# Asymmetric rear wheels.

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**Asymmetric rear wheels.**

you like? yeah or nay?

At first I though they looked goofy but then I rode an Ultegra wheel and pretty much feel like I can't live without it. Less flexy & more stable feeling.

At first I though they looked goofy but then I rode an Ultegra wheel and pretty much feel like I can't live without it. Less flexy & more stable feeling.

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Would more symmetrical spoke bracing angles possibly result in more lateral stiffness?

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Actually the asymmetric rims lower lateral stiffness some. They lower the bracing angle of the NDS and make them more similar to the DS spokes. You may be calling that more "symmetrical" angles. So in your terminology the asymmetric rims have more symmetrical angles. Weird, huh. But the result is lower stiffness. The DS angles don't increase, yet the NDS angles do decrease. For stiffness it is a slight loss.

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The DS bracing angles don't increase? How come?

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I have two bikes with Velocity Synergy O/C, and one with Velocity Aerohead O/C.

On both models of rim,

On both models of rim,

*all*the holes are offset.
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Yes, you are right. My mistake. What I should have said is this: the stiffness input from the DS side does increase a little and the stiffness from the NDS side decreases. The net change is negative because the relationship between bracing angle and stiffness is a squared function, not linear. So the small change in the lower angle side, the DS, produces less increase in stiffness than the small change in the larger angle side, the NDS.

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Really it depends on how wide the hub flanges are. I don't know how you can say it is less stiff by making the angles more even. More even = more even tension across the wheel making it stronger. Stronger can only be stiffer.

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But a simple explanation is this. Lateral wheel stiffness increases with the square of the sine function of the bracing angle. Put another way, stiffness is proportional to the square of the sine of the bracing angle (and the number of spokes, and the spoke gauge, etc.). When you make the NDS bracing angle smaller it removes stiffness from the wheel. When you make the DS bracing angle larger it adds stiffness to the wheel. But the key is the squared function. It emphasizes the decrease in the larger NDS angle more than the increase in the smaller DS angle. Think about how 2^2=4, but 10^2=100. The effect of squaring is much greater as numbers get larger.

Here is an example. Supposing you have a wheel that has an NDS bracing angle of 7° and a DS bracing angle of 4°. That would be fairly common. The sine of 7°= 0.12 and the square of that is 0.0148. The sine of 4° = 0.07 and the square of that is 0.0049. Now lets make that wheel asymmetrical by moving the spokes to the left enough so that the bracing angles on the two sides are equal, say 5.5° on each side. Making a long story short the square of the sines of the two equal angles is 0.0092. Now to get numbers that are proportional to the total lateral stiffness of the two wheel cases, just add up the NDS and DS values (squared sines) for each case. The symmetrical case gives you 0.0197 while the asymmetric case gives you 0.0184. That lower number means the asymmetric wheel is about 7% less stiff laterally. Not much, but the result is real. And it will always be that way no matter what the real values are. It is just a fact of life.

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But a simple explanation is this. Lateral wheel stiffness increases with the square of the sine function of the bracing angle. Put another way, stiffness is proportional to the square of the sine of the bracing angle (and the number of spokes, and the spoke gauge, etc.). When you make the NDS bracing angle smaller it removes stiffness from the wheel. When you make the DS bracing angle larger it adds stiffness to the wheel. But the key is the squared function. It emphasizes the decrease in the larger NDS angle more than the increase in the smaller DS angle. Think about how 2^2=4, but 10^2=100. The effect of squaring is much greater as numbers get larger.

Here is an example. Supposing you have a wheel that has an NDS bracing angle of 7° and a DS bracing angle of 4°. That would be fairly common. The sine of 7°= 0.12 and the square of that is 0.0148. The sine of 4° = 0.07 and the square of that is 0.0049. Now lets make that wheel asymmetrical by moving the spokes to the left enough so that the bracing angles on the two sides are equal, say 5.5° on each side. Making a long story short the square of the sines of the two equal angles is 0.0092. Now to get numbers that are proportional to the total lateral stiffness of the two wheel cases, just add up the NDS and DS values (squared sines) for each case. The symmetrical case gives you 0.0197 while the asymmetric case gives you 0.0184. That lower number means the asymmetric wheel is about 7% less stiff laterally. Not much, but the result is real. And it will always be that way no matter what the real values are. It is just a fact of life.

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I don't know much about all the physics behind it, but I am liking my Pacenti Forza very much.

