This is a much easier problem to visualize. It would actually work with 2 spokes as long as the bike is stationary and not moving, and if the rim itself is strong enough. However with only 2 spokes as you say, you add load to the hub and HALF of this load goes to the upper spoke and the OTHER HALF goes to the lower spoke. Also note that the internal forces will ALWAYS balance. (If you have unbalanced forces, then you have acceleration.) Instead of 2 spokes make it 32, but still with a very stiff rim. Now you will have a number of spokes at the top that combine to carry HALF of the load, and an equal number of spokes at the bottom that combine to carry the OTHER HALF. Take that same system and make the rim thin and flexible and as the rim gets thin enough to deform at the bottom where it contacts the pavement it will concentrate the forces at the bottom over a lesser number of spokes.

If you are interested in more on the "2 spoke" system, you can read my explanation in the other thread about a keyring and 2 rubber bands.


The easiest thing to do is to just cut them with some heavy duty wire cutters. And while you are riding your bike always remember to keep this part of the wheel on the bottom.

No, not half, all applied load is going through the top spoke. the bottom reduces the same amount which is why all forces always ballance.

Im not suggesting a 2 spoke wheel will work

shakes head and gives up completely and for ever

So do you just sit stationary on your bike?

Good. That is the best thing for you to do.

OK, bicycle is sitting untouched, top spoke has 200 pounds tension, bottom spoke has 200 pounds tension. Hub is being pulled both directions the same amount.

Now a 180 pound person sits on the bike, most of their weight is on the back wheel, maybe 120 pounds. According to your theory: upper spoke increases by the entire 120 pounds and becomes 320 pounds in tension. The lower spoke decreses by 120 pounds and becomes 80 pounds in tension. In engineering terms, summation of vertical forces must equal zero for static equilibrium. I.e. all of the forces on the hub must add up to equal zero. The hub has 3 forces acting on it- 320 pounds up from the upper spoke, 120 pounds down from the rider plus another 80 pounds down from the lower spoke. These forces are not balanced.

In reality here is what would happen in this simplified model: Rider applies 120 pounds to the rear wheel. Top spoke increases tension by half this, or 60 pounds, and now its tension is 260 pounds. Bottom spoke decreases by the other half of the load, or 60 pounds, and becomes 140 pounds. Again, summation of vertical forces must equal zero. The rider pushes down on the hub by 120 pounds, the lower spoke pulls down on the hub by 140 pounds, this is 260 pounds down. The upper spoke pulls up by 260 pounds which exactly balances the forces pulling down.

I could prove this with a finite element model but it is so obvious as to not require a computer for the solution.

It would work when the spokes were vertical but would not work when rotated 90 degrees. Also this is essentially radial spoking so it wouldn't work for the drive wheel. Now if you use at least 3 spokes, it would come closer to theoretically working for real light riders because no matter how the wheel is turned there are always vertical components to some of the spoke tensions to hold up the rider. It would take a very strong rim though. I wonder what the fewest spokes are that people have used for radial front wheels.

I make motor noises too.

So its not that only half goes to the upper spokes. The upper spokes get the full load, the lower spokes give up a balancing amount of tension to make things equal out again.
EDIT: NO, This isnt right I need to chew on this one more.

so no matter what the size of the load the lower spokes will automaticly give up enough ension to ballance the tension in the wheel? I mean until we load it so much that it fails right?

wow this is a tough concept to grasp. I can see the simplified math but the physical propertiesand the actions are hard to imagine

Erm no, thats not what I said. Also i didnt mean to talk about a 2 spoke wheel, rather 2 of the spokes from a multi spoke average wheel.

unloaded spoke tension = 200 pounds(say)
then apply 120 pounds to hub
Lower spoke becomes 80 pounds because rim deflects and effectively shortens the spoke
top spoke no longer needs to resist lower spoke 200 pounds it needs to resist only 80 from the lower spoke.
top spoke however now has to resist applied 120 + lower spoke 80, 120 + 80 = 200 = original tension, hub is still the same distance from top rim and central front to back, but closer to bottom rim.
so we have 200 pounds up force against 80 spoke tension down plus rider 120 acting down = 200 down
so forces ballance. Internal spoke tensions are out of ballance but system ballances as a whole due to externally applied load

If you think forces split equally check out previous FE posted on page 2 i think. almost all tension change is in the bottom due to rim deflection at bottom, very little rim deflection occurs anywhere else and the hub remains central.

