Re: I fixed a broken spoke!

12-23-2006, 04:22 PM

[Only registered and activated users can see links. ] wrote:
> On Sat, 23 Dec 2006 11:16:54 -0500, Peter Cole
> <[Only registered and activated users can see links. ]> wrote:
>
>> OK, I had some time this morning.
>>
>> I took an old (rear) wheel I have -- Mavic Reflex, 36 hole, rear, 2mm
>> spokes.
>>
>> 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.
>>
>> This was about what I expected, a hard (50-60lb) squeeze will give me
>> the nominal 50% over-tension I want in stress relieving. I didn't expect
>> 5:1, that's just what Phil H reported. My rims aren't as stiff, so that
>> would explain at least some of the difference. Parenthetically, this
>> also suggests my wheel was a little under-tensioned, something I don't
>> doubt, since as I recall, this rim replaced an identical prior rim that
>> cracked at the spoke holes (both anodized), so I think I took the
>> tension only as far as I needed to without the spokes unscrewing when
>> ridden.
>
> Dear Peter,
>
> Sorry, no tension gauge means no reasonably accurate measurement.
> You're missing repeated explanations of what's wrong with approach
> that you're using.
>
> A bicycle wheel is worthless as a makeshift tension gauge because all
> tension gauge calculations depend on the two support points being as
> rigidly fixed as possible.
>
> A mere quarter turn of a spoke nipple will visibly deform the rim, so
> squeezing four spokes twists the rim so much out of shape that the
> calculations are pointless.
>
> Applying large forces to an aluminum hoop braced by a few steel wires
> is simply not the same as using a massively braced tension gauge that
> applies small forces.
>
> The Park gauge braces its two posts with 3mm thick aluminum against a
> spring that applies a "weight" so small that you can put your finger
> between the spoke and the hammer applied by the spring. The whole idea
> is to avoid deforming the test rig.
>
> You measured one spoke's deflection at 12 mm for 49 pounds.
>
> Look at these results:
>
> http://home.comcast.net/~carlfogel/d...d/newspok2.jpg
>
> I measured the gap between the two squeezed spokes narrowing from ~82
> mm to ~47 mm, about ~35 for 2 spokes, or an ~18 mm single spoke
> deflection with a 100 pound weight hung.
>
> God knows what absurd imaginary tension increases might be calculated
> from this deflection, but the measured real tension increase was only
> about 90 pounds.
>
> Using a tension gauge is like using a rope stretched between a pair of
> oak trees and measuring the deflection caused by a 5 pound weight.
>
> Measuring the deflection of a spoke on an aluminum hoop under a 49
> pound weight is like watching a hammock sag between two saplings. Any
> calculations according to the bend are worthless.
>
> Buy a tension gauge and let us know what the actual tensions are. The
> Park gauge works fine and is about $50 to $75. Afterward, you can sell
> it on eBay to save money, but it would be a useful tool.
>
> If you're not willing to buy a tension gauge for $50, see if you can
> borrow one from a friend or a local bike shop.
>
> Whatever you do to obtain a tension gauge, I predict that it will not
> show even a 90 pound tension increase for a 60 pound squeeze force.
>
> But we're making progress. Your estimates are still high, but you're
> now aware that your wheels are nowhere near as stiff as you thought
> they were.
>
> Like you, I thought that the high tension increase claims were
> plausible until I started actually measuring things. Once I slapped a
> tension gauge on an actual spoke, I stopped relying on theoretical
> calculations based on theoretical wheels that theoretically were so
> stiff that they couldn't possibly be trued.
>
> Cheers,
>
> Carl Fogel

Ah, thanks for the concern about the mote in my eye -- shall we now turn
to the beam in yours? (double pun intended). I'm not sure if it's worth
the bother, as it does really require a grasp of high school level
physics -- but here goes:

First, to the theory (physics->vectors).

The angle of deflection with a known load does *exactly* determine the
tension. Your examples of sturdy oaks and tender saplings are exactly
equivalent. The angle *must* reveal the exact ratio of the tension force
to the load force (unless Newton was wrong, of course). Since you know
the load, and you know the *exact* ratio, you know the *exact* tension.

