Velocity Fusion Max Tension?

1234 writes:

If you thing it's over-tensioned, stress-relieve the wheel by
putting it on the floor and pressing down on opposite sides of the
rim.

This is not stress relieving.

Correct, the rim is quite flexible and deforms easily, so spoke
tension doesn't increase enough.

Stress relieving is not accomplished by pushing the rim parallel to
the axle.

That's always true, especially when you push with two hands (as Jeff
said, "on opposite sides of the wheel"). But there's at least one
exception: if the rim is pushed all around its circumference, not
just on two opposite sides.

The Rolf stressor box does just this: the hole in the table supports
the rim while the cup presses the hub down. This tensions all the
lower side spokes simultaneously. Rolf avoids wheel collapse by
reducing the tension in the top side spokes at the same time.

What is less apparent is that doing this by hand and doing it
incrementally at different places around the rim does not accomplish
what is needed. In the tensor box, tension of the down-side spokes is
doubled while up-side tension goes to zero, thus not overloading the
rim in circumferential compression.

Besides, doing this manually is hard to accomplish on wheels with
fewer than a 24 spokes, the requires force being greater than hands
can readily exert and for excessive contact pressure.

Jobst Brandt

Thanks Jeff

I still didn't see any tension specs. But I don't think they publish
them, or did I miss them also?

John

wrote:
Michael Press wrote:

"JeffWills" wrote:


If you thing it's over-tensioned, stress-relieve the wheel by putting
it on the floor and pressing down on opposite sides of the rim.

This is not stress relieving.


Correct, the rim is quite flexible and deforms easily, so spoke tension
doesn't increase enough.

how does the rim know whether tension originates with hand squeezing of
the spoke or from lateral deflection? the spoke can only be tensioned
if the rim resists.



Stress relieving is not accomplished by pushing the rim
parallel to the axle.


That's always true, especially when you push with two hands (as Jeff
said, "on opposite sides of the wheel"). But there's at least one
exception: if the rim is pushed all around its circumference, not just
on two opposite sides.

how so? all pushing all around the circumference achieves is stressing
all the spokes at once, not just the ones where the operators hands are
in the "push" method. otherwise it's identical.


The Rolf stressor box does just this: the hole in the table supports
the rim while the cup presses the hub down. This tensions all the lower
side spokes simultaneously. Rolf avoids wheel collapse by reducing the
tension in the top side spokes at the same time.

Jasper Janssen wrote:
On 13 Nov 2005 09:18:36 -0800, "Ed" wrote:

Do tell why it is impossible to create sufficient tension by
pushing parallel to the axle.

Only a few mm of hub to rim 'axial' displacement are required.
This does not seem out of reach for mans current ability with
machines.

What do machines have to do with doing this by hand, exactly?

Jasper

doing this 'by hand' often involves machines. The leverage of a
spoke wrench or the crank arm stress relief lever of Sheldon Brown
are examples.

Even if JeffWills' method of' 'pushing on the rim' worked, I'd rather
have something like turns of a nut to help me control the process.
I'm not talking hydraulics, here.

Michael Press' statement was not limited to hand operation.

(guessing at your meaning)
Ed

john wrote:
Thanks Jeff

I still didn't see any tension specs. But I don't think they publish
them, or did I miss them also?

John

you need to call them. theoretically, it's specialist info.

In article
om,
"Ed" wrote:

Michael Press wrote:
In article
.com,
"JeffWills" wrote:

Ben Munson wrote:

Thanks in advance,
Ben "I may have overtensioned" Munson

If you thing it's over-tensioned, stress-relieve the wheel by putting
it on the floor and pressing down on opposite sides of the rim. If it
"sproings" into a potato-chip shape, it's overtensioned. Back off the
tension two turns and start truing again. Final tension should be less
than the amount that potato-chipped the wheel.(This is Jobst's method-
it works for me.)

This is not stress relieving. The `sproings' sound is the
sound made by spokes untwisting that were wound up when
turning the nipples with the spoke wrench.

Stress relieving is not accomplished by pushing the rim
parallel to the axle. Not enough additional tension can be
put on the spokes in this way.

Do tell why it is impossible to create sufficient tension by
pushing parallel to the axle.

Only a few mm of hub to rim 'axial' displacement are required.
This does not seem out of reach for mans current ability with
machines.

Let's consider change in spoke tension from lateral
displacement of the rim.

According to Damon Rinard's measurements of lateral
stiffness a gravitational load of 11 kg deflects the
wheel by only 0.002 m. And according to the graph in
_The_Bicycle_Wheel a rim displacement of 0.002 m
corresponds to a negligible increase in spoke tension.
A 200 kg man putting half his weight on one side of a
rim and half on the other will increase the rim
displacement to about 0.02 m. This is enough to
increase spoke tension by 50 N. Still far short of the
spoke squeezing method.

It would take a rim displacement of about 0.03 m to
increase spoke tension by 200 N. The force and
displacement is far greater than I would care to apply
to _my_ wheel.


Let's do a calculation. We squeeze two spokes together
at their centers and calculate the increase in tension.

Spring constant for a spoke:
k = 1.23 x 10^6 N/m

Length of a spoke:
L = 0.3 m

Lateral displacement of center of a squeezed spoke:
s

Stretched length of a spoke from lateral displcement:
2 * sqrt((L/2)^2 + s^2) = L * sqrt(1 + (2s/L)^2) ~ L( 1 + 4s^2/(2L^2))

Change in length of the spoke from lateral displacement:
2s^2/L

Increase in spoke tension from lateral displacement:
2s^2/L k = 2 / 0.3 * (1.23 x 10^6) s^2 = 8.2 x 10^6 s^2

So if we squeeze a spoke by 5 mm = 0.005 m at the
center the increase in spoke tension is 200 N.