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But your forgetting about the transitional equivalency factor where you divide the angle of the stiffer side by the weaker side which is 1.75 and multiply that factor by the new sine value and you get .0322 effective stiffness as it is no longer prone to flex to one side or the other for a net improvement of almost 39% stiffness increase.

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I thought it was generally accepted that the butt dyno is never wrong. If a new part is installed, and the butt dyno determines that the change is positive, regardless of any actual numbers on paper, said part is an improvement. Once you start bringing math into it, the fun has most certainly died. Also, not being a math guy, I'd like to toss out there the fact that perception is reality-- if the OP feels like the new wheels are more laterally stiff, they are. No one else here can perceive his reality, we can only analyze our perceptions of his reality.

Or more likely, the Ultegra wheels are simply stiffer than what he had before, asymmetry notwithstanding.

Or more likely, the Ultegra wheels are simply stiffer than what he had before, asymmetry notwithstanding.

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Or more likely, the Ultegra wheels are simply stiffer than what he had before, asymmetry notwithstanding.

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But a simple explanation is this. Lateral wheel stiffness increases with the square of the sine function of the bracing angle. Put another way, stiffness is proportional to the square of the sine of the bracing angle (and the number of spokes, and the spoke gauge, etc.). When you make the NDS bracing angle smaller it removes stiffness from the wheel. When you make the DS bracing angle larger it adds stiffness to the wheel. But the key is the squared function. It emphasizes the decrease in the larger NDS angle more than the increase in the smaller DS angle. Think about how 2^2=4, but 10^2=100. The effect of squaring is much greater as numbers get larger.

Here is an example. Supposing you have a wheel that has an NDS bracing angle of 7° and a DS bracing angle of 4°. That would be fairly common. The sine of 7°= 0.12 and the square of that is 0.0148. The sine of 4° = 0.07 and the square of that is 0.0049. Now lets make that wheel asymmetrical by moving the spokes to the left enough so that the bracing angles on the two sides are equal, say 5.5° on each side. Making a long story short the square of the sines of the two equal angles is 0.0092. Now to get numbers that are proportional to the total lateral stiffness of the two wheel cases, just add up the NDS and DS values (squared sines) for each case. The symmetrical case gives you 0.0197 while the asymmetric case gives you 0.0184. That lower number means the asymmetric wheel is about 7% less stiff laterally. Not much, but the result is real. And it will always be that way no matter what the real values are. It is just a fact of life.

Hmmm ... .

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Here's another example: Let's suppose you have a wheel where the DS hub flange is in the same plane as the spoke holes so that the DS bracing angle is 0° and therefore the NDS bracing angle is 11°. The sine of 0° = 0.00 and the square of that is 0.0000. The sine of 11° = 0.191 and the square of that is 0.0364. The sum of the squared sines is 0.0364, which is almost twice the value of the two wheels above! The sum of the lateral forces from the DS spokes (spoke tension times sine of the bracing angle times number of spokes) is balanced by the lateral forces from the NDS spokes. Therefore the tension in the NDS spokes is 0.000 and as zero tension members can be removed from the structure without any loss of integrity.

Hmmm ... .

Hmmm ... .

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I believe all of you math may be irrelevant due to the fact that spokes can and will flex. Either way I think it's time to whip out a tension gauge.

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I haven't used asymmetrical rims but I have a lot of miles on zero dish or very low dish fix gear wheels. For my road wheels, I use DT Competitions on the DS, DT Revolutions on the NDS. My fix gears wheels I lace DT Revolutions both sides. My fix gear wheels feel far stiffer and stronger than my geared rear wheels. This is despite the spokes of both sides being of less effectiveness that the NDS of my geared wheels.

I find my geared wheels almost feeling scary after spending so much time on the rock solid fix gear wheels. I find it easy to believe that easing the really steep spoke angles of the geared DS could bring about big gains in how the wheel feels.

And to rpenmanparker (post#17) - you certainly can have a lateral tension of zero from those 11° NDS spokes. Don't tighten the nipples - at all! Then as gl981154 stated, the DS and NDS lateral forces are equal. As far as structural integrity, well there's a nice symmetry here. For acceptable deflection, the side force applied (in turns or from other sources) can be as high as the initial opposing side forces from the spoke tensions and angles. (Side force = spoke tension X the sine of the angle from vertical = DS spoke tension X sin(0) which also = NDS spoke tension X sin(11). NDS tension = 0. So DS spoke tension X sin(0) or NDS spoke tension X sin(11) or DS spoke tension X 0 = 0 X sin(11). Yup, it all works. A supper flimsy wheel stays as true as the rim was drawn - if you apply no load to it. (Stay off the bike.)

