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When you are starting the ride, I potentially believe ("know?") that with a smaller plate(cog with fewer teeth in the rear cassette), you obtain more velocity than with a bigger plate(cog with more teeth, in the rear cassette), applying same human power consumption on both cases. In first case (the correct way of starting the ride), the chosen plate radius is small, so it weights less and offers less resistance.

Once bike is running at a normal speed on flat terrain and gear's start resistance has been overcome, I want to know if I will get more speed with a smaller tooth rear cog than with a bigger cog, applying same human power consumption on both cases. Changing relations would of course mean I should ride faster of slower in order to waste my same power in all cases, basically because the different resistances of the plate, which depends on its chosen radius mostly.

So let's say I want to be using 200 Watts. Would different gear relation setups affect the output speed?

If my statements above are wrong in first please, correct me.

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    *If my statements above are wrong in first place, please, correct me. – Adrián Aug 22 '17 at 17:54
  • Using standard English words for different parts of the bike would make the question more understandable. – ojs Aug 22 '17 at 18:00
  • I get English may be a second language but this makes little to no sense. VTC – paparazzo Aug 22 '17 at 18:11
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    The efficiency of your muscles varies with the speed at which they're working. This is known as cadence in the sense of how fast you're pedalling. To a decent approximation, a bike's drivetrain efficiency is unaffected by the gear ratio (which I think makes the answer to your question no). Hopefully this may give you some hints as to how to make your question clearer. – Chris H Aug 22 '17 at 18:17
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    @paparazzi you may well be right. I'm more inclined to VTC than to answer but as the OP is new I thought I'd give them some hints and a chance to improve it. – Chris H Aug 22 '17 at 19:31
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Yay! Another physics question.

Restating the question: if the rider power input is constant, does the bicycle velocity change if the drive-train gear ratio is changed?

Let's assume the road surface, wind speed, gravity etc. all remain constant. Bike and rider are on a surface with no gradient.

Short answer is that constant power results in constant velocity regardless of gear ratio.

Power at the rear wheel = power provided by rider - power loss in drive-train.

If the power loss in the drive-train does not change with gear ratio selected, the power at the rear wheel does not change and hence velocity does not change.

Of course, the power loss in the drive-train will depend on the gear ratio selected to some degree. I believe the chain travelling more slowly around a larger sprocket consumes less power (I'm not sure on this so somebody please correct me if I'm wrong).

However the power needed to overcome air resistance and wheel rolling resistance is much greater than drive-train losses (if the velocity is not very low), so a change in drive-train losses will not effect velocity very much.


Update: addressing this statement from the OP (edited), as I think it's crux of the misunderstanding.

...with a smaller [rear sprocket] you obtain [higher] velocity than with a bigger [rear sprocket], applying same human power...

That would be true if human power was replaced with pedaling rotational speed.

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"Does changing gear relation produce different speed at same applied power?

Short answer, no.

Torque is the force we put to the end of a lever that is, say, about six inches long, the pedal crank arm.

Torque in English units of old was expressed in foot-pounds. It is a force, not a power, until movement occurs.

How fast goes the movement, times the force behind that movement, equals the power developed.

I am expressing these things in homely words. You will now understand to look upon the gearing not as wheels of a radius, but levers of a length, and when one is levering against another, the common denominator by which to find the force is the effective lever arm length working to move the load.

The gearing transforms torque but cannot change power. Power is the product of torque and speed.

Our human muscle force is at its greatest at zero velocity. Put a suitable length of lever under the foot for the job at hand: to start, perhaps a rather long lever to get us going promptly, say when starting a hill. Then, as speed is achieved, the leg's up and down stroke, which does not produce as much torque at higher rates of reciprocation as when it is stationary but pushing its hardest, is put back into the zone of power efficiency potential of the human machine by gearing up, using a shorter lever (so to speak), the smaller diameter driven plate, as you would call it.

The FORCE is what gets us going. Our force is at basis a reciprocation of one leg pushing, then another, much like the piston of an engine. Our power is that force times its rate of repetition. Our rpm (reciprocations per minute, you might say) range is limited. We use gearing as you know, to optimize the power potential of the human engine by keeping it working within its effective range of rpms. Our torque varies with the pedal position much as does the piston connecting rod of an internal combustion engine.

The gear ratio does not in itself produce different speed at the same applied power, period.

Can defense of the bold statement be simplified? "Gear inches"...copying now from Wikipedia: "Gear inches is one of [the] several relative measures of bicycle gearing, giving an indication of the mechanical advantage of different gears. Values for 'gear inches' typically range from 20 (very low gearing) via 70 (medium gearing) to 125 (very high gearing); as in a car, low gearing is for going up hills and high gearing is for going fast.

'Gear inches' is actually the diameter in inches of the drive wheel of a penny-farthing bicycle with equivalent gearing..."

Back in the day of the Ordinary bicycle a rider desiring speed rode as big a wheel as his inseam would allow. He literally geared UP. Could he go faster on a 60-inch-tall wheel than he could on a 52? Probably, but only because he could not spin the 52" wheel at a sufficient rate by which to develop the power he could develop at lower rpms on the literally-tall geared 60" bike (remember, our human torque/push pressure falls off with increasing rpms). But, if he could spin fast, he could go as fast on the 52" wheel as on the 60" wheel. Other factors being disregarded for this hypothetical example from 1880, you will see that both wheels (bikes were called wheels, then) would travel at precisely the same speed if equal power were applied.

  • Power is work per unit time not torque per unit time. – paparazzo Aug 23 '17 at 1:31
  • If force is N = kg.m.s^-2, then torque is N.m = kg.m^2.s^-2 and torque/time is N.m.s^-1 = kg.m^2.s^-3 = W is the same as power. Looks ok, although the original text is imprecise. – Useless Aug 23 '17 at 9:06
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    @Useless Same unit of measure does not make them the same – paparazzo Aug 23 '17 at 11:27

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