Spoke tension and buckling
Ben C
Manuals".
This is what I've gathered so far, which might be wrong:
1. High spoke tension decreases propensity to buckling because so long
as no spokes go slack, the wheel will hold its shape. Higher tension
means higher loads before it goes slack, where "loads" includes
hitting bumps in the road.
2. High spoke tension increases propensity to buckling. What's the
explanation here?
benc> But does it increase resistance to buckling or not? If not, why
benc> not?
jim beam> michael press seems to have the best handle on this.
jim beam> increasing tension increases propensity to buckle. the only
jim beam> wheel i've ever had spontaneously taco on me was a jobst-tight
jim beam> wheel.
Michael Press explained how in the buckled wheel, tension is thought to
be reduced in all spokes and the wheel is in a lower energy state. It
was thought that probably true and tacoed were both local energy minima
with true being higher energy than tacoed. But I'm not sure anything was
said about whether high tension increases propensity to buckling.
We know that if you tension your MA-2 too high then it buckles when you
stress-relieve it. Then you reduce the tension a bit. In theory hitting
bumps isn't supposed to raise the tension, so you should be able to ride
around safely at just below buckle-tension. But perhaps nasty bumps do
sometimes raise the tension a bit?
If true is a local energy minimum, one question is how steep are the
sides of the energy function at that point, with lower or higher spoke
tension. In other words, how much deformation does it take to snap you
into a buckle, and high vs low tension?
Certainly a wheel doesn't buckle when stress-relieved unless tension is
very high.
What does the graph of buckle-propensity against tension look like? Does
it increase monotonically with tension, or does it go downwards up to
some quite high tension and then suddenly shoot upwards at the end?
jobst....@stanfordalumni.org
> I think it's time to start a new thread. This came from "Maintenance
> Manuals".
> This is what I've gathered so far, which might be wrong:
> 1. High spoke tension decreases propensity to buckling because so
> long as no spokes go slack, the wheel will hold its shape.
> Higher tension means higher loads before it goes slack, where
> "loads" includes hitting bumps in the road.
You should define "buckling" first. While truing and tensioning a
wheel, rim buckling (becoming untrue in a slight taco shape) is a sign
of excessive tension. Collapsing a wheel from overload occurs when
spoke become sack and no longer furnish lateral stability to the wheel
in conjunction with lateral force.
> 2. High spoke tension increases propensity to buckling. What's the
> explanation here?
You haven't followed the discussion, possibly because there were rude
distractions. The buckling effect has been discussed for some time
now under the previous heading.
>>> But does it increase resistance to buckling or not? If not, why
>>> not?
>> michael press seems to have the best handle on this. increasing
>> tension increases propensity to buckle. the only wheel i've ever
>> had spontaneously taco on me was a jobst-tight wheel.
> Michael Press explained how in the buckled wheel, tension is thought
> to be reduced in all spokes and the wheel is in a lower energy
> state. It was thought that probably true and tacoed were both local
> energy minima with true being higher energy than tacoed. But I'm
> not sure anything was said about whether high tension increases
> propensity to buckling.
I explained that in previous responses. I can't imagine how you
missed that. Press also mentioned the same characteristic in that
there are fairly slack spokes in a forcefully pretzeled wheel. This
is not bucking from truing and tensioning. The two effects not being
parallel.
> We know that if you tension your MA-2 too high then it buckles when
> you stress-relieve it. Then you reduce the tension a bit. In
> theory hitting bumps isn't supposed to raise the tension, so you
> should be able to ride around safely at just below buckle-tension.
> But perhaps nasty bumps do sometimes raise the tension a bit?
Don't confuse that effect with a "tacoed" wheel.
> If true is a local energy minimum, one question is how steep are the
> sides of the energy function at that point, with lower or higher
> spoke tension. In other words, how much deformation does it take to
> snap you into a buckle, and high vs low tension?
You are mixing to different effects.
> Certainly a wheel doesn't buckle when stress-relieved unless tension
> is very high.
How "very"? That is a way of gauging whether a wheel is too tight and
it is dependent on the rim and number of spokes as I discussed at
length here and in "the Bicycle Wheel". It is not a single value for
different numbers of spokes.
> What does the graph of buckle-propensity against tension look like?
> Does it increase monotonically with tension, or does it go downward
> up to some quite high tension and then suddenly shoot upward at the
> end?
It is a single value so there is no curve. If you mean how does the
number of spokes affect it, it is not a linear effect because fewer
spokes act as fewer guy wires to keep the rim straight while at the
same time require more tension for the wheel to carry the same load
(there being fewer spokes in the load affected zone.
Jobst Brandt
jim beam
no mention of the wrong type of rims jobst? it makes all your dodging
and weaving about quantification much more amusing.
Ben C
>> Michael Press explained how in the buckled wheel, tension is thought
>> to be reduced in all spokes and the wheel is in a lower energy
>> state. It was thought that probably true and tacoed were both local
>> energy minima with true being higher energy than tacoed. But I'm
>> not sure anything was said about whether high tension increases
>> propensity to buckling.
>
> I explained that in previous responses. I can't imagine how you
> missed that. Press also mentioned the same characteristic in that
> there are fairly slack spokes in a forcefully pretzeled wheel.
Are you saying that it _follows_ from the fact that a buckled wheel has
looser spokes that spokes going slack in use leads to buckling, and
therefore that over-tight spokes does not?
If so, I did miss that part. It's not clear to me that that connection
is so easy to make.
[...]
>> What does the graph of buckle-propensity against tension look like?
>> Does it increase monotonically with tension, or does it go downward
>> up to some quite high tension and then suddenly shoot upward at the
>> end?
>
> It is a single value so there is no curve.
I think you're saying pretzeling (as opposed to buckling) is a single
value at a high tension. But collapsing from slack spokes presumably is
some sort of curve, with propensity to collapse going down as tension is
increased?
> If you mean how does the number of spokes affect it, [...]
No, not asking about number of spokes, just how tension affects
propensity to buckle.
jobst....@stanfordalumni.org
If there are no significant side loads, even wheels with slack spokes
do not collapse (unless greatly overloaded). Lateral collapse occurs
from too loose spokes when leaning the bicycle in the standing
position while pedaling hard, or most commonly after a crossing a
patch of poor traction that induces side slip. If traction recovers
after a slide slip, wheels are loaded laterally and a loosely spoked
wheels more readily fold. I don't call that buckling because buckling
is, in engineering terms, is lateral bowing of a column from
compression and is not a result of plastic deformation (see pole
vault). It is an elastic phenomenon that if carried far enough will
cause yield. That is not what causes most wheel collapse but rather
side loading.
Jobst Brandt
jim beam
jeepers jobst!! have you no shame? you blithely dismiss /any/
possibility that lateral loads are /any/ factor in wheel application
because it would contradict your entrenched position that spoke tensions
never increase, only decrease. and yet, here you are resorting to
lateral loading because it could contribute to lateral collapse. could
it be that you're trying to wriggle and squirm away from the fact that
such a collapse is much more likely the result of excess spoke tension?
> If traction recovers
> after a slide slip, wheels are loaded laterally and a loosely spoked
> wheels more readily fold.
no, excessively compressed rims more readily fold.
> I don't call that buckling because buckling
> is, in engineering terms, is lateral bowing of a column from
> compression and is not a result of plastic deformation (see pole
> vault).
different material - false analogy.
> It is an elastic phenomenon that if carried far enough will
> cause yield. That is not what causes most wheel collapse but rather
> side loading.
excess rim compression caused by excess spoke tension causes rim buckling.
Ben C
[...]
> no, excessively compressed rims more readily fold.
Suppose the rim is highly compressed by high spoke tension.
Then I hit a bump. The compression from the bump isn't just added to the
compression from the spoke tension, because as the rim starts to
compress, those spokes start to go slack. So the compression on the rim
stays about the same until the spokes have gone completely slack.
Is this right/wrong/too simplistic?
jim beam
the hoop compression is a function of spoke tension. i doubt locally
slackening spokes will reduce the compression of the whole because it's
an equilibrium structure and the bump load has to be taken up by the
whole. as far as local bending is concerned, it's a bending beam, with
a neutral plane, and compressive and tensile regions accordingly.
Peter Cole
> Ben C wrote:
>> On 2007-10-09, jim beam <spamv...@bad.example.net> wrote:
>> [...]
>>> no, excessively compressed rims more readily fold.
>>
>> Suppose the rim is highly compressed by high spoke tension.
