I fixed a broken spoke!

jim beam said:

imagine this:
[fixed width font]

v
concrete|---------|concrete

and
v
|--------------------|
| |
| |
| |<thin flexy pole
| |
//////////////////////////

now, which one do /you/ think is doing to show the greatest tension
increase when deflecting the wire at point "v"?

If the horizontal member is a string (i.e. no bending stiffness), and
you deflect point "v" until the string is bent at the same angle in both
cases, the tension is the same.

--
Dave
dvt at psu dot edu

dvt said:

jim beam said:

imagine this:
[fixed width font]

v
concrete|---------|concrete

and
v
|--------------------|
| |
| |
| |<thin flexy pole
| |
//////////////////////////

now, which one do /you/ think is doing to show the greatest tension
increase when deflecting the wire at point "v"?

If the horizontal member is a string (i.e. no bending stiffness), and
you deflect point "v" until the string is bent at the same angle in both
cases, the tension is the same.

Dear Dave,

The angle should work as you describe, so I think that you're right
and that I misunderstood things (Ron Ruff was patient enough to clear
things up elsewhere.)

But there's a practical problem worth describing here.

Mostly we're not measuring angles.

Instead, we use the almost equivalent equation T = (F x L) / (4 x D)
equation, which will be within two pounds or so of the angle result
for typical spoke-level measurements.

Measuring L and D seems simple. But in the examples, the L width
reduces as the springy ends come together, and the deflection D must
be measured from the new L position, so it's easy to mismeasure the
miserable things.

The concrete is easily described because the ends of L don't move:

123456789012345
<------L------>
concrete|------D------|concrete D1
. D . D2
' D ' D3

L measures 15 units across, D measures 3 units down.

The springy pole (archery bow, other flexible things) takes two crude
ASCII diagrams to show that the same force and angle changes L's width
and moves the L-end of D to a new position:

123456789012345
<-original L-->
springy |-------------|1
poles | |2
| |3
| |4
| |5

springy <-original L-->1
poles /LLLDLLLL\ 2 D1 new D is only 2
/ ' D ' \ 3 D2
/ \4
| |5
1234567890
new L is only 10

Same down force, same midspan bend angle, same resulting tension, but
easy to miscalculate if we don't measure L and D from where they end
up, not from where they started.

The angle would be a nice way to calculate the tension, but it's hard
to measure the angle with enough accuracy on a spoke in a wheel--not
that the displacement is much easier.

My present scheme is limited to checking the accuracy of my Park
tension gauge at even higher tension.

My Park gauge indicated 187 lbs tension with ~190 pounds of weights
and water buckets hanging from a ceiling spoke, but I need to find out
how close it is with 300 pounds hanging from the spoke, which involves
building a wooden platform to hold weights and a trash barrel filled
with water, with everything raised and lowered with a floor jack.

I've found what may be the ultimate stiff spoked wheel for testing,
but the Park gauge measures a rise to about 300 pounds of tension on a
squeezed spoke, while my displacement measurements indicate 500 pounds
of tension.

The 300 versus 500 pound difference is far too much to be just minor
measuring errors.

So maybe my Park gauge is accurate at 190-lb tension to within a few
pounds, but quickly becomes wildly inaccurate and claims that 500
pounds is only 300. (Unlikely, but I won't know until I check. And it
could be just my Park gauge, somehow abused or damaged, not the
typical Park gauge.)

Or maybe my method for measuring displacement (which I thought was
really simple, clever, and amazingly accurate) is somehow off.

Here's a picture of the kind of measurement that I'm getting:

http://i11.tinypic.com/487m1au.jpg

Click on the lower right in Explorer to see full-size. Filmed through
the valve hole of an impressive spoked wheel, the tip of zip tie fixed
to the wheel hub indicates 263 mm on a DT spoke ruler, which is
hanging from very close to the bend of a spoke.

In theory, I just look at the zip tie with the spoke at ~200 lbs of
tension by Park gauge and then look again after I hang ~55 lbs of
weights from the spoke.

The measurement is nicely repeatable, coming to 263 mm again and
again, but two such measurements produce a displacement D that
indicates 500 pounds of tension when my Park gauge indicates only 300
pounds.

So I could be goofing up or missing something when I measure
displacement, or my Park gauge could be wildly off. I can't see
anything wrong with how I'm measuring displacement, so the obvious
thing to do is to hang 300 lbs from a spoke and check my Park gauge.

