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It is from My personal experience I am saying that it takes considerable effort to move uphill on a bicycle while pedalling actively compared to simply pushing you bicycle along. But my questions are:

  • Is this real (i.e. its not just some kind of psychological effect)

  • If it is What is the reason for this

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From the energy balance there should be no difference, but firstly, if you don't have a special gear, you need a lot of force on a bike to reach the torque needed, and secondly you might not be trained for the specific movement that is needed, and that makes it hard. Thirdly the bike pedal pushes back against your foot (because the bike wants to roll downhill), even if you just try to take a rest, and that is unpleasant.

There is certainly an effect that makes it awkward going up a hill with a bike for majority of us, and that is that your bike goes slow, so it has little stabilization (which comes from the spin at normal speed), and you have to steer a lot while at the same time work hard in the pedals. You can call that psychological if you want.

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It's a matter of what constitutes "effort". If you consider proceeding uphill at the same rate in both cases, then the power requirements are roughly the same. However, depending on your gearing the force requirements may be a lot higher when riding the bicycle. Of course, this depends on the bicycle; some mountain bikes have very low gears available, which will allow you to ride up steep hills with little effort. In principle, you can have gearing such that riding uphill can require less "effort" than pushing the bike uphill, but at some point the difficulty of keeping the bicycle balanced that was mentioned by the other poster comes into play.

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  • $\begingroup$ This answer is factually incorrect; $P = \vec F \cdot \vec v$ so if $P$ and $\vec v$ are held constant then $|\vec F|$ cannot be made to vary by a simple gearing switch (which does not change any of the directions involved). $\endgroup$ – CR Drost Jan 8 '17 at 17:02
  • $\begingroup$ No, the answer is correct. You're forgetting that the $F$ in your equation is the force at the wheel. The force I am talking about is the one that is relevant to the rider, namely the force that s/he must exert on the pedals. The whole point of bicycle gearings (and transmissions in general) is to change the ratio of pedal force to force at the wheel. Of course, in order to do so the velocity of the pedals must change, so the product of torque times rotational speed remains the same for wheel and crankset. $\endgroup$ – Pirx Jan 8 '17 at 17:05
  • $\begingroup$ Ah, I see. You're pointing out that the gear makes your legs move faster/slower, which indeed changes the forces involved. $\endgroup$ – CR Drost Jan 8 '17 at 21:26
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It depends on the way you are applying the force. While pushing the cycle upwards, you are just applying the force linearly parallel to the motion, but during pedalling you are actually applying a tangential force that is being converted to linear motion.

Also while pushing the cycle, the force that you are applying is being acted upon the wheels and since the wheels have a large radius, the torque is much less as compared to when pedalling when you are producing the same linear motion as earlier but requiring a greater torque.

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The bicycle drive train is not a perfect force converter of the linear pushing of the pedals to the circular drive gears/wheel-it consumes some of the energy. In addition to the force required to overcome the friction of the chain & gears (which actually increases as a function of movement at lower gear ratios) there is energy expended slightly torquing the frame as the power of the rider is made to 'turn a corner' at the crank arm.

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