Is the conventional wisdom about hub maintinance wrong?
#26
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It can, in a very narrow range and at considerable force for a hardened steel axle. If you have an Instron UTS machine sure, but by hand.....uhhh no. Much like the argument made by expert nitpickers that a fly landing an a steel bridge cause it to flex to some degree.
Seriously, you think that compressing steel by hand is a more logical conclusion than the fact that there are multiple thread interfaces in the axle assembly and the force applied by the QR takes up the slop in the threads? Really?
Seriously, you think that compressing steel by hand is a more logical conclusion than the fact that there are multiple thread interfaces in the axle assembly and the force applied by the QR takes up the slop in the threads? Really?
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It can, in a very narrow range and at considerable force for a hardened steel axle. If you have an Instron UTS machine sure, but by hand.....uhhh no. Much like the argument made by expert nitpickers that a fly landing an a steel bridge cause it to flex to some degree.
Seriously, you think that compressing steel by hand is a more logical conclusion than the fact that there are multiple thread interfaces in the axle assembly and the force applied by the QR takes up the slop in the threads? Really?
Seriously, you think that compressing steel by hand is a more logical conclusion than the fact that there are multiple thread interfaces in the axle assembly and the force applied by the QR takes up the slop in the threads? Really?
Steel will stretch or compress an equal amount for an equal force pulling or pushing.
The amount of compression we're talking about is 1/16th turn on a 24tpi threaded axle, so about .0026" - the thickness of paper. Yes, I think a hollow axle could elastically compress 4" of hardened axle under the force of a cammed QR that much. A QR has enough leverage to deform mild steel dropouts with knurling marks.
Last edited by Kontact; 07-16-18 at 12:28 PM.
#28
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Steel can compress, but that's not what happens in this case. The proof against it compressing is that it relaxes when the QR is released. Compressed steel (plastic deformation) doesn't spring back to it's original shape.
Does the axle flex (elastic deformation)? If it does, that doesn't give me a lot of confidence in the axle integrity.
I would say the most likely culprit is: Thread tolerances of the cones, lock nuts and axle
Does the axle flex (elastic deformation)? If it does, that doesn't give me a lot of confidence in the axle integrity.
I would say the most likely culprit is: Thread tolerances of the cones, lock nuts and axle
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Are you a millwright or an engineer? This guy was a mechanical engineer. Wheel Bearing adjustment by Jobst Brandt
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
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My number is on speed dial for Aerospace engineers around the world.
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
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#32
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My number is on speed dial for Aerospace engineers around the world.
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
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Here you go sonny, read and learn...if that's possible for you
https://www.hexagon.de/rs/engineering%20fundamentals.pdf
https://www.hexagon.de/rs/engineering%20fundamentals.pdf
#34
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This would explain things -- there's no way you can get the clamping force of an internal-cam QR skewer with a bolt-on setup. So for your application, you are correct. Use the same adjustment, but with a QR skewer and the adjustment will be too tight.
#35
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Here you go sonny, read and learn...if that's possible for you
https://www.hexagon.de/rs/engineering%20fundamentals.pdf
https://www.hexagon.de/rs/engineering%20fundamentals.pdf
I am violating rule number 2.
#36
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My number is on speed dial for Aerospace engineers around the world.
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
I know that the cross sectional density of a QR skewer is significantly less than the cross sectional density of an axle, if your mechanical engineer friend would like to explain how tightening the skewer will compress a steel axle that has 800% larger cross section, I'm all ears.
If what you are trying to say is that the cross section area of the axle is 8 times the cross section area of the skewer, then the answer is that the axle will compress about 1/8 as much as the skewer will lengthen, assuming that the Youngs modulas of the axle is about the same as that of the skewer and that the axle and skewer are about the same length. This isn't rocket surgery, you know.
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What's your objective?
When I finish overhauling the hub I want the wheel to spin freely when clamped in the dropouts but not rock side-to-side. When I achieve both at the same time, I'm done.
When I finish overhauling the hub I want the wheel to spin freely when clamped in the dropouts but not rock side-to-side. When I achieve both at the same time, I'm done.
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What are the units of "cross sectional density?"
If what you are trying to say is that the cross section area of the axle is 8 times the cross section area of the skewer, then the answer is that the axle will compress about 1/8 as much as the skewer will lengthen, assuming that the Youngs modulas of the axle is about the same as that of the skewer and that the axle and skewer are about the same length. This isn't rocket surgery, you know.
If what you are trying to say is that the cross section area of the axle is 8 times the cross section area of the skewer, then the answer is that the axle will compress about 1/8 as much as the skewer will lengthen, assuming that the Youngs modulas of the axle is about the same as that of the skewer and that the axle and skewer are about the same length. This isn't rocket surgery, you know.
