Spoke delta tension?
jra...@njackn.com
With all the discussion in this group regarding spoke tension, I've been
wondering what are the various contributions to dynamic changes in
spoke tension due to the rider's force on the pedals. Of course there is
some change due to the torque on the rear wheel, but there must be a
number of others as well. Thoughts anyone?
-Jim Adney-
___ Blue Wave/QWK v2.12
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Jim Adney Melissa Kepner
jra...@njackn.com mgke...@facstaff.wisc.com
Laura Kepner-Adney
Madison, Wisconsin
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Jobst Brandt
> With all the discussion in this group regarding spoke tension, I've
> been wondering what are the various contributions to dynamic changes
> in spoke tension due to the rider's force on the pedals. Of course
> there is some change due to the torque on the rear wheel, but there
> must be a number of others as well. Thoughts anyone?
I can think of many effects and, having done so, analyzed them as you
probably know, being a longtime wreck.bicycles reader and contributor.
The results are collected and displayed in word, graph, and numerical
tables in "the Bicycle Wheel". I can recommend it.
Jobst Brandt <jbr...@hpl.hp.com>
Sheldon Brown
>
>
>With all the discussion in this group regarding spoke tension, I've been
>wondering what are the various contributions to dynamic changes in
>spoke tension due to the rider's force on the pedals. Of course there is
>some change due to the torque on the rear wheel, but there must be a
>number of others as well. Thoughts anyone?
>
Consider, for instance, torsional loading while climbing a hill in a 1:1
granny gear. Let us suppose that I am standing on a pedal with my whole
weight, the bike has low flange, 32 spoke hubs.
I am applying my 100 kGf on a 170mm moment arm (the crank). The 1:1 chain
drive transmits this force to the rear hub, which has a radius of 22 mm.
This translates into a tangential force at the hub of 770 kGf. If we assume
that the wheel is laced in a pattern that puts the spokes in line with the
tangent of the hub flange, we can divide the 770 kGf by 32 spokes, coming up
with an average tension change of 24 kGf per spoke. The trailing spokes will
each get 24 kGf tighter, the leading spokes will each get 24 kGf looser. 24
kGf is one quarter of the usual rule-of-thumb tension recommended for a good
quality wheel.
In the case of a typical dished drive wheel, one would suppose that the right
side spokes will do most of the work.
While standing on the pedal, I am probably also leaning the bike sideways,
introducing a substantial lateral load, which can cause serious instantaneous
tension changes.
On top of all that, I might hit a bump, causing a momentary stress to the
wheel of several times my weight. This will considerably loosen the spokes
closest to the ground, and somewhat tighten those immediately adjacent to the
impact area as they strain to hold the rim against bulging out and allowing
the impact area to collapse inward.
In the days when carbon steel spokes were common, the common failure mode was
breakage of the right-side trailing spokes as a result of overstress.
Now we use stainless steel, which has a higher tensile strength, but a
greater vulnerability to fatigue. This is exacerbated by the greater amount
of dish in many newer wheels. Now, the predominant cause of spoke breakage
is fatigue failure.
Fatigue occurs when the spoke is caused to go through cycles of tension and
slackness. That is, if a spoke is sometimes under high tension, sometimes
under slight tension, that is ok. If a spoke is sometimes under moderate
tension, but sometimes is allowed to go completely slack, that is where the
fatigue sets in.
Bicycles look simple, but the stresses on them are complex. They work as
well as they do, not because engineers calculated how strong all the parts
had to be, but because mechanics and framebuilders have tried everything over
the years.
Sheldon "By The Way, Is A kGf The Same As A Newton? They Didn't Talk About
Newtons In My High School Physics Class In 1962, And They Don't Define
Newtons In Any Of The Dictionaries Or Encyclopedias That I Have At Hand"
Brown
kGfville? Massachusetts
+-------------------------------------------------------+
| Je n'ai fait cell-ci plus longue que parce que |
| je n'ai pas eu le loisir de la faire plus courte. |
| (I have only made this so long because I did |
| not have the time to make it shorter.) |
| --Pascal |
+-------------------------------------------------------+
Jobst Brandt
> Now we use stainless steel, which has a higher tensile strength, but
> a greater vulnerability to fatigue. This is exacerbated by the
> greater amount of dish in many newer wheels. Now, the predominant
> cause of spoke breakage is fatigue failure.
