Forces on spokes

Maybe you should pluck spokes at various locations around the wheel
and nor which ones (by change in tone) are affected by placing a load
on the wheel. Let me tell you in advance what you will find (for pure
vertical loading). The only spokes affected by the load will be the
three or four spokes at the bottom directed at the road from the hub.

Those spokes are affected much more than the others by the load, I
agree. They undergo a dramatic loss of pretensioning.

If the spokes at the top are supporting the wheel, as you propose,
then they would be affected by the load, but they are not. I think
you are, as many others, not visualizing these things algebraically.
The problem is much like adding debits and credits to a bank account.

Here is my best explanation: The downward force on the axle due to the
weight of the rider must must be countered by an upward force of equal
magnitude exerted by the spokes on the hub. Thats just elementary
statics, what I think you are calling "debits and credits"

Lets divide the spokes into three somewhat imprecise categories:
1. Spokes at the bottom, underneath the axle.
2. Spokes to the side of the axle (mostly horizontal)
3. Spokes above the axle (mostly vertical)

How spokes in each category help to exert an upward force on the axle?

Category 1 would have to "push" upward on the axle from below. This
would put them in compression, which cannot, and does not, happen.

Category 2 cannot push up or down on the axle very much because they
are oriented mostly sideways to it.

Category 3 would have to pull upward on the axle, requiring them to be
in tension, which is what spokes are designed to do.

Conclusion, the load is supported by the spokes above the wheel. This
does not mean that their natural pretension has to increase very much,
or at all, when the weight is applied, since they no longer have to
support the pretensioning of the bottom spokes. But they are the ones
supporting the wheel by keeping the hub from dropping toward the
ground. You could remove the spokes at the bottom and at the sides
from a loaded wheel at rest and the wheel would not collapse.

Jeff

On 28 Aug 2006 14:41:50 -0700, "Jeff"
wrote:

Forces on a pre-tensioned wheel loaded at the axle:

http://www.astounding.org.uk/ian/wheel/index.html

Cheers,

Carl Fogel

An interesting and thorough analysis at that link, and yet I am not
sure that I believe
the final result. He concludes that the load is supported almost
exclusively by the
bottom few spokes (the ones pointing down toward the road) which are
strongly in
compression. However, long slender members such as spokes cannot
support large compressive loads because of their tendency to buckle
(bend). Also, much of the
strength of a wheel comes from the fact that all the spokes contribute
to the load at
all times. I suspect that he has not accounted fully for the
pretensioning of the spokes.
Jeff

Dear Jeff,

Actually, Ian's whole article is about accounting fully for the
pre-tensioning of the spokes.

True - his problem is not ignoring the pretensioning, sorry.

It's a subject that's been covered repeatedly. That's the nicest
online, detailed explanation that I know of.

You can find pretty much the same engineering analysis and conclusions
in "The Bicycle Wheel" by Jobst Brandt, any edition.

I certainly hope not.


And you can see experimental strain gauge confirmation in figures 10
and 11 Professor Gavin's paper he

http://www.duke.edu/~hpgavin/papers/...heel-Paper.pdf

The icicle-shapes on the graphs show the pre-tensioned spoke losing
and then regaining a large amount of tension as it rolls under the
loaded axle.

Yes, but this does not indicate that they are supporting the load at
that point, but rather that they are *not* supporting the load at that
point. The "icicles" are the spokes at the bottom getting shorter (a
strain gage measures distance) as they lose their pretensioning. The
load is being supported by all of the other spokes *except* for the few
spokes at the bottom that go slack.

Until all the pre-tension is used up, even a string will "support" a
compressive load,

I don't know what you mean by that sentence. A spoke that is
underneath the axle can't support the load whether it is pretensioned
or not because to oppose the load would require it to go into
compression.

which is why emergency repair spokes can be made of
kevlar string and why whole wheels can and have been made of them.

