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Maximum Human Torque

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Old 07-12-06 | 07:58 PM
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Originally Posted by Ideologue
I read that the maximum amount of torque momentarily produced by a professional (racing) cyclist is in the region of 1,000 Inch Pounds. That is 83.3r Foot Pounds, or 112.984 829 333 Newton Meters.

A 250cc motorcycle I used to ride only produced around 20 Newton Meters, so is the 112+Nm figure really an accurate one for the maximum torque a human can produce on a bicycle? It seems rather high to me. But then considering a rider with a mass of around 100kg standing up on the pedals and allowing his entire weight to push down on 175mm cranks (with a turning circle of 1099.6mm). This rider without even trying, and merely allowing gravity to pull his mass down and turning the cranks in the process, will exert approximately 539.35Nm of force through the bottom bracket spindle.

What do you make of this?
Well, it's really pretty simple. Your motorcycle produced 20 NM, but it did it at 7500 RPM, while the human produced it at 75 RPM. To get 75 RPM out of the motorcycle, you'd need to gear it down 100:1, and that would multiply the torque by 100, giving you 2000 NM.

That's why higher RPMs are sought after in engine design. If you can double the RPM range of an engine without lowering the torque (you can't, but bear with me..., then you could gear the engine down 2:1 and get double the torque.
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Old 07-12-06 | 10:22 PM
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Hmm, I really need to study this subject some more. Can anyone recommend a decent tutorial site on torque forces, power, etc.?

The interesting question is: Cog and sprocket design? I thought that these components are rated for torque, not power. I wonder what the maximum torque rating is for various bicycle sprocket sizes?
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Old 07-12-06 | 10:41 PM
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Originally Posted by Ideologue
Mine was a 4-stroke (KLR 250), and it certainly did like to rev a lot (4,000 RPM = 40mph). Its maximum power output was around 21 - 27hp.
Same as my riding lawn mower!
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Old 07-13-06 | 01:21 AM
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Originally Posted by Ideologue
Hmm, I really need to study this subject some more. Can anyone recommend a decent tutorial site on torque forces, power, etc.?
One thing to remember is your basic calculus. Power is torque integrated with respect to time. Which goes back to the gearing and RPM ideas. If you keep torque constant, but spin it twice as fast, you'll get twice the power. That's why the fastest sprinters spin such high-RPMs; there's a physiological limit to how much force & torque the human body can exert on the cranks. But if you spin it faster, you'll get more power & speed. Here's some auto sites that discuss torque & HP:
https://www.v8914.com/Horsepower-v-torque.htm
https://www.epi-eng.com/BAS-PwrTrq.htm
https://www.team-integra.net/sections...?ArticleID=467
https://www.howstuffworks.com/question381.htm

Originally Posted by Ideologue
The interesting question is: Cog and sprocket design? I thought that these components are rated for torque, not power. I wonder what the maximum torque rating is for various bicycle sprocket sizes?
It's not so much the torque, but the linear force in shear at the teeth that'll break them off. Calculate cross-sectional area of teeth at the contact point and figure breaking force at yield-strength. Divide by number of teeth and divide torque through radius of gears and see which is greater. Power is just how fast you spin the gear, so it won't hurt them that much to spin it faster... until you get into 200,000rpms+ where inertial forces starts coming into play.
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Old 07-13-06 | 02:56 AM
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Note the relation between torque and crank length.
Does higher torque at the BB require a stiffer frame/BB assembly?
Should big frames for big riders/long cranks be much stiffer than small frames for small riders /short cranks?
Are small bikes made too stiff and big bikes too floppy ?
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Old 07-13-06 | 03:34 AM
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Originally Posted by Mothra
It's not so much the torque, but the linear force in shear at the teeth that'll break them off. Calculate cross-sectional area of teeth at the contact point and figure breaking force at yield-strength. Divide by number of teeth and divide torque through radius of gears and see which is greater. Power is just how fast you spin the gear, so it won't hurt them that much to spin it faster... until you get into 200,000rpms+ where inertial forces starts coming into play.
I would think that friction heating would become a problem fairly early on at high power levels.
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Old 07-13-06 | 11:11 PM
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Originally Posted by Mothra
One thing to remember is your basic calculus. Power is torque integrated with respect to time. Which goes back to the gearing and RPM ideas. If you keep torque constant, but spin it twice as fast, you'll get twice the power. That's why the fastest sprinters spin such high-RPMs; there's a physiological limit to how much force & torque the human body can exert on the cranks. But if you spin it faster, you'll get more power & speed. Here's some auto sites that discuss torque & HP:
https://www.v8914.com/Horsepower-v-torque.htm
https://www.epi-eng.com/BAS-PwrTrq.htm
https://www.team-integra.net/sections...?ArticleID=467
https://www.howstuffworks.com/question381.htm

