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x-post: Bike Biz: Wheel ejection theory goes legal
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#273
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x-post: Bike Biz: Wheel ejection theory goes legal
On 2007-02-12, Tim McNamara wrote:
In article , Ben C wrote: [...] I think front mounted calipers is overkill, and has the drawback of putting the mountings in tensile fatigue as you've explained. And yet it is done successfully on other vehicles, so that is a surmountable problem and a straw man raised by jim. It was also interesting to read the reaction of Ben Micklem's frame builder to the suggestion of front-mounting the caliper. They didn't think it sounded like a good idea. But moving the dropout angle forwards and the caliper upwards a bit, as some designs already do-- wouldn't that solve the theoretical and/or real problems? It could. Look at motorcycle disk brakes, which often place the caliper up tight against the fork leg. The caliper ends up nearly at the top of the disk. The vector of the reaction force from braking would be in a much more benign direction. As I said in another thread, if the difference in direction is 45 degrees or better, I don't think you're going to get ejection. Unless my calculations are still wrong (it's been known...) Ben Micklem has a 54 degree difference with a 2:30 caliper and 20 degree forwards dropout. Should be perfectly safe. Placing the caliper in from of the fork would result in the reaction force driving the axle into the dropout and eliminating the ejection force altogether. Of course true, but although not impossible, undesirable. The 2007 range of "Ben C" mountain bikes will use rear calipers mounted at 2:30 and slightly forward-opening dropouts. And no tapering on the forks. |
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x-post: Bike Biz: Wheel ejection theory goes legal
A Muzi wrote:
p.s. - Jobst makes a good point about hub brakes (disc, drum whatever) on undersized/lightweight fork blades or stays. The occasional tinkerer discovers this with great surprise. Imagine what would happen if someone Chalo-sized had to do hard braking on this bike: Skinny fork, brake calipers about 90 degrees from the dropout opening http://www.hampsten.com/Tournesol/po...teur05_pop.htm No lawyer lips that I can see, but maybe they are really small http://www.hampsten.com/Tournesol/po...teur04_pop.htm Hole midway up the fork on the brake side http://www.hampsten.com/Tournesol/po...teur02_pop.htm |
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x-post: Bike Biz: Wheel ejection theory goes legal
On Mon, 12 Feb 2007 23:27:33 -0600, Tim McNamara wrote:
Other pictures on other pages of that site work correctly on my browser, just not that page. Odd. Do a "view image", which will get the small picture with a URL like http://materials.open.ac.uk/mem/imag...s_cc/ccf7r.jpg then take the "r" out of the name of the jpeg. http://materials.open.ac.uk/mem/imag...es_cc/ccf7.jpg Something is borked in their javascript, which is completely unnecessary for a simple link to a picture. Mike |
#276
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x-post: Bike Biz: Wheel ejection theory goes legal
jim beam wrote:
but the retention force exceeds payload by at least 3 times worst case scenario. Worst case? -- Dave dvt at psu dot edu |
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x-post: Bike Biz: Wheel ejection theory goes legal
On Mon, 12 Feb 2007 16:23:21 -0600, Ben C wrote:
Here is my working in the form of a GNU Octave script if anyone has the energy to check it. I haven't gone through it thoroughly (downloading Octave now), but there are a couple of assumptions that I think are incorrect: % Its magnitude is scaled by difference in radius between disk and tyre, % and multiplied again by 2 since the disk is only on one side of the axle. This factor is only relevant when the weight of bike + rider is included, which isn't part of this calculation. Instead of doubling the braking load I think you should halve the gravitational load (when it's used), and if you will grant some stiffness in the hub and axle that transfers part of the braking load to the other side, reduce the braking load by some percentage. % Add the braking force and the contact force (friction coefficient is 1.0) % to give the net force f on the wheel under braking. Once the rear wheel has lifted a greater friction coefficient has no effect, and I think you'll find that the actual value required for maximum braking is between 0.6 and 0.65 (Wilson gives 0.56 but I think he is assuming a crouched rider for his CoG location). I have a program that allows some experiment with friction and CoG position, but it's pretty crude and written in Awk, so I might rewrite it in Octave as a way of learning something about Octave. If you want to see the Awk version send me an email. Mike |
#278
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x-post: Bike Biz: Wheel ejection theory goes legal
On Feb 12, 4:06 pm, Tim McNamara wrote:
