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Let me preface by saying that this may not be the right SE. I considered asking on the Physics SE, but I thought I might try here first. If it is wrong, I am not against it being migrated.

From basic principles of Physics, power is computed as Work/time. So consider a rider and bike system climbing a hill. The work done is the difference in potential from bottom to top, and obviously the time will be climb-time.

Now, my question is:

Given the same rider, same bike weight, and same climb-time, does your gearing affect power? Also assume that the climb is efficient, no slipping tires, normal pedaling, etc.

From a physical perspective, I expect the answer is no. Same difference in potentials, same time, same power. However, from the rider perspective, I know that it sure feels like a lot more power is being used to climb with a harder ratio.

I expect that the answer is that the disparity comes from idealizing the system. If we consider the bike a closed system, we expect that all the energy put into the bike carries it up the hill, but this is not the case. Moreover I think the inefficiency of the human body will be relevant. However, I still cannot resolve the question.

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I also was at a loss for tags here. –  BBischof Apr 21 '11 at 8:10
    
We don't have a migration path from here to Physics, unfortunately. However, this question can really be asked here or on Physics but I think you'll get a better answer there. (I'd be fascinated to see the answer.) If you post it there as well, please post a link here as well. Cross-site collaboration will likely produce the best answer to this question. –  Neil Fein Apr 21 '11 at 14:27
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For now I will leave it here, if I don't get the answer I am looking for, I will try to dupe it over there. –  BBischof Apr 21 '11 at 14:37
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I suspect you mean efficiency rather than power, otherwise the question makes no sense. You're lifting the same weight over the same distance in the same time, so the power is the same. From a competition point of view, you probably either want to got faster (more power) for the same effort, or use less energy for the same climb. So you're looking at efficiency. –  Мסž Apr 22 '11 at 2:52
    
For just the bike, no, it does not affect power. The bicycle is a rather simple and linear system, it's pretty much power in == power our. However, the human body powering the thing, isn't even remotely linear. –  whatsisname Apr 22 '11 at 18:11

5 Answers 5

Maybe it's the difference between, what do you call it, 'isotonic' versus 'isometric' work?

What I mean is that, for example, it takes a human a lot of effort (force, power, or work) to try to move an immovable object: to push against a wall or something.

In too high a gear you push and push and go nowhere (lots of power to go nowhere => 0% efficiency).

In too low a gear it's too easy: you spin against no resistance; your spin rate is limited to ~120 RPM or so, i.e. can't increase infinitely; therefore (low force and limited RPM) you're limited in the amount of power you put out (it's less than your theoretical maximum power).

Possibly there's an efficient 'cadence' (perhaps 90 RPM) which you might want to use on all terrain (up, down, level), and the right thing (the right way to use your gears) is to continually adjust the gearing for the terrain in order to: a) maintain some constant, efficient cadence (e.g. 90 RPM); b) maintain a high enough force/power output at that cadence (e.g. if it seems too easy then switch to a higher gear, or if it's too difficult then switch to a lower gear, to maintain the cadence).

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The articles which are linked in the References section of the Wikipedia article about Cadence talk more: about performance, optimal cadence, gearing, etc. –  ChrisW Apr 21 '11 at 10:40
    
There's a similar effect with car engines and their gears: when the RPM is too low or too high then the engine's torque is low; you can graph the torque versus RPM, find a range of RPM at which the engine has the most torque (and I guess that 'power output' equals 'torque multiplied by RPM'). It's similar but not the same because human muscles aren't the same as internal combustion engines: e.g. an ICE can't do isometric work, and muscles' work apparently depends on factors like fast-twitch versus slow-twitch fibres, lactic acid buildup, etc. –  ChrisW Apr 21 '11 at 11:53
    
Thank you for this answer, I will read the articles and get back to you. –  BBischof Apr 21 '11 at 19:21

Given the same rider, same bike weight, and same climb-time, does your gearing affect power? Also assume that the climb is efficient, no slipping tires, normal pedaling, etc.

Well, that depends on which "power" you are measuring :-).

Obviously, the power exercised by the bicycle as a whole is the same - if it's moving at the same speed, it's the same power.

However, the power that your body exerts may well be different, for a variety of reasons:

  • Muscles probably have a speed and force level where they are most efficient, so the chemical energy/power your body must exert to produce muscle movement will be different.
  • The various energy loss processes due to flexing, friction etc. will probably be different depending on gearing. E.g. in lower gears there will be faster chain movement (thus more friction), on the other hand chain tension will be lower, which probably reduces friction. Also, in lower gears flexing of the frame in response to chain forces will probably be lower.

My impression is (though I don't have sources to back me up) that generally the human system is most power-efficient (i.e. best ration of pedal power to exertion) at cadences around 90-100 RPM, so that's what a cyclist should strive for.

Interestingly, the best cadence for maximum power is apparently much lower, that's why pro cyclists will use high gears and low cadences for sprints - however this is much more tiring than higher cadences, so inefficient over long distances.

