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#1
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Maximum torque on the crank?
Could anyone tell me the maximum torque that is applied to the bottom
bracket spindle of a road bicycle? Assume worst case scenarios, i.e. Very strong rider, very steep hill, etc. I would prefer a value in foot-pounds. Thanks, Bob |
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#2
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Maximum torque on the crank?
"Earls61" wrote in message ups.com... Could anyone tell me the maximum torque that is applied to the bottom bracket spindle of a road bicycle? Assume worst case scenarios, i.e. Very strong rider, very steep hill, etc. I would prefer a value in foot-pounds. Thanks, Bob That is so simple that I maight take a shot at it. The left pedal is the only one that puts torque on the spindle. The right pedal cannot put torque on the spindle. The crank arm is usually right at 7 inches long. Very rarely is a rider going to put more than 200 pounds of force on a single pedal. However, if the rider weighs, say, 300 pounds and then bounces on the pedal while the crank arm is parallel to the ground, he might get 350 pounds on it. Thus 350*7/12 is about 200 foot pounds. And the more reasoonable rider does 200*7/12=117 foot pounds. It probably does not matter much how strong the rider is or how steep the hill is. Further, if the chain is in a relatively small chain ring, say one with a radius of 2 1/2 inches, the force on the chain from the above heavy rider is 200*12/2.5 which is about 960 pounds. |
#3
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Maximum torque on the crank?
On 6 Aug 2005 19:25:10 -0700, "Earls61"
wrote: Could anyone tell me the maximum torque that is applied to the bottom bracket spindle of a road bicycle? Assume worst case scenarios, i.e. Very strong rider, very steep hill, etc. I would prefer a value in foot-pounds. Max-torque on the spindle would probably be achieved when doing a "trackstand" with the cranks nearly horizontal, in which the rider's weight is applied about equally to both pedals. Given a crank of 175mm in length and a rider weight of 350 lbs (not common, but certainly within the realm of what exists), I get 200 ft/lbs of torque applied to the spindle. I doubt that this would be exceeded when pedalling in most cases, though it might on hard climbs if the rider is both standing and applying additional force by pulling up on the bars. The difference is probably small, however...and with the exception of one of the more heroically-sized denizens of this group, I would not expect to find a 350lb rider trying that tactic very often. -- Typoes are a feature, not a bug. Some gardening required to reply via email. Words processed in a facility that contains nuts. |
#4
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Maximum torque on the crank?
On 6 Aug 2005 19:25:10 -0700, "Earls61"
wrote: Could anyone tell me the maximum torque that is applied to the bottom bracket spindle of a road bicycle? Assume worst case scenarios, i.e. Very strong rider, very steep hill, etc. I would prefer a value in foot-pounds. Thanks, Bob Dear Bob, If a 200-lb rider stands on a horizontal 175mm crank and doesn't go anywhere because the hill is too steep for the gearing, then he's applying . . . Let's see, 175mm / (25.4 mm/inch) is about 6.9 inches. So you'd have 200 lbs on a lever 6.9 inches long, and we want to convert it to pounds on a longer 12-inch lever . . . 200 lbs * (6.9 inches / 12 inches/foot ) 200 * 0.575 = 115 lb-feet So it looks as if the motionless 200-lb rider balancing on one horizontal pedal is applying the equivalent of 115 lbs on a 12-inch lever, or 115 ft-lbs of torque. If he pulls down hard on the handlebars, he might raise his 200-lb effect on the pedal to say 300 pounds--which gives 172.5 ft-lbs of torque. Heavier or stronger riders could raise the figure. Basically, stand on one foot on a bathroom scale, then pull up on a stout railing with both hands--the weight that you reach is then applied to a 6.9 inch lever, so multiply it by 0.575 to get foot-lbs. Carl Fogel |
#5
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Maximum torque on the crank?