In article .com, "Ed Pirrero" wrote: On Feb 11, 8:55 pm, "G.T." wrote: Ed Pirrero wrote: On Feb 11, 8:37 am, jim beam wrote: Tim McNamara wrote: snip underinformed opinion 1. there are no reported accidents that can be definitely distinguished from user error. Exactly. Tim like to throw bombs around about these reported incidents, but never once has anyone proven that it was actually the forces in question vs. user error. So because it hasn't been proven yet you guys are 100% certain that it's been user error? Strawman. No, it looks like just a poorly formed question. "A straw man argument is a logical fallacy based on misrepresentation of an opponent's position. To "set up a straw man" or "set up a straw-man argument" is to create a position that is easy to refute, then attribute that position to the opponent. A straw-man argument can be a successful rhetorical technique (that is, it may succeed in persuading people) but it is in fact a misleading fallacy, because the opponent's actual argument has not been refuted." Given the question mark at the end of Greg's post, I think it is reasonable to give him the benefit of the doubt. Were it a sentence, then I would agree that it is a straw man. However, you do appear to simply disregard any evidence that contradicts your theory. That may not be an accurate assessment of your thinking process but it is the impression I get from reading your posts. Your logic- and jim's- looks like this to me: A. It has been postulated that disk brake can cause wheel ejection. B. Uncertainty can be cast upon the evidence that A is true. C. Therefore A is false. That may or may not be what you intend. On my screen, that's how it reads. A reasonable reply! I am surprised, and pleased. OK, here's how it goes: A. Yes, I agree. B. Sort of - I think that because there is a force, that the possibility exists that a wheel could be ejected. C. No, not true at all. I'm still an agnostic on whether this "problem" is real, or just theoretical. Because, as Frank writes, so many conditions have to line up right for ejection to occur, it gets into the realm of doing a fix on something that's not broken. As I have said, to you and to others, I am willing to look at real data, and evaluate it for what it is. I don't dismiss data out of hand, I dismiss conclusions based on opinion rather than hard data. While to you the difference may be subtle, for me it makes all the difference in the world. Unfortunately, I don't consider the anecdotes "data", because the initial conditions are unknown. It is *possible* that the initial conditions are normal, but it is possible that they are not. Not exactly a dismissal, but not exactly full acceptance as data points. I consider any attempt to put me in the position of defending some absolute denial as strawman logical fallacy, for all of these reasons. I hope this clears things up. E.P. |
#279
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x-post: Bike Biz: Wheel ejection theory goes legal
On Feb 12, 3:49 pm, "G.T." wrote:
"Ed Pirrero" wrote in message ups.com... On Feb 12, 3:02 pm, "G.T." wrote: "Ed Pirrero" wrote in message groups.com... On Feb 12, 2:27 pm, "G.T." wrote: "Ed Pirrero" wrote in message roups.com... On Feb 11, 7:54 pm, Gary Young wrote: This is a variant of the my-uncle-was-a-smoker-and-he-lived-until-95 argument. Except for the small details that smoking will most definitely cause some harm, and, so far, disk brakes have caused none due to the ejection force being present. None? You're sure about that? Greg The answer to both questions is in the part you trimmed. "(Qualifier: if some harm has occurred, it certainly hasn't been distinguished from user error.)" So now you're omniscient? Strawman. If you've got any, and I mean ANY, credible data that any of the incidents involving wheel ejection have been proven as disk-brake caused, go ahead and cite it. It's sad that you and jb are such untrusting fools. Logical fallacy - ad hominem. E.P. |
#280
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x-post: Bike Biz: Wheel ejection theory goes legal
On 2007-02-13, Mike Causer wrote:
On Mon, 12 Feb 2007 16:23:21 -0600, Ben C wrote: Here is my working in the form of a GNU Octave script if anyone has the energy to check it. I haven't gone through it thoroughly (downloading Octave now), but there are a couple of assumptions that I think are incorrect: % Its magnitude is scaled by difference in radius between disk and tyre, % and multiplied again by 2 since the disk is only on one side of the axle. This factor is only relevant when the weight of bike + rider is included, which isn't part of this calculation. Yes, exactly right, and I realized this and pointed it out in an earlier post. You need to get rid of the factor of two. Instead of doubling the braking load I think you should halve the gravitational load (when it's used), and if you will grant some stiffness in the hub and axle that transfers part of the braking load to the other side, reduce the braking load by some percentage. I'm just considering the forces on the wheel: road force (horizontally backwards), ground reaction force (vertically upwards, same magnitude-- friction coefficient of 1.0), and caliper force. Add those three up and you get the net force on the wheel. Then you can see if its direction is within 45 degrees of the way out. The caliper force is computed so as to give zero net torque on the wheel. Now I'm thinking there's another error there. Ground reaction force should not be included, as it is opposed by the weight of the rider/bike pushing down on the wheel. This means I'm right back to my original version of the calcuation (no ground reaction force or factor of two) and I get 27 degrees from vertical for the caliper at 2:30 (which is 47 deg from the dropout if the dropout is 20 deg forwards), and 13 deg from vertical for caliper at 3:00 (or 14 deg if the disk/rim ratio is exactly 4 instead of 4.2). % Add the braking force and the contact force (friction coefficient is 1.0) % to give the net force f on the wheel under braking. Once the rear wheel has lifted a greater friction coefficient has no effect, and I think you'll find that the actual value required for maximum braking is between 0.6 and 0.65 (Wilson gives 0.56 but I think he is assuming a crouched rider for his CoG location). 1.0 was meant to be sort of "worst case", although actually the friction coefficient can get slightly higher than 1.0. I have a program that allows some experiment with friction and CoG position, but it's pretty crude and written in Awk, so I might rewrite it in Octave as a way of learning something about Octave. If you want to see the Awk version send me an email. I might do that. Thanks for looking through my version. |
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