Spokes can be made out of anything that is strong in tension. The fact
that they can be made out of string nicely illustrates the point that
spokes are never in compression, and shows why the spokes at the bottom
of the wheel don't support any of the load.

Jeff

Dear Jeff,

Loss of tension is the same as as compression. The spokes are strong
in tension, but weak in compression, so the trick is to make them work
while still in tension. Until a spoke loses all tension, it isn't
going to bend and fail in compression. (And when it does, it just
rattles a bit, since the nipple isn't fixed to the rim.)

It's the reverse of the trick used in pre-stressed concrete. Concrete
is strong in compression, but weak in tension--a concrete pillar
resists squashing nicely, but will pull apart comparatively easily. So
steel rods and other tricks are used to pre-compress a piece of
concrete that will be used in tension. If it's pre-compressed to a
thousand pounds, you have pull on it with a thousand-pound force
before the pre-compression is used up and the concrete begins to act
in its normal, uncompressed, feeble fashion.

The load on the axle is obviously supported by the spokes--nothing
else joins them to the rim.

Theory predicts and experiment confirms that the spokes under the axle
react to the load by losing a large amount of tension. Engineers using
the same tools keep coming up with the same figures--Jobst, Ian, and
others.

The 5 spokes right under the axle in Ian's example lose a large amount
of tension, just as if they were stiff wooden spokes--loss of tension
is the same as compression in the engineering world.

The other 31 spokes all gain some tension.

A common mistake is to think that the all the small tension increases
in other 31 spokes must add up to a large total.

They don't.

The reason is that they gain tension almost all the way around the
clock, so to speak--many of the spokes whose tension increases are
pulling the axle down, not pulling it up. You can only calculate the
lift (pure vertical force) of a 12 newton tension increase if you also
know the angle at which it's pulling:

"The tension is taken from the analysis. Multiplying the axial force
by the total angle factor gives the vertical component of the force.
This is the lift that one particular spoke is contributing to the hub.
If we add them up, we should get 1000, and luckily we do."

"Then, I've split the lift forces into two columns, depending upon
whether the spoke force was tensile or compressive. This is to see if
the hub hangs from tensile spokes, or stands on compressive ones."

"There are 31 tensile spokes. On average they contribute 1.436 N (0.14
kg, just under a third of a pound) each to holding up the hub.
There are 5 compressive spokes. On average they contribute 191.097N
(19 kg, just over 42 lbs) each to holding up the hub."

"Put it another way - the average compressive spoke contributes 133
times as much lift force as the average tensile spoke."

http://www.astounding.org.uk/ian/wheel/index.html

As Ian's table shows, the lift from the 5 bottom spokes is 955.5
newtons, while the 31 other spokes contribute a whopping 44.5 newtons
of lift.

Until you understand that loss of tension and compression are the same
thing, the whole analysis is really, really annoying and ridiculous. I
started out with even stronger feelings than you that the engineers'
analysis was insane.

Cheers,

Carl Fogel

Jeff Thomas writes:

Maybe you should pluck spokes at various locations around the wheel
and nor which ones (by change in tone) are affected by placing a
load on the wheel. Let me tell you in advance what you will find
(for pure vertical loading). The only spokes affected by the load
will be the three or four spokes at the bottom directed at the road
from the hub.

Those spokes are affected much more than the others by the load, I
agree. They undergo a dramatic loss of pretensioning.

If the spokes at the top are supporting the wheel, as you propose,
then they would be affected by the load, but they are not. I think
you are, as many others, not visualizing these things
algebraically. The problem is much like adding debits and credits
to a bank account.

Here is my best explanation: The downward force on the axle due to
the weight of the rider must must be countered by an upward force of
equal magnitude exerted by the spokes on the hub. Thats just
elementary statics, what I think you are calling "debits and
credits"

Lets divide the spokes into three somewhat imprecise categories:
1. Spokes at the bottom, underneath the axle.
2. Spokes to the side of the axle (mostly horizontal)
3. Spokes above the axle (mostly vertical)

How spokes in each category help to exert an upward force on the
axle?