Cheers,

Carl Fogel

dvt said:

jim beam said:

imagine this:
[fixed width font]

v
concrete|---------|concrete

and
v
|--------------------|
| |
| |
| |<thin flexy pole
| |
//////////////////////////

now, which one do /you/ think is doing to show the greatest tension
increase when deflecting the wire at point "v"?

If the horizontal member is a string (i.e. no bending stiffness), and
you deflect point "v" until the string is bent at the same angle in both
cases, the tension is the same.

only if the length remains the same. for the first example, deflection
can only be accommodated by elongation of the string as tension
increases. with the second, the accommodation comes from the flexy
poles, not string elongation.

jim beam said:

dvt said:

jim beam said:

imagine this:
[fixed width font]

v
concrete|---------|concrete

and
v
|--------------------|
| |
| |
| |<thin flexy pole
| |
//////////////////////////

now, which one do /you/ think is doing to show the greatest tension
increase when deflecting the wire at point "v"?

If the horizontal member is a string (i.e. no bending stiffness), and
you deflect point "v" until the string is bent at the same angle in both
cases, the tension is the same.

only if the length remains the same. for the first example, deflection
can only be accommodated by elongation of the string as tension
increases. with the second, the accommodation comes from the flexy
poles, not string elongation.

Dear Jim & Dave,

Rubber band attached to the tops of two flexy poles:
http://i13.tinypic.com/4cug5yo.jpg

Weight hung:
http://i11.tinypic.com/4d4edqr.jpg

Rubber band and weight moved to stiffer sections of flexy poles:
http://i14.tinypic.com/2hwk9c9.jpg

With the rubber band at the top, the spokes bow inward, the new L is
reduced, D is large, and the angle is deeper.

With the rubber band moved down, the spokes bow inward less, the new L
remains closer to the original, D is small, and the angle flattens.

As I understand things, the weight produces the same tension on the
rubber band at both heights because D and L vary inversely.

For anyone curious, the easy equation is . . .

tension = (weight * length) / (4 * deflection)

.. . . where length is the current distance between the two end points,
and deflection is the distance from a line drawn between them to the
bend. For practical purposes, it works as well as the precise equation
of tension = force / 2 x sine(half-the-bend-angle).

Here's an onscreen ruler and protractor if anyone wants to measure:

http://www.markus-bader.de/MB-Ruler

I'm not sure if this will address Jim's elongation point, but maybe
the pictures will lead to some better examples.

Cheers,

Carl Fogel

Quoted message said:

jim beam said:

dvt said:

jim beam wrote:
> imagine this:
> [fixed width font]
>
> v
> concrete|---------|concrete
>
> and
> v
> |--------------------|
> | |
> | |
> | |<thin flexy pole
> | |
> //////////////////////////
>
> now, which one do /you/ think is doing to show the greatest tension
> increase when deflecting the wire at point "v"?
If the horizontal member is a string (i.e. no bending stiffness), and
you deflect point "v" until the string is bent at the same angle in both
cases, the tension is the same.

only if the length remains the same. for the first example, deflection
can only be accommodated by elongation of the string as tension
increases. with the second, the accommodation comes from the flexy
poles, not string elongation.

Dear Jim & Dave,

Rubber band attached to the tops of two flexy poles:
http://i13.tinypic.com/4cug5yo.jpg

Weight hung:
http://i11.tinypic.com/4d4edqr.jpg

Rubber band and weight moved to stiffer sections of flexy poles:
http://i14.tinypic.com/2hwk9c9.jpg

With the rubber band at the top, the spokes bow inward, the new L is
reduced, D is large, and the angle is deeper.

With the rubber band moved down, the spokes bow inward less, the new L
remains closer to the original, D is small, and the angle flattens.

As I understand things, the weight produces the same tension on the
rubber band at both heights because D and L vary inversely.

For anyone curious, the easy equation is . . .

tension = (weight * length) / (4 * deflection)

. . . where length is the current distance between the two end points,
and deflection is the distance from a line drawn between them to the
bend. For practical purposes, it works as well as the precise equation
of tension = force / 2 x sine(half-the-bend-angle).

Here's an onscreen ruler and protractor if anyone wants to measure:

http://www.markus-bader.de/MB-Ruler

I'm not sure if this will address Jim's elongation point, but maybe
the pictures will lead to some better examples.

Cheers,

Carl Fogel

Dave was wrong.

The angle gives the ratio of load to tension.

In these scenarios, the load is the same, the angle different, the
tension different. What's so hard?