Second, to the measurements.

Does the deflection amount *exactly* determine the angle? Despite those
nasty rim deflections (seemingly moving the goal posts, as it were)? In
a word, yes.

As you may recall from middle school trigonometry, we can determine the
angle of a right triangle from the hypotenuse and the opposite side.

In this case, the hypotenuse is the spoke itself, unchanged in length
from its unloaded state except for an ignorable (much less than 1%)
amount of stretch. The opposite side is the measured deflection. Note
that the other side of the triangle (distance to rim) doesn't enter into
the calculation of angle.

Third, to the calculations.

I know you consider this "theoretical", but I assure you your
handy-dandy tensiometer relies on the same theory.

Tension = Force/2*sin(theta) (into the book directly from Newton).

alternately, Tension = Force*Length/4*Deflection (close approximation).

From the measurements you gave, I compute a final tension of 395lb with
your 100lb load and 17.5mm deflection, or a ratio of load to tension
increase of about 1.5.

Fourth, to the issue of "a wheel as a tensiometer".

The reason tensiometers (most, anyway) use small loads is so that they
won't be measuring the extra tension they create. Unfortunately, this
means deflections are small, requiring precision in displacement
measurement and also makes them sensitive to friction. These issues, in
turn, present challenges to manufacture at a palatable consumer price
point. In other words, a $50 tensiometer is probably not a very good
tensiometer.

A wheel, in comparison, makes a considerably better tensiometer, at
least in this rather unusual application. We get to use very large
forces and measure over the whole spoke length, making displacement
measurements relatively easy with cruder references. Since the forces
are large, errors from friction are also less likely to throw things off.

Fifth, "what about rim stiffness"?

Rim stiffness affects the angle formed by the load. In doing that, it
determines the ratio between the load and the tension. Consider the
extremes: if your saplings were very flexible, the angle could go to 90.
In that case, the tension would be 1/2 the load (the minimum). On the
other hand, consider your oaks to be infinitely strong and the rope
between them incapable of stretching in the slightest -- in that case, a
butterfly landing on it would create infinite tension (with an angle of
zero).

The reason I measured a force to tension ratio of about 2, and you about
1.5 (despite what your blue-handled marvel says) is only because my rim
is a bit stiffer.

If you still can't get you head around all this, send me your brain.
I'll do what I can and pay the return postage.

12-23-2006, 04:51 PM

On Sat, 23 Dec 2006 20:22:09 -0500, Peter Cole
<[Only registered and activated users can see links. ]> wrote:

>[Only registered and activated users can see links. ] wrote:
>> On Sat, 23 Dec 2006 11:16:54 -0500, Peter Cole
>> <[Only registered and activated users can see links. ]> wrote:
>>
>>> OK, I had some time this morning.
>>>
>>> I took an old (rear) wheel I have -- Mavic Reflex, 36 hole, rear, 2mm
>>> spokes.
>>>
>>> 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.
>>>
>>> This was about what I expected, a hard (50-60lb) squeeze will give me
>>> the nominal 50% over-tension I want in stress relieving. I didn't expect
>>> 5:1, that's just what Phil H reported. My rims aren't as stiff, so that
>>> would explain at least some of the difference. Parenthetically, this
>>> also suggests my wheel was a little under-tensioned, something I don't
>>> doubt, since as I recall, this rim replaced an identical prior rim that
>>> cracked at the spoke holes (both anodized), so I think I took the
>>> tension only as far as I needed to without the spokes unscrewing when
>>> ridden.
>>
>> Dear Peter,
>>
>> Sorry, no tension gauge means no reasonably accurate measurement.
>> You're missing repeated explanations of what's wrong with approach
>> that you're using.
>>
>> A bicycle wheel is worthless as a makeshift tension gauge because all
>> tension gauge calculations depend on the two support points being as
>> rigidly fixed as possible.
>>
>> A mere quarter turn of a spoke nipple will visibly deform the rim, so
>> squeezing four spokes twists the rim so much out of shape that the
>> calculations are pointless.
>>
>> Applying large forces to an aluminum hoop braced by a few steel wires
>> is simply not the same as using a massively braced tension gauge that
>> applies small forces.
>>
>> The Park gauge braces its two posts with 3mm thick aluminum against a
>> spring that applies a "weight" so small that you can put your finger
>> between the spoke and the hammer applied by the spring. The whole idea
>> is to avoid deforming the test rig.
>>
>> You measured one spoke's deflection at 12 mm for 49 pounds.
>>
>> Look at these results:
>>
>> http://home.comcast.net/~carlfogel/d...d/newspok2.jpg
>>
>> I measured the gap between the two squeezed spokes narrowing from ~82
>> mm to ~47 mm, about ~35 for 2 spokes, or an ~18 mm single spoke
>> deflection with a 100 pound weight hung.
>>
>> God knows what absurd imaginary tension increases might be calculated
>> from this deflection, but the measured real tension increase was only
>> about 90 pounds.
>>
>> Using a tension gauge is like using a rope stretched between a pair of
>> oak trees and measuring the deflection caused by a 5 pound weight.
>>
>> Measuring the deflection of a spoke on an aluminum hoop under a 49
>> pound weight is like watching a hammock sag between two saplings. Any
>> calculations according to the bend are worthless.
>>
>> Buy a tension gauge and let us know what the actual tensions are. The
>> Park gauge works fine and is about $50 to $75. Afterward, you can sell
>> it on eBay to save money, but it would be a useful tool.
>>
>> If you're not willing to buy a tension gauge for $50, see if you can
>> borrow one from a friend or a local bike shop.
>>
>> Whatever you do to obtain a tension gauge, I predict that it will not
>> show even a 90 pound tension increase for a 60 pound squeeze force.
>>
>> But we're making progress. Your estimates are still high, but you're
>> now aware that your wheels are nowhere near as stiff as you thought
>> they were.
>>
>> Like you, I thought that the high tension increase claims were
>> plausible until I started actually measuring things. Once I slapped a
>> tension gauge on an actual spoke, I stopped relying on theoretical
>> calculations based on theoretical wheels that theoretically were so
>> stiff that they couldn't possibly be trued.
>>
>> Cheers,
>>
>> Carl Fogel
>
>Ah, thanks for the concern about the mote in my eye -- shall we now turn
>to the beam in yours? (double pun intended). I'm not sure if it's worth
>the bother, as it does really require a grasp of high school level
>physics -- but here goes:
>
>First, to the theory (physics->vectors).
>
>The angle of deflection with a known load does *exactly* determine the
>tension. Your examples of sturdy oaks and tender saplings are exactly
>equivalent. The angle *must* reveal the exact ratio of the tension force
>to the load force (unless Newton was wrong, of course). Since you know
>the load, and you know the *exact* ratio, you know the *exact* tension.
>
>Second, to the measurements.
>
>Does the deflection amount *exactly* determine the angle? Despite those
>nasty rim deflections (seemingly moving the goal posts, as it were)? In
>a word, yes.
>
>As you may recall from middle school trigonometry, we can determine the
>angle of a right triangle from the hypotenuse and the opposite side.
>
>In this case, the hypotenuse is the spoke itself, unchanged in length
>from its unloaded state except for an ignorable (much less than 1%)
>amount of stretch. The opposite side is the measured deflection. Note
>that the other side of the triangle (distance to rim) doesn't enter into
>the calculation of angle.
>
>Third, to the calculations.
>
>I know you consider this "theoretical", but I assure you your
>handy-dandy tensiometer relies on the same theory.
>
>Tension = Force/2*sin(theta) (into the book directly from Newton).
>
>alternately, Tension = Force*Length/4*Deflection (close approximation).
>
> From the measurements you gave, I compute a final tension of 395lb with
>your 100lb load and 17.5mm deflection, or a ratio of load to tension
>increase of about 1.5.
>
>Fourth, to the issue of "a wheel as a tensiometer".
>
>The reason tensiometers (most, anyway) use small loads is so that they
>won't be measuring the extra tension they create. Unfortunately, this
>means deflections are small, requiring precision in displacement
>measurement and also makes them sensitive to friction. These issues, in
>turn, present challenges to manufacture at a palatable consumer price
>point. In other words, a $50 tensiometer is probably not a very good
>tensiometer.
>
>A wheel, in comparison, makes a considerably better tensiometer, at
>least in this rather unusual application. We get to use very large
>forces and measure over the whole spoke length, making displacement
>measurements relatively easy with cruder references. Since the forces
>are large, errors from friction are also less likely to throw things off.
>
>Fifth, "what about rim stiffness"?
>
>Rim stiffness affects the angle formed by the load. In doing that, it
>determines the ratio between the load and the tension. Consider the
>extremes: if your saplings were very flexible, the angle could go to 90.
>In that case, the tension would be 1/2 the load (the minimum). On the
>other hand, consider your oaks to be infinitely strong and the rope
>between them incapable of stretching in the slightest -- in that case, a
>butterfly landing on it would create infinite tension (with an angle of
>zero).
>
>The reason I measured a force to tension ratio of about 2, and you about
>1.5 (despite what your blue-handled marvel says) is only because my rim
>is a bit stiffer.
>
>If you still can't get you head around all this, send me your brain.
>I'll do what I can and pay the return postage.