Ben

I find my geared wheels almost feeling scary after spending so much time on the rock solid fix gear wheels. I find it easy to believe that easing the really steep spoke angles of the geared DS could bring about big gains in how the wheel feels.

And to rpenmanparker (post#17) - you certainly can have a lateral tension of zero from those 11° NDS spokes. Don't tighten the nipples - at all! Then as gl981154 stated, the DS and NDS lateral forces are equal. As far as structural integrity, well there's a nice symmetry here. For acceptable deflection, the side force applied (in turns or from other sources) can be as high as the initial opposing side forces from the spoke tensions and angles. (Side force = spoke tension X the sine of the angle from vertical = DS spoke tension X sin(0) which also = NDS spoke tension X sin(11). NDS tension = 0. So DS spoke tension X sin(0) or NDS spoke tension X sin(11) or DS spoke tension X 0 = 0 X sin(11). Yup, it all works. A supper flimsy wheel stays as true as the rim was drawn - if you apply no load to it. (Stay off the bike.)

Ben

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I find my geared wheels almost feeling scary after spending so much time on the rock solid fix gear wheels. I find it easy to believe that easing the really steep spoke angles of the geared DS could bring about big gains in how the wheel feels.

And to rpenmanparker (post#17) - you certainly can have a lateral tension of zero from those 11° NDS spokes. Don't tighten the nipples - at all! Then as gl981154 stated, the DS and NDS lateral forces are equal. As far as structural integrity, well there's a nice symmetry here. For acceptable deflection, the side force applied (in turns or from other sources) can be as high as the initial opposing side forces from the spoke tensions and angles. (Side force = spoke tension X the sine of the angle from vertical = DS spoke tension X sin(0) which also = NDS spoke tension X sin(11). NDS tension = 0. So DS spoke tension X sin(0) or NDS spoke tension X sin(11) or DS spoke tension X 0 = 0 X sin(11). Yup, it all works. A supper flimsy wheel stays as true as the rim was drawn - if you apply no load to it. (Stay off the bike.)

Ben

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Since you know the lateral tension on the NDS side is not zero, the structure cannot be created. If you try to tension the NDS spokes, even in order to provide just radial tension, they will pull the rim off center. That puts lateral tension into the DS spokes. You cannot Fool Mother Nature.

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And also a wheel cannot be stiff due to the tension in the spokes if they have no tension in them. The principles my calculation was based on is that lateral stiffness is another way of saying the tensile modulus of the spoke as reduced by the sine of the angle it is acting through. Tensile modulus is defined and measured in tension. Without tension, there is no stiffness. This is by definition which trumps further calculations. Your thought experiment is cute, but fatally flawed.

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tandems are 145 to 160 wide o.l.d. axles, to have the spoke bracing symmetrical , But..

the flange width , the base of the triangle is still narrow ...

Unlike Internal gear hubs that have wide hub shells and 1 cog , not needing to make compromises to fit 10 or 11.

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no, it is not. You are dealing with a limiting case, like dividing by zero. You think you are doing it, but it is not possible by definition. By definition the spokes of a bicycle wheel are in tension. Any tension on an angled spoke will cause the opposite spoke to assume a non-zero bracing angle.

And also a wheel cannot be stiff due to the tension in the spokes if they have no tension in them. The principles my calculation was based on is that lateral stiffness is another way of saying the tensile modulus of the spoke as reduced by the sine of the angle it is acting through. Tensile modulus is defined and measured in tension. Without tension, there is no stiffness. This is by definition which trumps further calculations. Your thought experiment is cute, but fatally flawed.

And also a wheel cannot be stiff due to the tension in the spokes if they have no tension in them. The principles my calculation was based on is that lateral stiffness is another way of saying the tensile modulus of the spoke as reduced by the sine of the angle it is acting through. Tensile modulus is defined and measured in tension. Without tension, there is no stiffness. This is by definition which trumps further calculations. Your thought experiment is cute, but fatally flawed.

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And also a wheel cannot be stiff due to the tension in the spokes if they have no tension in them. The principles my calculation was based on is that lateral stiffness is another way of saying the tensile modulus of the spoke as reduced by the sine of the angle it is acting through. Tensile modulus is defined and measured in tension. Without tension, there is no stiffness. This is by definition which trumps further calculations. Your thought experiment is cute, but fatally flawed.

your math is cute but it negates the fact of the overall structure of the wheel. An equal triangle will be harder to tip over than an unequal offset triangle. Your math may make sense but when you are apply drive side load combined with lean angle of a bike you will be applying many different forces that your fancy math can never account for.