Peter Cole said:

Quoted message said:

On Thu, 04 Jan 2007 22:31:28 -0800, jim beam
<[email hidden]> wrote:

Quoted message said:

Dave was wrong.

The angle gives the ratio of load to tension.

In these scenarios, the load is the same, the angle different, the
tension different. What's so hard?

As the OP, I think I should point out that this thread may have
exceeded the length of most helmet war threads.

Quoted message said:

For anyone curious, the easy equation is . . .

tension = (weight * length) / (4 * deflection)

. . . where length is the current distance between the two end points,
and deflection is the distance from a line drawn between them to the
bend. For practical purposes, it works as well as the precise equation
of tension = force / 2 x sine(half-the-bend-angle).

They are (or should be) identical; however, you wrote the wrong sine
formula. Correct is

sin(a) = deflection/hypotensus

where a is

-----------
\ a | a/
\ | /
\ | /
\ | /
\|/

I'm not sure what you mean by "half-the-bend-angle".

--
Joe Riel

dgk said:

Peter Cole said:

Quoted message said:

On Thu, 04 Jan 2007 22:31:28 -0800, jim beam
<[email hidden]> wrote:

Quoted message said:

Dave was wrong.

The angle gives the ratio of load to tension.

In these scenarios, the load is the same, the angle different, the
tension different. What's so hard?

As the OP, I think I should point out that this thread may have
exceeded the length of most helmet war threads.

Apparently, most of the length (and tedium) was due to a basic
misunderstanding of elementary physics. The worst thing about these
threads is that noobs come along regularly to begin the whole process
anew. It's a bore, but sometimes it's harder to let the blather stand
unchallenged.

What appears to be the underlying agenda in these threads is
"anti-engineering", or even "anti-science". It's common for engineering
to clash with intuition, but some people can't accept that.

Joe Riel said:

Quoted message said:

For anyone curious, the easy equation is . . .

tension = (weight * length) / (4 * deflection)

. . . where length is the current distance between the two end points,
and deflection is the distance from a line drawn between them to the
bend. For practical purposes, it works as well as the precise equation
of tension = force / 2 x sine(half-the-bend-angle).

They are (or should be) identical; however, you wrote the wrong sine
formula. Correct is

sin(a) = deflection/hypotensus

where a is

-----------
\ a | a/
\ | /
\ | /
\ | /
\|/

I'm not sure what you mean by "half-the-bend-angle".

Dear Joe,

Sorry, just my late-night incomprehensible goof for the
angle-over-there-that's-not-the-bend, the "a" in your diagram.

My understanding of the tension is that it simply doesn't matter how
the ends of the string, spoke, or rubber band end up at "a".

The ends of the string, spoke, or rubber band don't know whether
they're attached to stiff concrete or quivering poles. They only know
that they've stopped at those two points.

So the rubber band tension is the same at the top two points, which
are close together, whether the two points are anchored by bending
spokes or by concrete walls. The rubber band elongates the same amount
under the weight, no matter what is supplying the force to keep its
ends that far apart.

Similarly, the rubber band tension is the same at the two midpoints of
the spokes, which are further apart, no matter what happens to be
holding the ends of the rubber band that far apart. The rubber band's
tension is now greater than it was at the top, but only because the
ends are farther apart, not because of the nature of the material
holding the ends fixed in space.

So I think that Dave is right, and that Jim is mistaken. With a known
weight, we can ideally determine the tension from a picture that shows
only the vee of the rubber band--whatever supports the two ends of the
rubber band can be ignored.

But practically, we have to be able to determine the two current
endpoints that define the current L, the weight has to be right at the
midpoint of L, the force has to act at right angles to L, and the bend
has to be an idealized point instead of the flattening curve that it
actually is.

It's impossible, for example, for the force to act at right angles to
L when a pair of "parallel" spokes are squeezed (unless the squeezer
also exerts considerable force toward the rim, trying to slide his
hand outward from the hub).

A weight hung from a spoke will suffer the same loss of effective
force unless the two points determining the current L are dead level.

The squeezer's hand, to point out another problem, must flatten the
bend. The two other angles remain the same, but the measurement of D
will be wrong--D to the flattened bend must always be shorter than D
to the ideal bend:

Same bend angle, same L, same tension (I think), but different D:

\ / versus \__/
\/

Another potential problem occurs to me when I look at the alligator
clips on the rubber bands. They seem to tilt away from the spokes
instead of staying parallel as I naively expected.

Does this affect things? Where exactly do we measure from?

That is, do we measure from the bend up to the top of the rubber band
at the spoke, to the rubber band's midpoint, or the bottom of the
rubber band?

