wphamilton

I know there are some engineers in this group. I have an idle question should it pique your interest.

I was wondering how the tension on a given spoke in the rear wheel would change when you are applying force through the hub. If you were able to filter out the sine wave as it revolves around, and suppose a power input of say 200 watts, pick a given number and pattern of spokes, is the change enough to potentially measure? Is it linear with force?

corrado33

Well, here's a plot of the tension on a spoke during "normal" riding.

I believe the tension change due to the wheel turning (under some weight) would hide the tension change due to torsional stress. Even IF you could remove the tension change due to the radial stresses, you'd probably still have low signal to noise. I would guess that increasing the power delivered to the hub would simply apply an offset to the baseline. (Whether that baseline shift would be linear or else, I have no idea.) I don't think the tension would change based on its radial position when you remove the radial stress.


This image came from this report on Duke's website.
https://people.duke.edu/~hpgavin/pape...heel-Paper.pdf

wphamilton

Nice chart, it's always helpful to have a visualization. Two questions: isn't change in radial stress what I'm looking for, so I don't understand why we want to remove it *? And would you expect the tension change due to the wheel turning to be a smooth periodic function?

*or did that refer out out of plane loads?

OK I read the paper which cleared things up a bit. It also had this (analytic) graph of a strain from distributed load, such as sitting on the bike, which answers my second question above.


The Duke paper summarizes "spoke strains respond to lateral (cornering), tangential (accelerating and
braking), as well as radial (gravity) loads on the rim. " It seems to me that the load from pedaling against the various drag forces, even when not accelerating would also, would also be simply part of the tangential load, but wasn't mentioned by the authors since it doesn't really matter much to their analysis.

As a tangential load it makes sense as you said that you'd see it as a baseline shift in amplitude. I'm having a hard time visualizing why that should be swamped by the noise unless it's just that small a change ... but a low-pass filter of some sort?. Maybe another study has put some numbers to that.

Looigi

Basically, when applying torque to the hub via the chain and cog, that torque is transferred to the rim via the spokes. The trailing spoke tension increases and the leading spoke tension decreases. This is added to the effects of other forces placed on the wheel. Beyond that what do you want to know?

wphamilton

I don't follow that at all Looigi. If it's a tangential load, why the difference in trailing and leading spokes? Wouldn't it be more or less distributed among all of the spokes? I can't see the tension change in the horizontal spokes without a radial load at those points. Help me out there!

baron von trail 

I'd actually be surprised if hub-torque is the leading factor in spoke tension and compression. I would say it is the weight of the massive rider and the rest of the bike pushing down on the bottom spokes and pulling on the top spokes which loads them the most. The torque transfer is probably relatively low (as compared to weight bearing load) throughout most of the ride, barring the acceleration points like starts and climbs of course.

corrado33

This is what I was getting at. But, provided we can remove the stresses caused by weight, would the remaining stresses caused by the acceleration show measureable effects on the tension of the spokes.

wphamilton

Baron why is the torque transfer greater on a climb than over level road? Either way it's tangential force at the contact point vs force at the axle, right? This is evidently more complicated than I expected.

wphamilton

Succinctly put. Thank you.

Biker395 

Gotta love this group.

Digital Gee

I have to admit, I didn't understand ONE WORD of the technical comments on this thread.

osco53

Same here,, way way above my pay grade..

Personally this Old Fart In Training just doesn't worry about such things,, If I find a loose spoke I deal with it..

Biker395 

I'm like Salieri to Mozart. Talented enough to understand, not talented enough to do.

Carbonfiberboy 

Acceleration of gravity is 1g. You're not going to be hitting any 1g acceleration due to pedaling forces. They're tiny. You're probably not going to be hitting 1g with braking forces, either: that would have you coming to a complete stop from 15 mph in 1 second. Not happening with my bikes. That's endo country.