>>
>> Then I hit a bump. The compression from the bump isn't just added to the
>> compression from the spoke tension, because as the rim starts to
>> compress, those spokes start to go slack. So the compression on the rim
>> stays about the same until the spokes have gone completely slack.
>>
>> Is this right/wrong/too simplistic?
>
> the hoop compression is a function of spoke tension. i doubt locally
> slackening spokes will reduce the compression of the whole because it's
> an equilibrium structure and the bump load has to be taken up by the
> whole.
In other words, no, it's not "too simplistic" a bump won't significantly
change the rim compression, which is why you can operate a wheel within
a fairly small safety margin of compression buckling. The limit is more
what happens to wheel shape when you break a spoke.
> as far as local bending is concerned, it's a bending beam, with
> a neutral plane, and compressive and tensile regions accordingly.
A beam which is supported by the spokes, with the consequence that the
stiffness dramatically lowers when the support is removed.
The net is you want to have tight spokes to prevent excessive (damaging)
deflection of the rim under overload.
Ben C
> jim beam wrote:
>> Ben C wrote:
>>> On 2007-10-09, jim beam <spamv...@bad.example.net> wrote:
>>> [...]
>>>> no, excessively compressed rims more readily fold.
>>>
>>> Suppose the rim is highly compressed by high spoke tension.
>>>
>>> Then I hit a bump. The compression from the bump isn't just added to the
>>> compression from the spoke tension, because as the rim starts to
>>> compress, those spokes start to go slack. So the compression on the rim
>>> stays about the same until the spokes have gone completely slack.
>>>
>>> Is this right/wrong/too simplistic?
>> as far as local bending is concerned, it's a bending beam, with
>> a neutral plane, and compressive and tensile regions accordingly.
>
> A beam which is supported by the spokes, with the consequence that the
> stiffness dramatically lowers when the support is removed.
Sticking to local bending, _which_ spokes is the region of rim hitting
the bump supported by?
If it's the ones running from the hub to the bump, then those spokes
going slack would reduce the stiffness as you say.
But if it's all the other spokes, which aren't going slack, providing
the support, then the stiffness of the rim (locally) wouldn't change
much at the point where you hit the bump. Or it might even get stiffer
as the spoke goes slack (see below).
If the spokes weren't pretensioned and this was a cart wheel, it would
seem quite intuitive that it's the spokes near the bump that are
propping up the rim at that point. And people keep saying that a
pretensioned wheel behaves the same as a cart wheel.
But does it in every detail? It seems like it should be the other way
round. The effective stiffness of the rim at the bump should _increase_
as those spokes above it go slack, reaching its maximum stiffness when
they are completely slack.
Suppose you have a helical compression coil spring held in compression
with a rubber band around it. The spring holds the band in tension and
the band holds the spring in compression. They're in equilibrium. Both
obey Hooke's Law.
What's the response of the whole thing to compression and tension? If
you compress it, you're squashing the spring, but releasing the band a
bit. If you stretch it, you're stretching the band, and relaxing the
spring a bit.
If K is the spring constant of the spring, and L the spring constant of
the band, I would expect the spring constant in compression (for small
deviations from the equilibrium position) to be K - L and in tension
(for small deviations from the equilibrium position) to be L - K.
(But I would expect the system as a whole not to obey Hooke's Law-- it
should be like a spring that gets a bit stiffer the further you stretch
or compress it as the force contribution from the other element gets
less the further you go).
So, if the stiffness of the rim on its own is K and that of the spoke L,
then squashing the rim in (locally) should react like something with a
stiffness of K - L. When eventually L goes to zero (slack spoke) the
stiffness is K, which is higher than K - L.
Where have I gone wrong here?
jobst....@stanfordalumni.org
> [...]
> > no, excessively compressed rims more readily fold.
> Suppose the rim is highly compressed by high spoke tension.
I don't understand what you mean by this. If the wheel can withstand
hard rim braking it is not over structurally tensioned although rims
without sockets can fail by cracking (reducing tension in those
spokes) with use.
> Then I hit a bump. The compression from the bump isn't just added
> to the compression from the spoke tension, because as the rim starts
> to compress, those spokes start to go slack. So the compression on
> the rim stays about the same until the spokes have gone completely
> slack.
I think the deflection diagrams in "the Bicycle Wheel" make that
clearer as was mentioned about diagram being worth many words.
> Is this right/wrong/too simplistic?
I think you can test this on your own wheels (if they don't have fewer
than 20 spokes) by having someone pluck spokes while you stand on one
pedal and lean the bicycle with both feet in the pedals as in riding
while standing. You'll notice that the bottom spokes of the wheels
lose tension unequally, the "upperside" ones losing more tension than
the "underside" spokes. You'll need to lean the bicycle more than
what occurs while riding to get any other response. That would be at
an angle that one cannot pedal the bicycle.
Jobst Brandt
Joe Riel
> Suppose you have a helical compression coil spring held in compression
> with a rubber band around it. The spring holds the band in tension and
> the band holds the spring in compression. They're in equilibrium. Both
> obey Hooke's Law.
>
> What's the response of the whole thing to compression and tension? If
> you compress it, you're squashing the spring, but releasing the band a
> bit. If you stretch it, you're stretching the band, and relaxing the
> spring a bit.
>
> If K is the spring constant of the spring, and L the spring constant of
> the band, I would expect the spring constant in compression (for small
> deviations from the equilibrium position) to be K - L and in tension
> (for small deviations from the equilibrium position) to be L - K.
The spring and band are in parallel. The combined stiffness is K + L,
regardless whether the system is compressed or stretched.
--
Joe Riel
Peter Cole
> On 2007-10-09, Peter Cole <peter...@comcast.net> wrote:
>> jim beam wrote:
>>> Ben C wrote:
>>>> On 2007-10-09, jim beam <spamv...@bad.example.net> wrote:
>>>> [...]
>>>>> no, excessively compressed rims more readily fold.
>>>> Suppose the rim is highly compressed by high spoke tension.
>>>>
>>>> Then I hit a bump. The compression from the bump isn't just added to the
>>>> compression from the spoke tension, because as the rim starts to
>>>> compress, those spokes start to go slack. So the compression on the rim
>>>> stays about the same until the spokes have gone completely slack.
>>>>
>>>> Is this right/wrong/too simplistic?
> [...]
>>> as far as local bending is concerned, it's a bending beam, with
>>> a neutral plane, and compressive and tensile regions accordingly.
>> A beam which is supported by the spokes, with the consequence that the
>> stiffness dramatically lowers when the support is removed.
>
> Sticking to local bending, _which_ spokes is the region of rim hitting
> the bump supported by?
The spokes that are right there. Think railroad track & ties.
> If it's the ones running from the hub to the bump, then those spokes
> going slack would reduce the stiffness as you say.
>
> But if it's all the other spokes, which aren't going slack, providing
> the support, then the stiffness of the rim (locally) wouldn't change
> much at the point where you hit the bump. Or it might even get stiffer
> as the spoke goes slack (see below).
It's not.
> If the spokes weren't pretensioned and this was a cart wheel, it would
> seem quite intuitive that it's the spokes near the bump that are
> propping up the rim at that point. And people keep saying that a
> pretensioned wheel behaves the same as a cart wheel.
Because it does. And if those spokes were springy, how would the
combination of spoke spring and rim spring combine?
> But does it in every detail? It seems like it should be the other way
> round. The effective stiffness of the rim at the bump should _increase_
> as those spokes above it go slack, reaching its maximum stiffness when
> they are completely slack.
You are confusing stiffness with force. Stiffness is the change in force
with displacement. The force increases, the stiffness does not change
(up to the point where spokes go slack).
> Suppose you have a helical compression coil spring held in compression
> with a rubber band around it. The spring holds the band in tension and
> the band holds the spring in compression. They're in equilibrium. Both
> obey Hooke's Law.
>
> What's the response of the whole thing to compression and tension? If
> you compress it, you're squashing the spring, but releasing the band a
> bit. If you stretch it, you're stretching the band, and relaxing the
> spring a bit.
This is a bit non-intuitive, draw the graphs for each, paying attention
to sign, and you'll see that the slope (force over displacement) of the
combination is greater than either alone. The stiffnesses add.
> If K is the spring constant of the spring, and L the spring constant of
> the band, I would expect the spring constant in compression (for small
> deviations from the equilibrium position) to be K - L and in tension
> (for small deviations from the equilibrium position) to be L - K.