At least a spoke doesn't thin out at the midspan like a rubber band.

But I wonder if the bends at the endpoints of L on a spoke complicate
matters. We treat the spoke ends as idealized thin lines, but they're
actually 2 mm thick. If they don't behave ideally, the displacement D
can be off more than we expect--and we're measuring displacements of
only around 5 to 15 mm about 140 mm to one side.

Cheers,

Carl Fogel

Peter Cole said:

Quoted message said:

Quoted message said:

dvt wrote:
> jim beam wrote:
>> imagine this:
>> [fixed width font]
>>
>> v
>> concrete|---------|concrete
>>
>> and
>> v
>> |--------------------|
>> | |
>> | |
>> | |<thin flexy pole
>> | |
>> //////////////////////////
>>
>> now, which one do /you/ think is doing to show the greatest tension
>> increase when deflecting the wire at point "v"?

Quoted message said:

Quoted message said:

Quoted message said:

> If the horizontal member is a string (i.e. no bending stiffness),
> and you deflect point "v" until the string is bent at the same angle
> in both cases, the tension is the same.

Quoted message said:

Dave was wrong.

The angle gives the ratio of load to tension.

In these scenarios, the load is the same, the angle different, the
tension different. What's so hard?

You're right. I goofed. It's the ratio of load to tension, not only the
tension that is fixed by the angle. I'm embarrassed.

--
Dave
dvt at psu dot edu

Everyone confesses that exertion which brings out all the powers of body
and mind is the best thing for us; but most people do all they can to
get rid of it, and as a general rule nobody does much more than
circumstances drive them to do. -Harriet Beecher Stowe, abolitionist and
novelist (1811-1896)

dvt said:

Peter Cole said:

Quoted message said:

> dvt wrote:
>> jim beam wrote:
>>> imagine this:
>>> [fixed width font]
>>>
>>> v
>>> concrete|---------|concrete
>>>
>>> and
>>> v
>>> |--------------------|
>>> | |
>>> | |
>>> | |<thin flexy pole
>>> | |
>>> //////////////////////////
>>>
>>> now, which one do /you/ think is doing to show the greatest
>>> tension increase when deflecting the wire at point "v"?

Quoted message said:

Quoted message said:

>> If the horizontal member is a string (i.e. no bending stiffness),
>> and you deflect point "v" until the string is bent at the same
>> angle in both cases, the tension is the same.

Quoted message said:

Dave was wrong.

The angle gives the ratio of load to tension.

In these scenarios, the load is the same, the angle different, the
tension different. What's so hard?

You're right. I goofed. It's the ratio of load to tension, not only the
tension that is fixed by the angle. I'm embarrassed.

It's OK, statics confounds intuition. I still wince when I recall the
hilarity some of my contributions brought to my class. Apparently our
biology doesn't hard-wire a sense of vectors. I don't know if that was
your problem or you just misspoke (heh), but it does seem to be the
underlying issue with this thread. I'm just trying to keep it on track.

Peter Cole said:

dgk said:

Peter Cole said:

[email hidden] wrote:
> On Thu, 04 Jan 2007 22:31:28 -0800, jim beam
> <[email hidden]> wrote:

Quoted message said:

Dave was wrong.

The angle gives the ratio of load to tension.

In these scenarios, the load is the same, the angle different, the
tension different. What's so hard?

As the OP, I think I should point out that this thread may have
exceeded the length of most helmet war threads.

Apparently, most of the length (and tedium) was due to a basic
misunderstanding of elementary physics. The worst thing about these
threads is that noobs come along regularly to begin the whole process
anew. It's a bore, but sometimes it's harder to let the blather stand
unchallenged.

What appears to be the underlying agenda in these threads is
"anti-engineering", or even "anti-science". It's common for engineering
to clash with intuition, but some people can't accept that.

eh? from where i'm sitting, the agenda seems to be "engineers" having a
hard time coming to terms their grasp of reality. "oh, this piece of
metal is springing back after i bent it - that must be residual stress"
for example.

jim beam said:

Peter Cole wrote:

Quoted message said:

Quoted message said:

Apparently, most of the length (and tedium) was due to a basic
misunderstanding of elementary physics. The worst thing about these
threads is that noobs come along regularly to begin the whole process
anew. It's a bore, but sometimes it's harder to let the blather stand
unchallenged.

What appears to be the underlying agenda in these threads is
"anti-engineering", or even "anti-science". It's common for
engineering to clash with intuition, but some people can't accept that.

eh? from where i'm sitting, the agenda seems to be "engineers" having a
hard time coming to terms their grasp of reality.

