Re: I fixed a broken spoke! - Cycling Mob: Road Cycling, Mountain Biking and More

12-23-2006, 09:01 PM

[Only registered and activated users can see links. ] wrote:
> The diagrams and data are in radial and tangential dimensions. Spoke
> tension changes can be derived from cross section and length change,
> all of which are given.

Length changes are not given... at least not in the 2nd edition...
unless you mean the radial deflection of the nodes. The tangential
deflection adds a lot to this, but it is easy enough to calculate.

> This is about effects not absolute quantities. As I mentioned the
> vectors shown are all the same size, braking, torque, and radial load.
> I think the data and diagrams are clearly marked, considering all the
> reviews by engineers that were made before publishing.

I must be missing the cheat sheet...

Now that I know that "torque = 50" means a 50kg (?) load applied
tangent to a 300mm radius rim, and that "spoke angle = 5" means that
the spokes are angled 5 degrees from perpendicular to the rim, I have
enough info to calculate the tension change. A 5 degree angle gives an
effective hub radius of 26.5mm. So plugging in the numbers I get the
following. The first numbers use your torque values directly, and the
second converts to an equivalent force at the crank with a 2 to1 gear
ratio:

Effective flange radius (mm) 26.5 26.5
Crank length (mm) 300 170
Front sprocket teeth 1 2
Rear sprocket teeth 1 1
Crossed spokes on drive side 36 36
Force exerted on crank (kg) 50 176.5
Spoke diameter (mm) 1.6 1.6
Spoke length (mm) 300 300
Spoke area (mm^2) 2.01 2.01

Spoke tension change +-(kg) 15.7 15.7
Spoke length change +-(mm) 0.112 0.112

Note that this is an average tension change. Since the rim distorts,
the peak variation is somewhat larger... I calculated a maximum length
change from your FEM values of 0.147mm, or about 31% higher.

These *are* pretty small values... but they don't represent a "worst
case" for the effect of torsion on spoke tension. For one thing, to
have 36 spokes involved equally in resisting torque you would need a 36
hole hub (rare these days) and an infinitely stiff hub body
(impossible). In your book you include an analysis at the end where you
calculate that the left side of the hub (even with 3x) contributes only
12% to the torsional stiffness. Most modern hubs are stiffer than
this... but radial lacing on the NDS is also common. Radial spokes on
the NDS will transfer essentially zero torque when the DS is crossed.

There are plenty of rear wheels with 20 spokes and half radial... some
are even more extreme:

[Only registered and activated users can see links. ]

And obviously highly valued... even with a dent.

Also, an effective flange radius of 26.5 is a bit larger than
average... ~20mm is more normal and is par for Shimano hubs with 3x.

Again, if I plug in the numbers for a hypothetical yet perfectly
"normal" wheel (these days at least), I get:

Effective flange radius (mm) 20
Crank length (mm) 175
Front sprocket teeth 34
Rear sprocket teeth 25
Crossed spokes on drive side 10
Force exerted on crank (kg) 150
Spoke diameter (mm) 1.60
Spoke length (mm) 280
Spoke area (mm^2) 1.89

Spoke tension change +-(kg) 96.5
Spoke length change +-(mm) 0.661

And this isn't trivial at all. That is very near what spokes are
tensioned at to start with. In your book (2nd edition) you show charts
(fig 69) indicating that DT butted 1.8/1.6mm spokes will clearly yield
at ~175kg. So, if you build a wheel like the one shown above with 95kg
tension on the drive side, you can yield the trailing spokes and send
the leading ones to zero tension via torque alone.

I'll admit that the case shown above is rather extreme... but it was my
intention to show an extreme case (though not the most extreme case
that is out there). Some riders may never subject their equipment (and
their bodies) to a full power uphill sprint in a low gear... but I do
that occasionally. If you put your bike in the lowest gear at a
stoplight and take off with your full weight on the crank, then you
will achieve half the torque I've modeled here (if you weigh 75kg that
is). I think that a 150kg max force is very reasonable.

The wheels I am presently riding are 32, 3x, with a fairly large
diameter hub body, so if I assume that the left flange contributes 25%
to stiffness, I essentially have 1.33*16= 21.3 spokes involved in
torsion, dropping the spoke tension change to 45.3 kg instead of 96.5
kg. That is a value I feel a little more comfortable with.

Conclusion? Torsion should not be ignored when designing and building
wheels. There are plenty of wheel configurations out there now that are
pushing and exceeding the limit of what the spokes can take... and it
is no mystery that failures occur.