It's not so much the torque, but the linear force in shear at the teeth that'll break them off. Calculate cross-sectional area of teeth at the contact point and figure breaking force at yield-strength. Divide by number of teeth and divide torque through radius of gears and see which is greater. Power is just how fast you spin the gear, so it won't hurt them that much to spin it faster... until you get into 200,000rpms+ where inertial forces starts coming into play.

Thanks for your post and the links, they look pretty good to me!
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Old 07-13-06 | 11:29 PM
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Originally Posted by MichaelW
Note the relation between torque and crank length.
I am yet to study the above links thoroughly but it seems to me, and using the equation posted early on in this very thread, that the longer the crank arms the more leverage, ergo more torque. But the cadence would not be so high, unless the rider was very strong (assuming a single gear) with very long legs, so the rider’s knees might suffer after a while! The ride would be more anaerobic than aerobic too, and perhaps a more stressful experience!
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Old 07-14-06 | 09:33 AM
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Originally Posted by Mothra
One of the reasons you can exert more force than your body-weight is that you'll pulling up on one side while pushing down on the other. The back-muscles is also used when pulling up on the bars to counteract the upward movement of the body. So you won't automaticaly launch yourself off the bike if you apply more than your body-weight. Of course, it takes years, decades of training in order to get the brain/muscle-coordination to spin smoothly while sprinting.

The other thing you have to consider is the gearing. It's really torque at the rear wheels that makes the bike move. Actually thrust at the contact patch is what counts. So take torque at the crank, run it through the gear-ratios, divide out radius of rear-wheel and you get linear thrust at the contact patch. You'll notice that the bike's gearing divides (reduces) the torque, while a motorcycle's gearbox multiplies (increases) the torque from the crank.

Finally, it's power that's the real determination of speed. Which is torque X RPM, so the faster you spin, the more power you make. A motorcycle's torque multiplied by 9000rpms is gonna generate a heck of a lot more power than a human at 90rpms. You can take McEwen's 1700-watt output and work backwards through the gearing & RPMs to arrive at his torque@crank and realise the guy's a monster!
+1

My guess is that the torque numbers from the pros isn't all that much higher(and may be less) than that of the general public. But, since they can maintain high rpms all day, they have much better total power. What I think would be a real interesting experiment would be to see how a racer would perform if they carried/pulled a significant load vs someone who rides loaded as a matter of routine, or a real in-shape clydesdale(250 lb with low bodyfat). My guess(and again, it's a guess) is that as the pulled load increases, the clyde would perform better in comparison, as maximum torque becomes more important than maximum power.
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Old 07-14-06 | 10:59 AM
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It's power that's doing the work. Torque is simply a substitue for gearing. With proper gearing, power always wins over torque.
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Old 07-14-06 | 03:45 PM
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Power & torque are directly related through time & RPM. It's like speed vs. acceleration, there's an extra integration step with respect to time. When it comes down to it, at any given speed, 30, 60, 80kph, there's a certain amount of power required. In order to generate that power, you can use high-torque/low-RPM or low-torque/high-RPM.