Think about it, when you squash the spring, the band eases, "helping"
you less, making the spring "seem" stiffer. You really have to draw it
out I think to fully grasp it. The spring constants add.
> (But I would expect the system as a whole not to obey Hooke's Law-- it
> should be like a spring that gets a bit stiffer the further you stretch
> or compress it as the force contribution from the other element gets
> less the further you go).
>
> So, if the stiffness of the rim on its own is K and that of the spoke L,
> then squashing the rim in (locally) should react like something with a
> stiffness of K - L. When eventually L goes to zero (slack spoke) the
> stiffness is K, which is higher than K - L.
It obeys Hooke's Law, it must, it's linear, so you just superimpose
forces. Plot combined force against displacement. There is no need to
think about equilibrium or small displacements, assume perfect springs
of arbitrary constants. Don't think about what it "should be".
> Where have I gone wrong here?
Try drawing it out, it's hard to do in your head.
Ben C
> Ben C wrote:
>> On 2007-10-09, Peter Cole <peter...@comcast.net> wrote:
>>> jim beam wrote:
>>>> Ben C wrote:
>>>>> On 2007-10-09, jim beam <spamv...@bad.example.net> wrote:
>>>>> [...]
>>>>>> no, excessively compressed rims more readily fold.
>>>>> Suppose the rim is highly compressed by high spoke tension.
>>>>>
>>>>> Then I hit a bump. The compression from the bump isn't just added to the
>>>>> compression from the spoke tension, because as the rim starts to
>>>>> compress, those spokes start to go slack. So the compression on the rim
>>>>> stays about the same until the spokes have gone completely slack.
>>>>>
>>>>> Is this right/wrong/too simplistic?
>> [...]
>>>> as far as local bending is concerned, it's a bending beam, with
>>>> a neutral plane, and compressive and tensile regions accordingly.
>>> A beam which is supported by the spokes, with the consequence that the
>>> stiffness dramatically lowers when the support is removed.
>>
>> Sticking to local bending, _which_ spokes is the region of rim hitting
>> the bump supported by?
>
> The spokes that are right there. Think railroad track & ties.
Yes, that was what I was thinking of.
[...]
>> But does it in every detail? It seems like it should be the other way
>> round. The effective stiffness of the rim at the bump should _increase_
>> as those spokes above it go slack, reaching its maximum stiffness when
>> they are completely slack.
>
> You are confusing stiffness with force. Stiffness is the change in force
> with displacement. The force increases, the stiffness does not change
> (up to the point where spokes go slack).
No, whatever I'm confusing, it's not stiffness and force. See [1] below.
>> Suppose you have a helical compression coil spring held in compression
>> with a rubber band around it. The spring holds the band in tension and
>> the band holds the spring in compression. They're in equilibrium. Both
>> obey Hooke's Law.
>>
>> What's the response of the whole thing to compression and tension? If
>> you compress it, you're squashing the spring, but releasing the band a
>> bit. If you stretch it, you're stretching the band, and relaxing the
>> spring a bit.
>
> This is a bit non-intuitive, draw the graphs for each, paying attention
> to sign, and you'll see that the slope (force over displacement) of the
> combination is greater than either alone. The stiffnesses add.
If you pay attention to the sign, yes. But I will draw some more graphs.
>> If K is the spring constant of the spring, and L the spring constant of
>> the band, I would expect the spring constant in compression (for small
>> deviations from the equilibrium position) to be K - L and in tension
>> (for small deviations from the equilibrium position) to be L - K.
>
> Think about it, when you squash the spring, the band eases, "helping"
> you less, making the spring "seem" stiffer. You really have to draw it
> out I think to fully grasp it.
[1] That's exactly what I thought. Which is why I think the whole
assembly gets more stiff the further you compress (or extend) it, as the
other element helps you less.
Anyway, thanks for your input, I think I can work this out.
Ben C
> Ben C? writes:
>
>> [...]
>> > no, excessively compressed rims more readily fold.
>
>> Suppose the rim is highly compressed by high spoke tension.
>
> I don't understand what you mean by this.
Just that the spokes are pulling the rim inwards compressing it.
[...]
>> Is this right/wrong/too simplistic?
>
> I think you can test this on your own wheels (if they don't have fewer
> than 20 spokes) by having someone pluck spokes while you stand on one
> pedal and lean the bicycle with both feet in the pedals as in riding
> while standing. You'll notice that the bottom spokes of the wheels
> lose tension unequally, the "upperside" ones losing more tension than
> the "underside" spokes.
Interesting, but my question was about something else-- how spoke
tension affects what happens to the rim when you hit a bump.
jobst....@stanfordalumni.org
>>> [...]
>>>> no, excessively compressed rims more readily fold.
>>> Suppose the rim is highly compressed by high spoke tension.
>> I don't understand what you mean by this.
> Just that the spokes are pulling the rim inward compressing it.
> [...]
>>> Is this right/wrong/too simplistic?
>> I think you can test this on your own wheels (if they don't have
>> fewer than 20 spokes) by having someone pluck spokes while you
>> stand on one pedal and lean the bicycle with both feet in the
>> pedals as in riding while standing. You'll notice that the bottom
>> spokes of the wheels lose tension unequally, the "upperside" ones
>> losing more tension than the "underside" spokes.
> Interesting, but my question was about something else-- how spoke
> tension affects what happens to the rim when you hit a bump.
As was mentioned in another thread, a rim is an elastically supported
beam and acts like a railroad track supported by elastic cross ties.
By performing a linear mental development of the circular wheel into a
linear one, some of these effects are more readily apparent.
Consider the spokes as not becoming slack and that they have an axial
elasticity. The deflections (changes in tension and length) require
forces that make rim shape mainly irrelevant. Deflections are on the
scale of 0.001's of inches as you can see in the diagrams of FEA
analysis. You can easily deform a bare rim radially with thumb
pressure but you can't do that on a complete wheel where spoke
stiffness is the measure. Please look at the deflection diagrams that
make this clear.
Jobst Brandt
carl...@comcast.net
Dear Jobst,
How much does a bare deep modern rim deform under thumb pressure?
I used to think that bare bicycle rims with no spokes flattened quite
easily, but posts on RBT and turn-of-the century factory claims that a
man could stand on a bare no-spoke highwheeler rim have made me wonder
about this.
I don't have any bare rims, so I'm curious if anyone can put a 10 or
20 pound weight on some aero-style hoops and report what happens.
I expect that bare no-spoke rims deform more than a tensioned wheel,
but I have to admit that I've never tried the experiment and don't
know what to expect--a tenth of an inch, an inch, two inches, a bent
rim?
Cheers,
Carl Fogel
jobst....@stanfordalumni.org
> Dear Jobst,
> How much does a bare deep modern rim deform under thumb pressure?
Oh several times as much as a standard cross section rim 36 spoked wheel,
0.050" at least. As I said, the RIM.
> I used to think that bare bicycle rims with no spokes flattened quite
> easily, but posts on RBT and turn-of-the century factory claims that a
> man could stand on a bare no-spoke highwheeler rim have made me wonder
> about this.
They lied. Even though those were steel rims, many were U-shaped flat
material. You can sit on a bare MA-2 but it deforms enormously
(about 1.5").
> I don't have any bare rims, so I'm curious if anyone can put a 10 or
> 20 pound weight on some aero-style hoops and report what happens.
I don't have any either, but its not anything like a spoked wheel,
whose radial rigidity is almost entirely the support element
elasticity (the spokes). Didn't you recently assess what that was?
> I expect that bare no-spoke rims deform more than a tensioned wheel,
> but I have to admit that I've never tried the experiment and don't
> know what to expect--a tenth of an inch, an inch, two inches, a bent
> rim?
Well, my MA-2 did not take a set after I sat on it.
Jobst Brandt
jobst....@stanfordalumni.org
> Dear Jobst,
> How much does a bare deep modern rim deform under thumb pressure?
Oh several times as much as a standard cross section rim 36 spoked wheel,
0.050" at least. As I said, the RIM.
> I used to think that bare bicycle rims with no spokes flattened quite
> easily, but posts on RBT and turn-of-the century factory claims that a
> man could stand on a bare no-spoke highwheeler rim have made me wonder
> about this.
They lied. Even though those were steel rims, many were U-shaped flat
material. You can sit on a bare MA-2 but it deforms enormously
(about 1.5").
> I don't have any bare rims, so I'm curious if anyone can put a 10 or
> 20 pound weight on some aero-style hoops and report what happens.