With previous hubs like the old Campagnolo Record road hubs, very
little torque was transmitted to the left side spokes and still this
does not cause the right spokes to fail more often than the left and
less stressed spokes. The cause of failure is the residual overlaying
stress from manufacture and lacing. Today's stainless spokes from the
leading makers are far more fatigue resistant than earlier non
stainless spokes.
> Fatigue occurs when the spoke is caused to go through cycles of
> tension and slackness. That is, if a spoke is sometimes under high
> tension, sometimes under slight tension, that is OK. If a spoke is
> sometimes under moderate tension, but sometimes is allowed to go
> completely slack, that is where the fatigue sets in.
The principal fatigue failures occur from cyclic stress near the yield
stress. This is caused by residual stresses overlayed on cyclic
loading, not by stress reversal because there is no reversal in a
spoke. It cannot support compression.
> Sheldon "By The Way, Is A kGf The Same As A Newton? They Didn't Talk About
> Newtons In My High School Physics Class In 1962, And They Don't Define
> Newtons In Any Of The Dictionaries Or Encyclopedias That I Have At Hand"
> Brown
A newton is the force of 1 kg accelerated by 1 g. A g=9.8m/sec/sec,
therefore one kg force is about 10 Newtons or 1 kg / 9.8 m/sec/sec = 1 N.
Jobst Brandt <jbr...@hpl.hp.com>
Jobst Brandt
> Well...... There's the specific question you pose, and then there
> are other questions you don't bring up, for reasons I don't know. So
> I'll bring them up briefly myself!
>
> 1. How does the change in spoke tension relate to spoke failure?
> It's obvious that, all else being equal, the spoke with the greatest
> tension variation will break soonest, or otherwise would have to be
> made the strongest. So it does make sense to redesign rims, hubs,
> and lacings to reduce spoke deltaT.
In theory that might seem to be the reason for spoke dimensions but it
isn't so. If you tried to build wheels with some of the spokes on the
fringe of standard dimensions, you will find that they twist off when
made thinner, that they come loose if made fatter than 2.0mm, that
even spokes without elbows break at the head and threads and that
there is reason to make "conventional" hubs with the flanges that we
see most often.
Spokes break in use mainly from failure to stress relieve. You can
assure yourself of this if you review the margin from in use tension
to the yield stress. I am still riding the spokes on the same hubs
that I spoked up when you and I first met... many years ago. I weigh
185 lbs and ride in the roughest terrain with my 120 psi road tires.
I haven't broken a spoke in years except one that got nicked when a
chain got dropped into the spokes. Essentially, spokes do not break
on well built wheels.
> 2. What relation does pedaling force bear to spoke deltaT? Spoke
> tensions are affected by many different loads at the ground. Radial
> force.... lateral force..... and tangential force if either brakes
> or drive are applied.
As you can measure and compute, torque does not add substantially to
tension, up or down. The margin to the yield point is still enormous.
> The biggest deltaT may arise from hitting large bumps, which
> potentially relaxes taut spokes almost completely. But lateral
> loads, whether from 1. bike lean 2. violent maneuvers 3. off-center
> bumps or ledges probably leads to tension levels well above the
> norm. (I also have a fairly dim memory of very high tensions near
> the load, when the wheel is loaded radially to near the point of
> damage). Normal pedaling torques in high road gearing don't make
> for much tension change, but granny gears and steep inclines may be
> another matter. Brake forces are large, but are not derived from hub
> torques unless a disk brake is used.
These values can all be computed and if you look in "the Bicycle
Wheel" they are shown for the standard unit load of 200 lbsf on the
level, up a vertical wall, and for braking straight down, assuming the
wheel does not slip. The values are insignificant, and my experience
with the wheels on which I have put about 10000 miles per year for
eons supports these findings.
> But your specific topic was spoke deltaT from pedaling forces......
