Re: Archery test

Carl Fogel said:

Like you, I now expect that if D is measured accurately from the
correct starting position (somewhere in the thin air between the two
current end points of L) to the bend, then the equation should agree
quite closely with an accurate tension gauge.

Why mess with measuring d and L when you can just measure the angle of the
bend in the spoke? The sine of half that angle is what you trying to get
with d and L anyway.

I measured the angle of the bend in the lower of the two "squeezed" spokes
in your picture http://i16.tinypic.com/33uygk1.jpg off of my screen to be
about 12.65 degrees (probably off a bit because of parallax). The equation
for tension reduces down to T = F/2sin(theta) where theta is 1/2 of the bend
angle. If you have F= 60 lbs on that lower spoke, the tension works out to
60/2sin(6.325) = 272 lbf or 123.5 kgf. I used the lower spoke because the
tensiometer on the upper spoke is in the way. This doesn't take into
account the stiffness of the spoke, but should be in the ballpark.

Andrew Lee

Andrew Lee said:

This doesn't take into account the stiffness of the spoke, but should be
in the ballpark.

I should have just left it at "it doesn't take into account the stiffness of
the spoke". Because some of the force was put into bending the spoke, the
actual tension is less. I'm sure someone else can calculate how much
less...

On Tue, 2 Jan 2007 20:27:37 -0900, "Andrew Lee"

whatsupandrewathotmaildotcom said:

Carl Fogel said:

Like you, I now expect that if D is measured accurately from the
correct starting position (somewhere in the thin air between the two
current end points of L) to the bend, then the equation should agree
quite closely with an accurate tension gauge.

Why mess with measuring d and L when you can just measure the angle of the
bend in the spoke? The sine of half that angle is what you trying to get
with d and L anyway.

I measured the angle of the bend in the lower of the two "squeezed" spokes
in your picture http://i16.tinypic.com/33uygk1.jpg off of my screen to be
about 12.65 degrees (probably off a bit because of parallax). The equation
for tension reduces down to T = F/2sin(theta) where theta is 1/2 of the bend
angle. If you have F= 60 lbs on that lower spoke, the tension works out to
60/2sin(6.325) = 272 lbf or 123.5 kgf. I used the lower spoke because the
tensiometer on the upper spoke is in the way. This doesn't take into
account the stiffness of the spoke, but should be in the ballpark.

Andrew Lee

Dear Andrew,

Hmmm . . . the idea of using the camera is appealing, though slow.

How much will it matter that the force applied to the lower bend is
not bisecting the angle? Regrettably, you can't squeeze a pair of
spokes in real life and get right angles to bends.

As for the relative accuracy . . .

For a 60lb force, a 290 mm L after the bend is made (close to what
you're using as an example) my spreadsheet indicates these figures:

tension tension
displacement degrees f*2(sine) (f*l)/(4*d)
15.1 5.945 289.6 288.1
16.1 6.336 271.8 270.2
17.1 6.726 256.1 254.4

An error of +/ 0.4 degrees is as much as an error of +/- 1 mm on
displacement.

I'm more confident of measuring a displacement in millimeters than I
am of measuring a camera distorted angle to within 0.4 degrees.

Right now, I'm working on an even stiffer rig than a pipe-clamp that
lets me measure displacement and apply a Park gauge, but the first
results left me baffled when I downloaded the pictures and read the
displacement off the ruler.

Possibly the tension on my new rig really does climb to 500 pounds,
which is what my displacement measurement indicates.

If so, my Park gauge is off by 200 pounds, because it indicates only
300 pounds of tension.

Reconciling the two results requires either an incredibly inaccurate
tension gauge (unlikely) or a displacement of 8 mm that's 5 mm too
small (just as unlikely--I was expecting maybe 0.5 mm, not 5 mm), so
I'm hoping to figure out that I'm doing something incredibly stupid
when I measure displacement.

Cheers,

Carl Fogel

Carl Fogel said:

Dear Andrew,

Hmmm . . . the idea of using the camera is appealing, though slow.

How much will it matter that the force applied to the lower bend is
not bisecting the angle?

That there is a curve rather than a point change in direction is more the
reason to measure the angle rather than the displacement. The formula we
are using, F/2sin(theta), where theta is the difference in angle on either
side of the load between the unloaded spoke and the same spoke when loaded,
is based on vectors, so the angle is what matters. The forumla that you
are plugging your displacement data in, F*L/4d, is just a simplied version
of F/2sin(theta) where we assume that theta is small, so tan(theta) is about
the same as sin(theta). Tan(theta) is 2d/L and is substituted for
sin(theta) in the approximation.