#39
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The point remains a QR skewer absolutely will not compress a steel axle. The link I supplied clearly explains how threaded fasteners work.
Oh and by the way, you can thank me for being part of the team that provided a metallurgical solution to prevent another Deep Hole Horizon disaster. That may be almost impressive as your ability to tighten a bolt.
Last edited by desmodue; 07-21-18 at 05:10 PM.
#40
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Funny how the scientific method is ignored in this thread by supposed scientists in favor of calculation. Yeah, the skewer does tighten the bearings. My opinion is that the amount of axle compression which must happen to effect the small perceptible change in bearing slop is very, very small. However that may be, it's certainly noticeable in old-style cup-and-cone bearings. One can test this supposition more quickly than reading this thread.
My WAG is that slop at rim before the skewer should be on the order of .005", but that's going to depend on the axle. I'll let someone else do the calculations.
My WAG is that slop at rim before the skewer should be on the order of .005", but that's going to depend on the axle. I'll let someone else do the calculations.
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Last edited by Carbonfiberboy; 07-21-18 at 08:50 PM.
#41
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Your thinking is highly flawed in that you assume that the axle and skewer are both limited to 2 axis, in reality what happens is the skewer being much thinner than the axle will flex instead of compressing. Simply put, its a freaking piece of crap aluminum hub not stress rupture machine. For all the idiots that think a steel axle compresses, here's a simple test" get a .500" thick piece of plate steel. Drill a 9/32' hole in it 1/2' from the edge. Measure the plate thickness with a caliper. Put a 1/4' x 20 tpi grade 8 bolt in it with washers and a nut tighten the nut and bolt as hard as you dare without twisting it off. Measure the plate thickness again. Not that the dimension remans the same. That's all I have to say about that.
Let's see how this works out. A QR skewer is about 4.8 mm in diameter. That's about 18 X 10^-6 m^2
The axle is about 10mm in diameter with about a 5mm hole in the center. That's about 60 X 10^-6 m^2.
So, the cross section of the axle is about 3.3 times the cross section of the skewer, not the 8 times (or 9 times) that you proposed.
So, how much tension should it be reasonable for a QR skewer to experience? I don't know the mechanical properties of the QR skewer , but I don't know the mechanical properties of a 1.8 mm diameter spoke, either. What I do know, is that a 1.8 mm spoke and a QR skewer probably have about the same yield strength. So, it is quite obviious that a QR skewer can handle the stress that a 1.8 mm spoke can handle. A 1.8 mm spoke can easily handle in excess of 1300 N. A 4.8 mm QR skewer, therefeore, can easily handle about 9000 N as the area of the skewer is about 7 times the area of the spoke.
So, if you take a 130 mm axle and apply 9000 N compressive force, how much does the axle compress? That is clearly explained by the well known stress/strain relationship.
Stress/strain = Young's modulus, which for various steels is about 200 X 10^9 N/m^2
So, how much does a steel axle compress if a compressive tension of 9000 N is applied?
strain = stress/Young's modulus X axle length
strain (i.e. compression) = (9X 10^3 N / 60 X 10^-6 m^2) / (200 X 10 ^ 9 N/m^2) X (130 X 10 ^-3 m)
If you calculate that out, you come up with a compression of .0975 mm -- a little less than .004".
It was suggested above by Kontact that the compression would be (1/24)/16". About .003"
Seems like it's reasonable for a QR skewer to cmpress an axle by at least that much.
I did these calculations with a slide rule and I'm assuming that a QR and spoke have about the same yield strength and I'm assuming that the very high tension that can be exerted with a QR eccentric is about 9000 N. If my calculations are off, or my assumptions are unwarranted, please explain. Else, go back to serving coffee.
#42
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So here's a strange twist: I've since serviced a few more wheel sets, and I noticed a strange thing. I'll get the cone adjustment just right, and then test it again AFTER I put the wheel back on the bike. The results? It's either exactly the same, OR there is play that I couldn't feel when I had the wheel off the bike.
I'll say that again. When I install the wheel back onto the bike, sometimes there is play that I couldn't feel when I was adjusting the cones.
Weird, huh?
I'll say that again. When I install the wheel back onto the bike, sometimes there is play that I couldn't feel when I was adjusting the cones.
Weird, huh?
#43
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dear Gormless Gob****e:
Let's see how this works out. A QR skewer is about 4.8 mm in diameter. That's about 18 X 10^-6 m^2
The axle is about 10mm in diameter with about a 5mm hole in the center. That's about 60 X 10^-6 m^2.
So, the cross section of the axle is about 3.3 times the cross section of the skewer, not the 8 times (or 9 times) that you proposed.