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I have some confusions here. First, the bike doesn't exert and power, the only force it exerts is friction. Your two points are exactly the things that I had in mind but I was having trouble expressing them. Thank you. But I dont really understand your last two paragraphs at all. What do you mean here by power-efficient vs max-power. And why does this difference follow from gear size. Sorry for my confusion. –  BBischof Apr 21 '11 at 19:25
    
As to the power thing: What I meant to say was that it depends on where you measure the power. If you measure power at the wheel, it will always be the same for the same speed and terrain. However, the power that the human has to exert may be different. –  sleske Apr 21 '11 at 21:38
    
As to max vs. efficient: Most efficient would be like a car - the speed/cadence which lets you travel the longest distance before you run out of fuel. Like a car, that is not necessarily the speed/cadence which gives you maximum power. And higher cadence generally means lower gears. –  sleske Apr 21 '11 at 21:41

I suspect you mean efficiency rather than power.

In my opinion the main trade-off is between increased biomechanical losses at higher rpm (basically muscle friction) and decreased blood flow with higher forces at lower rpm. The balance depends both on the rider and the duration.

In the IHPVA Journal of Human Power, Issue 45(pdf, index here) is a paper called Maximum Human Power where they talk about Tyler Hamilton winning the Mt Washinton climb in 51 minutes:

"he rode much of the climb, however, in the 23-tooth cog, and made several surges in the 21." If he had 700- mm wheels, as seems likely, his average cadence would have been 63 RPM.

The whole article is worth reading, and it might pay to browse the index for similar papers.

The flip side is that top sprinters often go well over 150rpm in the final sprint. At that point they're trading biomechanical efficiency for peak power. I used to peak at over 900W for 10 seconds (>8W/kg) at around 130rpm, but my hour performance of around 350W used a cadence of about 80-90rpm.

The real answer is specific to you. It will depend on your body shape, muscle type, fitness and more transient factors. It's also a question best answered by experiment, and should be part of your tr4aining schedule if you're competing. If not, I suggest finding a climb you ride regularly and keep a training diary.

There has also been much discussion over the hydration for long climbs. Is it better to hydrate and start heavier, or run slightly dehydrated so you weigh less? IIRC the conclusion was that hydration was better, but I can't find the reference.

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No, gear and gain ratios do not affect power. While you are correct in assuming that it would feel different to the rider, if the other three variables are equal, then the power rate will be the same. In this case, in an "easier" gear ratio, the cadence would require a significant increase to maintain the same climb time (speed) and if the rider is identical, then the work rate is identical. The increase in speed of pedaling makes up the difference in wattage expenditure compared to the "harder" gear at a lower cadence.

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I sort-of-disagree. While this is absolutely true in theory, realistically we have a cadence/force band where we emit optimal power. At very high cadences we don't apply enough force to match the same power, and vice versa for high power with low cadence. –  Stephen Touset Jan 26 '12 at 16:35
    
@StephenTouset That cadence/force band isn't constant -- it varies according to conditions. See here or here for a discussion of cadence, pedal force, and "optimal" power. –  R. Chung Jan 26 '12 at 18:40
    
@StephenTouset: That's what I meant when I said that this assumes all three variables are equal. If you change the force of pedaling, then the power will vary, but that is not the gearing, it is a limitation of the rider. If your cadence becomes so high that the mechanics of your body can't sustain it, the your power rate will decrease. But it will not be because of the gearing. –  zenbike Jan 28 '12 at 10:31

To my understanding, it shouldn't. The simplest explanation is that power out is equal to power in times efficiency (efficiency being the energy loss due to friction, air resistance, rolling resistance, heat, etc.). Changing gears doesn't change the power in (that part is all on you), nor does it change the mechanical efficiency. Therefore, power output doesn't change.

For a little bit deeper, power is the total work performed over the total time (P_avg = ΔW/Δt). In this case, we're considering it over identical durations, so Δt is constant. In a rotational context, W is the torque (rotational force) exerted times angular velocity (rotational speed), or W = τθ. A gear will only change the ratio between torque and angular velocity while maintaining a constant work output. In other words, going to a higher gear might require twice as much torque, but the pedals will spin half as fast. A lower gear might let you spin twice as fast, but you'll be using half the torque. Since the work output is the same, the power output is the same.

How does this affect wheel-speed? Well, the same W = τθ affects your wheels, too, but in reverse (your wheels see it backwards: imagine if you were pedaling on your cog, and the wheels were attached to the bottom bracket). A lower gear will put more torque on the wheels (enabling high acceleration), but have correspondingly low angular velocity (rotational speed). A higher gear won't put much torque on the wheels (which is why it's so hard to accelerate), but will make them spin like mad. So ideally, being in as high gear as possible would give you the greatest speed.

However, that's where the human body comes into play. We have two complementary systems for generating power: the cardiovascular system, which produces less power but for very long durations, and the muscular system, which excels at producing high power, but only for a short period of time. Ideally, when not sprinting, you want both systems to be producing as much power as they can sustain. The sum of that power (minus efficiency losses) will be your total power output, and the elevation change, rolling resistance, and your aerodynamics will determine what proportion of that output will be ultimately used for torque versus distance (and thus your gear ratio).

Hope that helps.

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