Dear Peter,

There's not much point in discussing things with someone who'll do
anything but measure the actual tension under discussion.

I predict that you won't see 90 pounds of tension rise for a squeeze
force of 60 pounds if you ever actually measure the spoke tension
instead of insisting that you can calculate it correctly.

Your rim is unlikely to be anywhere near as stiff in regard to the
tensions applied as a steel pipe clamp.

Since it's easy to do, your reluctance to confirm your claim to
implausibly high tension rises by a simple direct measurement will be
the last word on the matter.

Cheers,

Carl Fogel

12-23-2006, 05:23 PM

Peter Cole wrote:
> [Only registered and activated users can see links. ] wrote:
>> On Sat, 23 Dec 2006 11:16:54 -0500, Peter Cole
>> <[Only registered and activated users can see links. ]> wrote:
>>
>>> OK, I had some time this morning.
>>>
>>> I took an old (rear) wheel I have -- Mavic Reflex, 36 hole, rear, 2mm
>>> spokes.
>>>
>>> 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.
>>>
>>> This was about what I expected, a hard (50-60lb) squeeze will give me
>>> the nominal 50% over-tension I want in stress relieving. I didn't
>>> expect 5:1, that's just what Phil H reported. My rims aren't as
>>> stiff, so that would explain at least some of the difference.
>>> Parenthetically, this also suggests my wheel was a little
>>> under-tensioned, something I don't doubt, since as I recall, this rim
>>> replaced an identical prior rim that cracked at the spoke holes (both
>>> anodized), so I think I took the tension only as far as I needed to
>>> without the spokes unscrewing when ridden.
>>
>> Dear Peter,
>>
>> Sorry, no tension gauge means no reasonably accurate measurement.
>> You're missing repeated explanations of what's wrong with approach
>> that you're using.
>> A bicycle wheel is worthless as a makeshift tension gauge because all
>> tension gauge calculations depend on the two support points being as
>> rigidly fixed as possible.
>> A mere quarter turn of a spoke nipple will visibly deform the rim, so
>> squeezing four spokes twists the rim so much out of shape that the
>> calculations are pointless.
>>
>> Applying large forces to an aluminum hoop braced by a few steel wires
>> is simply not the same as using a massively braced tension gauge that
>> applies small forces.
>>
>> The Park gauge braces its two posts with 3mm thick aluminum against a
>> spring that applies a "weight" so small that you can put your finger
>> between the spoke and the hammer applied by the spring. The whole idea
>> is to avoid deforming the test rig.
>>
>> You measured one spoke's deflection at 12 mm for 49 pounds.
>>
>> Look at these results:
>>
>> http://home.comcast.net/~carlfogel/d...d/newspok2.jpg
>>
>> I measured the gap between the two squeezed spokes narrowing from ~82
>> mm to ~47 mm, about ~35 for 2 spokes, or an ~18 mm single spoke
>> deflection with a 100 pound weight hung.
>> God knows what absurd imaginary tension increases might be calculated
>> from this deflection, but the measured real tension increase was only
>> about 90 pounds.
>>
>> Using a tension gauge is like using a rope stretched between a pair of
>> oak trees and measuring the deflection caused by a 5 pound weight.
>> Measuring the deflection of a spoke on an aluminum hoop under a 49
>> pound weight is like watching a hammock sag between two saplings. Any
>> calculations according to the bend are worthless.
>>
>> Buy a tension gauge and let us know what the actual tensions are. The
>> Park gauge works fine and is about $50 to $75. Afterward, you can sell
>> it on eBay to save money, but it would be a useful tool.
>> If you're not willing to buy a tension gauge for $50, see if you can
>> borrow one from a friend or a local bike shop.
>>
>> Whatever you do to obtain a tension gauge, I predict that it will not
>> show even a 90 pound tension increase for a 60 pound squeeze force.
>>
>> But we're making progress. Your estimates are still high, but you're
>> now aware that your wheels are nowhere near as stiff as you thought
>> they were.