Problem is you max out how much torque a human can generate since the leg muscles can only exert so much force on the crank. Then in order to get more power and more speed at 100% leg-exertion, the only thing you can do is increase RPMs once you're pushing as hard as you can. So as john bono said, the pros aren't that much stronger, but they're more coordinated and smoother so they can exert higher average torque through a wider percentage of the 360-degree pedal-stroke than us mere mortals....and do it at higher RPMs... resulting in more power & speed.

Last edited by Mothra; 07-15-06 at 08:36 PM.
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Old 07-14-06 | 07:58 PM
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Originally Posted by CdCf
It's power that's doing the work. Torque is simply a substitue for gearing. With proper gearing, power always wins over torque.

Power amplifies torque.
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Old 07-15-06 | 12:31 AM
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Originally Posted by Ideologue
Power amplifies torque.
It's gearing that amplifies torque. Power is just a measurement of how quickly you are generating that torque. Imagine torque as single punches in a bar fight, each one has a certain "hit". The brawler who generates more power is the one that can land 5-blows/sec vs. 3-blows/sec for the other guy. So even though each punch is exactly the same force, the one landing more blows per second generates more power.

So it's actually the torque that is measured. If you were to measure anything on the bike, you'd measure torque at the crank or at teh rear wheel. Then combine that with a timer to measure RPMs and power is then calculated. Power is basically torque per unit time.
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Old 07-15-06 | 02:38 PM
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That is what I meant. Poor semantics, perhaps!
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Old 07-17-06 | 03:45 PM
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An interesting comparison of torque vs. HP comes in looking at cars while examining the weight, HP and torque. Check out performance figures between the '86 Ferrari Testarossa vs. the '86 Ferrari 288GTO... and compare to their weight, HP and torque.
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Old 07-17-06 | 04:01 PM
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The maximum moment on one crank is more than body weight. The force on that crank can be increased significantly by pulling up on the handle bar. Meanwhile the other leg can be pulling up on the other crank. I suspect that the true maximum can be somewhere in the area of twice body weight.
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Old 07-17-06 | 04:40 PM
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Originally Posted by CTAC
His *total* power output could be 650W, but only 240W are used for *wheels motion*. He's not 100% efficient. Do you agree?
I'm no engineer, so I probably don't belong here, but wouldn't you need to find the point in the arc where torque is maximum? Couldn't that just be a hundredth of a second, or less? The original post was about MAXIMUM torque, wasn't it?

Bike designers and manufacturers have had engineers looking at the same principles and came up with something very similar to what we're all using now, which came along in the 50's.
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Old 07-02-13 | 11:04 AM
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Originally Posted by ericgu
Well, it's really pretty simple. Your motorcycle produced 20 NM, but it did it at 7500 RPM, while the human produced it at 75 RPM. To get 75 RPM out of the motorcycle, you'd need to gear it down 100:1, and that would multiply the torque by 100, giving you 2000 NM.