> This divides up into hub torque from the chain varying in-plane
> loading from the rider bobbing (or 'leaping') up and down side
> forces from bike lean or steering waver modest side forces from the
> dynamic imbalance of one thigh ascending while the other descends
> (large at high cadence).
At about 15 degrees lean when standing, tension just begins to
increase and by a tangent function that is typically flat for small
angles. Only the lateral component couples into an increase in
tension while the radial load dominates compression, the principal
effect.
> It's up to you whether you want to count the leaning, swerving,
> bobbing, and imbalance as related to pedal force.
It's not hard to overlay these effects and they are easily evaluated
individually.
> Strictly speaking, a rider can exert pedal torque without any of
> those added loads, but in road riding at least, the resulting deltaT
> is not the main one.
I'll second that contention.
> Jobst has a planar finite element analysis of this loading in his
> book, and Pippard has a similar result (continuum approximation)
> from the 30's. I recently saw an analysis of 12-spoke wheels from
> the 50's, but can't recall if they included hub torque (reacted by
> ground traction).
There is another publication in which Karl Wiedemer, Kraftverteilung
am Speichenrad, in "Konstruktion" 14. Year(1962) Heft 2, S. 64-66,
clearly shows the compression nature of a spoked wheel, superimposed
on its pretensioning. His work escaped discovery because it was not
related to bicycles, but rather to industrial hand carts using such
wheels. This becomes apparent because nowhere are bicycles mentioned
and the analyzed wheel has 42 spokes, a number not found on bicycles.
The paper shows the radial displacement, spoke loads, and rim bending
moment.
It is the only work I could find that clearly shows an awareness of
the compressive nature of wire spoked wheels. Pippard hypothesized
about what the forces might be but never discovered that wooden wagon
wheels and prestressed wire wheels carry loads in the same manner,
i.e. by compressing the bottom spokes. He wrote equations that could
not be solved without computers, and therefore never saw the results.
Jobst Brandt <jbr...@hpl.hp.com>
jra...@njackn.com
Ne> From: Jim Papadopoulos <bicyc...@delphi.com>
Ne> <jra...@njackn.com> writes:
>With all the discussion in this group regarding spoke tension, I've been
>wondering what are the various contributions to dynamic changes in
>spoke tension due to the rider's force on the pedals. Of course there is
>some change due to the torque on the rear wheel, but there must be a
>number of others as well. Thoughts anyone?
Ne>
Ne> Well...... There's the specific question you pose,
Ne> and then there are other questions you don't bring up,
Ne> for reasons I don't know. So I'll bring them up
Ne> briefly myself!
<Various intesesting comments snipped>
Ne> Spoke tensions are affected by many different loads
Ne> at the ground. Radial force.... lateral force..... and tangential
Ne> force if either brakes or drive are applied.
This is starting to go where I was interested.
Ne> But your specific topic was spoke deltaT from pedalling forces......
Ne> This divides up into
Ne> hub torque from the chain
Ne> varying in-plane loading from the rider bobbing
Ne> (or 'leaping') up and down
Ne> side forces from bike lean or steering waver
Ne> modest side forces from the dynamic imbalance of one
Ne> thigh ascending while the other descends (large at high cadence).
I think there's more here.
Consider a bike that just has "footpegs" (no cranks). If I apply an
impulse=1_pedal_stroke to the right footpeg this causes a torque along
the bike's roll axis that is clockwise if viewed from behind. If this
impulse is applied by a rider, then the torque is at least partially
countered by forces applied to the handlebars. I suspect that the torque
cancellation isn't perfect within each stroke. This leads to lateral
forces at the contact points of the tires with the ground which force the
roll center to be along the contact line rather than thru the bike/rider
center of mass. (Assume, for convenience, that the bike tracks a
perfectly straight line.)
The question then becomes, "How large are these forces and what is their
effect on spoke delta tension?" While the lateral forces due to this
effect may be small, the spoke/wheel geometry may still make the spoke
deltaT significant.
jra...@njackn.com
Ne> From: pe...@hpl.hp.com (Peter Webb)
Ne> Jobst Brandt (jbr...@hpl.hp.com) wrote:
Ne> : A newton is the force of 1 kg accelerated by 1 g. A g=9.8m/sec/sec,
Ne> : therefore one kg force is about 10 Newtons or
1 kg / 9.8 m/sec/sec = 1 N.
m * a = F This should read: 1 kg * 9.8 m/sec/sec = 10 N.