Category 1 would have to "push" upward on the axle from below. This
would put them in compression, which cannot, and does not, happen.

That's where the problem lies. Is "pulling down less" the same as
pushing up more? That's an algebraic sum.

Category 2 cannot push up or down on the axle very much because they
are oriented mostly sideways to it.

Category 3 would have to pull upward on the axle, requiring them to
be in tension, which is what spokes are designed to do.

Conclusion, the load is supported by the spokes above the wheel.
This does not mean that their natural pretension has to increase
very much, or at all, when the weight is applied, since they no
longer have to support the pretensioning of the bottom spokes. But
they are the ones supporting the wheel by keeping the hub from
dropping toward the ground. You could remove the spokes at the
bottom and at the sides from a loaded wheel at rest and the wheel
would not collapse.

Your conclusion is based on that semantic problem and one of statics
(forces). I take it you are not a structural engineer and therefore
have not been involved with superposition of forces. This is such a
case and the net tension in the wheel does not alter its function. It
only absolves the wires of buckling.

"the Bicycle Wheel" has been in print since 1981 and has been reviewed
by many scientists among which we find Karl Wiedemer, professor of
mechanical engineering who published a paper on this item that year.
You are not the first to see it differently but in that, disagree with
experts in the art of structural analysis who agree with the work.

Jobst Brandt

On 28 Aug 2006 13:50:54 -0700, wrote:

That article makes the simplifying assumption that if you can hang from
a rope, you can sit on it.

A foolish linearity is the hobgoblin of little minds.

Dear Kendall,

Cut a rubber band, tie it to something that weighs a few pounds, and
take it to the post office.

Note the weight on the digital scale.

Use one finger to pull up on the rubber band until half the weight
disappears from the scale.

Push your finger down a bit with your other hand.

The digital scale will indicate that you are pushing down on it
through the stretched rubber band.

The scale will keep showing how hard you push until all the rubber
band's pre-tension is used up.

A string will do the same thing, but its range of elasticity is so
small that we can't see what happens--and the range of elasticity of
steel spokes is even smaller.

I agree that the ability of a pre-tensioned member to function in
compression is a very annoying principle.

Cheers,

Carl Fogel

Unfortunately, I missed this sentence which redefines the variables:

"This is a change from the unloaded state, so compression doesn't
actually mean compression, it means reduction in tension."

This is a perturbation in force.

In article , says...
On 28 Aug 2006 13:50:54 -0700, wrote:

That article makes the simplifying assumption that if you can hang from
a rope, you can sit on it.

A foolish linearity is the hobgoblin of little minds.

Dear Kendall,

Cut a rubber band, tie it to something that weighs a few pounds, and
take it to the post office.

Note the weight on the digital scale.

Use one finger to pull up on the rubber band until half the weight
disappears from the scale.

Push your finger down a bit with your other hand.

The digital scale will indicate that you are pushing down on it
through the stretched rubber band.

The scale will keep showing how hard you push until all the rubber
band's pre-tension is used up.

A string will do the same thing, but its range of elasticity is so
small that we can't see what happens--and the range of elasticity of
steel spokes is even smaller.

I agree that the ability of a pre-tensioned member to function in
compression is a very annoying principle.

Cheers,

Carl Fogel

I think the issue here is more in the semantics of the term "function in compression" than in any mis-understanding of
the underlying physics.

The problem comes as follows. Assume the parcel has weight W, and you stretch the rubber band until the scale registers
a weight of W/2. Note that the band now has tension W/2, but as this is your equilibrium starting point you ignore it.
Now you press down on your hand with a force f W/2 and the weight registers by the scale increases by precisely the
same amount f. In effect, the force of your hand on your finger appears to be transferred through the rubber band and,
as you are pressing down on it, this must be a compressive force. At this point it is reasonable to claim that the
rubber band is "function[ing] in compression" . But now, increase the force until f = W/2. According to the "function
in compression" arguement the rubber band is now passing a compressive load of W/2 to the parcel to increase the scale
reading from the initial W/2 equilibrium to W. But if your helpful postmaster now leans across the bench wielding a
pair of scissors, he/she can now cut the rubber band and the weight registered by the scales remains unchanged. So now
where does this mysterious extra W/2 on the scales come from. It can no longer be considered as a compressive force
acting through the rubber band because the band is no longer in the picture.

