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#151
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well, as i understand the argument, it goes like this: resolved "ejection" force is postulated to be 1825N, and iso pullout spec is alledged to be only 500N, therefore the sky will fall tomorrow. what's being ignored is that a conservative estimate for pullout force based on just part of one face of a serrated axle [what all disk brake hub manufacturers afaik use] is in excess of 5000N. and that's /not/ including any effect of serrated skewer nut facings either. if that's not ignoring a critical part of the analysis, i don't know what is. I'm just trying to resolve a force of 5000N through small ridges on an alloy fork, coupled with what is known about the properties of the alloy to which it is clamped and not getting deformation of the fork dropout faces. Look at almost any bike fork and you will see marks from the skewer nuts on the outside and marks from the axle on the inside. That is even before placing braking stresses on it. It doesn't take a rocket scientist to work out that if you are putting a system like that where you are relying on ductile metals to hold something in place under strong vibrational loads, then it can potentially deform and work loose. I think jim's arguement can be stated in "If it is properly tightened then it can't work loose". And the observation being that the wheel was properly tightened but did work loose. The point in question could be the timescale between being verifiably correctly fastened and the skewer allegedly working loose. I am not terribly much in favour of the static models produced so far. It doesn't take a great leap of imagination to determine realistic scenarios where there is no loading on the handlebars but a maximal braking force on the front wheel, potentially a pull-out force before the brakes even come on. A more realistic test would be the following: Hold the forks rigid and put the brakes on. monitor slippage with respect to the torsional force on the wheel. Do we get slippage before the wheel collapses? At what force do we get slippage? With that data we have to ask how representative of real life are these forces. The moral of the story is that when riding any reasonable definition of rough terrain one should always reseat the q/r (undo and retighten). Ideally I'd like to see manufacturers moving towards the better style of axle attachment for off-road machines. Having said that, I cant remember the last time I checked the q/r on my bikes but I don't ride o/r very often, and when I do it is hardly severe stuff. ...d |
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#152
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On Tue, 28 Sep 2004 20:40:35 -0700, jim beam wrote:
what's being ignored is that a conservative estimate for pullout force based on just part of one face of a serrated axle [what all disk brake hub manufacturers afaik use] is in excess of 5000N. and that's /not/ including any effect of serrated skewer nut facings either. if that's not ignoring a critical part of the analysis, i don't know what is. Not including effect of serrations, so just friction? Well, steel-on-steel coefficient of friction is in teh order of 0.1 to 0.3, so ignoring the fact that you claim a conservative estimate, and aiming for an optimistic estimate, that assumes a normal force of 5000/0.3, or 16.7E3 N. Suppose a typical skewer is 4mm diameter, that's a CSA of 12.5mm2. Thus we require teh skewer to be working at 1326 N/mm2, or about 5 times the yield stress of mild steel. Admittedly, it's possible to get steel with those sort of yields, but I'd question whether skewers are made of them. When we return to your claim of 'conservative' estimate, I think I'd need to allow for a low coefficient of friction, say 0.1, requiring a stress in teh skewer of about 4000 N/mm2, which I think exceeds any sensible steel. All of which ignores the fact that this force has to be exerted by teh person closing the QR - a quick work calculation, based on teh assumption that teh skewer shortens by say 2mm while the lever moves through 90 degrees and has ane effective length 50mm, so 90mm stroke, so 45:1, so it would (using teh conservative friction figure) require a closing force of aboiut 1100 N, or 110kg. Do you really claim that QRs should be adjusted such that it takes a 110kg force on teh lever to close them? I'd be interested to see your conservative estimate that gets 5000N pull-out. regards, Ian SMith -- |\ /| no .sig |o o| |/ \| |
#153
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On Wed, 29 Sep 2004 02:10:12 +0100, Tony Raven
wrote: The only thing I can't recall seing come loose is a Nord-Lock washer set, and I bet even they will in the right (or wrong) circumstances. That still works because serrations on the top of the top washer lock it to the nut rotation. i.e. it uses serrations to work just like a Shimano skewer. Sure. But the serrations are much deeper, and the torque is typically much greater. All this is speculation. Why have the people with the wherewithal to actually test the hypothesis (coincidentaly the people who make money selling the product) not done so? Guy -- May contain traces of irony. Contents liable to settle after posting. http://www.chapmancentral.co.uk 88% of helmet statistics are made up, 65% of them at Washington University |