That's why higher RPMs are sought after in engine design. If you can double the RPM range of an engine without lowering the torque (you can't, but bear with me..., then you could gear the engine down 2:1 and get double the torque.
I think Eric's got it right, on an average a human produces around 30-40 N-m of torque at 80-90 rpm at the crank, if you have 46 teeth on your crank and 26 on your wheel, you have a ratio of 0.59, and therefore the torque at the wheel is 30x0.59=17.7 N-m, and 80/0.59 = 135 rpm at the wheel. And this the first gear of your mountain bike, as you shift down the ratio changes, 0.41, 0.33, 0.24 etc.
The scooter engine on the other hand produces 20 N-m at lets say 1000 rpm, therefore you'll need a different gear ratio, something like 2-3, to see an rpm of 300-500 at the wheel, the torque in which case becomes 20x2=40 N-m or 20x3=60 N-m at the wheel.
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Old 07-02-13 | 04:24 PM
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Originally Posted by MichaelW
Note the relation between torque and crank length.
Does higher torque at the BB require a stiffer frame/BB assembly?
Should big frames for big riders/long cranks be much stiffer than small frames for small riders /short cranks?
Are small bikes made too stiff and big bikes too floppy ?
I will have to go looking for it, and maybe this will jog someone's memory, since I think I found the link on this forum originally, but there is a research paper that compared power, torque, and RPMs for something like 30 riders, and the numbers aren't as straightforward as one might think. The basic takeaway was: find the crank length and RPM that is most comfortable for you.
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Old 07-02-13 | 05:07 PM
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Originally Posted by operator
Sprinters are putting out what.. 2000w in the TDF sprints? Something around that neighbourhood?
No. The top track sprinters can put out around 2100w for a few seconds. But they are too heavily muscled to be remotely competitive in a road race. Cavendish was an endurance rider on the track, not a sprinter. He's a little guy who generates his speed by a very aero position and great technique, his estimated power in a sprint at the end of a stage is only about 1200w. Greipel is more powerful, estimates have him at maybe 1400-1500w.
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Old 07-02-13 | 05:20 PM
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You can't compare pedaling a bicycle to doing 400 lb leg presses, since you're not pushing against a padded steel frame. The only reaction force that allows you to push against the cranks is the force produced by your arms pulling against the handlebars plus your body weight. Most do not have the upper body strength to pull up with that much force.
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Old 07-02-13 | 05:55 PM
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Y'all do know that this 'debate' had been slumbering peacefully for nearly 7 years, right?
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Old 07-02-13 | 06:05 PM
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Originally Posted by no1mad
Y'all do know that this 'debate' had been slumbering peacefully for nearly 7 years, right?
Nonsense! A good torque debate never goes out of style. Has there been some new technological advance in that time that has nullified the amount of torque applied to the cranks?
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Old 07-02-13 | 06:22 PM
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Hi,

Static torque is a force like tightening a bolt. However is rotating systems
it is the equivalent static force that is transferring the power in each part.

A person on a bike can produce the most torque at the crank by standing
on the pedals, and the most torque at the rear wheel in the lowest gear
when not moving, an engine clearly can't do that at all.

If you compare a motorcycle torque and power output curve at various
rpm's to a human they couldn't be more different. As an engine goes
faster its gets more powerful and typically has a fairly flat torque curve,
though it always rises quite a lot from low revs to sensible revs.

A humans torque curve drops off the faster you pedal, starting at maximum.
As rpm increases it drops off slowly so power output increases, but you then
hit a region where power output remains ~ constant over a range of cadences,
the torque curve must be downtilted in this region, i.e. you can't pedal fast hard.

rgds, sreten.
.
Theoretically any person can produce any arbitrary torque with a big enough lever.

Anyone who is 140 lbs with 7" cranks sets off with 1000 lbs inches of torque.

Producing that torque at speed in an entirely different matter.

Last edited by sreten; 07-02-13 at 06:37 PM.
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Old 07-05-13 | 08:19 AM
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This is an interesting thread. Some posters touched on what I question. On a DF if a rider is standing and his or her full weight is on one pedal, some of the numbers quoted seem low. Now if a 200 pound rider is pushing down his weight, plus the force added by pulling up on the handle bars for maybe 90 degrees the torque would seem to me would be around 250 foot pounds. That would be repeated with the other leg, for a total of 180 degrees. Then divide that by 2, because of a full 360 degree circle. Then drop that by the gearing of say 42 by 15 and you would get the torque actually applied to the ground. So to me you would have 125 times .3 would equal 37.5 foot pound of torque. Those are approx numbers for a 200 pound person and would be reduced by less weight and less pulling up on the handle bars. That would seem to me to be the max, but again that would be reduced by mechanical efficiencies. It also occurs that that would be the torque on the cassette. It would again have to be reduced by the relationship of the sprocket to the wheel depending on the size. I believe that is why my Tailwind bent would take off like a scalded rabbit with its 20 inch rear wheel. Comments?

Last edited by rydabent; 07-07-13 at 07:41 AM.
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