=537... times 15g mass = 8N.

No, neeb is not right. That's not the force through the tyre.

Centripetal acceleration is acceleration towards the centre of circular motion. So that equation calculates a force toards the wheel hub, through the spoke. Without that force the wheel would fly off the hub. Your equation is relevant when talking about the force needed to increase rotational velocity of the wheel (well, a derivation from it, the moment of inertia of the wheel), but not when discussing the force through the tyre.

The force through the tyre is the same as the force exerted up into the tyre by the road, otherwise the tyre would fall through the tarmac (ref Newton's first law of motion). The tarmac doesn't know the wheels are spinning. In fact, when the tyre comes into contact with the road, the part in contact is not spinning. It's stationary. Therefore the force through the tyre is a simple vertical force. That vertical force is calculated as F=ma (Newton's second law), force = mass*acceleration. The acceleration is the gravitational constant, or 9.81m/s^2. The mass in question is 15g, so 0.015kg. So the reduction in force through the tyre caused by the 15g lighter wheel valve is approximately 0.015*10 = 0.15N.

Boom baby.

imposter2.0

drlodge wrote:

I haven't ready all the pages either! All things being equal lighter wheels are obviously an advantage when hill climbing for 2 reasons:
- less weight to lug up the hill (same applies to all aspects of the rider and bike)
- easier to accelerate the bike on each pedal stroke and/or if trying to accelerate the bike uphill due to less rotating mass.

The relative advantages of the above can be debated, depending on the amount of weight at issue. And all things are not always equal e.g. lighter wheel might mean weaker wheel so there is more energy being lost as the wheel distorts more.

Maybe you should read the other pages. What you have just said has already been debunked as nonsense elsewhere in the thread - apart from the less overall weight bit, obviously...

smidsy

drlodge wrote:

smidsy wrote:

Riding up hills is essentially constant acceleration as you need to combat gravity.

Er, no it isn't. Going uphill at a constant pace means zero acceleration. There is, however, a constant force (gravity) that is trying to pull you down the hill and you have to work against that. Which I guess is a bit like constantly accelerating but its not the same thing

Well perhaps I used poor language.

Going up hills requires constant hard effort (otherwise you slow very rapidly). To me it is like acceleration over a prolonged period, which is why I said 'essentially'.

Anyway I have nothing constructive to add so will dip out now.

drlodge

Imposter wrote:

Maybe you should read the other pages. What you have just said has already been debunked as nonsense elsewhere in the thread - apart from the less overall weight bit, obviously...

May be you and others should consider this...think about the wheel weight at extremes to make the point:
- how easy would going up the hill be if the wheels weighed nothing?
- how easy would going up the hill be if the wheels weighed a ton?

imposter2.0

drlodge wrote:

Imposter wrote:

Maybe you should read the other pages. What you have just said has already been debunked as nonsense elsewhere in the thread - apart from the less overall weight bit, obviously...

May be you and others should consider this...think about the wheel weight at extremes to make the point:
- how easy would going up the hill be if the wheels weighed nothing?
- how easy would going up the hill be if the wheels weighed a ton?

You are still missing the point - which various people have tried to make time and again on this thread. If the wheels weighed 'nothing', the bike would obviously be lighter. Lighter bike = better climbing. If the wheels weighed 'a ton' (for the sake of argument), then the bike would obviously be heavier. Heavier bike = harder climbing.

Nobody has ever disputed that a lighter bike would not be better uphill. But your point on wheel weight and how it relates to 'improved acceleration' is misinformed, from a climbing persective.

neeb

Slo Mo Jones wrote:

Bar Shaker wrote:

Er, Neeb is right. That isn't the formula for centripedal acceleration, it is a=v2/r or Force = mass times v2/r

At 30 mph, v=13.333m/s r=.33m so a=537... times 15g mass = 8N.

No, neeb is not right. That's not the force through the tyre.

Centripetal acceleration is acceleration towards the centre of circular motion. So that equation calculates a force toards the wheel hub, through the spoke. Without that force the wheel would fly off the hub. Your equation is relevant when talking about the force needed to increase rotational velocity of the wheel (well, a derivation from it, the moment of inertia of the wheel), but not when discussing the force through the tyre.