#154
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Ian Smith wrote:
On Tue, 28 Sep 2004 20:40:35 -0700, jim beam wrote: what's being ignored is that a conservative estimate for pullout force based on just part of one face of a serrated axle [what all disk brake hub manufacturers afaik use] is in excess of 5000N. and that's /not/ including any effect of serrated skewer nut facings either. if that's not ignoring a critical part of the analysis, i don't know what is. Not including effect of serrations, so just friction? no, _including_ effect of serrations - which are evident on all forks/hubs that i've seen for disk brakes. Well, steel-on-steel coefficient of friction is in teh order of 0.1 to 0.3, so ignoring the fact that you claim a conservative estimate, and aiming for an optimistic estimate, that assumes a normal force of 5000/0.3, or 16.7E3 N. Suppose a typical skewer is 4mm diameter, that's a CSA of 12.5mm2. Thus we require teh skewer to be working at 1326 N/mm2, or about 5 times the yield stress of mild steel. Admittedly, it's possible to get steel with those sort of yields, but I'd question whether skewers are made of them. When we return to your claim of 'conservative' estimate, I think I'd need to allow for a low coefficient of friction, say 0.1, requiring a stress in teh skewer of about 4000 N/mm2, which I think exceeds any sensible steel. All of which ignores the fact that this force has to be exerted by teh person closing the QR - a quick work calculation, based on teh assumption that teh skewer shortens by say 2mm while the lever moves through 90 degrees and has ane effective length 50mm, so 90mm stroke, so 45:1, so it would (using teh conservative friction figure) require a closing force of aboiut 1100 N, or 110kg. Do you really claim that QRs should be adjusted such that it takes a 110kg force on teh lever to close them? I'd be interested to see your conservative estimate that gets 5000N pull-out. regards, Ian SMith here's my original post. http://groups.google.com/groups?selm...&output=gplain |
#155
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in message , Just zis Guy,
you know? ') wrote: On Wed, 29 Sep 2004 02:10:12 +0100, Tony Raven wrote: The only thing I can't recall seing come loose is a Nord-Lock washer set, and I bet even they will in the right (or wrong) circumstances. That still works because serrations on the top of the top washer lock it to the nut rotation. i.e. it uses serrations to work just like a Shimano skewer. Sure. But the serrations are much deeper, and the torque is typically much greater. All this is speculation. Why have the people with the wherewithal to actually test the hypothesis (coincidentaly the people who make money selling the product) not done so? If you refer to Cannondale in this, may I again point out to you that in Cannondales 2005 lineup you will find at most two models which combine disk brakes with front quick-releases? Cannondale did the test. Whatever they thought of the results which they submitted back to the consumer protection people, they clearly decided that building bikes which combined disk brakes with front quick releases was not a business they really wished to be in. -- (Simon Brooke) http://www.jasmine.org.uk/~simon/ ;; no eternal reward will forgive us now for wasting the dawn. ;; Jim Morrison |
#156
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Slacker wrote:
[disc brake ejection] Because you're sucking valuable brandwidth. Yes! It must be saved for the endless rbt US politics flamewar. God forbid anyone talk about bicycles... -- David Damerell flcl? |
#157
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On Wed, 29 Sep 2004 06:01:30 -0700, jim beam wrote:
Ian Smith wrote: Not including effect of serrations, so just friction? no, _including_ effect of serrations - which are evident on all forks/hubs that i've seen for disk brakes. I'd be interested to see your conservative estimate that gets 5000N pull-out. here's my original post. http://groups.google.com/groups?selm...&output=gplain You maintain that fork legs are always soft and axles always embed in them. Possibly you have in mind that fork dropouts are always aluminium, but that's not true. Having maintained that the drop-outs are soft enough that embedment occurs, you then decide that they require 200 N/mm2 to shear (maintaining that this is a conservative figure), but assuming a Tresca yield criterion, this requires a material with a uniaxial yield stress in excess of 400 N/mm2, which is high for aluminium, and moderate for steel (well above mild steel). You go from a 94mm2 gross contact area, to a 25mm2 area of sheared material. I'm not sure how you'd predict the area of material sheared by a given embedment of a given serration pattern, but I'll try and work out how you came up with that. Suppose a 3mm seration spacing, implies 31mm length of contact in teh area. Given your 25mm2, that in turn suggests a shearing plane width of 0.8mm at each serration, which if we assume a 45 degree plane, suggests teh serrations must bite into teh dropout by about 0.6mm. Is that compatible with your assumptions? It seems excessive to me to assume that every time you tighten teh QR, you have to embed teh serations 0.6mm into teh material of teh fork dropout. Can you justify that? I think after a few weeks use, fork ends would be so chewed to bits you'd need new ones. regards, Ian SMith -- |\ /| no .sig |o o| |/ \| |