>> Like you, I thought that the high tension increase claims were
>> plausible until I started actually measuring things. Once I slapped a
>> tension gauge on an actual spoke, I stopped relying on theoretical
>> calculations based on theoretical wheels that theoretically were so
>> stiff that they couldn't possibly be trued.
>>
>> Cheers,
>>
>> Carl Fogel
>
> Ah, thanks for the concern about the mote in my eye -- shall we now turn
> to the beam in yours? (double pun intended). I'm not sure if it's worth
> the bother, as it does really require a grasp of high school level
> physics -- but here goes:
>
> First, to the theory (physics->vectors).
>
> The angle of deflection with a known load does *exactly* determine the
> tension. Your examples of sturdy oaks and tender saplings are exactly
> equivalent. The angle *must* reveal the exact ratio of the tension force
> to the load force (unless Newton was wrong, of course). Since you know
> the load, and you know the *exact* ratio, you know the *exact* tension.
>
> Second, to the measurements.
>
> Does the deflection amount *exactly* determine the angle? Despite those
> nasty rim deflections (seemingly moving the goal posts, as it were)? In
> a word, yes.
>
> As you may recall from middle school trigonometry, we can determine the
> angle of a right triangle from the hypotenuse and the opposite side.
>
> In this case, the hypotenuse is the spoke itself, unchanged in length
> from its unloaded state except for an ignorable (much less than 1%)
> amount of stretch. The opposite side is the measured deflection. Note
> that the other side of the triangle (distance to rim) doesn't enter into
> the calculation of angle.
>
> Third, to the calculations.
>
> I know you consider this "theoretical", but I assure you your
> handy-dandy tensiometer relies on the same theory.
>
> Tension = Force/2*sin(theta) (into the book directly from Newton).
>
> alternately, Tension = Force*Length/4*Deflection (close approximation).
>
> From the measurements you gave, I compute a final tension of 395lb with
> your 100lb load and 17.5mm deflection, or a ratio of load to tension
> increase of about 1.5.
>
> Fourth, to the issue of "a wheel as a tensiometer".
>
> The reason tensiometers (most, anyway) use small loads is so that they
> won't be measuring the extra tension they create. Unfortunately, this
> means deflections are small, requiring precision in displacement
> measurement and also makes them sensitive to friction. These issues, in
> turn, present challenges to manufacture at a palatable consumer price
> point. In other words, a $50 tensiometer is probably not a very good
> tensiometer.
>
> A wheel, in comparison, makes a considerably better tensiometer, at
> least in this rather unusual application. We get to use very large
> forces and measure over the whole spoke length, making displacement
> measurements relatively easy with cruder references. Since the forces
> are large, errors from friction are also less likely to throw things off.
>
> Fifth, "what about rim stiffness"?
>
> Rim stiffness affects the angle formed by the load. In doing that, it
> determines the ratio between the load and the tension. Consider the
> extremes: if your saplings were very flexible, the angle could go to 90.
> In that case, the tension would be 1/2 the load (the minimum). On the
> other hand, consider your oaks to be infinitely strong and the rope
> between them incapable of stretching in the slightest -- in that case, a
> butterfly landing on it would create infinite tension (with an angle of
> zero).
>
> The reason I measured a force to tension ratio of about 2, and you about
> 1.5 (despite what your blue-handled marvel says) is only because my rim
> is a bit stiffer.
>
> If you still can't get you head around all this, send me your brain.
> I'll do what I can and pay the return postage.
>

wow dude, you're deeply afflicted with nonsensicitis, with a large dose
of dontgethebasics stubbornius thrown in. really, that has to be one of
the most willfully obtuse responses i've ever seen on r.b.t., and that's
saying something.

this stuff is real basic.