I don't have any deep-V rims either, but its not anything like a
spoked wheel, whose radial rigidity is almost entirely the support
element elasticity (the spokes). Didn't you recently assess what that
was?
> I expect that bare no-spoke rims deform more than a tensioned wheel,
> but I have to admit that I've never tried the experiment and don't
> know what to expect--a tenth of an inch, an inch, two inches, a bent
> rim?
Well, my MA-2 did not take a set after I sat on it.
Jobst Brandt
jim beam
> jim beam wrote:
>> Ben C wrote:
>>> On 2007-10-09, jim beam <spamv...@bad.example.net> wrote:
>>> [...]
>>>> no, excessively compressed rims more readily fold.
>>>
>>> Suppose the rim is highly compressed by high spoke tension.
>>>
>>> Then I hit a bump. The compression from the bump isn't just added to the
>>> compression from the spoke tension, because as the rim starts to
>>> compress, those spokes start to go slack. So the compression on the rim
>>> stays about the same until the spokes have gone completely slack.
>>>
>>> Is this right/wrong/too simplistic?
>>
>> the hoop compression is a function of spoke tension. i doubt locally
>> slackening spokes will reduce the compression of the whole because
>> it's an equilibrium structure and the bump load has to be taken up by
>> the whole.
>
> In other words, no, it's not "too simplistic" a bump won't significantly
> change the rim compression,
as a whole, but local elastic deformation can change local compression
significantly.
> which is why you can operate a wheel within
> a fairly small safety margin of compression buckling.
except that,... see above.
> The limit is more
> what happens to wheel shape when you break a spoke.
no, the limit is spoke tension sufficient to prevent spokes going slack.
>
>> as far as local bending is concerned, it's a bending beam, with a
>> neutral plane, and compressive and tensile regions accordingly.
>
> A beam which is supported by the spokes, with the consequence that the
> stiffness dramatically lowers when the support is removed.
>
> The net is you want to have tight spokes to prevent excessive (damaging)
> deflection of the rim under overload.
except that it has to be balanced against increasing propensity to taco
and rim cracking. spokes only need to be tight enough to not go slack.
no more.
frkr...@gmail.com
I don't have any high-wheeler rims nor any deep-section aero rims.
But I just proved to my satisfaction that 180 pounds will flatten an
ancient, bare Rigida AL-320 (I think that's the model. It's not quite
legible.)
10 pounds produced almost exactly 0.1" elastic deflection. The rim is
19 mm wide, 15 mm deep, and weighs about one pound (453 grams).
I guess I was saving that rim for just this occasion.
- Frank Krygowski
carl...@comcast.net
Dear Frank,
Thanks!
I'm surprised that 10 pounds produced only a tenth of an inch of drop.
Now I gotta think of something to send you to pay for that old rim.
Cheers,
Carl Fogel
jim beam
nor did mine. which leads me to ask, why do you claim that wheels
"collapse" if some spokes go slack? clearly the rim won't.
carl...@comcast.net
Dear Jobst,
Sigh . . . What a nice post, except for where you call people you
don't know liars about something that you've never tested.
Here's what Rudge wrote about their 1887 highwheeler:
"The FELLOES are of the Clement hollow pattern, consisting of one
piece of steel tubing rolled and pressed into the proper shape, and
capable of bearing a weight of two hundred pounds, even before a
single spoke has been inserted."
http://tallbike.com/tall/87rudge.html
Why not consider the possibility that you may not know everything?
And why not consider what readers will think when you call Rudge liars
when I mention that their bare highwheel would support the weight of a
man, but then go on to state that your aluminum MA2 supported your 180
pound weight without taking a set, dropping a whole inch and a half
when you sat on it?
For all I know, you might even be right, but it's awfully hard to
credit you when you take such a bizarre approach.
Cheers,
Carl Fogel
Tom Sherman
> ...
> Dear Jobst,
>
> Sigh . . . What a nice post, except for where you call people you
Not that "Dear Carl" would ever call someone a liar based on NO
evidence. [end sarcasm].
--
Tom Sherman - Holstein-Friesland Bovinia
Beer - It's not just for breakfast anymore!
Ben C
Yes, you're right, and so is Peter Cole.
I have drawn some more graphs and thought about it some more and am
now convinced of this fact.
The difficulty is getting the signs right and consistent so you're
adding the correct two functions together.
It feels OK intuitively now too, although it's not that easy to explain,
and there's probably not much point since it's only myself I need to
explain it to.
Anyway, this means Peter Cole is also right that stiffness is higher
when the spokes aren't slack than when they are (railroad ties etc.).
This means that high tension spokes mean more deformation is possible
before you reach the transition point where the rim responds with K
alone rather than with K+L.
I was on my own tangent there, but now I don't understand why jim beam
says that higher spoke tension (higher than needed to prevent slack
spoke flexing fatigue) achieves nothing, or why he says it makes it more
likely that the rim will dent when you hit a bump.
jim beam
simple. look at a stress/strain graph. all you're doing is moving
along the straight line region and approaching the yield point. that is
true of any loaded point in the structure. the only way "increasing"
anything can mitigate a load is if it has an opposite sign to the load.
increasing tension is great for spokes, for exactly that reason. but
it's terrible for rims as loading increase and tension increase are the
same sign!
Ben C
> along the straight line region and approaching the yield point. that is
> true of any loaded point in the structure. the only way "increasing"
> anything can mitigate a load is if it has an opposite sign to the load.
> increasing tension is great for spokes, for exactly that reason. but
> it's terrible for rims as loading increase and tension increase are the
> same sign!
So, the high-tension spokes are compressing the rim, and the bump you
hit compresses it even more? So by pre-loading it, you're bringing it
nearer to yield?
If this is what you're saying, then the part I was having trouble with
was this: how is it that rim compression from spoke tension and from the
bump are added together, since as you hit the bump, the spoke becomes
slacker.
But actually I now think that's exactly the same confusion I had about
the railroad ties/my compression spring wrapped up in a rubber band.
Until the spoke is completely slack, then the rim _is_ being compressed
both by the bump and by the spoke.
The fact that the spoke is there makes the whole assembly stiffer, but
also means that the rim is operating at closer to yield.
So, assuming Hooke's Law for spokes and rim (but assuming that spokes
don't work in compression):
1. Having a spoke that isn't slack supports the rim and makes the whole
assembly stiffer.
2. Making the tension even higher doesn't make it any _more_ stiff.
3. But it does bring the rim closer to yield making it easier to
flat-spot.
4. But it does keep it stiffer for longer-- the stress it can take
before the spokes go completely slack and the assembly loses
stiffness is greater.
There's some critical value of spoke tension such that the spoke goes
slack just as the rim reaches yield stress.
More tension than that critical value and you're definitely reducing rim
strength without any benefit.
Less tension than that and the stress level at which you lose spoke
support is reduced. But a stiffer rim doesn't need as much spoke
support.
And then there's rim fatigue cracking, which may be a problem at a lower
tension than at the "critical value" mentioned above.
Peter Cole
> On 2007-10-10, jim beam <spamv...@bad.example.net> wrote:
> [...]
>> simple. look at a stress/strain graph. all you're doing is moving
>> along the straight line region and approaching the yield point. that is
>> true of any loaded point in the structure. the only way "increasing"
>> anything can mitigate a load is if it has an opposite sign to the load.
>> increasing tension is great for spokes, for exactly that reason. but
>> it's terrible for rims as loading increase and tension increase are the
>> same sign!
>
> So, the high-tension spokes are compressing the rim, and the bump you
> hit compresses it even more? So by pre-loading it, you're bringing it
> nearer to yield?
>
> If this is what you're saying, then the part I was having trouble with
> was this: how is it that rim compression from spoke tension and from the
> bump are added together, since as you hit the bump, the spoke becomes
> slacker.
Good question.
>
> But actually I now think that's exactly the same confusion I had about
> the railroad ties/my compression spring wrapped up in a rubber band.
Should be a sign.
> Until the spoke is completely slack, then the rim _is_ being compressed
> both by the bump and by the spoke.
No, just the spokes.
> The fact that the spoke is there makes the whole assembly stiffer, but
> also means that the rim is operating at closer to yield.
The wheel is "closer to yield", in the sense that rim compression
eventually will cause a buckle.
> So, assuming Hooke's Law for spokes and rim (but assuming that spokes
> don't work in compression):
>
> 1. Having a spoke that isn't slack supports the rim and makes the whole
> assembly stiffer.