Ne> 1 N is the force required to accelerate 1 kg at 1 m/s/s. The
Ne> following sentence, 1 kg force is about 10 N, is correct.
Forgive me if this seems pedantic, but I have never been comfortable
with anything that encourages confusion of units. The kilogram is a
unit of mass. Introducing the KgF as a unit of force may be technically
correct, but it starts us down a very confusing road.
Dave Blake
>
[snip]
>
> Forgive me if this seems pedantic, but I have never been comfortable
> with anything that encourages confusion of units. The kilogram is a
> unit of mass. Introducing the KgF as a unit of force may be technically
> correct, but it starts us down a very confusing road.
>
> -Jim Adney-
Unfortunately, such units are sneaking into the literature in
various fields. In my neurophysioloogical field, grams-force is
a common unit defined as one gram in 1 g. In many cases units like
this are simply more understandable than newtons.
Dave Blake
dbl...@bme.jhu.edu
Jim Papadopoulos
> Consider a bike that just has "footpegs" (no cranks). If I apply an
> impulse=1_pedal_stroke to the right footpeg this causes a torque along
> the bike's roll axis that is clockwise if viewed from behind. If this
> impulse is applied by a rider, then the torque is at least partially
> countered by forces applied to the handlebars. I suspect that the torque
> cancellation isn't perfect within each stroke. This leads to lateral
> forces at the contact points of the tires with the ground which force the
> roll center to be along the contact line rather than thru the bike/rider
> center of mass. (Assume, for convenience, that the bike tracks a
> perfectly straight line.)
had BETTER be very good, or the bicycle will go spinning into oblivion.
It is perfectly possible to pedal (seated OR standing) such that the frame
does not lean from side to side. In that case the net torque about the
track was zero. If the bicycle leans (5 deg. or 10 deg, say) with
a given frequency = CADENCE, then we can calculate the sinusoidal
torque which must have been applied, and it's not very big (even granting
that a little extra must be supplied to balance the moment due to
gravity when leaned).
Your point, that even a small lateral force may cause large spoke stresses,
is basically valid. But I think that severalfold larger wheel sideforces arise
from frame tilt (with the rider's c.m. still
substantially over the track line).
OK, here's how I look at wheel sideforces (and I'm ignoring other lateral
physics such as precessional stresses and moments):
First of all, when the rider forms a rigid body with the frame,
all pedalling forces are reacted internally and cause no support
reactions. But the rider may steer slightly.... acceleration of
the bike about vertical and horizontal axes (to put it simplistically)
gives rise to (or is caused by!) horizontal forces at each wheel contact.
But horizontal forces are not wheel sideforces, you also have to
recognise that the bike is tipping back and forth, and the
vertical support force also generally has a component normal to the plane
of the wheel. The condition for no wheel sideforces to occur in
this scenario I think is that the rider should be a point mass.....
in an inverted simple pendulum, the support reaction has zero moment
about the c.m.
But now think about a bicycle steering straight, but with teh
rider not rigid nor fastened rigidly to the frame. Lateral wheel
forces, large ones, now become more likely (In the above case
with steering only, I think large lateral forces only arise in
rapid maneuvers).
The relation of body english to sideforce can be seen in no-hands riding
when you want to swerve a little..... (oops, I guess that's not
strictly true, but my editor doesn't let me return! We can get into it
later.....)
OK, if you accelerate your body rotationally relative to the frame,
about either vertical or horizontal axes, you'll get side forces.
When standing and rocking the frame, it's the light structure which
gets all the acceleration, so the force is small.... but when seated and
bendign at the waist, or when seated and twisting your forward-leaned
torso left or right, some real forces are generated.
But not only are horizontal forces generated, it's the vertical forces
acting on a tipped wheel which I have usually seen as the largest
related to pedalling activity.
When you stand and rock the frame it's clear what the wheel sideforce
is, as long as you travel straight. When you sit, and bend sideways
at the waist, there are (quasi) static and dynamic aspects.