1. we have theory and we have practice. if the two do not accord,
either the theory is incomplete or the practice is wrong. do you deny that?

2. since the measuring instrument is a commercial off the shelf tool
available to us all, with presumably reasonable calibration and
consistency from instrument to instrument, that leaves measuring discord
in the territory of fogel fraud. are you accusing him of that?

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.

12-23-2006, 11:13 PM

[Only registered and activated users can see links. ] wrote:
> Since it's easy to do, your reluctance to confirm your claim to
> implausibly high tension rises by a simple direct measurement will be
> the last word on the matter.

I can't offer an explanation of why yours and Peter's results differ,
but the way he is determining the spoke tension (*with* the load
applied) truly is independant of the stiffness of the rim. It is simple
trig. *If* we can assume that a spoke is as flexible as a thread (more
plausible over long spans), and if we know the length, deflection, and
force applied... then the tension is absolutely defined. I know it
doesn't seem intuitively correct, but you can move the end points
around all you want... as long as you measure L, d, and F accurately,
then you *can* calculate T.

This doesn't tell us what the initial tension was... only the tension
with the applied load.

I don't know if the Park tensiometer is any good but I picked one up
from Performance when they were having their $20 off $50 with free
shipping sale, and I'm having fun with it. It seems to be pretty
consistent at least, if not very precise... ie a single digit on the
scale can represent a 4 to 20 kg variation in tension. The force it
exerts on the spoke is not that small either... considering that it
takes ~3lb to squeeze the handle and a leverage ratio of ~8 to 1 gives
a force of ~24lb. Supposedly their conversion table compensates for
this...

12-24-2006, 06:08 AM

Ron Ruff wrote:
> [Only registered and activated users can see links. ] wrote:
>> Since it's easy to do, your reluctance to confirm your claim to
>> implausibly high tension rises by a simple direct measurement will be
>> the last word on the matter.
>
> I can't offer an explanation of why yours and Peter's results differ,
> but the way he is determining the spoke tension (*with* the load
> applied) truly is independant of the stiffness of the rim. It is simple
> trig. *If* we can assume that a spoke is as flexible as a thread (more
> plausible over long spans), and if we know the length, deflection, and
> force applied... then the tension is absolutely defined.

no because the math assumes the ends are fixed in space. it doesn't
explicitly state that, but it's the assumption nevertheless. so peter
doesn't have a result, he has a calculation which doesn't reflect
reality. carl otoh has an actual repeatable consistent measurement.

> I know it
> doesn't seem intuitively correct, but you can move the end points
> around all you want... as long as you measure L, d, and F accurately,
> then you *can* calculate T.

if the length of the wire remains the same, and there's deflection, then
by definition, the end points have moved. and that is precisely the
problem with the calculation.

>
> This doesn't tell us what the initial tension was... only the tension
> with the applied load.
>
> I don't know if the Park tensiometer is any good but I picked one up
> from Performance when they were having their $20 off $50 with free
> shipping sale, and I'm having fun with it. It seems to be pretty
> consistent at least, if not very precise... ie a single digit on the
> scale can represent a 4 to 20 kg variation in tension. The force it
> exerts on the spoke is not that small either... considering that it
> takes ~3lb to squeeze the handle and a leverage ratio of ~8 to 1 gives
> a force of ~24lb. Supposedly their conversion table compensates for
> this...
>

don't blame the tool.