Yes.
> 2. Making the tension even higher doesn't make it any _more_ stiff.
Yes.
> 3. But it does bring the rim closer to yield making it easier to
> flat-spot.
No, just the opposite. The deflection for a given load (over the slack
point) is the load to slack times the stiffness of rim and spokes, plus
the load above slack times the stiffness of the rim alone. The earlier
the spoke slacks, the greater the total deflection for a given load. You
should also consider that there may be more than one spoke slacking.
> 4. But it does keep it stiffer for longer-- the stress it can take
> before the spokes go completely slack and the assembly loses
> stiffness is greater.
Yes.
> There's some critical value of spoke tension such that the spoke goes
> slack just as the rim reaches yield stress.
In this case, "yield" means bending beyond the elastic limit of the rim.
In the reference FEA model in Jobst's book (or Ian's online -- largely
the same), the displacement is only 0.75mm before slacking. That's not
going to create much of a flat spot, even if it's plastically deformed
the rim. Your "critical value" might be well above practical spoke
tension, that would be my guess, it would be an interesting computation.
If you observe that 0.75mm corresponds to a load 5x greater than
nominal, you realize the limit is really a factor in impulse loads, not
hauling a heavy backpack. In this kind of scenario, it's more a matter
of absorbing the energy in the impact with a hole or debris. The wheel
has more impact tolerance.
Ben C
>> Until the spoke is completely slack, then the rim _is_ being compressed
>> both by the bump and by the spoke.
>
> No, just the spokes.
Hell, now I'm confused again :)
If it was just the spokes, then the rim wouldn't move. As you hit the
bump, you've got the load from the bump plus the load from the spokes.
The load from the spokes has gone down a bit, because they are going a
bit slack, but the total load has gone up on the rim, or it wouldn't
deform.
But if the yield point of the rim is above the spokes-going-slack point,
then it doesn't matter that you're bringing the rim closer to yield--
It's not going to yield until after the spokes have gone slack anyway.
[...]
>> 3. But it does bring the rim closer to yield making it easier to
>> flat-spot.
>
> No, just the opposite. The deflection for a given load (over the slack
> point) is the load to slack times the stiffness of rim and spokes, plus
> the load above slack times the stiffness of the rim alone. The earlier
> the spoke slacks, the greater the total deflection for a given load.
Yup, I think I get that. (3) _is_ true if tension is higher than the
"critical value" though.
[...]
>> There's some critical value of spoke tension such that the spoke goes
>> slack just as the rim reaches yield stress.
>
> In this case, "yield" means bending beyond the elastic limit of the rim.
> In the reference FEA model in Jobst's book (or Ian's online -- largely
> the same), the displacement is only 0.75mm before slacking. That's not
> going to create much of a flat spot, even if it's plastically deformed
> the rim. Your "critical value" might be well above practical spoke
> tension, that would be my guess, it would be an interesting computation.
Well I think that's the decider.
If spoke-slacking happens before rim-yielding then the extra tension
isn't a problem (from the point of view of flat-spots).
As for "modern" rims-- they're stiffer, so a bigger load before spoke
slacking. If they're not as much stronger as they are stiffer then the
critical value moves down.
But I still don't know if it comes into the picture. As you say 0.75mm
is not a big displacement.
Peter Cole
> On 2007-10-10, Peter Cole <peter...@comcast.net> wrote:
>> Ben C wrote:
> [...]
>>> Until the spoke is completely slack, then the rim _is_ being compressed
>>> both by the bump and by the spoke.
>> No, just the spokes.
>
> Hell, now I'm confused again :)
You're working hard to be.
> If it was just the spokes, then the rim wouldn't move.
Hold on, you said "compressed" above, by that I thought you meant
circumferential rim compression, not wheel radial compression.
> As you hit the
> bump, you've got the load from the bump plus the load from the spokes.
> The load from the spokes has gone down a bit, because they are going a
> bit slack, but the total load has gone up on the rim, or it wouldn't
> deform.
>
> But if the yield point of the rim is above the spokes-going-slack point,
> then it doesn't matter that you're bringing the rim closer to yield--
> It's not going to yield until after the spokes have gone slack anyway.
Yes, but until the spokes go slack the aggregate wheel has a much higher
spring constant, remember?
>
> [...]
>>> 3. But it does bring the rim closer to yield making it easier to
>>> flat-spot.
>> No, just the opposite. The deflection for a given load (over the slack
>> point) is the load to slack times the stiffness of rim and spokes, plus
>> the load above slack times the stiffness of the rim alone. The earlier
>> the spoke slacks, the greater the total deflection for a given load.
>
> Yup, I think I get that. (3) _is_ true if tension is higher than the
> "critical value" though.
No.
> [...]
>>> There's some critical value of spoke tension such that the spoke goes
>>> slack just as the rim reaches yield stress.
>> In this case, "yield" means bending beyond the elastic limit of the rim.
>> In the reference FEA model in Jobst's book (or Ian's online -- largely
>> the same), the displacement is only 0.75mm before slacking. That's not
>> going to create much of a flat spot, even if it's plastically deformed
>> the rim. Your "critical value" might be well above practical spoke
>> tension, that would be my guess, it would be an interesting computation.
>
> Well I think that's the decider.
"Decider" of what?
> If spoke-slacking happens before rim-yielding then the extra tension
> isn't a problem (from the point of view of flat-spots).
No value of spoke tension is a "problem" for flat spots.
> As for "modern" rims-- they're stiffer, so a bigger load before spoke
> slacking.
Yes, but still much less stiff than the spokes, the spokes still dominate.
> If they're not as much stronger as they are stiffer then the
> critical value moves down.
Aluminum will permanently bend at 0.2% deformation, whatever the rim.
> But I still don't know if it comes into the picture. As you say 0.75mm
> is not a big displacement.
You snipped the part where I said to think of it in terms of energy.
Whack a wheel with a baseball bat.
The tire will absorb some energy.
The spokes will absorb some energy.
The rim will absorb the rest.
The tighter the spokes, the more energy they can absorb (force x distance).
The rim will always absorb some energy, no matter how softly you hit the
wheel. The tires and spokes will stop absorbing at some level, and when
they do, the rim will start absorbing all the energy, and displace at a
faster rate (not as stiff). Tighter spokes just postpone that
transition. The same thing happens in lateral loading.
Ben C
>> then it doesn't matter that you're bringing the rim closer to yield--
>> It's not going to yield until after the spokes have gone slack anyway.
>
> Yes, but until the spokes go slack the aggregate wheel has a much higher
> spring constant, remember?
Yes I remember.
>> [...]
>>>> 3. But it does bring the rim closer to yield making it easier to
>>>> flat-spot.
>>> No, just the opposite. The deflection for a given load (over the slack
>>> point) is the load to slack times the stiffness of rim and spokes, plus
>>> the load above slack times the stiffness of the rim alone. The earlier
>>> the spoke slacks, the greater the total deflection for a given load.
>>
>> Yup, I think I get that. (3) _is_ true if tension is higher than the
>> "critical value" though.
>
> No.
It's true that the rim is brought closer to yield by higher spoke
tension.
The question is not whether it's true but whether it's a herring. If
tension is lower than this "critical value", then I think it is.
>> [...]
>>>> There's some critical value of spoke tension such that the spoke goes
>>>> slack just as the rim reaches yield stress.
>>> In this case, "yield" means bending beyond the elastic limit of the rim.
>>> In the reference FEA model in Jobst's book (or Ian's online -- largely
>>> the same), the displacement is only 0.75mm before slacking. That's not
>>> going to create much of a flat spot, even if it's plastically deformed
>>> the rim. Your "critical value" might be well above practical spoke
>>> tension, that would be my guess, it would be an interesting computation.
>>
>> Well I think that's the decider.
>
> "Decider" of what?
The factor that decides whether high spoke tension increases propensity
to flat-spot or not.
>> If spoke-slacking happens before rim-yielding then the extra tension
>> isn't a problem (from the point of view of flat-spots).
>
> No value of spoke tension is a "problem" for flat spots.
If tension were so high that hitting a bump meant the rim reached yield
before the spokes went slack then I think you would get flat spots at a
lower applied force than if tension weren't that high.
I know you said tension is most unlikely to be that high, and you are
probably right.
>> As for "modern" rims-- they're stiffer, so a bigger load before spoke
>> slacking.
>
> Yes, but still much less stiff than the spokes, the spokes still dominate.