If performed slowly, just look at the frame's angle of lean, and
assume the ground force is vertically upward. But if done quidkly,
a rapid bend at the waist actually moves the c.m. to one side of
the contact line. (That's why it's theoretically possible to
balance without moving, or to balance a chair on two legs.) So
we return to horizontal forces, and also the need to perform corrective
steering actions.
Boy, you don't know how good it feels to be pedantic like this --
I guess I miss being a professor. I'm sure the foregoing is
repetitious, vague, and disorganized, but I'll be happy to take up
individual points, and even get into estimates of dynamical
quantities..... Maybe I'll even be granted the pleasure of
learning that I've been missing the point all along?!
THAT'S what I'm looking for in rec.bicycles.tech.
Jim
jra...@njackn.com
Ne> From: bl...@bard.mb.jhu.edu (Dave Blake)
Ne> In article <951123231...@njackn.com>, jra...@njackn.com says...
>
> Forgive me if this seems pedantic, but I have never been comfortable
> with anything that encourages confusion of units. The kilogram is a
> unit of mass. Introducing the KgF as a unit of force may be technically
> correct, but it starts us down a very confusing road.
Ne> Unfortunately, such units are sneaking into the literature in
Ne> various fields. In my neurophysioloogical field, grams-force is
Ne> a common unit defined as one gram in 1 g. In many cases units like
Ne> this are simply more understandable than newtons.
Gack! Things are worse than I thought. Dynes and Newtons are less well
understood only because they are seldom used in everyday life. Thanks
for the feedback, but let's try to keep things on a higher plane if we
can.
Jobst Brandt
> Your point, that even a small lateral force may cause large spoke
> stresses, is basically valid. But I think that severalfold larger
> wheel sideforces arise from frame tilt (with the rider's c.m. still
> substantially over the track line).
Jim! Hold the phone. How about measuring these forces rather than
to conjecture about them. I have computed and measured them and even
with greater than average lean to the bicycle while climbing in the
standing position, the spokes in the load affected zone experience a
reduction in tension, albeit not as
great as when just rolling along.
Jobst Brandt <jbr...@hpl.hp.com>
jra...@njackn.com
Ne> From: Jim Papadopoulos <bicyc...@delphi.com>
Ne> <jra...@njackn.com> writes:
> Consider a bike that just has "footpegs" (no cranks). If I apply an
> impulse=1_pedal_stroke to the right footpeg this causes a torque along
> the bike's roll axis that is clockwise if viewed from behind.
Ne> But not only are horizontal forces generated, it's the vertical forces
Ne> acting on a tipped wheel which I have usually seen as the largest
Ne> related to pedalling activity.
I've been thinking about what you've written, and, while I'm not sure I
understand all of it, I think I agree with your basic contention here
that the normal component of the vertical force (weight) is the dominant
contributor to side force on the rim.
However (you were waiting for this, weren't you?) when I think about how
one actually rides it appears that the two effects actually add. Consider
a rider honking up a hill. Swinging the bike from side to side with the
cadence, note that as I stroke down with the right foot I am swinging the
bike to the left by the handlebars. Since this is simultaneous with my
right pedal stroke, I must conclude that the torque applied via the bars
is greater than that due to the impulse on the pedals. (Just talking
about torque about the roll axis here!)
A quick sketch here will show you that for the first half of this stroke
both the normal component of the weight and the swinging of the frame
require a force to the left applied from the road to the rim.
Since the whole point here is to determine how close to failure our wheels
are, it is clear that we have to consider the worst case. I'm not sure I
buy your "no side force" senario, but it's really not important.
What I'd really like to do now is raise the point of spoke angle, and what
price we pay for more cogs in the rear and the increased dish that comes
with them.
Jim Papadopoulos
>Jim! Hold the phone. How about measuring these forces rather than
>to conjecture about them. I have computed and measured them and even
>with greater than average lean to the bicycle while climbing in the
>standing position, the spokes in the load affected zone experience a
>reduction in tension, albeit not as
>great as when just rolling along.
are at my 'country home'.
