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#1 |
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Registered User
Join Date: Dec 2003
Location: Connecticut
Posts: 916
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anybody have any info on how much torque elite track sprinters do in standing starts(team sprint)?
i know they ride bigger chains and disk wheels so that the equipment won't break. but, i was doing some max torque efforts in my driveway today, and even at 1000 inch lbs, my road bike held up fine. (i am riding 32 spoke open pros, shimano 9speed chain) my chain has 10,000 miles on it. when i do standing starts in a big gear i only do half that torque.(FYI, my peak is about 400watts under Ryan Bayley's, so i am not weak) so what's all this i hear about track sprinters destroying road equipment? |
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#2 |
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Registered User
Join Date: Jul 2003
Posts: 76
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I don't know about the torque but if you really do have 10,000 miles on your chain you should replace it.
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#3 | |
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Registered User
Join Date: Apr 2004
Location: UK
Posts: 23
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Quote:
A couple of articles, one from the archives on here, see Ric Stern's comments...... http://www.cyclingforums.com/archiv...hp/t-74232.html & another from John Moss.... http://worldofendurance.com/article.asp?a_id=1013851 So somewhere between 1800 to 2400 watts ![]() |
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#4 |
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Registered User
Join Date: Jun 2004
Location: Jyvaskyla, Finland
Posts: 665
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Wavydavy, one of the links you gave was a thread started by velomanct. Nice going.
Velomanct is asking about the peak torque generated by top level track sprinters, and torque is a measure of the twisting or rotational force about a point. In this case the point is the crank of a racing bicycle. Velomanct, I don't know the answer to your question, but your peak torque numbers sound really lowm, in fact I'm sure you're much higher than what you say. Estimating from the fact that I can partial deadlift 675 lbs at a rate of about 5 cm per second, I would estimate I can briefly generate over 5000 inch lbs (420ft lbs) of torque at a foot speed of a few cm per sec. Doesn't matter though as my torque drops off very rapidly at higher foot speeds and I get tired very quickly at max power. I've improved my endurance quite a bit though with hard 15-20 km efforts and my sprinting speed has improved without doing much sprinting. I myself am not interested in peak torques, as they typically peak at very low speeds. More important is your ability to generate torque at a much higher foot velocity like at 100-140 rpm. In fact I would probably say that my back and leg strength (and grip!!) is far better than even elite track sprinters, it's the power production at higher velocities that eludes me. As for destroying equipment, I used to ride in the highest gear my old mountainbike had, a 48 x 12 all the time. I broke the crank twice within 6 months and the same with the bike I had before that. The young woman in the bike store said I was riding my bike wrong. Probably. |
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#5 | |
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Registered User
Join Date: Dec 2003
Location: Connecticut
Posts: 916
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Quote:
i am using a powertap to measure my watts and torque. it is a lot easier to put out high amounts of torque when lifting weights, compared to pushing crank arms on a bike. i am interested in torque, because that is what breaks equipment, not power. i would estimate, based on my own performance that elite track sprinters hit around 800 inch lbs of torque at the start on the team sprint. i suppose its possible for them to peak over 1000 inch lbs, which would explain why they break road equipment. you are right, peak torques don't really mean a lot concerning performance, watts are what matter. bikeguy, it sounds you are a really strong rider. you should work on your speed and i bet you could do a really good sprint. i am just getting into wieght training and i hope i can improve my sprint a lot. (many people call me skinny and underweight for height) you should try to get yourself tested, using a powermeter to see what kind of watts you can produce in a sprint. from reading other posts, it sounds like you are already a fast sprinter. |
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#6 |
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Registered User
Join Date: Jun 2004
Location: Jyvaskyla, Finland
Posts: 665
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Velomanct,
I just realised that the foot position on the bike doesn't lend itself very well to maximal force transfer as the heel is floating in the air. All I know is I can make my bike creak on a hill if I sprint with the handlebars bending as I pull up on them hard to generate added force on the pedals. I've been considering a wingate test, but I'm out of commission for at least a couple of weeks because I got pericarditis. I'm still getting getting pain in the chest occassionally so there will be no cycling of any kind for quite a while let alone a maximal wingate. |
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#7 |
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Registered User
Join Date: May 2003
Posts: 25
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Here's a good estimate based on my standing start performance. I can do a standing 100m in a 48x14 in around 9.5 seconds. Here's the quick way to calculate torque based on my numbers using the laws of physics:
Newton's 2nd law states Force = mass * acceleration or F=ma. (* is multiplication) My average acceleration for the 100m start can be obtained by using the equation x=1/2a*t^2. (x is distance in meters and t^2 is time in seconds "squared".) x=100m, t=9.5s, so a=2.21m/s/s My mass with bicycle is around 90kg. Using these numbers I can solve for the force. This will be force at the tire/track contact point. F=90kg*2.21m/s/s F=200N To convert force to torque, we need to know the radius of the tire. It's around 340mm (0.340m) if I recall correctly. Torque(T) is T=F*r where r is the tire radius. That means my axle torque is: T=200N * 0.340m = 68Nm (or around 50ftlb.) This is NOT at the crank though. At the crank there is more torque by a factor of the gear ratio. The ratio is 48/14=3.43. So, at the crank I'm making on average over a 100m standing start: 68Nm * 3.43 = 233Nm torque. Humans do not produce constant torque around the pedal stroke. Plotted out it would look much like a series of humps with flat spots of near-zero torque when the cranks are near the top and bottom of the stroke. This implies that the 233Nm average is a consequence of peaks rather larger than that, averaged together with flat spots of near-zero torque. Using a common RMS(root mean square) estimation, I'm making peak torque of 233Nm * 1.41 = 329Nm (or 244lbft) of torque at the cranks. If this sounds crazy, think about it this way... the pedal force would me a max of only 431lbs. Just standing on the crank I'm getting 180lbs (my bodyweight) into it. That means that the combination of my 400lb-squat-capable front leg and my rear leg pulling upwards while accelerating is making an additional 251lbs of force. The peak axle torque by the same argument would be around 50*1.41=70.5ftlbs. This being averaged over 100m, I'm not sure how much the best stroke among those strokes would have attained. I'd think it'd be fairly constant once underway, but perhaps a bit higher at the first few strokes. Last comment...promise. This all is a slight underestimate also since it neglects wind resistance (my time in the very first calculation would have been faster without wind). I figure it's within 10% though. A standing 50m time would be better, but I don't know mine. Hope this helps. Last edited by Budarz : 29-11.-2004 at 02:48 PM. |
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#8 |
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Registered User
Join Date: Aug 2004
Posts: 40
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Budarz, you're forgetting energy expended to rolling resistance and air drag. Factor those in and you should be spot on.
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#9 |
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Registered User
Join Date: May 2003
Posts: 25
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I realize this. That's why my last paragraph reads:
"Last comment...promise. This all is a slight underestimate also since it neglects wind resistance (my time in the very first calculation would have been faster without wind). I figure it's within 10% though." True I should have written rolling resistance in there too. I have written complete cycling simulations that predict my kilo times every half lap to within 0.1s of the real splits by including max force develpment, max power output, an analytic function which accounts for diminished power output over time, air drag, rolling resistance, etc. I was just intending to give velomanct a little help. |
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