>
>> If they're not as much stronger as they are stiffer then the
>> critical value moves down.
>
> Aluminum will permanently bend at 0.2% deformation, whatever the rim.
OK, thanks. So if they're stiffer, they're also stronger.
>> But I still don't know if it comes into the picture. As you say 0.75mm
>> is not a big displacement.
>
> You snipped the part where I said to think of it in terms of energy.
>
> Whack a wheel with a baseball bat.
>
> The tire will absorb some energy.
> The spokes will absorb some energy.
> The rim will absorb the rest.
>
> The tighter the spokes, the more energy they can absorb (force x distance).
Thinking of it in terms of energy is interesting too. But my current
concern is with strength. It's not the whole story, but I'm just trying
to evaluate the question of whether or why high spoke tension might
increase the propensity of the rim to flat-spot.
Peter Cole
> On 2007-10-10, Peter Cole <peter...@comcast.net> wrote:
>> Ben C wrote:
> [...]
>>> But if the yield point of the rim is above the spokes-going-slack point,
>>> then it doesn't matter that you're bringing the rim closer to yield--
>>> It's not going to yield until after the spokes have gone slack anyway.
>> Yes, but until the spokes go slack the aggregate wheel has a much higher
>> spring constant, remember?
>
> Yes I remember.
>
>>> [...]
>>>>> 3. But it does bring the rim closer to yield making it easier to
>>>>> flat-spot.
>>>> No, just the opposite. The deflection for a given load (over the slack
>>>> point) is the load to slack times the stiffness of rim and spokes, plus
>>>> the load above slack times the stiffness of the rim alone. The earlier
>>>> the spoke slacks, the greater the total deflection for a given load.
>>> Yup, I think I get that. (3) _is_ true if tension is higher than the
>>> "critical value" though.
>> No.
>
> It's true that the rim is brought closer to yield by higher spoke
> tension.
There are 2 "yields" here, first is rim circumferential compression from
spoke tension (buckle), second is rim bending from load/impact. They are
entirely separate.
>
> The question is not whether it's true but whether it's a herring. If
> tension is lower than this "critical value", then I think it is.
>
>>> [...]
>>>>> There's some critical value of spoke tension such that the spoke goes
>>>>> slack just as the rim reaches yield stress.
>>>> In this case, "yield" means bending beyond the elastic limit of the rim.
>>>> In the reference FEA model in Jobst's book (or Ian's online -- largely
>>>> the same), the displacement is only 0.75mm before slacking. That's not
>>>> going to create much of a flat spot, even if it's plastically deformed
>>>> the rim. Your "critical value" might be well above practical spoke
>>>> tension, that would be my guess, it would be an interesting computation.
>>> Well I think that's the decider.
>> "Decider" of what?
>
> The factor that decides whether high spoke tension increases propensity
> to flat-spot or not.
>
>>> If spoke-slacking happens before rim-yielding then the extra tension
>>> isn't a problem (from the point of view of flat-spots).
>> No value of spoke tension is a "problem" for flat spots.
>
> If tension were so high that hitting a bump meant the rim reached yield
> before the spokes went slack then I think you would get flat spots at a
> lower applied force than if tension weren't that high.
No, flat spots are from bending, the rim will bend the same no matter
what the tension is, as long as the spokes don't go slack, if they do,
it will bend sooner.
> I know you said tension is most unlikely to be that high, and you are
> probably right.
>
>>> As for "modern" rims-- they're stiffer, so a bigger load before spoke
>>> slacking.
>> Yes, but still much less stiff than the spokes, the spokes still dominate.
>>
>>> If they're not as much stronger as they are stiffer then the
>>> critical value moves down.
>> Aluminum will permanently bend at 0.2% deformation, whatever the rim.
>
> OK, thanks. So if they're stiffer, they're also stronger.
>
>>> But I still don't know if it comes into the picture. As you say 0.75mm
>>> is not a big displacement.
>> You snipped the part where I said to think of it in terms of energy.
>>
>> Whack a wheel with a baseball bat.
>>
>> The tire will absorb some energy.
>> The spokes will absorb some energy.
>> The rim will absorb the rest.
>>
>> The tighter the spokes, the more energy they can absorb (force x distance).
>
> Thinking of it in terms of energy is interesting too. But my current
> concern is with strength. It's not the whole story, but I'm just trying
> to evaluate the question of whether or why high spoke tension might
> increase the propensity of the rim to flat-spot.
It's the same as asking whether high tire pressure will make the wheel
more likely to flat spot. Changing the circumferential rim compression
doesn't change the chance of flat spotting, changing the tension does.
You are confusing rim circumferential compression with rim bending --
the rim won't buckle from compression under radial loads.
jobst....@stanfordalumni.org
>>> Dear Jobst,
> Dear Jobst,
> Sigh... What a nice post, except for where you call people you don't
> know liars about something that you've never tested.
This is a deception of the second kind. As I said most high wheelers
used U-shaped rims of flat sheet metal. That you cannot crush a
deep-V rim with body weight is likewise true except that hollow cross
section penny-farthing bicycle were uncommon and therefore do not
represent the norm. In that respect, the claim is false and as you
pointed out, exceptions don't make the rule.
> Here's what Rudge wrote about their 1887 highwheeler:
> "The FELLOES are of the Clement hollow pattern, consisting of one
> piece of steel tubing rolled and pressed into the proper shape, and
> capable of bearing a weight of two hundred pounds, even before a
> single spoke has been inserted."
> http://tallbike.com/tall/87rudge.html
> Why not consider the possibility that you may not know everything?
Why not consider that Rudge was not the maker of most high wheels and
that even these were out of the ordinary.
> And why not consider what readers will think when you call Rudge liars
> when I mention that their bare highwheel would support the weight of a
> man, but then go on to state that your aluminum MA2 supported your 180
> pound weight without taking a set, dropping a whole inch and a half
> when you sat on it?
Where did I mane Rudge as liars? I find your application of
translation to what others post unseemly and an obvious personal
attack. I also notice that you don't write such responses to those
who pass disinformation to the newsgroup in rudest manner.
> For all I know, you might even be right, but it's awfully hard to
> credit you when you take such a bizarre approach.
What is bizarre is your take on these things. The information is in
what I wrote but you choose to cite an exception in spite of my
explanation to what rims I was referring. I sense that you have a bit
of jim beam disorder at work from your harsh comments to most anything
I write.
Jobst Brandt
carl...@comcast.net
Dear Jobst,
Er, don't you remember writing "They lied"?
Cheers,
Carl Fogel
jim beam
no. the rim material doesn't care /how/ it exceeds yield, only that it
will plastically deform if applied stress exceeds that level.
>
>>
>> The question is not whether it's true but whether it's a herring. If
>> tension is lower than this "critical value", then I think it is.
>>
>>>> [...]
>>>>>> There's some critical value of spoke tension such that the spoke goes
>>>>>> slack just as the rim reaches yield stress.
>>>>> In this case, "yield" means bending beyond the elastic limit of the
>>>>> rim. In the reference FEA model in Jobst's book (or Ian's online --
>>>>> largely the same), the displacement is only 0.75mm before slacking.
>>>>> That's not going to create much of a flat spot, even if it's
>>>>> plastically deformed the rim. Your "critical value" might be well
>>>>> above practical spoke tension, that would be my guess, it would be
>>>>> an interesting computation.
>>>> Well I think that's the decider.
>>> "Decider" of what?
>>
>> The factor that decides whether high spoke tension increases propensity
>> to flat-spot or not.
>>
>>>> If spoke-slacking happens before rim-yielding then the extra tension
>>>> isn't a problem (from the point of view of flat-spots).
>>> No value of spoke tension is a "problem" for flat spots.
>>
>> If tension were so high that hitting a bump meant the rim reached yield
>> before the spokes went slack then I think you would get flat spots at a
>> lower applied force than if tension weren't that high.
>
> No, flat spots are from bending,
no, the combined effects of bending /plus/ hoop compression.
the rim will bend the same no matter
> what the tension is, as long as the spokes don't go slack, if they do,
> it will bend sooner.
except that the hoop compression decreases the available loading capacity.
no it's not.
> Changing the circumferential rim compression
> doesn't change the chance of flat spotting, changing the tension does.
er, increasing the circumferential hoop compression most definitely
/does/ increase the chance of flat spotting. if yield is [pick a
number] 300MPa, and hoop compression is 200MPa, that's only 100MPa of
available for additional compressive loading. shirley it's obvious that
lower hoop compression leaves more available load compression capacity?
