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I know that the physics of disc brakes are different than rim brakes, but the differences are not intuitive to me so I want to double-check this. I understand that one fail point of single walled rims is the spoke hole; that they are prone to cracking. Will this tendency be exacerbated by a disc brake (compared to a rim brake)?

FWIW, this is the front wheel I am thinking of.

EDIT: if the answer to the question is no, what do I do with the various yes answers that I find? For example from Park Tool's website:

A similar torsion or twisting load is applied through the spokes by rotor disc braking.... The rim caliper brakes, however, do not apply braking force through the spokes in the same way.
http://www.parktool.com/blog/calvins-corner/wheel-tension-balance-web-app

Am I confusing the discussion of load with the forces that cause a spoke nipple to crack the rim at the hole? Or is it, as @Rider_X says, that the forces are different but the differences are negligible?

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    @Mσᶎ I looked at that question. I couldn't identify a discernible answer that I can apply here. – jqning Nov 30 '15 at 3:46
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    I agree that the answers are vague, but your question is one of the unfortunate yes/no ones. So the answer there as far as you're concerned is "no". My reading of that question is that there's much discussion of possible mechanisms but no-one managed to explain how a wheel would be more likely to fail. – Móż Nov 30 '15 at 4:15
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    Given that a wheel works by being in tension, and it has to sustain your weight plus impacts when you hit something like a pot hole, I doubt that disc brake would exacerbate it much beyond what it already sees. – Rider_X Nov 30 '15 at 8:30
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    No the physics are not different. You have a given amount of inertia / momentum (mass * velocity) and that force goes thru the spokes. Rim pulling on hub or hub pulling on rim it is the same force / work. – paparazzo Nov 30 '15 at 12:57
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Well, let's compute the forces. For example's sake, I'm assuming 80kg of bike + rider, all braking force on front wheel and deceleration of 5 m/s^2 (extreme, but possible). This equals 400N at contact surface. The radius of 559mm rim and tire is something like 320mm, giving 128Nm of torque. The spoke hole is at 22mm radius, and cross 3 lacing places the spokes roughly tangential to hub flange. Using the axle as pivot and spoke attachment to hub as lever, we can calculate that we need force of 5818N to produce the torque.

Spread over 32 spokes, that is 181 newtons per spoke. Since spokes in a properly built wheel are already at 1000N or more, the increase is around 15-20%. This requires a rim with stronger spoke holes, but not by that large margin.

Keep in mind, that in many double wall rims the spokes attach to outer wall. In this case, the extra wall doesn't help at all.

The case with radial spokes on both sides is different: The spokes exit the hub at zero angle, there is no leverage and if the spokes wouldn't stretch and let the hub twist, the force would be infinite. Calculating the actual force is left as an exercise to the reader :)

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  • Loads (e.g., braking, bumps, etc) will not load all spokes equally at any given instantaneous moment. Peak force will be the most important when talking about fatigue. – Rider_X Dec 1 '15 at 3:24
  • True. Impacts at rim unload the spokes at contact point and only slightly increase tension elsewhere, so I didn't try anything more exact. The case with evenly spread load gets close enough for me. Let's see if anyone has time to play around with FEM. – ojs Dec 1 '15 at 17:52

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