Is there any formula for figuring out how much torque is needed to spin a crank?
 

A formula where you would plug in your weight, the gears that you are in, the speed your are traveling at, and maybe the incline of the hill you want to go up.

So I want to just compare how much torque is needed to spin a bike up a 5 degree incline at 5mph in the gear selection of say a 26t rear cog and a 42t front cog with a 200 pound rider. Maybe put in the weight of the bike as well, not sure if that would matter all that much though.

I think its as simple as the amount of force needed to lift the total mass up the slope via newtonian physics, then just multiply by your gear ratio and by the wheel radius divided by crank arm length. and fudge everything for friction

Total weight x cos (incline angle) x gravitational acceleration (32ft/s/s) gives you the force down the slope.
Force down the slope has to be opposed by your force which is the torque like "pierce" says.
Speed does not enter the maths until you account for wind and friction, so if you like you can ignore those.
Thus you have the force x the distance = the work done to get up the hill. The work done is the same whether you go slow or fast if there's not much wind.
Power = force x velocity. So although the work done is the same, you will require more power to go up the hill faster.
If you are only interested in the torque, forget all this talk about work and power.:)

Wouldn't this only work for a bicycle being ridden by a perfectly spherical cow? ;)

No, it will also apply to an elliptical cow.

I understand the question, and the answers are correct, but fail to account for the effects of gearing, though that's also easy enough. A more sophisticated analysis could also factor in wind drag, and the parasitic drag of the system.

But all of this doesn't matter, because what you'll get is a number, and it won't be useful because the input force, namely the leg strength isn't calibrated.

Figuring gears, torques, and all other date may be somewhat interesting, but useless when riding. You're equipped with an excellent bio-feedback system, use it to advantage by listening to your legs, and selecting gearing accordingly.

That said, here's a link that might be exactly what Bob is looking for. It calculates cadence, power, torque, etc. values for various speeds and grades. Have fun.

I'm with Frank. I and others here could relatively easily derive the equations, put in a corrector for rolling resistance drag and even for slope and air drag. I could program it in Excel. But first, what use would it really be? Bobotech, to what use would you put those numbers? How often?

Second, there is a German site on analytical solutions to cycling mechanics and understanding cycle performance, and they actually have nearly the model you would need on line, and have had for a number of years. For a little while a number of the people here and on Internet BOB used it to argue about wheel efficiency and other more subtle performance factors, but ultimately you ride a bike rather than validate calculations about a bike.

I just wonder if your real question could be answered some other way.

It more for curiosity than anything else. I really don't need an exact number, but a way of comparing how much force is needed for one sized person compared is another. I have often wondered if I were to suddenly lose 200 pounds within like 6 months, what kind of bike rider would I end up being?

Depending on where you're starting, either real happy or dead.

Seriously, If you have anywhere near a potential 200#s to lose, I commend the effort and wish you all the success possible, though don't a quick miracle. One irony of weight loss by bicycle is that after an initial drop, the rate of loss slows down significantly.

That's for 2 reasons, the first is you become more efficient, so need ever longer, harder workouts to maintain progress. The second is good news, while you're losing fat you're also gaining muscle, so the weight loss understates the real improvement in your condition. You can have a radical improvement, with a large reduction in things like waist size, yet the scale may not reflect it honestly, so don't get discouraged.

If you want a real sense of what the weight loss may mean, try making up a 20# payload. When doing simulations, I usually use potatoes, but anything will do. When you see how much harder just 20#s can make hill climbing, you can imagine what an improvement 10 times that may be like. From time to time, you can add back the 20#s as a reminder of the progress you're making.

What affect would bio pace chainrings have on the ellpitcal cow?

Also, don't forget to visit the clydes http://www.bikeforums.net/forumdispl...00-lb-91-kg%29 Plenty of people in that forum have lost a real 200 pounds, or close to it.

Awesome!

Obviously, don't answer if you find this question a bit personal or whatever, and I assure you I mean no disrespect, but - how much do you weigh, dude, and how tall are you?

Is there any formula for figuring out how much torque is needed to spin a crank?
 

Bobotech, have you looked at www.analyticcycling.com? This is the one I would play with, given I had your question.

Oh i have never hid my weight here. I do go into the clyde subforum once in a while but I am somewhat north of 350 pounds. A big guy. I mainly ride my Specialized Sequoia Elite with a triple or my Motobecane Fantom Cross Team Ti with a triple. When I got the Cross Team, it was just a bare frameset so I built it up as a triple instead of a compact double that it originally came with. I have often wondered what kind of rider I would end up being if I lost 200 pounds very fast. Would I just breeze up hills that now challenge me to nearly passing out? Would I be able to average 15-17mph on a longer ride instead of barely making 11mph on a 5 mile ride?

Hi,

In simplistic terms the force needed to hold a rider static on a hill
is simply a proportion of the riders + bike weight. The "torque" at
any point is the drive train is simply the effective leverage of that
point regarding the the static force due to the incline.

When you start moving things change. Friction, rolling resistance
and drag come into play, overcoming gravity is only important on
steep hills, for modest hills drag can still dominate effort for speed.

Still, essentially power to weight ratio determines hill climbing ability.
Torque determines power at a given rpm. If you lost weight but your
power and optimum cadence remained the same, then you'd simply
use a higher gear to go up the hill faster.

work = force x distance, work per second = power. Lose weight and
the force drops, for the same power you will go near proportionately
faster, assuming the speed is low and that gravity is very dominating.

rgds, sreten.

You'll find that the terrain definitely gets flatter. You'll probably see close to a 50% increase in average speed, not because losing weight makes flat riding that much easier, but because your body gets more efficient. Likewise your range will expand materially, and distances of 30-50 or miles will be within your easy reach.

It isn't digital, you don't have to lose all the weight to see the improvement, just losing 10% or 35#s will already start making a big difference. Even before you lose much weight, the exercise will improve your heart and lungs and you'll find everything, like climbing stairs easier.

Problem is it's a bit of a catch-22, since the weight makes it hard for you to ride long enough to burn off much. You need to find a way to push at the fringes of your capacity, and expand the range slowly.