Do wheels hang from or stand on their spokes?

[dkahn400]

The tug-o-war analogy can also be used to show me 'pushing' via a rope.

Imagine two matched people. The rope is static, they are pulling at each other with the same force. You agree this is tension, yes?

I walk along, and push one person towards the other person. Neither alters their effort. However, I've just introduced a new force. What will happen? What will you call that force?

Illustrated above already, by me.  The problem with your push is that the illusion of a push/compression disappears the moment the person at the other end stops pulling.  It's not appropriate to call this compression, because there is only ever tension or slack in the rope.  Compare that with a tightrope analogy, except using a suitably large diameter steel tube instead of the rope, where you can now describe compression.

You keep changing the ball game. Firstly a tug-of-war is a fairly poor analogy to a bicycle wheel in any case. Secondly, the moment one of the people stops pulling the analogy fails completely. It's like the "cut away the spokes and see what happens" argument. If you alter the structure it's not surprising if you also alter its physical properties.

[Wendy]

You keep saying that, but what you really mean is that your description fails in many similar models.  You shouldn't be blaming the model, you should be blaming your description.  I'll say it again, there is no compressive load, because there is still actual tension.  You've consistently failed to explain this.

[rogerzilla]

Wheel stresses (Jobst Brandt)

[andygates]

All these two-directional larks miss the load on all the other spokes, don't they?  And in a tangentially-spoked wheel there are no direct up and down loads ever. 

My vote is for the hub squatting in a tensioned web, cackling to itself.

[Air Dancer]

You keep saying that, but what you really mean is that your description fails in many similar models.  You shouldn't be blaming the model, you should be blaming your description.  I'll say it again, there is no compressive load, because there is still actual tension.  You've consistently failed to explain this.

Because there is a downward force at the axle due to the proportion of the rider plus bike weight and a corresponding equal reaction from the ground, the wheel between these points is under a compressive load.

The spokes in the wheel still remain in tension because this compressive load is not great enough to over come this, but will cause slight deformation of the wheel.

[Wendy]

Because there is a downward force at the axle due to the proportion of the rider plus bike weight and a corresponding equal reaction from the ground, the wheel between these points is under a compressive load.

The spokes in the wheel still remain in tension because this compressive load is not great enough to over come this, but will cause slight deformation of the wheel.

I agree, but you can't replace the bolded bit above with "the spoke", which is how I perceive Dave's explanation.

[Ian H]

Assuming that Engineers' English departs from regular usage, I still find it hard to understand how it can be useful to refer to a reduction in tension as 'compression'. If you cut the bottom spoke of a normal, loaded cycle wheel, it would still go bang as tensile forces pulled it apart.

[dkahn400]

Assuming that Engineers' English departs from regular usage, I still find it hard to understand how it can be useful to refer to a reduction in tension as 'compression'. If you cut the bottom spoke of a normal, loaded cycle wheel, it would still go bang as tensile forces pulled it apart.

A reduction in tension is not necessarily a net compression but it is an increase in compression.

[mattc]

A reduction in tension is not necessarily a net compression but it is an increase in compression.

He doesn't give up, does he?

[Ian H]

A reduction in tension is not necessarily a net compression but it is an increase in compression.

Yebbut...to compression is surely a force acting inwards, not a force acting outwards even if lessened.

[dkahn400]

A reduction in tension is not necessarily a net compression but it is an increase in compression.

Yebbut...to compression is surely a force acting inwards, not a force acting outwards even if lessened.

Compression is simply negative tension. If you sit on top of your wheel you compress it, but the spokes remain under tension.

[mrcharly-YHT]

A reduction in tension is not necessarily a net compression but it is an increase in compression.

Yebbut...to compression is surely a force acting inwards,

Correct.
There is more than one force acting on each spoke. You have to add them up to see what is happening, not consider each in isolation.
(damn, I said to myself I wasn't going to join in anymore!)

[GruB]

Got the wall mounted installed today.  My mistake - it holds onto the rim, not a spoke 

[dkahn400]

Because there is a downward force at the axle due to the proportion of the rider plus bike weight and a corresponding equal reaction from the ground, the wheel between these points is under a compressive load.

The spokes in the wheel still remain in tension because this compressive load is not great enough to over come this, but will cause slight deformation of the wheel.

I agree, but you can't replace the bolded bit above with "the spoke", which is how I perceive Dave's explanation.

And what parts of the wheel are there between these two points?