Conclusions
Your FE model would be worthless, what you describe is the results of an analysis which does not allow for rim deflection at the bottom or for the wheel to move down as a whole as the rim deflects. Otherwise I would agree. Models need to represent what they are analysing

By the way the example above is not quite correct because FE does show some small increase in top spoke tension due to the wheel distortion.

The applied load hangs from the spokes, no one has been able to show an alternative load path. Its really not difficult.

Yes, does this make item 7 work then? I'll remember that next time i put it on the scales........jeez

Nothin like a good spirited debate!

The FE model includes all of the spokes, and a rim with realistic properties so it deforms at the bottom. The deformation of the rim is why there is more change in the group of fewer spokes in the bottom. (If the rim was very stiff then there would be equal change in forces top and bottom.) However we were talking about a system with only 2 spokes, which would behave differently. With only 2 spokes you have the same number of spokes acting above and below, so the change in stress is the same in each spoke, i.e. the forces split evenly.

I'm sitting at my desk in front of my workstation computer, with lots of different FE software loaded. I can analyze both models and support my claims.

You can? This is not a difficult thing for you to do?

I am still obsessing over the butted (or swaged) vs straight spoke question. In post #129 above, markymark69 referred to a paper by Henri Gavin and a link which will get you the pdf. ( I tried posting that link but for some reason when I did it, it brought an error rather than the pdf.) Here's the thing: in your brief discussion about this earlier, you compared the performance of butted vs straight gage spokes when they each see the same decrease in length (i.e. compression -- I will use "compression" to mean decrease in length) -- that is, when the rim would deform the same for each wheel. For the same rim deformation you get the same compression of the spokes and, in that case, straight spokes will lose more tension than will butted spokes. Perhaps to the point that the straight spokes become slack. But, that's not what happens when two wheels are subjected to the same load.

In the Gavin paper (figure 5), radial wheel stiffness is plotted against bending moment of inertia of the rim for various crossing patterns, spoke diameters, and two different analytic models. There is a metric buttload (or, swageload, if you prefer) of information presented in that figure but I think there's something that is pretty clear: Radial wheel stiffness in response to a point load on the rim is less for a wheel built with thinner spokes. That is, when you push on the rim at a point (like the ground contact point), a wheel built with thinner spokes will deform more than a wheel built with thicker spokes. So, it is true that for the same compression a thinner spoke will lose less tension than will a thicker spoke, but for the same wheel loading, thinner spokes will compress more than will thicker spokes. So, the question is not how different spokes behave when compressed equally, it is, what is the difference in tension that occurs in thin spokes as opposed to thick ones when the rim is subjected to the same load. A possible complicating factor is noted in the Gavin paper: "At higher radial loads, spokes lose tension at a decreasing rate per unit radial load. As spokes lose tension, loads are carried to a greater extent by moments in the rim." The effect of the tire is also a complicating factor.

Perhaps the answer to the question is contained in Gavin's presentation of the data in his figure 5, but if it is, I am not sufficiently clever to find it. Is this an easy question for you to answer with your FEA programs? Is a wheel built with thinner (or butted/swaged) spokes actually less likely to experience complete loss of spoke tension in some spokes under normal riding conditions?