> You are confusing rim circumferential compression with rim bending --
> the rim won't buckle from compression under radial loads.
it will if the hoop compression is excessive - at the absolute tipping
point, the slightest thing can cause the instability that will cause the
rim to suddenly pretzel.
jim beam
> On 2007-10-10, Peter Cole <peter...@comcast.net> wrote:
>> Ben C wrote:
> [...]
>>> Until the spoke is completely slack, then the rim _is_ being compressed
>>> both by the bump and by the spoke.
>> No, just the spokes.
>
> Hell, now I'm confused again :)
peter cole is the one confused, not you.
>
> If it was just the spokes, then the rim wouldn't move. As you hit the
> bump, you've got the load from the bump plus the load from the spokes.
> The load from the spokes has gone down a bit, because they are going a
> bit slack, but the total load has gone up on the rim, or it wouldn't
> deform.
>
> But if the yield point of the rim is above the spokes-going-slack point,
> then it doesn't matter that you're bringing the rim closer to yield--
> It's not going to yield until after the spokes have gone slack anyway.
correct.
>
> [...]
>>> 3. But it does bring the rim closer to yield making it easier to
>>> flat-spot.
>> No, just the opposite. The deflection for a given load (over the slack
>> point) is the load to slack times the stiffness of rim and spokes, plus
>> the load above slack times the stiffness of the rim alone. The earlier
>> the spoke slacks, the greater the total deflection for a given load.
>
> Yup, I think I get that. (3) _is_ true if tension is higher than the
> "critical value" though.
that portion of the rim where the spokes are slack is less supported,
and therefore more elastic. but if it's not under high hoop
compression, it takes more deformation to yield.
>
> [...]
>>> There's some critical value of spoke tension such that the spoke goes
>>> slack just as the rim reaches yield stress.
>> In this case, "yield" means bending beyond the elastic limit of the rim.
>> In the reference FEA model in Jobst's book (or Ian's online -- largely
>> the same), the displacement is only 0.75mm before slacking. That's not
>> going to create much of a flat spot, even if it's plastically deformed
>> the rim. Your "critical value" might be well above practical spoke
>> tension, that would be my guess, it would be an interesting computation.
>
> Well I think that's the decider.
>
> If spoke-slacking happens before rim-yielding then the extra tension
> isn't a problem (from the point of view of flat-spots).
correct.
>
> As for "modern" rims-- they're stiffer, so a bigger load before spoke
> slacking.
correct.
jim beam
<snip crap>
> I sense that you have a bit
> of jim beam disorder at work from your harsh comments to most anything
> I write.
>
why is pointing out your mistakes and inconsistencies "jim beam
disorder"? why don't you simply correct your mistakes, then quit
repeating them?
wheels "collapse" when spokes go slack vs. "my MA-2 did not take a set
after I sat on it" for instance.
jobst....@stanfordalumni.org
> <snip crap>
>> I sense that you have a bit of jim beam disorder at work from your
>> harsh comments to most anything I write.
> why is pointing out your mistakes and inconsistencies "jim beam
> disorder"? why don't you simply correct your mistakes, then quit
> repeating them?
he disorder is qualified adequately in the sentence above.
> wheels "collapse" when spokes go slack vs. "my MA-2 did not take a
> set after I sat on it" for instance.
Please read the thread before jumping to conclusions. This is about
bare rims with no spokes. Collapsing wheels has been discussed at
length and the snippet you cite does not represent what was said.
Jobst Brandt
jim beam
> http://www.jimbeam.com/beam/default.aspx? writes:
>
>> <snip crap>
>
>>> I sense that you have a bit of jim beam disorder at work from your
>>> harsh comments to most anything I write.
>
>> why is pointing out your mistakes and inconsistencies "jim beam
>> disorder"? why don't you simply correct your mistakes, then quit
>> repeating them?
>
> he disorder is qualified adequately in the sentence above.
er, so /why/ don't you correct your mistakes rather than repeat them?
or is that "jobstian avoidance disorder"?
>
>> wheels "collapse" when spokes go slack vs. "my MA-2 did not take a
>> set after I sat on it" for instance.
>
> Please read the thread before jumping to conclusions. This is about
> bare rims with no spokes.
i /know/ it's a bare rim jobst - i've done the same experiment. fact
is, it's inconsistent with your previous statements about wheel
collapse. if you make two contradictory assertions, you must be
prepared to explain, not simply avoid.
> Collapsing wheels has been discussed at
> length and the snippet you cite does not represent what was said.
eh? you've never shown any collapsing wheels, only asserted that
supposition. the inconsistency is that you expect us to believe that a
wheel where only a few spokes are slack will "collapse", but then prove
that a bare rim with no spokes [and thus no bracing] somehow
mysteriously doesn't!
meanwhile...
http://www.flickr.com/photos/38636024@N00/417157612/
Joe Riel
> except that the hoop compression decreases the available loading capacity.
Yes, but the relevant question is, by how much? A back of the envelope
computation on a 36 spoke MA2 rim with 90 kgf tension per spoke gives
a compression of about 70 MPa. Any idea on the yield strength of the
rim alloy?
--
Joe Riel
Ben C
>> then it doesn't matter that you're bringing the rim closer to yield--
>> It's not going to yield until after the spokes have gone slack anyway.
>
> correct.
So do you think the yield point is below the spokes-slack point for an
MA-2 or for an Open Pro rim built to just-sub-taco tension?
[...]
> that portion of the rim where the spokes are slack is less supported,
> and therefore more elastic. but if it's not under high hoop
> compression, it takes more deformation to yield.
I think I get this. Support from the spokes make it stiffer but not
stronger. So without the support, same yield stress, but more
deformation before yield.
jim beam
i'd guess in the 200-250MPa range. it's 6106. should be able to find a
number for "full hard" extrusion or cold worked.
also, don't forget, we're looking at situations with /excess/ spoke
tension, not tensions in the normal range that you've just given.
jim beam
> On 2007-10-11, jim beam <spamv...@bad.example.net> wrote:
>> Ben C wrote:
> [...]
>>> But if the yield point of the rim is above the spokes-going-slack point,
>>> then it doesn't matter that you're bringing the rim closer to yield--
>>> It's not going to yield until after the spokes have gone slack anyway.
>> correct.
>
> So do you think the yield point is below the spokes-slack point for an
> MA-2 or for an Open Pro rim built to just-sub-taco tension?
not for the bulk, no, otherwise it would deform on build. however, at
high tensions, you can get local deformation around the eyelet, so that
is at yield.
>
> [...]
>> that portion of the rim where the spokes are slack is less supported,
>> and therefore more elastic. but if it's not under high hoop
>> compression, it takes more deformation to yield.
>
> I think I get this. Support from the spokes make it stiffer
well, not stiffer in that the modulus of any component doesn't change,
but it's in "high elasticity" mode slightly longer if the spokes aren't
slacking.
> but not
> stronger. So without the support, same yield stress, but more
> deformation before yield.
elastic deformation, yes. and it's only because the stress at any point
on the rim is lower.
bottom line, you want spoke tension as it /does/ brace the rim. and it
prevents spoke nipples unscrewing. but you don't want to crank it up to
the limit. i see it as the intersection of two lines on a graph, one
positive slope, the other negative [spoke load capacity vs. rim].
intersection is ideal spoke tension, not reducing capacity of the rim
simply to favor the spokes.
Peter Cole
My quick calculations agree with yours. From the data in Jobst's book, I
get around 20MPa for the stress from the nominal wheel load (50kg).
Since that load causes 0.15mm deflection, and the yield of the rim is
probably around 250MPa, there's roughly 10x deflection until yield, or
1.5mm. (250-70)/20 = 9.
Since the spoke slacks at 5x load (0.75mm), that's about half way to
yield distance wise, but if rim stiffness is 1/2 the spoke stiffness (by
my rough calc.), that's only an additional 1.5 load, so 6.5x load total
to start of rim yield. If the spokes were 50% tighter, you'd have 7.5x
load to slack, and 0.5x additional to yield, or an 8x total load to rim
yield.
Ben C
> Joe Riel wrote:
>> jim beam <spamv...@bad.example.net> writes:
>>
>>> except that the hoop compression decreases the available loading capacity.
>>
>> Yes, but the relevant question is, by how much? A back of the envelope
>> computation on a 36 spoke MA2 rim with 90 kgf tension per spoke gives
>> a compression of about 70 MPa. Any idea on the yield strength of the
>> rim alloy?