OK, perhaps you forgot where this started:

I said:

"I measured 7mm deflection with an initial 23lb, and 12mm with an
additional 26lb (49lb total). I used hung weights, so I'm sure of the
forces. I used a caliper to measure the deflections. I don't have a
tensiometer, so I don't know the initial tension, but the differential
should be reasonably accurate."

"By the formula in Jobst's book (T=Force*length/4*displacement), I
calculated a tension of 230lb for the first load, 285 for the second.
So, the additional 26lb increased tension 55lb, or a little over 2:1."

you replied:
"that's the problem! jobst's formula doesn't account for rim
distortion. you need to measure tension of the spoke directly. which
exactly what fogel did."

If you understood vectors, you'd have known that "L" included rim
deflection. If you understood wheels (and vectors), you'd realize that
even not taking that into account doesn't significantly affect the outcome.

All you need to know to accurately measure spoke tension is the load and
the angle.

Carl understands now, how about you?

Ready to retract?

Peter Cole said:

jim beam said:

Peter Cole wrote:

Quoted message said:

Quoted message said:

Apparently, most of the length (and tedium) was due to a basic
misunderstanding of elementary physics. The worst thing about these
threads is that noobs come along regularly to begin the whole process
anew. It's a bore, but sometimes it's harder to let the blather stand
unchallenged.

What appears to be the underlying agenda in these threads is
"anti-engineering", or even "anti-science". It's common for
engineering to clash with intuition, but some people can't accept that.

eh? from where i'm sitting, the agenda seems to be "engineers" having
a hard time coming to terms their grasp of reality.

OK, perhaps you forgot where this started:

I said:

"I measured 7mm deflection with an initial 23lb, and 12mm with an
additional 26lb (49lb total). I used hung weights, so I'm sure of the
forces. I used a caliper to measure the deflections. I don't have a
tensiometer, so I don't know the initial tension, but the differential
should be reasonably accurate."

"By the formula in Jobst's book (T=Force*length/4*displacement), I
calculated a tension of 230lb for the first load, 285 for the second.
So, the additional 26lb increased tension 55lb, or a little over 2:1."

you replied:
"that's the problem! jobst's formula doesn't account for rim
distortion. you need to measure tension of the spoke directly. which
exactly what fogel did."

If you understood vectors, you'd have known that "L" included rim
deflection. If you understood wheels (and vectors), you'd realize that
even not taking that into account doesn't significantly affect the outcome.

condescending drivel. /you/ are the guy that "forgot" to take L into
account. /you/ are the guy with the problem recognizing elasticity. as
i pointed out right at the start. now of course, you're trying to gloss
over all this, but hey peter, don't let little details get in the way of
your method when you "measure" spoke tension.

Quoted message said:

All you need to know to accurately measure spoke tension is the load and
the angle.

Carl understands now, how about you?

Ready to retract?

er, /you/ are the one that needs to eat crow here buddy. your unabashed
attempts to ignore the inconvenient stuff is quite ridiculous.

jim beam said:

condescending drivel. /you/ are the guy that "forgot" to take L into
account. /you/ are the guy with the problem recognizing elasticity. as
i pointed out right at the start. now of course, you're trying to gloss
over all this, but hey peter, don't let little details get in the way of
your method when you "measure" spoke tension.

I didn't "forget" to take "L" into account. What you (continue to) fail
to grasp is that rim deflection doesn't matter. The ratio of the force
to the tension is exactly equal to the ratio of the displacement to the
hypotenuse, which, in this case is "L", and doesn't change.

Your response (to my complete explanation) was:

"This stuff is real basic.

"3. if there is no fraud, then the math model is incomplete. and half a
moment's thought paying attention to the fact that we don't see rim
stiffness in the equation should make anyone pause. assumption that the
distance between the two ends of the wire therefore remains perfectly
fixed is incorrect."

"this stuff is real basic peter and it concerns me that you either don't
get this stuff or are prepared to deny it."

It is "real basic". Rim stiffness doesn't enter into it. Nobody is
assuming "the distance between the two ends of the wire remains
perfectly fixed". It doesn't matter. Everyone seems to "get this stuff"
by now except you.

Quoted message said:

er, /you/ are the one that needs to eat crow here buddy. your unabashed
attempts to ignore the inconvenient stuff is quite ridiculous.

Ridiculous, indeed. I think you're the one who needs to give it more
than "half a moment's thought".