The force through the tyre is the same as the force exerted up into the tyre by the road, otherwise the tyre would fall through the tarmac (ref Newton's first law of motion). The tarmac doesn't know the wheels are spinning. In fact, when the tyre comes into contact with the road, the part in contact is not spinning. It's stationary. Therefore the force through the tyre is a simple vertical force. That vertical force is calculated as F=ma (Newton's second law), force = mass*acceleration. The acceleration is the gravitational constant, or 9.81m/s^2. The mass in question is 15g, so 0.015kg. So the reduction in force through the tyre caused by the 15g lighter wheel valve is approximately 0.015*10 = 0.15N.

Boom baby.

OK, I'm no engineer or physicist, so I'm quite prepared to learn here if necessary, talk me though it...

The centripetal force (towards the hub) is theoretically equal and opposite to the centrifugal force, away from the hub, right? But doesn't assuming that they are completely equal imply that the hub is completely motionless, or rather that you are in a frame of reference in which it appears to be?

An unbalanced centrifuge will shake violently, I assume because the axle isn't being held completely stationary in relation to the circle described by the revolution of the unbalanced bit. The weight of the machine (a mass that requires force to be accelerated) will partially counteract the centrifugal force through centripetal force, but only in proportion to that mass. So in the frame of reference of someone looking at the centrifuge, the centrifugal force is causing the thing to shake because the centripetal force isn't matching it.. Or something like that... However it may be the case that in the frame of reference of the centrifuge itself (where the axle is assumed to be stationary) the centripedal force and centripetal force are completely equal.

Similarly when riding a bike, the centrifugal force created by the valve spinning could cause the hub to move slightly up and down, so that rather than this force being opposed by centripetal force towards a completely stationary hub, it is being opposed by the tyre deforming...

Probably got that at least partially wrong, in which case, enlighten me..

Bar Shaker

Slo Mo Jones wrote:

Bar Shaker wrote:

Er, Neeb is right. That isn't the formula for centripedal acceleration, it is a=v2/r or Force = mass times v2/r

At 30 mph, v=13.333m/s r=.33m so a=537... times 15g mass = 8N.

No, neeb is not right. That's not the force through the tyre.

Centripetal acceleration is acceleration towards the centre of circular motion. So that equation calculates a force toards the wheel hub, through the spoke. Without that force the wheel would fly off the hub. Your equation is relevant when talking about the force needed to increase rotational velocity of the wheel (well, a derivation from it, the moment of inertia of the wheel), but not when discussing the force through the tyre.

The force through the tyre is the same as the force exerted up into the tyre by the road, otherwise the tyre would fall through the tarmac (ref Newton's first law of motion). The tarmac doesn't know the wheels are spinning. In fact, when the tyre comes into contact with the road, the part in contact is not spinning. It's stationary. Therefore the force through the tyre is a simple vertical force. That vertical force is calculated as F=ma (Newton's second law), force = mass*acceleration. The acceleration is the gravitational constant, or 9.81m/s^2. The mass in question is 15g, so 0.015kg. So the reduction in force through the tyre caused by the 15g lighter wheel valve is approximately 0.015*10 = 0.15N.

Boom baby.

Your maths is correct... but only when the wheel is stationary.

In order to try and make things easier for you to understand, have a think about why your back gets wet when riding through a puddle. Then think about how this is affected by your speed. Why does water get flung up your back at 20mph but not at 5mph? Now see if your formula can describe what is going on.

Then try the same with my formula.

Slo Mo Jones

Bar Shaker wrote:

Slo Mo Jones wrote:

Bar Shaker wrote:

Er, Neeb is right. That isn't the formula for centripedal acceleration, it is a=v2/r or Force = mass times v2/r

At 30 mph, v=13.333m/s r=.33m so a=537... times 15g mass = 8N.

No, neeb is not right. That's not the force through the tyre.

Centripetal acceleration is acceleration towards the centre of circular motion. So that equation calculates a force toards the wheel hub, through the spoke. Without that force the wheel would fly off the hub. Your equation is relevant when talking about the force needed to increase rotational velocity of the wheel (well, a derivation from it, the moment of inertia of the wheel), but not when discussing the force through the tyre.

The force through the tyre is the same as the force exerted up into the tyre by the road, otherwise the tyre would fall through the tarmac (ref Newton's first law of motion). The tarmac doesn't know the wheels are spinning. In fact, when the tyre comes into contact with the road, the part in contact is not spinning. It's stationary. Therefore the force through the tyre is a simple vertical force. That vertical force is calculated as F=ma (Newton's second law), force = mass*acceleration. The acceleration is the gravitational constant, or 9.81m/s^2. The mass in question is 15g, so 0.015kg. So the reduction in force through the tyre caused by the 15g lighter wheel valve is approximately 0.015*10 = 0.15N.