Alfred Ryder wrote:
Very rarely is a rider going to put more than 200 pounds of force on a single pedal. It is actually easy to put in excess of 200 lbs of force onto the pedals when sprinting... particularly if it is a steep uphill. In this case you are using the momentum of your weight and the power of your leg to produce a force that is much higher than what you weigh... and you don't even need to be pulling up on the bars to do it. If you have a bathroom scale, try a little experiment. Stand on it with one leg. Then let your body drop down a little and push on it hard like you would when sprinting. When you let your body drop the scale goes close to zero, then it maxes out when you push down. Or you could think about jumping up and down. Your *average* weight pushing down on the floor will be whatever you weigh... but obviously there is no weight pushing down when you are in the air, so it must be a lot more than your weight in some point in the cycle. I frankly don't know what the peak torque would be, but I'd guess it make sense to look at a force of something around 2-4 times body weight in a sprint. Then there is the issue of landing from jumps... I wonder, does the OP want to know the total stresses in the bottom bracket spindle or the torque only? Torque is the greatest factor, but there are other forces besides torque. -Ron |
#6
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Maximum torque on the crank?
Your track stand analogy fits in with Carl Fogel's "stationary rider"
analogy somewhat. I do them all the time, however, and I know I don't put any where near 350 ft.-lbs. of torque on the spindle. I think this is because this is one of the few things I am actually good at! |
#7
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Maximum torque on the crank?
Alfred wrote:
"The left pedal is the only one that puts torque on the spindle. The right pedal cannot put torque on the spindle" That is a good point. I didn't think about that. the pedal force is applied directly to the chanring through the chainring bolts. |
#8
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Maximum torque on the crank?
Thanks for the clear explanation, Carl. I would think a stationary
rider would put more torque on the spindle as you say, rather than Ron Ruff's sprinting rider mentioned in his post. But for some reason, I was thinking that the torque would be larger 170-180ft-lbs. Basically, stand on one foot on a bathroom scale, then pull up on a stout railing with both hands--the weight that you reach is then applied to a 6.9 inch lever, so multiply it by 0.575 to get foot-lbs. Later this morning, I am going to take the bathroom scale out to the garage and do exactly that. The workbench out there is bolted securely to the floor, so I can use that to pull on. Bob |
#9
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Maximum torque on the crank?
Earls61 wrote:
Thanks for the clear explanation, Carl. I would think a stationary rider would put more torque on the spindle as you say, rather than Ron Ruff's sprinting rider mentioned in his post. You'd certainly get the most torque at lower revs, that's basic physics. Sprinting riders aren't using their maximum strength because they're pedalling too fast, but riders overgeared on a steep hill will be much closer. However, a stationary rider isn't putting in the same *peak* force as a moving one. Look at it like this: you've been cutting down a tree in the garden and now you have a branch you want to break. To put the most force into it, do you stand on it gently rocking back and forth, or do you jump up in the air and stamp down on it? So what you want for your worst-case scenario is: a big heavy rider (250lb) with big long cranks (180mm) grinding away at very low speed (so he puts maximal force into each pedal stroke) flinging his considerable weight into each pedal stroke (which multiplies his effective weight by some unknown factor) and pulling around on the bars (ditto). Unfortunately we've got two unknowns in there so you can't really get a good answer, but it's easy to calculate the torque of a stationary rider and then you've just got to guesstimate how much gets added by throwing weight around and hauling on bars. Just out of interest, why do you want to know? |
#10
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Maximum torque on the crank?
On 6 Aug 2005 19:25:10 -0700, Earls61 wrote:
Could anyone tell me the maximum torque that is applied to the bottom bracket spindle of a road bicycle? Assume worst case scenarios, i.e. Very strong rider, very steep hill, etc. I would prefer a value in foot-pounds. Petacchi is worth to produce a max of 130 kg during his fast sprint at Milano-Sanremo. With a crank of 175mm it means an impressive torque of 130*0.175 = 22.75 kgm. Assuming 1 kg = 2.20 pounds and 1 m = 3.2 feet, this should result in: 22.75 * 2.20 * 3.2 = 160.16 pounds*feet. I suppose this is the maximum power a man could produce on a bottom brack! -- Massimo Bacilieri AKA Crononauta Ravenna, Italy. |
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