12-23-2006, 09:05 PM

Ron Ruff wrote:
> [Only registered and activated users can see links. ] wrote:
>> The diagrams and data are in radial and tangential dimensions. Spoke
>> tension changes can be derived from cross section and length change,
>> all of which are given.
>
> Length changes are not given... at least not in the 2nd edition...
> unless you mean the radial deflection of the nodes. The tangential
> deflection adds a lot to this, but it is easy enough to calculate.
>
>> This is about effects not absolute quantities. As I mentioned the
>> vectors shown are all the same size, braking, torque, and radial load.
>> I think the data and diagrams are clearly marked, considering all the
>> reviews by engineers that were made before publishing.
>
> I must be missing the cheat sheet...

>
> Now that I know that "torque = 50" means a 50kg (?) load applied
> tangent to a 300mm radius rim, and that "spoke angle = 5" means that
> the spokes are angled 5 degrees from perpendicular to the rim, I have
> enough info to calculate the tension change. A 5 degree angle gives an
> effective hub radius of 26.5mm. So plugging in the numbers I get the
> following. The first numbers use your torque values directly, and the
> second converts to an equivalent force at the crank with a 2 to1 gear
> ratio:
>
> Effective flange radius (mm) 26.5 26.5
> Crank length (mm) 300 170
> Front sprocket teeth 1 2
> Rear sprocket teeth 1 1
> Crossed spokes on drive side 36 36
> Force exerted on crank (kg) 50 176.5
> Spoke diameter (mm) 1.6 1.6
> Spoke length (mm) 300 300
> Spoke area (mm^2) 2.01 2.01
>
> Spoke tension change +-(kg) 15.7 15.7
> Spoke length change +-(mm) 0.112 0.112
>
> Note that this is an average tension change. Since the rim distorts,
> the peak variation is somewhat larger... I calculated a maximum length
> change from your FEM values of 0.147mm, or about 31% higher.
>
> These *are* pretty small values... but they don't represent a "worst
> case" for the effect of torsion on spoke tension. For one thing, to
> have 36 spokes involved equally in resisting torque you would need a 36
> hole hub (rare these days) and an infinitely stiff hub body
> (impossible). In your book you include an analysis at the end where you
> calculate that the left side of the hub (even with 3x) contributes only
> 12% to the torsional stiffness. Most modern hubs are stiffer than
> this... but radial lacing on the NDS is also common. Radial spokes on
> the NDS will transfer essentially zero torque when the DS is crossed.
>
> There are plenty of rear wheels with 20 spokes and half radial... some
> are even more extreme:
>
> [Only registered and activated users can see links. ]
>
> And obviously highly valued... even with a dent.
>
> Also, an effective flange radius of 26.5 is a bit larger than
> average... ~20mm is more normal and is par for Shimano hubs with 3x.
>
> Again, if I plug in the numbers for a hypothetical yet perfectly
> "normal" wheel (these days at least), I get:
>
> Effective flange radius (mm) 20
> Crank length (mm) 175
> Front sprocket teeth 34
> Rear sprocket teeth 25
> Crossed spokes on drive side 10
> Force exerted on crank (kg) 150
> Spoke diameter (mm) 1.60
> Spoke length (mm) 280
> Spoke area (mm^2) 1.89
>
> Spoke tension change +-(kg) 96.5
> Spoke length change +-(mm) 0.661
>
> And this isn't trivial at all. That is very near what spokes are
> tensioned at to start with. In your book (2nd edition) you show charts
> (fig 69) indicating that DT butted 1.8/1.6mm spokes will clearly yield
> at ~175kg. So, if you build a wheel like the one shown above with 95kg
> tension on the drive side, you can yield the trailing spokes and send
> the leading ones to zero tension via torque alone.
>
> I'll admit that the case shown above is rather extreme... but it was my
> intention to show an extreme case (though not the most extreme case
> that is out there). Some riders may never subject their equipment (and
> their bodies) to a full power uphill sprint in a low gear... but I do
> that occasionally. If you put your bike in the lowest gear at a
> stoplight and take off with your full weight on the crank, then you
> will achieve half the torque I've modeled here (if you weigh 75kg that
> is). I think that a 150kg max force is very reasonable.
>
> The wheels I am presently riding are 32, 3x, with a fairly large
> diameter hub body, so if I assume that the left flange contributes 25%
> to stiffness, I essentially have 1.33*16= 21.3 spokes involved in
> torsion, dropping the spoke tension change to 45.3 kg instead of 96.5
> kg. That is a value I feel a little more comfortable with.
>
> Conclusion? Torsion should not be ignored when designing and building
> wheels. There are plenty of wheel configurations out there now that are
> pushing and exceeding the limit of what the spokes can take... and it
> is no mystery that failures occur.
>
very interesting - thanks ron!