Boom baby.

Your maths is correct... but only when the wheel is stationary.

In order to try and make things easier for you to understand, have a think about why your back gets wet when riding through a puddle. Then think about how this is affected by your speed. Why does water get flung up your back at 20mph but not at 5mph? Now see if your formula can describe what is going on.

Then try the same with my formula.

The tyre isn't in contact with the road when it flicks water up on to your back. The road doesn't know what's going on up there. My post was about the force through the tyre into the road, the force deflecting the tyre, which is clearly not 8N as my post clearly explains. 8N, through each tyre as suggested, would be like riding with an extra 1.6kg strapped to your back. You really think 15g on a valve is going to be the equivalent of that? Water flicking up onto your back - what has this got to do with the force through the tyre?

Neeb, I am an engineer. Or at least, I was. The term "centrifugal force" is a bit pants. There's not really any such thing. "The centripetal force (towards the hub) is theoretically equal and opposite to the centrifugal force, away from the hub, right?" No, this is an understandable, erm, misunderstanding. When you're on a waltzer at the fairground, and you think it's "centrifugal" force pushing you against the outside of the car you're in, what you're actually feeling is the centripetal force of the car pulling you towards the centre of the ride.

There is not an equal and opposite force to the centripetal force. How can that be when the rim remains 350mm from the hub? Wouldn't the rim be accelerating towards the hub, at F (centripetal force) / mass = acceleration? Well, as odd as it seems, At any point in time, assuming the wheels are turning, any point on the rim is accelerating towards the hub. At a = r.w^2, where w is the angular velocity, r being the radius of the wheel.

If there were an equal and opposite force to centripetal force, you would feel in equilibrium on the fairground waltzer, rather than feeling the outside of the waltzer car push on you until the ride stops.#

A boring post, I know, but I think it's important.

neeb

Slo Mo Jones wrote:

My post was about the force through the tyre into the road, the force deflecting the tyre, which is clearly not 8N as my post clearly explains. 8N, through each tyre as suggested, would be like riding with an extra 1.6kg strapped to your back. You really think 15g on a valve is going to be the equivalent of that?

Well, leaving the physics for later, firstly I don't get the "clearly not" bit - I've seen how much force half a gram in an unbalanced centrifuge can exert - OK, the speed of revolution is much more than a bicycle wheel, but it's enough to convince me not to take anything for granted where asymmetrical fast-spinning things are concerned.

Slo Mo Jones wrote:

The term "centrifugal force" is a bit pants. There's not really any such thing. "The centripetal force (towards the hub) is theoretically equal and opposite to the centrifugal force, away from the hub, right?" No, this is an understandable, erm, misunderstanding. When you're on a waltzer at the fairground, and you think it's "centrifugal" force pushing you against the outside of the car you're in, what you're actually feeling is the centripetal force of the car pulling you towards the centre of the ride.

Ok, I get this now. I understand that the centripetal force is effectively a continuous acceleration towards the center of rotation, because any point on the circumference needs to be continuously "pulled back" from a constant velocity straight line path. Remembering Newton's(?) diagrams of planetary elliptical motion now...

Slo Mo Jones wrote:

There is not an equal and opposite force to the centripetal force. How can that be when the rim remains 350mm from the hub? Wouldn't the rim be accelerating towards the hub, at F (centripetal force) / mass = acceleration? Well, as odd as it seems, At any point in time, assuming the wheels are turning, any point on the rim is accelerating towards the hub. At a = r.w^2, where w is the angular velocity, r being the radius of the wheel.
If there were an equal and opposite force to centripetal force, you would feel in equilibrium on the fairground waltzer, rather than feeling the outside of the waltzer car push on you until the ride stops.

Ok, I can accept and understand this as stated, but it seems to me you are still assuming a stationary hub and a symmetrical rim. Can you explain to me why it is NOT the case that an unbalanced bicycle rim (15g at one point on a rim that is otherwise balanced) spinning at the velocity it would be when the bike is going at 20mph, isn't going to act exactly like an unbalanced centrifuge? If you spin a wheel up to speed when the bike is on a stand, you will nearly always see/feel a fairly violent rhythmic shaking due to asymmetry on the rim/tyre combination. Yes, the rim is remaining 350mm from the hub, but the whole rim, wheel and hub (as well as most of the bicycle) is vibrating backwards and forwards with a force that I can easily believe is in the same ball park as the weight of 1.6kg under 1g. The only difference when the bike is being ridden is that this movement is being opposed by the ground, so presumably the proportion of the force that is no longer expressed as motion is being expended in deforming the tyre.