Have any of us *measured* the forces (or spoke tensions) while
actually riding? My measurements relate to spoke tension while
a bicycle is slowly rolled forward and an accomplice plucks
one spoke. A friend of mine strain gaged his wheels,
and sent the data to an FM tape recorder -- unfortunately
rode only upright, on smooth roads.
What method do you use?
As for calculations, I disagree with your
approach in The Book. That does not take into
account the finite size of the LATERAL load affected
zone (much bigger than radial load affected zone), which arisies
due to rim torsional(and to some extent, lateral bending)
flexibility. Spoke tensions go UP outside the radial
load affected zone... at least in my experiments.
Thanks for the comments, though.
Jim
Jobst Brandt
>> Jim! Hold the phone. How about measuring these forces rather than
>> to conjecture about them. I have computed and measured them and
>> even with greater than average lean to the bicycle while climbing
>> in the standing position, the spokes in the load affected zone
>> experience a reduction in tension, albeit not as great as when just
>> rolling along.
>
> You have me at a disadvantage, my 'spoke delta T' measurements are
> at my 'country home'. Have any of us *measured* the forces (or
> spoke tensions) while actually riding? My measurements relate to
> spoke tension while a bicycle is slowly rolled forward and an
> accomplice plucks one spoke. A friend of mine strain gaged his
> wheels, and sent the data to an FM tape recorder -- unfortunately
> rode only upright, on smooth roads. What method do you use?
Well I did my measurements at my Dacha and let my porters do the
riding and measuring.
I did my measurement statically by taking up a standing position with
the brakes on and plucked spoked to see if the tone was higher or
lower than for the unloaded wheel. I also look down at the ground
contact point while climbing and note that it seldom lies outside the
flanges. Gravity makes this assessment accurate because my CG must be
vertically over the contact patch and my nose is centered over this
point if I don't contort my body.
> As for calculations, I disagree with your approach in The Book. That
> does not take into account the finite size of the LATERAL load
> affected zone (much bigger than radial load affected zone), which
> arises due to rim torsional(and to some extent, lateral bending)
> flexibility. Spoke tensions go UP outside the radial load affected
> zone... at least in my experiments.
The graph in the book shows what effect lateral displacement of the rim
does to spoke tension. To compute the effect of leaning the bicycle
you need only draw a triangle of hub and spokes and swing the load vector
through the expected angles to see that you must be at fairly unusual
angles to increase tension.
The vector must be outside the lateral angle of the spokes to cause an
increase in tension because the rim has lateral bending stiffness that
involves more than just one cross section, but this a reasonable
approximation of what goes on judging from the acoustic test. Normal
lean of the bicycle while standing does not exceed this angle.
Jobst Brandt <jbr...@hpl.hp.com>
Jim Papadopoulos
>I did my measurement statically by taking up a standing position with
>the brakes on and plucked spoked to see if the tone was higher or
>lower than for the unloaded wheel. I also look down at the ground
valid. Dramatic tension rises were found only OUTSIDE the so-called
load affected zone.
>lower than for the unloaded wheel. I also look down at the ground
>contact point while climbing and note that it seldom lies outside the
>flanges. Gravity makes this assessment accurate because my CG must be
>vertically over the contact patch and my nose is centered over this
>point if I don't contort my body.
And it doesn't take much lean to exceed that angle......
but you are thinking one-dimensionally if that's your criterion
for tension generation.
>The graph in the book shows what effect lateral displacement of the rim
>does to spoke tension. To compute the effect of leaning the bicycle
motions of each point of the rim. But you don't..... I bet you
think the rim moves in the direction of the load vector.......
that would be a mistake.
>does to spoke tension. To compute the effect of leaning the bicycle
>you need only draw a triangle of hub and spokes and swing the load vector
>through the expected angles to see that you must be at fairly unusual
>angles to increase tension.
If you push the rear rim with an inplane force, the load affected
zone wants to move towards the freewheel side. (To put it another way,
if you move that section of rim radially, the lateral spoke forces
will be unbalanced.)
That unbalanced force is carried..... by spokes OUTSIDE the
radial-load-affected-zone. .... yielding a tension
increase...... and this is without any lean of the bike.