It's too bad that there is a curve in the bend, but that is spoke stiffness
coming into play. If you are measuring displacement from the side of the
spoke before loading to the same side of the spoke on the curve after
loading, then you are understating the actual angle theta.

Quoted message said:

Regrettably, you can't squeeze a pair of
spokes in real life and get right angles to bends.

I think you mean that the bend comes to a point, not a right angle.

Quoted message said:

As for the relative accuracy . . .

For a 60lb force, a 290 mm L after the bend is made (close to what
you're using as an example) my spreadsheet indicates these figures:

tension tension
displacement degrees f*2(sine) (f*l)/(4*d)
15.1 5.945 289.6 288.1
16.1 6.336 271.8 270.2
17.1 6.726 256.1 254.4

An error of +/ 0.4 degrees is as much as an error of +/- 1 mm on
displacement.

I'm more confident of measuring a displacement in millimeters than I
am of measuring a camera distorted angle to within 0.4 degrees.

I am about as confident of getting within 0.4 degrees as I would be
measuring displacement within 1 mm. This is true especially considering the
curve in the bend and that you are really measuring displacement to find the
angle change. Just make sure you take the photos level and perpendicular
with the spokes and back the camera away from the subject as much as
practible to minimize parallax error. If that haves you shooting telephoto,
that's good because there's less distortion in there in most zoom lenses. I
didn't measure the angle directly in my example. I lined up the edge of a
sheet of paper on the right side of the spoke and traced the left side of
the spoke onto the paper. I then calculated the angle by measuring the
sides of the resulting triangle and using arctangent. The point is to get
the angles in the straight sections outside of the curve caused by the load
interacting with spoke stiffness. Using length and displacement more
problematic with the spoke crossings affecting the effective L and problems
with measuring displacement that I mentioned above.

Quoted message said:

Right now, I'm working on an even stiffer rig than a pipe-clamp that
lets me measure displacement and apply a Park gauge, but the first
results left me baffled when I downloaded the pictures and read the
displacement off the ruler.

Possibly the tension on my new rig really does climb to 500 pounds,
which is what my displacement measurement indicates.

If so, my Park gauge is off by 200 pounds, because it indicates only
300 pounds of tension.

Reconciling the two results requires either an incredibly inaccurate
tension gauge (unlikely) or a displacement of 8 mm that's 5 mm too
small (just as unlikely--I was expecting maybe 0.5 mm, not 5 mm), so
I'm hoping to figure out that I'm doing something incredibly stupid
when I measure displacement.

Try measuring the angle... or try measuring displacement to the location
where the two straight sections of the spoke intersect. That might give you
a few more millimeters.

A proposal for how you might take into account spoke stiffness. See how
much force it takes to bend an unsecured/untensioned spoke to the same angle
that you get with your 60 lb load. Then subtract that value from 60 lbs
before plugging into T = F/2sin(theta). So if it takes 3 lbs to put that
12.65 degree bend in the spoke, then plug in T = 57/2sin(6.325) = 259 lbf.
I didn't think this through too much, so maybe this isn't a good
approximation...

Quoted message said:

Right now, I'm working on an even stiffer rig than a pipe-clamp that
lets me measure displacement and apply a Park gauge, but the first
results left me baffled when I downloaded the pictures and read the
displacement off the ruler.

Possibly the tension on my new rig really does climb to 500 pounds,
which is what my displacement measurement indicates.

If so, my Park gauge is off by 200 pounds, because it indicates only
300 pounds of tension.

Reconciling the two results requires either an incredibly inaccurate
tension gauge (unlikely) or a displacement of 8 mm that's 5 mm too
small (just as unlikely--I was expecting maybe 0.5 mm, not 5 mm), so
I'm hoping to figure out that I'm doing something incredibly stupid
when I measure displacement.

"When you have eliminated all which is impossible, then whatever
remains, however improbable, must be the truth." -- Sherlock Holmes,
"The Blanched Soldier"

Andrew Lee said:

I measured the angle of the bend in the lower of the two "squeezed" spokes
in your picture http://i16.tinypic.com/33uygk1.jpg off of my screen to be
about 12.65 degrees (probably off a bit because of parallax). The equation
for tension reduces down to T = F/2sin(theta) where theta is 1/2 of the bend
angle. If you have F= 60 lbs on that lower spoke, the tension works out to
60/2sin(6.325) = 272 lbf or 123.5 kgf. I used the lower spoke because the
tensiometer on the upper spoke is in the way. This doesn't take into
account the stiffness of the spoke, but should be in the ballpark.