12-24-2006, 06:49 AM

jim beam wrote:
> Ron Ruff wrote:
>> [Only registered and activated users can see links. ] wrote:
>>> Since it's easy to do, your reluctance to confirm your claim to
>>> implausibly high tension rises by a simple direct measurement will be
>>> the last word on the matter.
>>
>> I can't offer an explanation of why yours and Peter's results differ,
>> but the way he is determining the spoke tension (*with* the load
>> applied) truly is independant of the stiffness of the rim. It is simple
>> trig. *If* we can assume that a spoke is as flexible as a thread (more
>> plausible over long spans), and if we know the length, deflection, and
>> force applied... then the tension is absolutely defined.
>
> no because the math assumes the ends are fixed in space. it doesn't
> explicitly state that, but it's the assumption nevertheless. so peter
> doesn't have a result, he has a calculation which doesn't reflect
> reality. carl otoh has an actual repeatable consistent measurement.
>
>> I know it
>> doesn't seem intuitively correct, but you can move the end points
>> around all you want... as long as you measure L, d, and F accurately,
>> then you *can* calculate T.
>
> if the length of the wire remains the same, and there's deflection, then
> by definition, the end points have moved.

on re-reading, i need to qualify that:
if the length of wire remains the same, and there's deflection, then, by
definition, two possible things have happened.

1. if the end points are not affixed to something perfectly rigid, then
the two ends have moved closer together.

2. if the end points /are/ perfectly rigid, then the wire has become
longer by elasticity, and will experience a tension rise accordingly.

reality is, there is a mix of the two. there will be /some/ tension
rise, but that is defined by the rigidity of the mountings. as an
illustration, have two kids hold a piece of rope, then pull in the
middle of the rope. how much additional tension can the kids exert?
not much! therefore to accommodate the the deflection, the kids get
pulled closer together. the rope experiences /some/ tension increase,
but the majority of geometry change is accommodated by movement of the
two anchor points.

pretty basic. as is the concept that rims are far from rigid. anyone
that's ever trued a wheel knows that.

> and that is precisely the
> problem with the calculation.
>
>>
>> This doesn't tell us what the initial tension was... only the tension
>> with the applied load.
>>
>> I don't know if the Park tensiometer is any good but I picked one up
>> from Performance when they were having their $20 off $50 with free
>> shipping sale, and I'm having fun with it. It seems to be pretty
>> consistent at least, if not very precise... ie a single digit on the
>> scale can represent a 4 to 20 kg variation in tension. The force it
>> exerts on the spoke is not that small either... considering that it
>> takes ~3lb to squeeze the handle and a leverage ratio of ~8 to 1 gives
>> a force of ~24lb. Supposedly their conversion table compensates for
>> this...
>>
>
> don't blame the tool.

12-24-2006, 09:20 AM

jim beam wrote:
> no because the math assumes the ends are fixed in space. it doesn't
> explicitly state that, but it's the assumption nevertheless.

Let me try to re-explain. At first I thought the same as you... it is
intuitively obvious that if the end points are not rigid (they come
together some amount), you'll be able to deflect the spoke farther with
the same force... and this true! But at the same time it doesn't
matter. The calculation of tension via T= F*L/4/d is pure trig (force
balance perpendicular to the spoke) and doesn't care if the end points
have moved at all. The spoke is modeled as thread with zero bending
stiffness; we don't care about it's modulus or anything. You only need
to know those quanities, F, L, and d fairly accurately to determine the
tension in the spoke. Note that the "initial" spoke tension or
conditions is not a part of this at all... we are just looking at the
tension *after* the side load (F) is applied. This is a key point in
understanding I think.

It is true that if you have perfectly rigid endpoints a given F will
result in a smaller d, and T will be higher. But the equation is not
invalidated if the ends come together a little. Then d will be greater
and you will calculate a smaller T.

I can't say if Peter made an error in measurement or not, but his
method is sound.

12-24-2006, 03:02 PM

On 24 Dec 2006 10:20:42 -0800, "Ron Ruff" <[Only registered and activated users can see links. ]>
wrote:

>
>jim beam wrote:
>> no because the math assumes the ends are fixed in space. it doesn't
>> explicitly state that, but it's the assumption nevertheless.
>
>Let me try to re-explain. At first I thought the same as you... it is
>intuitively obvious that if the end points are not rigid (they come
>together some amount), you'll be able to deflect the spoke farther with
>the same force... and this true! But at the same time it doesn't
>matter. The calculation of tension via T= F*L/4/d is pure trig (force
>balance perpendicular to the spoke) and doesn't care if the end points
>have moved at all. The spoke is modeled as thread with zero bending
>stiffness; we don't care about it's modulus or anything. You only need
>to know those quanities, F, L, and d fairly accurately to determine the
>tension in the spoke. Note that the "initial" spoke tension or
>conditions is not a part of this at all... we are just looking at the
>tension *after* the side load (F) is applied. This is a key point in
>understanding I think.
>
>It is true that if you have perfectly rigid endpoints a given F will
>result in a smaller d, and T will be higher. But the equation is not
>invalidated if the ends come together a little. Then d will be greater
>and you will calculate a smaller T.
>
>I can't say if Peter made an error in measurement or not, but his
>method is sound.

Dear Ron,

If the method is sound in this specific situation, then it would be
easy to confirm its soundness with a tension gauge.

For fun, I'm going to see about hanging a weight around 200 pounds
from a spoke and applying a tension gauge. I'll be very surprised if
it is the 50% off suggested elsewhere.

(The immediate problem is that the only handy and easily mounted ~200
weight in my basement will be busy trying to apply the tension gauge,
and it's too dark and cold to try to hoist a motorycle into the air
out on the car port.)

As an analogy, some people are claiming that measuring the depth of
two footprints in a muddy field lets them determine that whoever
walked through the field and left those tracks must have weighed
either 300 pounds or 1,000 pounds--the muddiness makes no difference
and they scoff at the notion that the soil might vary.

But everyone who actually stood on a bathroom scale at the far side of
the muddy field seemed to weigh around 200 pounds.

Why not step on the scale?

RBT has a bad habit of debating theories in a vacuum instead of trying
to test whether the actual situation involves what the theory
requires.

Think of the fun if I report that my Park tension gauge is wildly
inaccurate!

Cheers,

Carl Fogel

12-24-2006, 03:10 PM

Ron Ruff wrote:
> jim beam wrote:
>> no because the math assumes the ends are fixed in space. it doesn't
>> explicitly state that, but it's the assumption nevertheless.
>
> Let me try to re-explain. At first I thought the same as you... it is
> intuitively obvious that if the end points are not rigid (they come
> together some amount), you'll be able to deflect the spoke farther with
> the same force... and this true! But at the same time it doesn't
> matter. The calculation of tension via T= F*L/4/d is pure trig (force
> balance perpendicular to the spoke) and doesn't care if the end points
> have moved at all.

yes it does. try unanchoring one end of the wire. it can still be
deflected, but see what happens to the tension.

> The spoke is modeled as thread with zero bending
> stiffness; we don't care about it's modulus or anything. You only need
> to know those quanities, F, L, and d fairly accurately to determine the
> tension in the spoke. Note that the "initial" spoke tension or
> conditions is not a part of this at all... we are just looking at the
> tension *after* the side load (F) is applied. This is a key point in
> understanding I think.
>
> It is true that if you have perfectly rigid endpoints a given F will
> result in a smaller d, and T will be higher. But the equation is not
> invalidated if the ends come together a little. Then d will be greater
> and you will calculate a smaller T.
>
> I can't say if Peter made an error in measurement or not, but his
> method is sound.
>
his calculated values don't agree with measured values for exactly that
reason - he didn't measure, he just calculated! and the calculation he
performed did not allow for the changes in length where the end anchors
are not rigid.

12-24-2006, 10:54 PM

jim beam wrote:
> yes it does. try unanchoring one end of the wire. it can still be
> deflected, but see what happens to the tension.

Or... you can anchor both ends... or not... it doesn't matter. I think
you are still stuck on the idea that the for a given force, the
deflection (and tension) will vary a lot depending on how rigid the
ends are fixed... and you are absolutely correct! If you apply a 50 lb
load to spoke that has rigidly fixed ends, the tension rise will be
much higher than if the ends flex a large amount... but in *both* cases
you'll still be able to determine the *resulting* tension if you can
measure F, L, and d!

I had a block about this too at first... I even started to write Peter
a post telling him he was wrong... then I went back and looked at how
the equation was derived... It's just vectors and force balance.

It's trickier to tell what the initial tension was though... this only
gives you the tension after the load is applied.