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PostPosted: Sat Mar 17, 2012 12:32 am 
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RollingGoat wrote:
....


I have witten the equilibrium equations and I'm right. You can see the calculations here:

http://cds-0.blogspot.fr/2012/03/rim-an ... tical.html

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Posted: Sat Mar 17, 2012 12:32 am 


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PostPosted: Sat Mar 17, 2012 3:07 am 
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You're missing the component of the braking force perpendicular to the seatstay for the disk brake.

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PostPosted: Sat Mar 17, 2012 11:14 am 
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The angle between the speed direction of the disc in the point of contact with the pads and the seatstays is very small. Anyway, I have made the calculation for that case too and the moment M is Mbrake(L1+L2(cos(alpha)-1))/(L1+L2) with alpha the angle between the speed direction of the disc and the seatstays, so even smaller than for the case with the disc force acting parallel to the seatstays.

Are you convinced now?

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PostPosted: Sat Mar 17, 2012 11:17 am 
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Guys no offence but can you hold this discussion with PM's, please?

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PostPosted: Sat Mar 17, 2012 4:33 pm 
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No, because a) you're missing reaction forces for both cases and b) the speed direction of the disk is very nearly perpendicular to most seatstays, and is usually completely perpendicular for the chainstays.
There's a professor at my school who's written a few papers on bicycle dynamics. I'll email him and see what he thinks.

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PostPosted: Sat Mar 24, 2012 12:26 pm 
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not this year.

http://velonews.competitor.com/2012/03/ ... 013_210440


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PostPosted: Sat Mar 24, 2012 1:16 pm 
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This is actually a very, very good thing. 2012 is already 25% over now. Nobody even expected the UCI to consider discs this year. However, the article states that the UCI equipment commission will review disc brake usage for road competition this summer for possible approval as early as 2013!

:D


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PostPosted: Sat Mar 24, 2012 8:43 pm 
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My apologies if this has already been covered, but I was thinking about this during today's ride:

We've discussed how disc brakes shift the braking force from the rim to the hub and how rims may get heavier/need more spokes. But how does this force compare to the power put through the hub during a sprint? A pro sprinter will pump out some serious power through the hub which is then carried through the spokes to the rim. In a sense, this is the opposite of slowing with a disc brake. How do these forces compare in magnitude?

I have no engineering background, but would love to hear what you guys think.


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PostPosted: Sat Mar 24, 2012 9:09 pm 
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Braking far more powerful than any sprint.

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PostPosted: Sat Mar 24, 2012 9:13 pm 
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First, pedaling force comes through rear wheel which currently has more spokes to share the load than the front one.
Second, pedaling force is several times less than what can come from braking. Just compare what time/distance you'll have to travel to speed yourself up from 0 to say 40 km/h with how fast you can slow down from 40 km/h to full stop by braking.

edit: it's obvious that when you speed up you work against the air and when you slow down it helps you, so it's not a perfect example, but it's still holds true since the difference will be far larger than what can be accounted to aerodynamics...


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PostPosted: Sat Mar 24, 2012 9:23 pm 
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Emergency braking of a 90kg rider is 100Nm of torque in the disc fixing points
Chris Hoy sprinting with a 52/13 ratio puts 150Nm of torque in the freehub pawls

For a strong amateur cyclist, the sprinting torque in the freehub pawls would be 70-80Nm

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PostPosted: Sat Mar 24, 2012 10:31 pm 
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Cool--thanks for the details. Out of curiosity, where did you find these numbers?

So it's possible that rear wheel lacing won't have to change considerably. Front wheels will definitely need some beefing up, but maybe only to a level found on rear wheels today.


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PostPosted: Sat Mar 24, 2012 11:37 pm 
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arbitrage wrote:
Cool--thanks for the details. Out of curiosity, where did you find these numbers?


Own calculations

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PostPosted: Sat Apr 07, 2012 4:29 pm 
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Image

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Posted: Sat Apr 07, 2012 4:29 pm 


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PostPosted: Sat Apr 07, 2012 6:24 pm 
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OMG I want one of those :shock:

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