So in a statically loaded bicycle wheel the question of which spokes are responding to the force applied by the rider
depends on the starting point. Considering a tensioned wheel as the starting point it is reasonable to claim that the
bottom spoke acts in compression. Considering the individual untensioned spokes and rim as the starting point it is
reasonable to claim that the spokes act in concert to relieve the load - with the bottom spoke under the least tension
and the lateral spokes under the greatest tension. Neither view is more right or wrong than the other.

Regards,
Mike

Mike writes:

So in a statically loaded bicycle wheel the question of which spokes
are responding to the force applied by the rider depends on the
starting point. Considering a tensioned wheel as the starting point
it is reasonable to claim that the bottom spoke acts in
compression. Considering the individual untensioned spokes and rim
as the starting point it is reasonable to claim that the spokes act
in concert to relieve the load - with the bottom spoke under the
least tension and the lateral spokes under the greatest
tension. Neither view is more right or wrong than the other.

But one may be more useful than the other. Note that these are static
analyses. Consider a buoy anchored to the sea floor via a rope.
Tapping on rope at the bottom transmits a compressional wave through
the rope that vibrates the buoy. It's clear that the rope is
transmitting the force wave, not the ocean.

--
Joe Riel

On Tue, 29 Aug 2006 16:09:19 +1200, Mike
wrote:

In article , says...
On 28 Aug 2006 13:50:54 -0700, wrote:

That article makes the simplifying assumption that if you can hang from
a rope, you can sit on it.

A foolish linearity is the hobgoblin of little minds.

Dear Kendall,

Cut a rubber band, tie it to something that weighs a few pounds, and
take it to the post office.

Note the weight on the digital scale.

Use one finger to pull up on the rubber band until half the weight
disappears from the scale.

Push your finger down a bit with your other hand.

The digital scale will indicate that you are pushing down on it
through the stretched rubber band.

The scale will keep showing how hard you push until all the rubber
band's pre-tension is used up.

A string will do the same thing, but its range of elasticity is so
small that we can't see what happens--and the range of elasticity of
steel spokes is even smaller.

I agree that the ability of a pre-tensioned member to function in
compression is a very annoying principle.

Cheers,

Carl Fogel

I think the issue here is more in the semantics of the term "function in compression" than in any mis-understanding of
the underlying physics.

The problem comes as follows. Assume the parcel has weight W, and you stretch the rubber band until the scale registers
a weight of W/2. Note that the band now has tension W/2, but as this is your equilibrium starting point you ignore it.
Now you press down on your hand with a force f W/2 and the weight registers by the scale increases by precisely the
same amount f. In effect, the force of your hand on your finger appears to be transferred through the rubber band and,
as you are pressing down on it, this must be a compressive force. At this point it is reasonable to claim that the
rubber band is "function[ing] in compression" . But now, increase the force until f = W/2. According to the "function
in compression" arguement the rubber band is now passing a compressive load of W/2 to the parcel to increase the scale
reading from the initial W/2 equilibrium to W. But if your helpful postmaster now leans across the bench wielding a
pair of scissors, he/she can now cut the rubber band and the weight registered by the scales remains unchanged. So now
where does this mysterious extra W/2 on the scales come from. It can no longer be considered as a compressive force
acting through the rubber band because the band is no longer in the picture.