The spoke squeezing method is simpler
and less likely to damage a wheel
than pushing the rim laterally.

--
Michael Press

Michael Press wrote:
In article
om,
"Ed" wrote:

Michael Press wrote:
In article
.com,
"JeffWills" wrote:

Ben Munson wrote:

Thanks in advance,
Ben "I may have overtensioned" Munson

If you thing it's over-tensioned, stress-relieve the wheel by putting
it on the floor and pressing down on opposite sides of the rim. If it
"sproings" into a potato-chip shape, it's overtensioned. Back off the
tension two turns and start truing again. Final tension should be less
than the amount that potato-chipped the wheel.(This is Jobst's method-
it works for me.)

This is not stress relieving. The `sproings' sound is the
sound made by spokes untwisting that were wound up when
turning the nipples with the spoke wrench.

Stress relieving is not accomplished by pushing the rim
parallel to the axle. Not enough additional tension can be
put on the spokes in this way.

Do tell why it is impossible to create sufficient tension by
pushing parallel to the axle.

Only a few mm of hub to rim 'axial' displacement are required.
This does not seem out of reach for mans current ability with
machines.

Let's consider change in spoke tension from lateral
displacement of the rim.

According to Damon Rinard's measurements of lateral
stiffness a gravitational load of 11 kg deflects the
wheel by only 0.002 m.

Damon Rinard applied a load at one point on the rim which
was free to flex, all spokes could react. Dividing the load
and applying it to more points reduces the freedom, less
displacement at the point of the load results (see the posts
of Diane_1234, Jobst Brandt in this thread).

And according to the graph in
_The_Bicycle_Wheel a rim displacement of 0.002 m
corresponds to a negligible increase in spoke tension.

Sorry, but I don't see what you're referring to as I peruse
my copy.

using your 300mm spoke, "The Bicycle Wheel"'s 35.5mm
half flange width, simple case of radial spoking,
the rim to spoke hole radial distance is
sqrt(300.00^2 - 35.500^2) = 297.892mm

With an immobile rim, a pure axial hub displacement
of 2mm results in a length of
sqrt(297.892^2 + (35.500+ 2.000)^2) = 300.243mm
for an increased spoke length of 0.243 mm.

using your Spring constant for a spoke
the resulting tension change is
1.23 x 10^6 N/m * 0.243 x 10^-3 m = 299N


The axial force required (per spoke) would be about
spoke tension * 37.5 / 300.243
This is attainable in the home, but us mere 80kg folks
will need some mechanical advantage to handle
36 spoke wheels.

please check this, its late at night.


A 200 kg man putting half his weight on one side of a
rim and half on the other will increase the rim
displacement to about 0.02 m. This is enough to
increase spoke tension by 50 N.

Still far short of the
spoke squeezing method.

It would take a rim displacement of about 0.03 m to
increase spoke tension by 200 N. The force and
displacement is far greater than I would care to apply
to _my_ wheel.

I think this all comes from the Rinard data. I wouldn't
do it that way either.


snip spoke squeeze calc

The spoke squeezing method is simpler
and less likely to damage a wheel
than pushing the rim laterally.

The point of yours that I questioned was that it is impossible
to create sufficient tension by pushing parallel to the axle.

You have not addressed this other than for the limited two
point loaded case and using one point load data. Perhaps
that is the case to which you meant to confine your remark.

Thanks for your reply.

Ed

Ben Munson wrote:
On 2005-11-13 06:28:12 -0800, "Qui si parla Campagnolo"
said:


Ben Munson wrote:
I am building a non-dished track wheel with 700c Velocity Fusions. What
is the max spoke tension for this rim? Without heating any of the usual
strong personalities up to simmer, does anyone have knowledge of its
real-world limits? Is there a comprehensive place to get this
information online? If not, is there a reason nobody has started one?

Thanks in advance,
Ben "I may have overtensioned" Munson

100 kgf is what you ought to be looking for. For a symmetrical hub,
both sides. I have built hundreds of Fusions and that's what I use.

Thanks Peter! That's the kind of info I like. Is there much variance in
your tension between Aeroheads to Fusions to deep Vs? I am going to be
moving on to some Deep V mountain bike wheels for my pops after this
little learning experience. He is a heavyweight, so all of this gets
more critical.

Ben Munson

No, the same for all three..Using enough spokes, and not a too light
rim and proper tension is what makes wheels work. Nothing wrong with
36h wheels.

On 13 Nov 2005 20:44:05 -0800, "Ed" wrote:

Jasper Janssen wrote:
On 13 Nov 2005 09:18:36 -0800, "Ed" wrote:

Do tell why it is impossible to create sufficient tension by
pushing parallel to the axle.

Only a few mm of hub to rim 'axial' displacement are required.
This does not seem out of reach for mans current ability with
machines.

What do machines have to do with doing this by hand, exactly?

doing this 'by hand' often involves machines. The leverage of a
spoke wrench or the crank arm stress relief lever of Sheldon Brown
are examples.

And neither of them involve hub/rin axial displacement.

Even if JeffWills' method of' 'pushing on the rim' worked, I'd rather
have something like turns of a nut to help me control the process.
I'm not talking hydraulics, here.

The operation *as described* was putting the wheel on the floor and
pushing on the rim. Please tell me what kind of 'turns of the nut' Rube
Goldberg device you've got worked out for that.

Michael Press' statement was not limited to hand operation.

Yes, it was.

Jasper