The restoring force due to spoke stiffness is F = 48*E*I*w / L^3
where E is the modulus of steel (200 kN/mm^2), I is the moment
of inertia (about 0.79 mm^4 for a 2.0mm spoke), w is the displacement,
and L is the length of the beam. See my post in the "Spoke tension
deflection test" thread for more details. Take w=15mm and L=280mm.
Then the restoring force is about 5 N or 0.5 kgf, negligible compared
to the applied weight. Spokes are not very stiff in bending,
especially when measured over the whole length of the spoke.

This is, of course, one reason why we tension the spokes, and why
bicycle wheels aren't much like wagon wheels.

Ben

Quoted message said:

Andrew Lee said:

I measured the angle of the bend in the lower of the two "squeezed" spokes
in your picture http://i16.tinypic.com/33uygk1.jpg off of my screen to be
about 12.65 degrees (probably off a bit because of parallax). The equation
for tension reduces down to T = F/2sin(theta) where theta is 1/2 of the bend
angle. If you have F= 60 lbs on that lower spoke, the tension works out to
60/2sin(6.325) = 272 lbf or 123.5 kgf. I used the lower spoke because the
tensiometer on the upper spoke is in the way. This doesn't take into
account the stiffness of the spoke, but should be in the ballpark.

The restoring force due to spoke stiffness is F = 48*E*I*w / L^3
where E is the modulus of steel (200 kN/mm^2), I is the moment
of inertia (about 0.79 mm^4 for a 2.0mm spoke), w is the displacement,
and L is the length of the beam. See my post in the "Spoke tension
deflection test" thread for more details. Take w=15mm and L=280mm.
Then the restoring force is about 5 N or 0.5 kgf, negligible compared
to the applied weight. Spokes are not very stiff in bending,
especially when measured over the whole length of the spoke.

This is, of course, one reason why we tension the spokes, and why
bicycle wheels aren't much like wagon wheels.

Ben

Of course, after we tension the spokes, bicycle wheels are very much
like wagon wheels.

I just had to say it.

In article <[email hidden]>,

Andrew Lee whatsupandrewathotmaildotcom said:

A proposal for how you might take into account spoke stiffness. See how
much force it takes to bend an unsecured/untensioned spoke to the same angle
that you get with your 60 lb load. Then subtract that value from 60 lbs
before plugging into T = F/2sin(theta). So if it takes 3 lbs to put that
12.65 degree bend in the spoke, then plug in T = 57/2sin(6.325) = 259 lbf.
I didn't think this through too much, so maybe this isn't a good
approximation...

Dave (dvt at psu dot edu) has used this exact approach before.
If you follow this approximation through to its end, the effect of
spoke stiffness is to subtract a constant offset (dependant only on
stiffness and measurement span, but not on tension) amount from the
tension predicted from the flexible-spoke equation.

This is a step in the right direction, and for tensions of
interest, the constant offset of tension is pretty accurate as an
empircal model, but the offset is over twice what your/Dave's approach
predicts.

-Luns

In article <[email hidden]>,

Luns Tee said:

In article <[email hidden]>,

Andrew Lee whatsupandrewathotmaildotcom said:

A proposal for how you might take into account spoke stiffness. See how
much force it takes to bend an unsecured/untensioned spoke to the same angle
that you get with your 60 lb load. Then subtract that value from 60 lbs
before plugging into T = F/2sin(theta). So if it takes 3 lbs to put that
12.65 degree bend in the spoke, then plug in T = 57/2sin(6.325) = 259 lbf.
I didn't think this through too much, so maybe this isn't a good
approximation...

Quoted message said:

Dave (dvt at psu dot edu) has used this exact approach before.

Actually, I misspoke somewhat - Dave's approach was to subtract
the force from the T= FL/(4d) formula rather than the angle
expression. If you're bypassing the curve and taking the the angle
between the straight sections outside the curve, that's already taking
into account spoke stiffness.

-Luns

Luns Tee said:

Luns Tee said:

In article <[email hidden]>,

Andrew Lee whatsupandrewathotmaildotcom said:

A proposal for how you might take into account spoke stiffness. See how
much force it takes to bend an unsecured/untensioned spoke to the same
angle
that you get with your 60 lb load. Then subtract that value from 60 lbs
before plugging into T = F/2sin(theta). So if it takes 3 lbs to put that
12.65 degree bend in the spoke, then plug in T = 57/2sin(6.325) = 259
lbf.
I didn't think this through too much, so maybe this isn't a good
approximation...

Quoted message said:

Dave (dvt at psu dot edu) has used this exact approach before.

Actually, I misspoke somewhat - Dave's approach was to subtract
the force from the T= FL/(4d) formula rather than the angle
expression. If you're bypassing the curve and taking the the angle
between the straight sections outside the curve, that's already taking
into account spoke stiffness.

-Luns

You might be right. I think I'll sleep on it.

Goodnight.