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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. Gear ratios change angular velocity and torque but because the product of the two is constant power remains the same.

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.

  • tl;dr the answer is yes, then :) ! Here an example of power loss in the drive-train "The frictional losses of the system were highest for each chainring when the chain ran on the smallest cog. This is also where the difference in frictional losses were greatest between the two drivetrains: at the highest gear ratio (4.8) the 48 X 10-tooth combination consumed six watts more than the 53 X 11-tooth combination." ( bikeradar.com/news/… ) – EarlGrey Sep 19 at 13:38
  • Re. drivetrain friction and gearing. I believe that the more the chain bends (i.e. around smaller sprockets/chainrings), the more friction there is (hence why some racers are using oversize pulley wheels which save about 1-2 W). The more the chain is angled side to side (i.e. cross-chained), the more friction there is. I don't recall reading that it has anything to do with the speed the chain travels, and I think it was more about how much the chain is bent. – Weiwen Ng Sep 20 at 14:51
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You are mixing up pretty much everything here. The main misconception is about the role of the gearshift:

Gearing is about force (more precise: torque), not power. If you put 200W of power into any gear, you get 200W of power out of the gear (ignoring the losses which are small in relation). It is the torque on the wheel that changes.

Say, for instance, you are putting 250Nm torque on your pedal. Assuming a 10cm radius chainring, that translates to a force of 2500N on the chain. If the sprocket is only 5cm radius, you only get 125Nm torque on the wheel. But for each rotation of the cranks, the wheel rotates twice. Energy is torque times revolutions, and 250Nm * 1 revolution = 125Nm * 2 revolutions.

And that is the key point of gears: A low gear allows the rider to spin at 90 revolutions per minute while going up a hill, turning all the spinning into a lot of force to overcome gravity; and a high gear allows the rider to spin at 90 revolutions per minute while going down the same hill, turning all the force into a lot of spinning of the rear wheel. Without the gear shift, the rider would need to put unpractical force on the cranks while going up, and would not be able to follow his cranks with his feet while going down.


The second misconception is about the role of weight. Weight has no role in gearing. Period. Racers try to reduce weight because unsprung weight is an energy drain, but that effect is fully ignorant on whether you are pedaling or not.


Now, human power consumption is an aspect.

Try standing with a bent knee on anything solid. You are not doing work (your power output is zero), but it is straining your muscles and you are burning energy to keep your muscles in tension. All the energy you put in is wasted.

Now, try sitting on a bike with the rear wheel suspended in the air. Spin the pedals. Fast. Again, you are not producing any power, as there is no force you need to overcome. But the fast movement will likely get you hot quite quickly: Your body is again burning energy (you get hot), but all the energy you put in is wasted.

These are the two edge cases: Force without movement gives no power, and movement without force gives no power. Nevertheless your body burns energy in both cases due to its inefficiencies. Somewhere in between these extremes is a point where your body can produce the most power for the energy it burns. That point is somewhere in the vicinity of 90 pedal revolutions per minute.

And that is exactly the job of a gear shift: Keeping your legs spinning at their optimal speed, to get the most power out of your legs for the energy your body burns.

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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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I know this is an old question, but I couldn't help jumping in and offering a contrarian opinion.

Of course changing the gear changes the speed! Why do you think they have gears on a bike?! Watch the Tour de France, they all have gears on their bikes. They make their handlebars as thin as paperclips to save weight, but they leave the gears on because gears make them go faster. Period. Different gears can make you go faster or slower in different scenarios. The scenario including terrain, biker fitness, biker weight, bike lubrication etc.

You can try it yourself. Get going at a comfortable rate on a bike in flat terrain at a middle gear. Try switching down to your lowest gear while keeping your exertion the same. You will slow down. Try switching to your highest gear while keeping exertion the same. You will slow down. There is a single optimal gear (for that terrain, bike, and human body) which maximizes the power transfer from your body to the bike and thus leading to the highest speed.

The term for this in physics is impedance matching. These concepts are more discussed in electronics, but there are parallels in mechanical systems, with springs being similar to capacitance, and momentum being similar to inductance for example.

Basically, the idea is that all power transfer systems have some amount of resistance when transferring power from the source (your legs) to the load (rubber pushing on the cement). This is very intuitive when when considering the simple case of a constant torque source and resistance. However, your legs are not a constant torque source they are sinusoidal. And as it turns out, the resistance (technically impedance) of power transfer between the source and the load changes depending on the frequency. The gears effectively adjust the frequency of the power source (your legs) to maximize the transfer of power from your legs to the cement.

Since energy is always conserved, what happens to all the energy that your body generates if the impedance is mismatched? Well, some of it certainly is dissipated as heat from simple friction, but the real phenomenon at play here is reflection. Whenever there is a discontinuity in impedance, some of the energy is reflected back towards the source. When you pedal in the wrong gear, you are quite literally pedaling against yourself. Like pushing your hands together really hard, it will make you tired after a while, but you won't see much effect.

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    From physics standpoint, this didn't make any sense at all.The oscillations that happen in bike frame and drivetrain are all in range of several Hz upwards. Pedaling on the other hand is at most 3 Hz and often well below 2. – ojs Sep 19 at 11:42
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    Right conclusion for wrong reasons, Yes, human legs can product the most power in a range of cadence and that drops off if pedaling to fast or too slow. It it not due to 'impedance mismatch' between legs and drivetrain however. – Argenti Apparatus Sep 19 at 12:13
  • @ArgentiApparatus Thanks for the response! I've seen those plots of power verse cadence for motors before, but they seem kind of unsatisfying to me because it is an empirical observation and doesn't offer any insight into why they behave that way. When I hear frequency dependent system response, complex impedance seems like the obvious way to represent that. Do you see any other fundamental way of describing the phenomenon? It could just be that real portion of the source impedance is a function of frequency that happens to be the plot you mentioned. Just feels like a cop-out. – david11 Sep 19 at 22:36
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    @david11 lancet.mit.edu/motors/motors3.html - explanation of torque and power curves based on a simple model of internal resistance in a DC motor. – Argenti Apparatus Sep 20 at 0:08
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    If you want to understand why cycling, walking and running cadences are what they are, it's better to forget about electric motors and read about physiology instead. – ojs Sep 21 at 10:07

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