However . . . if you have rim brakes, the deceleration from braking will be similar to the acceleration of gravity, i.e. force transmitted from the hub to rim in the vector direction of the acceleration. But with disc brakes, the brake is connected to the hub and thus "winds up" the spokes as well as creating the force vector of the deceleration. Thus disc brakes create a good bit more stress on spokes and the rim bed.

leob1

Guys, go for a ride.

wphamilton

CFB, everything (almost) that you put into the pedal is applied through the rear wheel. It will not usually be as much force as your body weight, but "tiny" sounds wrong to me. Check me, how many watts would it take to accelerate 65 kg at one g for 2 seconds? distance for 2 seconds is 19.6 meters, then (e= mass x acceleration x distance) about 12,500 joule, or power of 6250 watts for the two seconds. Someone putting out 300 watts for those two seconds is about a twentieth of that. Much smaller, but not tiny.

Can we see that amount in spoke tension, and what would it look like? I suspect that you could, and it would be distributed according to the spoke pattern but otherwise symmetric with respect to rotation.

I might be looking at it wrong (which is why I started this thread) but as I see it all of the force propelling you forward is applied at the ground at the rear wheel, through the axle against the dropouts. That will be a torque, all of it transmitted through the spokes. Regardless whether the drag is from uphill, constant pace in flats, etc.

corrado33

I attempted to run a simulation in solidworks of a 10 N*m load (about the torque you'd put on a wheel with 200 W) on a bike wheel, unfortunately the model I downloaded wasn't great, and the simulation failed. (The model was made of surfaces, not solid bodies, so it just wasn't great.) I'm not good enough (it'd take far too long) to make my own correct model. (Especially since I have no idea how spokes are threaded.)

If anyone is good at cad and can make a solid body drawing of a correctly built wheel (spokes can be fused to the rim and hub, no need for spoke nipples and what not) I'd gladly run it through some simulations and have it calculate the tension of the spokes under load and weight.

But then that ruins this "thought experiment."

Stevie47

Try envisioning the spokes as rubber bands with a lot of give. When you crank on the pedals the rear hub will rotate forward more than the rim because the rubber bands will give. If you look at the angle of the spokes and picture this you will realize that half of the rubber bands (spokes) will stretch more than normal and half will be stretched less. Increased force in half and decreased force in half. I'm sure it could be measured. Strain gauge technology is very good. That's how power meters work (another discussion). And yes the change in strain should be linear with force, under normal circumstances (no deformation, etc)

RoadTire 

Yep, over my pay grade too, but maybe I can learn something here. Does this chart indicate a -650% change in spoke tension?

-275-50/50×100=-650

jppe

These were the types of problems we solved in our Dynamics classes in Engineering. At the time it seemed like "fun", so much so I took Dynamics as an elective. And we used slide rules instead of calculators. I take my slide rule with me to show kids when I talk to them about Engineering during Engineer's week. They really enjoy working simple problems with it.

wphamilton

The negative strain is where the spoke points straight down from the hub. They took only the smooth and periodic changes in spoke strain, and averaged them out over many revolutions to come up with this chart. The solid line is someone's mathematical model.

The way I imagine it, which is probably oversimplified, is that the weight on the hub hanging down on spokes from the rim wants to squash the rim. Pulling down at the top and spreading out front and back. The rim widening is opposed by the spokes everywhere except for spokes pointing up and down (in some range of angles). Spokes pointing more up support more weight directly and less against the rim being squashed, those pointing down do neither hence negative strain. What's interesting and kind of surprising to me is that the strain is mostly uniform through the rest of the revolution.

The authors don't account for rider input in their road tests. I don't know if it's because they deemed it irrelevant like many respondents here, or if they kept it constant.

RChung

Method and Apparatus for Measuring Torque Transmitted by Driven Wheel of a Cycle or the Like Vehicle - Patent application

wphamilton

That has a good examination of the forces I think. If you've looked at the patent filing, any estimate of the general magnitude of delta T1, delta T2 etc in figure 4?

osco53

Biker,, you shouldn't have retired..go to Nasa, I think they could use you

wphamilton

I only asked because he'd probably know or could guess, and if it's as small delta-t as I suspect then it's not something you could slap together with mail order electronic parts.