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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
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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

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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I'd drink ~300-500 metres before the start of the climb when its still flat, and I've got time to drink, stow the bottle, and burp before the rise. I also pull up long sleeves, all the better for cooling. You're going to carry the bottle up, just a matter of whether the water is in you or in the bottle. And many climbs lack a water source at the top. – Criggie Jan 7 at 5:51

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

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

Of course gear ratio affects the "potential" power that you can produce. Consider a maximum muscular effort to go up a steep hill. Neglecting chain friction and other secondary effects, you’ll go up the hill the fastest at the highest power that your muscles can produce. Note that power = k x torque x cadence (where k is a just a constant that determines the units of power (watts, horsepower, etc.). Say you're riding in too high a gear so you can't move forward on the hill (your cadence is 0). At 0 cadence your torque is at the maximum that it can be and your power is 0. As you increase your cadence (by lowering your gear ratio) your torque decreases. However, the product of torque and cadence (which is proportional to power) increases. As you keep increasing your cadence by lowering your gear ratio you will eventually reach the energetically optimal cadence (EOC). At the EOC, the power that your muscles can produce is at a maximum. Increasing cadence above EOC reduces your maximum potential power.

Bottom Line: Choose that gear ratio that allows you to spin as close as possible to the EOC. You’ll climb the steep hill fastest at this cadence.

Note: The power vs. cadence curve looks like an upside down parabola. It’s a direct result of the work done by Archibald Vivian Hill, who won a Nobel Prize for his work on this and many other topics in biophysics. Also note that maximum endurance probably occurs at a cadence less than EOC.

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Welcome to bicycles.stackexchange. Your answer applies only instantaneously. For climbs of nontrivial duration, power is limited by metabolic processes (mostly, aerobic). That is, what you've written is true for maximal instantaneous power but when climbing any nontrivial hill your power output will be decidedly submaximal. There are still boundary values on gearing that will limit power production but and long as you are reasonably far from those boundaries your limitation is metabolic, not either force-limited nor limited by the speed of msucular contractions. – R. Chung Jan 7 at 4:44
    
Gidday and welcome to SE Bicycles. Good first answer - have you got any further information or links on EOC? – Criggie Jan 7 at 5:47
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@R.Chung,you're right, but I reckon a sort of aerobically-stable EOC could be used. – Chris H Jan 7 at 8:26
    
This is basically the right answer. If it were not true that gear ratio affects the power available at the wheel the automobiles would have no need of multi-speed transmissions. – Daniel R Hicks Jan 7 at 13:06
    
@ChrisH One might think so, but empirically not. Many riders can produce > 1 kW for a few seconds in a narrow range of cadence but they might climb a nontrivial hill at, say, 200 - 250 watts. Empirical analysis of their cadence choices show that at a lower level of output they can and generally do produce fixed power over a wide range of cadence and crank torque. – R. Chung Jan 7 at 14:12

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

There are several factors involved here, so any answer is not simple. First, as Leon noted, you get zero power to the wheels when the gear is so hard that you can't move. And you get vanishingly small power to the wheels when the gear ratio is so easy that you're spinning at 200 RPM.

But more importantly, AVERAGE power over a period of time is highly dependent on details of how the muscles work. Primarily there is AEROBIC vs ANAEROBIC exercise. With the average rider, with normal blood sugar, any riding above about 80 RPM will be largely aerobic, and any (halfway-challenging) riding below about 60 RPM will have a large anaerobic piece. Aerobic exercise burns blood sugar, but anaerobic exercise burns glycogen stored in the muscles.

For short periods of time (how short depending on how intense the exercise and how much blood flow there is) muscles in good health can burn glycogen about as efficiently as blood glucose, but the amount of glycogen stored in the muscles is only sufficient for maybe 15-30 minutes of high-intensity exercise (though with training specifically targeted towards increasing the body's glycogen stores this can be increased to several hours).

Thus, riding in a "difficult" gear that produces a low RPM more rapidly exhausts muscle glycogen and leads to more rapid fatigue. And obviously, as you fatigue your power output drops. (And of course, riding with too "easy" a gear results excessively high RPMs, and the average rider's "optimal" RPM is generally below 100.) In between, you're trading off modest glycogen consumption vs a somewhat increased muscle power you can get by engaging the "slow twitch" muscles and some other factors. (Keep in mind that you need the glycogen for short, high-demand situations, such as climbing a short, steep hill without downshifting. You can actually injure your muscles in some circumstances if glycogen is totally exhausted.)

(And there is also the point to consider that in susceptible individuals one can cause knee injury by consistently using too difficult a gear.)

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I don't understand the link between "riding in a difficult gear" and "more rapidly exhausts muscle glycogen". Why is that? Surely your muscles will use glycogen or glucose depending on what's available, rather than being determined by that gear you're in? It'd be useful if you had references so I could read up on the mechanism. – Nuі Jan 8 at 7:18
    
@Nuі - When your muscles move slowly they burn more glycogen. There are two reasons: First, you're probably demanding more instantaneous energy from them, and, second, because of the lack of motion there's not as much blood flow in the legs (tight muscles restrict blood flow). – Daniel R Hicks Jan 8 at 13:37

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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there's a 404 error on that link. – OraNob Jan 7 at 13:27

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