Relating this specifically to the physics - presumably the centripetal force is greater for the point on the rim that is slightly heavier. The centripetal acceleration is the same for all points on the rim, but the force required to continuously accelerate the heavier part of the rim towards the hub is greater due to the mass. So the heavier part of the rim has a greater inwards momentum than the part opposite on the other side. I presume that this is what causes an unbalanced centrifuge (or an unbalanced wheel) to shake.

neeb

Putting all that another way - I understand that with the revolution of a planet around a star, it is actually more correct to think of the star and the planet revolving around each other - the force of gravity acts on both the planet and the star, although due its much greater mass the star only wobbles a tiny but. This is how planets around other stars are detected I believe, by looking for the "wobble" in the star.

In the same way, the centripetal force acting on the 15g of the valve weight will cause the hub to wobble slightly. The force causing this will increase with the speed of wheel rotation as well as with the weight of the valve, so I still don't see why my initial calculation wasn't correct.

Slo Mo Jones

I don't quite understand the post, but the greater force required to accelerate the heavier part of the rim towards the hub will manifest itself in the spoke, not on the ground.

Remember, the ground doesn't know that the wheel is spinning.

neeb

Slo Mo Jones wrote:

I don't quite understand the post, but the greater force required to accelerate the heavier part of the rim towards the hub will manifest itself in the spoke, not on the ground.

Remember, the ground doesn't know that the wheel is spinning.

Just for the sake of argument though, let's assume that the spoke is completely rigid - wouldn't this force then act to accelerate the bike and rider upwards slightly (and downwards, backwards and forwards etc as the wheel rotates), i.e the hub spindle would in effect describe a tiny circle? And wouldn't this in turn be partially smoothed out by tyre deformation? The tyre would actually be slightly more compressed when the 15g was at the TOP of the wheel and the centripetal force pushing the whole structure downwards slighty, and slightly less compressed when the 15g was nearest the ground with the centripetal force pushing up towards the hub...

Slo Mo Jones

neeb wrote:

Slo Mo Jones wrote:

I don't quite understand the post, but the greater force required to accelerate the heavier part of the rim towards the hub will manifest itself in the spoke, not on the ground.

Remember, the ground doesn't know that the wheel is spinning.

Just for the sake of argument though, let's assume that the spoke is completely rigid - wouldn't this force then act to accelerate the bike and rider upwards slightly (and downwards, backwards and forwards etc as the wheel rotates), i.e the hub spindle would in effect describe a tiny circle? And wouldn't this in turn be partially smoothed out by tyre deformation? The tyre would actually be slightly more compressed when the 15g was at the TOP of the wheel and the centripetal force pushing the whole structure downwards slighty, and slightly less compressed when the 15g was nearest the ground with the centripetal force pushing up towards the hub...

Erm, no. Centripetal force manifests itself as a tension in the spoke. And spokes are designed to be excellent at withstanding tension.

If you had the measuring equipment of NASA, you could say that the spoke, under tension, will undergo a miniscule elastic strain (i.e. extension), making the wheel deformed by an imperceptibly tiny amount, which would require a tiny, tiny, tiny weeny bit more energy to spin. But that would be guff.

The reduction in force exerted through the tyre into the road by using a valve 15g lighter really is only 0.15N.

Bar Shaker

Slo Mo Jones wrote:

The tyre isn't in contact with the road when it flicks water up on to your back. The road doesn't know what's going on up there. My post was about the force through the tyre into the road, the force deflecting the tyre, which is clearly not 8N as my post clearly explains. 8N, through each tyre as suggested, would be like riding with an extra 1.6kg strapped to your back. You really think 15g on a valve is going to be the equivalent of that? Water ******* up onto your back - what has this got to do with the force through the tyre?

Perhaps I misunderstood the initial point, which I thought was the force which a 15g valve exerted onto the tyre of wheel rotating at 20mph. The road had nothing to do with it.

1.6kg does not come into it as the forces exist all round the wheel, not just at the point where the valve is closest to the ground. We cannot talk about centripetal acceleration and stationary wheels. We cannot describe the forces going on in rotating wheels using gravity in the formula.