So in a statically loaded bicycle wheel the question of which spokes are responding to the force applied by the rider
depends on the starting point. Considering a tensioned wheel as the starting point it is reasonable to claim that the
bottom spoke acts in compression. Considering the individual untensioned spokes and rim as the starting point it is
reasonable to claim that the spokes act in concert to relieve the load - with the bottom spoke under the least tension
and the lateral spokes under the greatest tension. Neither view is more right or wrong than the other.

Regards,
Mike

Dear Mike,

If you cut a spoke or a rubber band that's in tension, it's no longer
a pre-tensioned structure. This is a common mistake in discussions of
the wheel's behavior.

The response of the pre-tensioned spokes on a bicycle wheel to a load
on the axle is predictable and measurable by experiment.

The spokes directly under the axle account for 95% of the change from
an unloaded state.

The other spokes gain a little tension, but their measured vertical
force (the result of the tension gain and its angle) supports only
about 5% of the load.

The behavior is not obvious, which leads to attempts to prove that
something else must happen, but no usable theory has ever replaced the
one that Jobst and Ian worked out. In engineering circles, it's about
as controversial as calculating the area of a circle.

The pre-tensioned spokes must somehow tranfer the load from the axle
to the ground. Their tension changes can be predicted, and the changes
are confirmed by strain gauge testing.

Objections to the model used by Jobst, Ian, and other engineers always
involve spoke strains that cannot be calculated and never show up in
tests.

Incidentally, the lateral spokes don't show significantly different
tension changes than the rest of the non-bottom spokes. Check Ian's
tables again:

http://www.astounding.org.uk/ian/wheel/index.html

Nor does testing show significant differences for the lateral spokes.
See Professor Gavin's test result graphs again, figures 10 and 11:

http://www.duke.edu/~hpgavin/papers/...heel-Paper.pdf

Cheers,

Carl Fogel

wrote:
Jeff Thomas writes:

Experiment seems to confirm theory.

The experiment confirms that the spokes do indeed go slack as they
pass under the hub. It doesn't in anyway prove that they are
supporting the wheel through compressive loading before they go
slack.

All the spokes are accounted for in both theory and experiment.

What else besides the spokes connects the wheel to the loaded axle?

If the forces don't show up anywhere else, what supports the load?

Ian's page goes through this in patient detail--the increase in
tension in the other spokes isn't anywhere near enough to support
the load.

Then his analysis must be wrong. The load must be supported by the
spokes that are not underneath the axle, because those spokes are
unable to push upward against the downward force exerted by the
axle. Remember, there are only a few spokes at the bottom, and some
30 spokes not at the bottom.

Maybe you should pluck spokes at various locations around the wheel
and nor which ones (by change in tone) are affected by placing a load
on the wheel. Let me tell you in advance what you will find (for pure
vertical loading). The only spokes affected by the load will be the
three or four spokes at the bottom directed at the road from the hub.

I've already done that experiment and reported the results a while back
on this ng, where the results were called "unexpected". The spokes
parallel to the ground had their pich rise significantly.

Joe Riel wrote:
writes:

wrote:
Until all the pre-tension is used up, even a string will "support" a
compressive load

A string may support a compressive load if it is pretensioned but if
there is nothing to support the string it doesn't matter. There is no
way for a spoke under compressive load to support anything except by
its nipple's friction with the spoke hole. Any compressive load will
try to push the spoke out the outside of the rim.

What would you say if the string were replaced with a chain that was
welded to the rim? How is the link to link interface of the chain any
different from the nipple to rim interface?

There is a major difference: the nipple to rim interface is essentially
nonexistent when the compressive force is applied perpendicular to it;
compressive force applied to a chain in the same direction is
perpendicular to the direction of force that would tend to buckle the
chain. Tensioning counteracts the tendency of the chain to buckle at
the links, as it counteracts the tendency of the spoke to buckle, but
the problem is not the spoke buckling, it's the spoke telescoping into
the spoke hole.

The point being, the pretension in the spoke acts on the nipple to rim
interface just as it does on the links (or string or spoke).

No, it doesn't.