>
> My quick calculations agree with yours. From the data in Jobst's book, I
> get around 20MPa for the stress from the nominal wheel load (50kg).
> Since that load causes 0.15mm deflection, and the yield of the rim is
> probably around 250MPa, there's roughly 10x deflection until yield, or
> 1.5mm. (250-70)/20 = 9.
I was going to try and attempt some calculations myself, but how do you
figure when the rim's going to yield? We can say we're loading the rim
with 250MPa, but actually it's bending, across the span of however many
spokes have gone slack. So it seems one would have to take into account
the length of the span and bending moments and things.
Peter Cole
> On 2007-10-11, Peter Cole <peter...@comcast.net> wrote:
>> Joe Riel wrote:
>>> jim beam <spamv...@bad.example.net> writes:
>>>
>>>> except that the hoop compression decreases the available loading capacity.
>>> Yes, but the relevant question is, by how much? A back of the envelope
>>> computation on a 36 spoke MA2 rim with 90 kgf tension per spoke gives
>>> a compression of about 70 MPa. Any idea on the yield strength of the
>>> rim alloy?
>> My quick calculations agree with yours. From the data in Jobst's book, I
>> get around 20MPa for the stress from the nominal wheel load (50kg).
>> Since that load causes 0.15mm deflection, and the yield of the rim is
>> probably around 250MPa, there's roughly 10x deflection until yield, or
>> 1.5mm. (250-70)/20 = 9.
>
> I was going to try and attempt some calculations myself, but how do you
> figure when the rim's going to yield? We can say we're loading the rim
> with 250MPa, but actually it's bending, across the span of however many
> spokes have gone slack. So it seems one would have to take into account
> the length of the span and bending moments and things.
Yeah, all that stuff.
jim beam
> Ben C wrote:
>> On 2007-10-11, Peter Cole <peter...@comcast.net> wrote:
>>> Joe Riel wrote:
>>>> jim beam <spamv...@bad.example.net> writes:
>>>>
>>>>> except that the hoop compression decreases the available loading
>>>>> capacity.
>>>> Yes, but the relevant question is, by how much? A back of the envelope
>>>> computation on a 36 spoke MA2 rim with 90 kgf tension per spoke gives
>>>> a compression of about 70 MPa. Any idea on the yield strength of the
>>>> rim alloy?
>>> My quick calculations agree with yours. From the data in Jobst's
>>> book, I get around 20MPa for the stress from the nominal wheel load
>>> (50kg). Since that load causes 0.15mm deflection, and the yield of
>>> the rim is probably around 250MPa, there's roughly 10x deflection
>>> until yield, or 1.5mm. (250-70)/20 = 9.
>>
>> I was going to try and attempt some calculations myself, but how do you
>> figure when the rim's going to yield? We can say we're loading the rim
>> with 250MPa, but actually it's bending, across the span of however many
>> spokes have gone slack. So it seems one would have to take into account
>> the length of the span and bending moments and things.
>
> Yeah, all that stuff.
ignorance bomber.
Ben C
> bottom line, you want spoke tension as it /does/ brace the rim. and it
> prevents spoke nipples unscrewing. but you don't want to crank it up to
> the limit. i see it as the intersection of two lines on a graph, one
> positive slope, the other negative [spoke load capacity vs. rim].
So on that graph, spoke load capacity is load-until-slack, and rim load
capacity is load-until-yield. So I think that intersection is the same
one I was talking about.
> intersection is ideal spoke tension, not reducing capacity of the rim
> simply to favor the spokes.
Yes, although it would be interesting to be able to estimate what the
ideal tension is for different rim/spoke count combinations. Needs an
FEA really I think.
And of course this isn't the only factor-- the ideal tension as given
by that intersection might be too high and lead to spoke-hole fatigue.
Maybe that's a "badly designed rim"-- spoke bed too crummy for available
rim strength.
I imagine aero rims would make the biggest difference in stiffness. MA-2
and Open Pro are roughly the same in cross-section I think. What's the
difference between them anyway (apart from anodizing, machined brake
track, the way they join up the hoop)? Do they cook the Open Pro more to
make it stiffer?
jim beam
> On 2007-10-11, jim beam <spamv...@bad.example.net> wrote:
> [...]
>> bottom line, you want spoke tension as it /does/ brace the rim. and it
>> prevents spoke nipples unscrewing. but you don't want to crank it up to
>> the limit. i see it as the intersection of two lines on a graph, one
>> positive slope, the other negative [spoke load capacity vs. rim].
>
> So on that graph, spoke load capacity is load-until-slack, and rim load
> capacity is load-until-yield. So I think that intersection is the same
> one I was talking about.
>
>> intersection is ideal spoke tension, not reducing capacity of the rim
>> simply to favor the spokes.
>
> Yes, although it would be interesting to be able to estimate what the
> ideal tension is for different rim/spoke count combinations. Needs an
> FEA really I think.
>
> And of course this isn't the only factor-- the ideal tension as given
> by that intersection might be too high and lead to spoke-hole fatigue.
> Maybe that's a "badly designed rim"-- spoke bed too crummy for available
> rim strength.
absolutely, that is a given. first though we need to establish the
mechanical principles, then we can refine for materials.
>
> I imagine aero rims would make the biggest difference in stiffness. MA-2
> and Open Pro are roughly the same in cross-section I think.
ma2 is about as shallow as you can get for a normal brake track width.
open pro is nearly twice as deep.
> What's the
> difference between them anyway (apart from anodizing, machined brake
> track, the way they join up the hoop)? Do they cook the Open Pro more to
> make it stiffer?
i don't know if mavic heat treat. i know some rims were - the rim in my
missing spoke photo has a "heat treated" label on it for instance, but
it's not mavic. and heat treatment doesn't make things stiffer, just
stronger [lengthens the straight line section of the stress/strain
graph, not changes the slope].
Ben C
>> and Open Pro are roughly the same in cross-section I think.
>
> ma2 is about as shallow as you can get for a normal brake track width.
> open pro is nearly twice as deep.
Sounds like that's the significant difference then.
http://www.astounding.org.uk/ian/wheel/sectprops.html
Ian's model used 25x10mm as the box section, Open Pro is 18.4x19.6mm.
That is about twice as deep, and "oversquare".
Peter Cole
As you've seen, a wheel is basically a 2 spring system, where the spring
constants add. The spoke "spring" drops out when the spoke goes slack,
causing the aggregate spring constant to drop suddenly at that point by
a factor of about 3. The stiffness of the rim "spring" varies inversely
with the cube of the length, so that spring constant goes down rapidly
as spokes become unloaded. Putting the spokes closer together and/or
increasing the initial spoke tension is much more effective than
increasing rim stiffness in preventing flat spotting.
Ben C
> open pro is nearly twice as deep.
I'm wondering now how they got it to be only 10 or 15g heavier. Perhaps
the walls are almost half the thickness.
jim beam
mine's about 25g heavier fwiw, but that would be my guess, yes - i've
not sectioned a ma2 so can't post a pic.
jim beam
increasing rim stiffness is achieved by bracing with spokes. however,
cranking spoke tension up so that the rim approaches yield decreases the
load the "spring" can take before plastic deformation.
Ben C
This page has cross-sections and information about various Mavic
products apparently from 1996:
http://www.bikepro.com/products/rims/mavicroad.html
The Open 4 looks similar to today's Open Pro. The rim they make now that
looks most like the MA-2 in cross-section is the Reflex Tubular.
jim beam
> On 2007-10-13, jim beam <spamv...@bad.example.net> wrote:
>> Ben C wrote:
>>> On 2007-10-12, jim beam <spamv...@bad.example.net> wrote:
>>> [...]
>>>> ma2 is about as shallow as you can get for a normal brake track width.
>>>> open pro is nearly twice as deep.
>>> I'm wondering now how they got it to be only 10 or 15g heavier. Perhaps
>>> the walls are almost half the thickness.
>> mine's about 25g heavier fwiw, but that would be my guess, yes - i've
>> not sectioned a ma2 so can't post a pic.
>
> This page has cross-sections and information about various Mavic
> products apparently from 1996:
>
> http://www.bikepro.com/products/rims/mavicroad.html
"The MA2 uses two-piece, double wall stainless steel eyelets"
that's not quite true. the "socket" is plated steel that rusts. the
"rivet" is the only stainless part.