That is one claim made for butted spokes -- they are less likely to completely lose tension in use. The other claim made for them is that they are less likely to experience fatigue failure in normal use. Although the topic of Gavin's paper is fatigue failure, he doesn't, as far as I can tell, address the question. He is only interested in the effect of crossing patterns on fatigue failure. I think all the data to answer the question was collected, but if the analysis was done and a conclusion reached, I don't see it.

Gavin, and I infer others, present cycles to fatigue failure as a function of stress range. When plotted log-log, it's presented as a linear relationship. He states, "Cycles with the largest spoke strain have the largest contribution to [probability of spoke failure]. If we are comparing the fat portion of a butted spoke to the same portion of a straight spoke (assuming, eg, 14/15/14 vs 14 gauge), simply comparing change in tension will give us both change in stress and strain. So, will the magnitude of tension cycles in thinner spokes be significantly less than the magnitude of tension cycles in thicker spokes when subjected to comparable loads? I think that answering this question with respect to straight 14ga spokes vs straight 15ga spokes will give us the answer for 14 ga vs 14/15/14ga spokes. Is this a hard question to answer?

I know that it is not your function in life to satisfy my curiosity, and I understand if you aren't as interested in this as I am. I appreciate your contributions in these threads, it's enlightening.

I'm thinking that I might try to contact Gavin to see if he has any thoughts that he would be willing to share. Of course, he also has a life independent of my curiosity.

There is an interplay of the different elements, those being spokes and rims, and their properties (which is why it is called an indeterminate system). Change the properties of different pieces and it affects stiffnesses, deflections and stresses. (As far as spoke fatigue is concerned it would be mainly stresses that you are interested in.) What we said before was that butted spokes can be stretched more during tensioning which makes their likelihood of stress reversal less. That is one affect. They make the wheel less stiff so the wheel deflects more at the bottom. This increases the stress range in the spokes. Another affect is that they make the stiffness of the spokes relative to the rim less, so the rim distributes its load to more spokes at the bottom. These last 2 things act against each other as far as the spokes are concerned, so which one is more important? I could make an educated guess but it would be interesting to do an analysis to confirm it.

"Imagine one spoke at the top of a wheel and one at the bottom, wheel unloaded"
How do you make this a 2 spoke wheel? I,m just simplifying by talking about only 2 spokes.
Spoke tension and where the load goes is so obvious it is the least interesting topic about wheels.
The butted spoke question is more interesting and can be simply explained by the relationship between stress and strain.

Why not analyse something useful like what the best course of action is after your chain goes into the wheel and bends a spoke? This is genuinely useful. If you bend a spoke and then get the wheel trued by a shop without replacing said spoke, just straighten the wheel up with the bent spoke still there, what is the effect?
I have produced finite element models which suggest a fatigue failure in the spoke NEXT TO the bent one could possibly occur soon after. From this I know to insist on having bent spokes replaced, rather than just having the wheel straightened. I'd be interested to know if anyone else came to that conclusion?

But having said that, what the hell do I know.

Action Plan to resolve this question:

1. measure spoke tension in all spokes of a wheel on your bike
2. have someone sit on your bike while you remeasure spoke tension
3. document and diagram changes in tension to conclusively describe the change in forces
4. change your name to Jobst Brandt and collect royalties on his book which describes all of this already

Yeah, I bet you could make a bunch of money with that plan. Probably hundreds of dollars -- every year.

edit: Perhaps I should mention that I'm just making a joke and mean no harm. It's just that a business plan which involves trying to make money from royalties from books which have a very narrow and small popular appeal is probably not a good one. It's sort of funny.

Already been done. http://www.bikeforums.net/showthread...1#post10796777
Except the researcher didnt change his name. He opted for tenure instead.

Actual strain in 2x, 3x, 4x lacing pattern (tension peaks at 120deg TDC..compression mostly occurs 40deg BDC):
http://img291.imageshack.us/img291/2...oketension.jpg

My head just exploded.

Huh?

It's just Euler's buckling equation. (pronounced "oiler")

ftfy