Slo Mo Jones wrote:

Neeb, I am an engineer. Or at least, I was. The term "centrifugal force" is a bit pants. There's not really any such thing. "The centripetal force (towards the hub) is theoretically equal and opposite to the centrifugal force, away from the hub, right?" No, this is an understandable, erm, misunderstanding. When you're on a waltzer at the fairground, and you think it's "centrifugal" force pushing you against the outside of the car you're in, what you're actually feeling is the centripetal force of the car pulling you towards the centre of the ride.

Centripetal acceleration (away from the centre of rotation) is matched by the force exerted on the spokes. If the centripetal acceleration in a disc exceeds the force capacity of the spokes, the wheel will explode. An unbalanced wheel doesn't exert more force on the spokes, it exerts more on what is holding the hub stationary. Fortunately we cannot ride fast enough for this to happen(!) but it is a factor for the rating of tyres on high performance cars.

Slo Mo Jones wrote:

There is not an equal and opposite force to the centripetal force. How can that be when the rim remains 350mm from the hub? Wouldn't the rim be accelerating towards the hub, at F (centripetal force) / mass = acceleration? Well, as odd as it seems, At any point in time, assuming the wheels are turning, any point on the rim is accelerating towards the hub. At a = r.w^2, where w is the angular velocity, r being the radius of the wheel.

If there were an equal and opposite force to centripetal force, you would feel in equilibrium on the fairground waltzer, rather than feeling the outside of the waltzer car push on you until the ride stops.#

A boring post, I know, but I think it's important.

Woohoo, I think you have it.

So back to the original point, what force is a 15g valve exerting on a tube/tyre in a wheel rotating at 20mph....?

It is being flung away from the hub with a force of 8N.

neeb

Slo Mo Jones wrote:

Erm, no. Centripetal force manifests itself as a tension in the spoke. And spokes are designed to be excellent at withstanding tension.

If you had the measuring equipment of NASA, you could say that the spoke, under tension, will undergo a miniscule elastic strain (i.e. extension), making the wheel deformed by an imperceptibly tiny amount, which would require a tiny, tiny, tiny weeny bit more energy to spin. But that would be guff.

The reduction in force exerted through the tyre into the road by using a valve 15g lighter really is only 0.15N.

But you are ignoring my assumption, i.e. that the spoke is completely rigid / inelastic - I know that's not true, but I just want to establish what the situation would be if it was, so that I can then meaningfully understand what is happening in terms of the elasticity of the spokes.

Slo Mo Jones

Bar Shaker wrote:

Slo Mo Jones wrote:

The tyre isn't in contact with the road when it flicks water up on to your back. The road doesn't know what's going on up there. My post was about the force through the tyre into the road, the force deflecting the tyre, which is clearly not 8N as my post clearly explains. 8N, through each tyre as suggested, would be like riding with an extra 1.6kg strapped to your back. You really think 15g on a valve is going to be the equivalent of that? Water ******* up onto your back - what has this got to do with the force through the tyre?

Perhaps I misunderstood the initial point, which I thought was the force which a 15g valve exerted onto the tyre of wheel rotating at 20mph. The road had nothing to do with it.

1.6kg does not come into it as the forces exist all round the wheel, not just at the point where the valve is closest to the ground. We cannot talk about centripetal acceleration and stationary wheels. We cannot describe the forces going on in rotating wheels using gravity in the formula.

Slo Mo Jones wrote:

Neeb, I am an engineer. Or at least, I was. The term "centrifugal force" is a bit pants. There's not really any such thing. "The centripetal force (towards the hub) is theoretically equal and opposite to the centrifugal force, away from the hub, right?" No, this is an understandable, erm, misunderstanding. When you're on a waltzer at the fairground, and you think it's "centrifugal" force pushing you against the outside of the car you're in, what you're actually feeling is the centripetal force of the car pulling you towards the centre of the ride.

Centripetal acceleration (away from the centre of rotation) is matched by the force exerted on the spokes. If the centripetal acceleration in a disc exceeds the force capacity of the spokes, the wheel will explode. An unbalanced wheel doesn't exert more force on the spokes, it exerts more on what is holding the hub stationary. Fortunately we cannot ride fast enough for this to happen(!) but it is a factor for the rating of tyres on high performance cars.