Finally, while tension increases occur both for small lean angles
(both wheels) and no lean angle (rear wheel) of course the biggest
changes come from larger leans, or swerves, or off-center rocks.
I routinely lean my bike more than you do, and many others do too.
(That is, when I ride....)
JIm
Jobst Brandt
>> I did my measurement statically by taking up a standing position with
>> the brakes on and plucked spoked to see if the tone was higher or
>> lower than for the unloaded wheel. I also look down at the ground
>
> OK, then I think you'll concede that my approach is at least as
> valid. Dramatic tension rises were found only OUTSIDE the so-called
> load affected zone.
Hold it. I found no such changes. An increase in tension is easily
detected in tone change and there was nothing like that in spokes to
either side of the load affected zone in the tests I performed.
>> lower than for the unloaded wheel. I also look down at the ground
>> contact point while climbing and note that it seldom lies outside the
>> flanges. Gravity makes this assessment accurate because my CG must be
>> vertically over the contact patch and my nose is centered over this
>> point if I don't contort my body.
>
> Kind of hard to look down at the freewheel side flange that way.
> And it doesn't take much lean to exceed that angle...... but you
> are thinking one-dimensionally if that's your criterion for tension
> generation.
When I stand, my weight is primarily on the front wheel so loading on the
rear is hardly significant. Besides, as I pointed out, spokes last for
more than 100000 miles on my wheels so the effect can't be significant,
front or rear.
> >The graph in the book shows what effect lateral displacement of the rim
> >does to spoke tension. To compute the effect of leaning the bicycle
>
> OK, all you have to do is KNOW the lateral (and radial) motions of
> each point of the rim. But you don't..... I bet you think the rim
> moves in the direction of the load vector....... that would be a
> mistake.
I don't know what you are suggesting but what are you getting at?
>> does to spoke tension. To compute the effect of leaning the bicycle
>> you need only draw a triangle of hub and spokes and swing the load vector
>> through the expected angles to see that you must be at fairly unusual
>> angles to increase tension.
>
> NO, NO, NO!
> If you push the rear rim with an in-plane force, the load affected
> zone wants to move towards the freewheel side. (To put it another
> way, if you move that section of rim radially, the lateral spoke
> forces will be unbalanced.)
>
> That unbalanced force is carried..... by spokes OUTSIDE the
> radial-load-affected-zone. ....yielding a tension increase...
> and this is without any lean of the bike.
What? This is getting weirder and weirderer, so to speak. How about
putting this on a FEM analysis and demonstrating these effects. I
certainly haven't qualitatively seen anything like this.
> Finally, while tension increases occur both for small lean angles
> (both wheels) and no lean angle (rear wheel) of course the biggest
> changes come from larger leans, or swerves, or off-center rocks.
>
> I routinely lean my bike more than you do, and many others do too.
Not often and not for any significant mileage. To do so requires
substantial gratuitous work that may be possible for a demonstration
but not for any length of hill climbing.
Jobst Brandt <jbr...@hpl.hp.com>
Eric Soroos
>> That unbalanced force is carried..... by spokes OUTSIDE the
>> radial-load-affected-zone. ....yielding a tension increase...
>> and this is without any lean of the bike.
>
>What? This is getting weirder and weirderer, so to speak. How about
>putting this on a FEM analysis and demonstrating these effects. I
>certainly haven't qualitatively seen anything like this.
>
Time to jump in. I have a 3d bike wheel model for a MATLAB based
structural analysis routine. (Structural analysis == spokes modeled as
truss elements, rim as frame elements) It includes dishing effects,
simplistic spoke crossing(i.e. you get a 3x wheel, the spokes just don't
contact each other in the middle), and varying number of spokes.
Unfortunately, I will be on a plane in a couple of hours, and I won't
really have access to a professional version of matlab for a couple of
weeks.
I will run some trials when I get back into town.
If you are impatient and have a version of matlab that will handle
~200x200 matricies, or if you would like to look at the code, it is at:
http://www.ce.washington.edu/~soroos/matlab/index.html
If you have good values the physical parameters of the rim, I would be
interested in seeing them. (ixx,iyy, J, E etc).
eric