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Puzzled about skewer tension

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Puzzled about skewer tension

Old 06-10-11, 10:02 PM
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Puzzled about skewer tension

I wanted to figure out if a Pitlock skewer would be OK for my horizontal rear dropout. (There are contradictory opinions: Pitlock says no, some folks here agree with that and some don't.) So I tried to get an idea of how the tension generated by the torque Pitlock says is their limit compares with what works with QR.

I've ended up baffled. Multiplying the mechanical gain factor for M5 bolts by the force Pitlock suggests gives a tension several times the strength of the bolt. Now, friction is a big factor - but in the sources I've seen, was only supposed to consume half or less of the input.

The numbers:
  • tensile strength of an M5 bolt is about 2500 pounds (1/40 of a square inch tensile stress area - that's cross-section when you take thread into account; steel strength around 100,000 pounds/square inch). Maybe more for deluxe steel, but not twice as much.
  • mechanical gain of screw: thread pitch is 0.8 mm - which is therefore the skewer motion produced by a full turn. If we presume a 6 cm wrench handle, tip travels 38 cm for a gain factor of 470.
  • Pitlock's 9 nm, at radius .06 m, is a force of 150 newtons. Multiplying by 470: 70,500 n / 15,800 pounds. DT Swiss also makes screw-tightened skewers; they say 15 nm: x 470 yields 117,500 n / 26000 pounds.
  • so even let friction consume half the input (one number I've heard; others are lower) - the DT Swiss spec still generates 6 tons on a rod good for 1.
Does anyone know the real story on this? Is friction really consuming 5/6 of the force?

Edit: OK - spotted the answer. Yes, friction does dominate - because the path the screw thread is dragged along is about 20 times longer than the movement in the direction of tension. Per revolution, thread is traveling roughly 15 mm against a frictional force of about 15% of tension - so thread-friction work is 15mm times .15 T, while tensioning work done is .8mm times T. And friction on the nut face is similar (a bit higher actually, due to greater radius). So roughly 5T frictional work to 0.8T tensioning work.

Last edited by Antifriction; 06-10-11 at 11:11 PM.
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Old 06-10-11, 10:56 PM
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I don't have time right now to go through it all, but it seems like your math is fishy. Do a unit check first and go from there.
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Old 06-10-11, 11:46 PM
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I'm too lazy to read all that. I have a simple test: is it a steel enclosed cam skewer? If yes, use it without hesitation in horizontal dropouts. If no, use it at your own risk and don't be surprised if it slips.

Sheldon Brown on QR skewers: https://www.sheldonbrown.com/skewers.html
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Old 06-11-11, 12:30 AM
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He's talking about locking skewers (which screw on with a special tool), not quick-release. I haven't used pitlocks, but I did use pinhead skewers with horizontal dropouts for a while. It's possible, but it can be difficult to get them sufficiently tight with the short key you use. I prefer using Sheldon Brown's locking method (u-lock around the wheel inside the rear triangle) with a quick-release on the rear and a locking skewer on the front.
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