#158
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Ian Smith wrote:
You maintain that fork legs are always soft and axles always embed in them. Possibly you have in mind that fork dropouts are always aluminium, but that's not true. Having maintained that the drop-outs are soft enough that embedment occurs, you then decide that they require 200 N/mm2 to shear (maintaining that this is a conservative figure), but assuming a Tresca yield criterion, this requires a material with a uniaxial yield stress in excess of 400 N/mm2, which is high for aluminium, and moderate for steel (well above mild steel). You go from a 94mm2 gross contact area, to a 25mm2 area of sheared material. I'm not sure how you'd predict the area of material sheared by a given embedment of a given serration pattern, but I'll try and work out how you came up with that. Suppose a 3mm seration spacing, implies 31mm length of contact in teh area. Given your 25mm2, that in turn suggests a shearing plane width of 0.8mm at each serration, which if we assume a 45 degree plane, suggests teh serrations must bite into teh dropout by about 0.6mm. Is that compatible with your assumptions? It seems excessive to me to assume that every time you tighten teh QR, you have to embed teh serations 0.6mm into teh material of teh fork dropout. Can you justify that? I think after a few weeks use, fork ends would be so chewed to bits you'd need new ones. The word "teh" appears seven times. What does "teh" mean? -- Tom Sherman |
#159
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Ian Smith wrote:
On Wed, 29 Sep 2004 06:01:30 -0700, jim beam wrote: Ian Smith wrote: Not including effect of serrations, so just friction? no, _including_ effect of serrations - which are evident on all forks/hubs that i've seen for disk brakes. I'd be interested to see your conservative estimate that gets 5000N pull-out. here's my original post. http://groups.google.com/groups?selm...&output=gplain You maintain that fork legs are always soft and axles always embed in them. suspension forks, yes. Possibly you have in mind that fork dropouts are always aluminium, but that's not true. correct, they're frequently magnesium alloy. i never stated aluminum, so please don't make assumptions for me. Having maintained that the drop-outs are soft enough that embedment occurs, you then decide that they require 200 N/mm2 to shear (maintaining that this is a conservative figure), but assuming a Tresca yield criterion, Tresca is a fudge. use von Mises instead. this requires a material with a uniaxial yield stress in excess of 400 N/mm2, which is high for aluminium, and moderate for steel (well above mild steel). eh? you'll have to explain that leap of logic for me. You go from a 94mm2 gross contact area, to a 25mm2 area of sheared material. I'm not sure how you'd predict the area of material sheared by a given embedment of a given serration pattern, but I'll try and work out how you came up with that. Suppose a 3mm seration spacing, implies 31mm length of contact in teh area. Given your 25mm2, that in turn suggests a shearing plane width of 0.8mm at each serration, which if we assume a 45 degree plane, suggests teh serrations must bite into teh dropout by about 0.6mm. Is that compatible with your assumptions? no. i have no idea where you're getting your shear analysis from. this is not cupping of a tensile sample, it's simple planar shear. if you want to get into an analysis based on serration orientation, differential embedding depths, etc., be my guest, but don't make assumptions like your others here; get yourself access to a mtb fork and make some measurements. your comment below seems to indicate that you don't have direct personal experience. It seems excessive to me to assume that every time you tighten teh QR, you have to embed teh serations 0.6mm into teh material of teh fork dropout. Can you justify that? I think after a few weeks use, fork ends would be so chewed to bits you'd need new ones. that guess makes no allowance for serration patterns to be persistent, which is what we see in practice. once serrations are made, fork & axle mesh in the same place time after time. http://home.comcast.net/~carlfogel/d...d/Img_3199.jpg this is my fork. that pic was taken somewhere north of the 2000 mile mark. persistent serrations are clearly evident. and the wheel gets taken out of this fork each and every ride, so it's got to be roughly 100 insertions you see there. not exactly "chewed to bits" is it? regards, Ian SMith |
#160
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Tom Sherman wrote in message ...
Ian Smith wrote: It seems excessive to me to assume that every time you tighten teh QR, you have to embed teh serations 0.6mm into teh material of teh fork dropout. Can you justify that? I think after a few weeks use, fork ends would be so chewed to bits you'd need new ones. The word "teh" appears seven times. What does "teh" mean? http://www.winternet.com/~mikelr/flame31.html twinks |
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