Slo Mo Jones wrote:

There is not an equal and opposite force to the centripetal force. How can that be when the rim remains 350mm from the hub? Wouldn't the rim be accelerating towards the hub, at F (centripetal force) / mass = acceleration? Well, as odd as it seems, At any point in time, assuming the wheels are turning, any point on the rim is accelerating towards the hub. At a = r.w^2, where w is the angular velocity, r being the radius of the wheel.

If there were an equal and opposite force to centripetal force, you would feel in equilibrium on the fairground waltzer, rather than feeling the outside of the waltzer car push on you until the ride stops.#

A boring post, I know, but I think it's important.

Woohoo, I think you have it.

So back to the original point, what force is a 15g valve exerting on a tube/tyre in a wheel rotating at 20mph....?

It is being flung away from the hub with a force of 8N.

Hehe, hang on a second, you getting sarcastic with me? Despite my clear and detailed description of why you were / are wrong, showing my workings and everything? And you still misunderstand my post and continue to show off your ignorance? wowzers.

To compound that you are again incorrect on a few points

The 15g valve is not being "flung" away from the away from the hub with a force of 8N. I've just explained that to you clearly. It is being pulled towards the hub, by the spokes, with a force of 8N.

The valve is exerting no force on the tyre or tube other than a vertical force of 0.15N on the tyre at the point of contact with the road.

Btw, the original point I was responding to was this:

neeb wrote:

That's 8 N squashing the tyre once every wheel revolution

which is still wrong.



My goodness. Study up son.

lawrences

Slo Mo Jones wrote:

My goodness. Study up son.

+1

neeb

Ok, let's reiterate / go back to basics...

Spin a bike wheel off the ground by turning the cranks. You will often experience a fairly violent vibration. Clearly this is due to the centripetal force of small imbalance(s) in rim weight, OK? This is the 8N or so under discussion. It's not so insignificant as to be undetectable, neither is it being absorbed much by spoke tension. The force imbalance may be acting inwardly rather than outwardly if you want to be technical about it, but it's still shaking a 7kg bike.

What happens to this same force when you are riding the bike? Is it ALL now being absorbed by spoke tension? If so, this must be because of the weight of the loaded bike compressing the wheel against the ground, thus preventing it from moving as it does when it is off the ground. So why would this not act to deform the tyre as well as tension/untension the spokes?

I'm quite prepared to accept I'm wrong if/when the situation is explained in a way that makes sense.

Slo Mo Jones

*sigh*

Neeb, when I spin the rear wheel off the ground, there is no violent vibration. The wheel spins smoothly. If there is a violent vibration when you do it, it is not caused by the valve. I would suggest that any violent vibration indicates that the bike is kaput.

Bar Shaker

Per rotation - agreed.

Missed that bit :oops:

neeb

Slo Mo Jones wrote:

*sigh*

Neeb, when I spin the rear wheel off the ground, there is no violent vibration. The wheel spins smoothly. If there is a violent vibration when you do it, it is not caused by the valve. I would suggest that any violent vibration indicates that the bike is kaput.

Perhaps you haven't tried it with very many bikes, or haven't spun the wheel fast enough (use the cranks). Of course if your wheel is very well balanced it won't happen. I can assure you however that with many bikes this does occur, and can only be down to slight weight differences at different points on the rim (including the tyre, tube, valve etc).

Why would it NOT occur, even with a 15g imbalance, given that this WILL produce 5 - 8N or so of centripetal force that isn't being opposed by the same force on the other side of the wheel? (as it would be if the wheel was balanced).

Do you accept that an unbalanced centrifuge vibrates violently?

pride4ever

Bottom line is a light wheelset weighing a kg or less/more is going to make bugger all difference in the long run. Just watch any pro race, some climbers are riding wheels that are basic skinny hoops and some are riding wheelsets with 20+ sections. Its all bunkum.

neeb

pride4ever wrote:

Bottom line is a light wheelset weighing a kg or less/more is going to make bugger all difference in the long run. Just watch any pro race, some climbers are riding wheels that are basic skinny hoops and some are riding wheelsets with 20+ sections. Its all bunkum.

I think this thread has been on a different subject for a while now..

FlacVest

Well I certainly learned a lot here; one man's quest to educate many and his fight against ignorance.

The power of one's ego in leading people to believe their assumptions are actually worth something.

And a nice refresher in classical physics. A great way to spend a Sunday night.

But for anybody joining who says, "TLDR", make an exception this time. It's worth it.