So, how much tension should it be reasonable for a QR skewer to experience? I don't know the mechanical properties of the QR skewer , but I don't know the mechanical properties of a 1.8 mm diameter spoke, either. What I do know, is that a 1.8 mm spoke and a QR skewer probably have about the same yield strength. So, it is quite obviious that a QR skewer can handle the stress that a 1.8 mm spoke can handle. A 1.8 mm spoke can easily handle in excess of 1300 N. A 4.8 mm QR skewer, therefeore, can easily handle about 9000 N as the area of the skewer is about 7 times the area of the spoke.
So, if you take a 130 mm axle and apply 9000 N compressive force, how much does the axle compress? That is clearly explained by the well known stress/strain relationship.
Stress/strain = Young's modulus, which for various steels is about 200 X 10^9 N/m^2
So, how much does a steel axle compress if a compressive tension of 9000 N is applied?
strain = stress/Young's modulus X axle length
strain (i.e. compression) = (9X 10^3 N / 60 X 10^-6 m^2) / (200 X 10 ^ 9 N/m^2) X (130 X 10 ^-3 m)
If you calculate that out, you come up with a compression of .0975 mm -- a little less than .004".
It was suggested above by Kontact that the compression would be (1/24)/16". About .003"
Seems like it's reasonable for a QR skewer to cmpress an axle by at least that much.
I did these calculations with a slide rule and I'm assuming that a QR and spoke have about the same yield strength and I'm assuming that the very high tension that can be exerted with a QR eccentric is about 9000 N. If my calculations are off, or my assumptions are unwarranted, please explain. Else, go back to serving coffee.
Let's see how this works out. A QR skewer is about 4.8 mm in diameter. That's about 18 X 10^-6 m^2
The axle is about 10mm in diameter with about a 5mm hole in the center. That's about 60 X 10^-6 m^2.
So, the cross section of the axle is about 3.3 times the cross section of the skewer, not the 8 times (or 9 times) that you proposed.
So, how much tension should it be reasonable for a QR skewer to experience? I don't know the mechanical properties of the QR skewer , but I don't know the mechanical properties of a 1.8 mm diameter spoke, either. What I do know, is that a 1.8 mm spoke and a QR skewer probably have about the same yield strength. So, it is quite obviious that a QR skewer can handle the stress that a 1.8 mm spoke can handle. A 1.8 mm spoke can easily handle in excess of 1300 N. A 4.8 mm QR skewer, therefeore, can easily handle about 9000 N as the area of the skewer is about 7 times the area of the spoke.
So, if you take a 130 mm axle and apply 9000 N compressive force, how much does the axle compress? That is clearly explained by the well known stress/strain relationship.
Stress/strain = Young's modulus, which for various steels is about 200 X 10^9 N/m^2
So, how much does a steel axle compress if a compressive tension of 9000 N is applied?
strain = stress/Young's modulus X axle length
strain (i.e. compression) = (9X 10^3 N / 60 X 10^-6 m^2) / (200 X 10 ^ 9 N/m^2) X (130 X 10 ^-3 m)
If you calculate that out, you come up with a compression of .0975 mm -- a little less than .004".
It was suggested above by Kontact that the compression would be (1/24)/16". About .003"
Seems like it's reasonable for a QR skewer to cmpress an axle by at least that much.
I did these calculations with a slide rule and I'm assuming that a QR and spoke have about the same yield strength and I'm assuming that the very high tension that can be exerted with a QR eccentric is about 9000 N. If my calculations are off, or my assumptions are unwarranted, please explain. Else, go back to serving coffee.
#44
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Discernible play in the axle may be be a hit or miss off the bike, your perception to detect when gripping at the axle ends on both sides, low leverage point.
However, on the bike you should be using the wheel rim for leverage, hence making it easier to detect the proper preload when at 45 degrees from full closure (some play) and then at full closure (no play). I mark my OR's for reference (Tape or Sharpe marker on the nut and lever closure side) so they end up in the same place every time when tightened (when removed and reinstalled), to give the same pressure on the preload every time.
Also it helps to look straight down at the area, aliening the reference marks with the center of the axle, to avoid parallax distortion. Hows that for type 2 OCD (Obsessive Cycling Desire). Really want to get my moneys worth out of those Grade 25 BB's.
However, on the bike you should be using the wheel rim for leverage, hence making it easier to detect the proper preload when at 45 degrees from full closure (some play) and then at full closure (no play). I mark my OR's for reference (Tape or Sharpe marker on the nut and lever closure side) so they end up in the same place every time when tightened (when removed and reinstalled), to give the same pressure on the preload every time.
Also it helps to look straight down at the area, aliening the reference marks with the center of the axle, to avoid parallax distortion. Hows that for type 2 OCD (Obsessive Cycling Desire). Really want to get my moneys worth out of those Grade 25 BB's.
Last edited by toddbiker; 07-24-18 at 03:31 AM.
#45
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Smart guy, this one^. Yeah. That's probably it.
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