[dkahn400]

Got the wall mounted installed today.  My mistake - it holds onto the rim, not a spoke 

And in answer to the original issue, it is perfectly OK to hang your bicycles by their rims.   

[dkahn400]

A reduction in tension is not necessarily a net compression but it is an increase in compression.

He doesn't give up, does he?

It's a simple concept; I can't understand why people are struggling with it. Think of your bank account if it helps. If it's in credit you can take money away from it and still leave it in credit. If it's in the red you can add money to it and still leave it in the red. If something is in tension you can compress it and still leave it in tension.

[Robbo4]

For some reason this thread reminds me of Humphrey Lyttelton introducing "one song to the tune of another".

[mattc]

A reduction in tension is not necessarily a net compression but it is an increase in compression.

He doesn't give up, does he?

It's a simple concept; I can't understand why people are struggling with it. Think of your bank account if it helps. If it's in credit you can take money away from it and still leave it in credit. If it's in the red you can add money to it and stil leave it in the red. If something is in tension you can compress it and still leave it in tension.

I understand it perfectly - I just think it's very unhelpful!
To pick at your analogy - when you deposit money in your savings account, do you call it
"reducing your overdraft".

(This is rather like your post on AUK-Yahoo - you were right about non-linear gear ratio steps, but not in a way that the layman would easily understand. )

[dkahn400]

I understand it perfectly - I just think it's very unhelpful!
To pick at your analogy - when you deposit money in your savings account, do you call it
"reducing your overdraft".

This is the difference between everyday language and technical language. If you prefer to say "reduction in tension" rather than "compression" that's fine by me. I don't agree that it's helpful though, as it complicates matters unnecessarily. For example, in the case of the cast moped wheel you don't actually know whether the spokes are in compression or tension to begin with. So you then have to say it supports a load at the hub by compression of the lower spokes, or possibly by a reduction in their tension, or maybe an initial reduction in tension followed by compression.

Or in Mike's case he would have to say it supports a load at the hub by compression of the lower spokes or, if the spokes happen to be in tension and remain in tension by its entire structure with the lower spokes reducing in tension but not actually contributing anything, or if the lower spokes start in tension but finish in compression then initially it supports a load by its entire structure with the lower spokes reducing in tension but not actually contributing anything then suddenly it works by an entirely different mechanism with the lower spokes supporting the load by compression.

Do you see why for a technical description of how a wheel supports a load at the hub it's best to say it does so by deformation of the bottom of the rim and compression of the lower spokes? This description applies equally to all spoked wheels regardless of the initial tensions.

(This is rather like your post on AUK-Yahoo - you were right about non-linear gear ratio steps, but not in a way that the layman would easily understand. )

My post? I thought I'd stayed out of that one.

[mattc]

(This is rather like your post on AUK-Yahoo - you were right about non-linear gear ratio steps, but not in a way that the layman would easily understand. )

My post? I thought I'd stayed out of that one.

You're right, my mistake - it was Rod Dalitz. I hope neither of you are offended.

[GruB]

There will be a fight tomorrow morning in the rear yard of Melksham Divisional HQ.
The four wall hooks have been installed.
The estates man has put signs up over each - with our names displayed !! 
First in best dressed and all that. 

[drossall]

I rather liked andrew_s's peace-making attempt on page 1, but everyone ignored it

And then:

Okay, this is very interesting.  How does the actual turning of the wheel effect all this?  Is there not a force created when it turns?

Er, no. There are (essentially balanced) forces in the wheel of course, as has been discussed endlessly. Turning the wheel means that a different part is deformed at the bottom at each moment, so the forces "move round" the wheel. However, turning does not create a force as such. Indeed, in most situations, if there is constant speed there is no acceleration, which implies that there are no forces being applied*.

*Sometimes, of course, there are balanced forces - you freewheel downhill at a constant speed because wind resistance balances gravity, not because there are no forces at all.

[sas]

Okay, this is very interesting.  How does the actual turning of the wheel effect all this?  Is there not a force created when it turns?

Er, no. There are (essentially balanced) forces in the wheel of course, as has been discussed endlessly. Turning the wheel means that a different part is deformed at the bottom at each moment, so the forces "move round" the wheel. However, turning does not create a force as such. Indeed, in most situations, if there is constant speed there is no acceleration, which implies that there are no forces being applied*.

If something's turning there must be a centripetal force, so in a balanced wheel all spokes will have an equal additional increase in tension. Unless you want to be really pedantic, in which case you have to take into account that the deformation of the rim at the bottom slightly decrease the radius, which therefore results in a slightly lower centripetal force.