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B Bicycle gears, force, torque

  1. Nov 12, 2017 at 3:56 PM #1
    Hello,

    I'm trying to get a conceptual understanding of why switching gears makes pedaling easier or harder. I think I understand what's happening with the rear gears, but I'm not sure I understand what's going on with the front gears. I will outline my thinking:

    Let's suppose our bicycle has one cog at the front, and two sprockets at the back. Since no changes can be made at the front, let's focus on the back: when the chain is on the largest sprocket, the moment arm is at its longest and therefore maximum torque is applied, requiring less force to complete one full rotation of the wheel but requiring a lot of angular displacement of the sprocket for that rotation to complete. This translates to "easier" pedaling, but lower speed (more crank turns required to turn the wheel fully once). If we switch to the second, smaller sprocket at the back, the moment arm is shortened, reducing torque. This translates to "harder" pedaling, and higher speed (fewer crank turns required to turn the wheel fully once). Please correct me if I am wrong on any of this.

    Now, let's magically add one extra cog to the front, making for two in total. This new cog is twice as large as the one we've been using so far. Without making any changes to chain position on the rear (let's assume the chain is on the largest rear sprocket), we switch from the smaller front cog, to the "new" larger front cog. What I am unclear on is why does this make it harder to pedal? Changing the front gear does not seem to impact torque on the rear wheel.

    My thinking so far is the following: because we are now using a frontal cog that is twice as large, we are effectively required to move twice as much chain with one full rotation of the crank (due to the 2x larger cog circumference). This means that, for each single rotation of the crank, we are now doing double the work on the rear sprocket, and therefore on the wheel.

    So essentially: changes at the back make it easier/harder to pedal due to change in torque. Changes at the front make it easier/harder to pedal due to increasing/reducing the number of sprocket rotations we achieve with a single rotation of the crank.

    I am not certain on my final conclusion. And even if my final conclusion is correct, I'm not 100% certain my train of thought that brought me there is accurate.

    Am I correct in my assumptions above, as well as my conclusion?

    Thanks.
     
    Last edited: Nov 12, 2017 at 4:09 PM
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  3. Nov 12, 2017 at 5:15 PM #2

    andrewkirk

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    What you wrote sounds broadly correct. The way I look at it is as follows. It is just a series of several applications of the rule:

    Torque = Force x Distance​

    • Rider applies a force ##F_{pedal}## to the pedal, which is at the end of a crank of length ##R_{crank}##. That imparts a torque of ##T_{crank} = F_{pedal}.R_{crank}## to the crankset
    • The radius of the chainring (front sprocket) bearing the chain is ##R_{chainring}##. The torque ##T_{crank}## translates into a force of ##F_{chain}=\frac{T_{crank}}{R_{chainring}}## on the chain, pulling it forward.
    • The radius of the rear sprocket bearing the chain is ##R_{sprocket}##. The force ##F_{chain}## is applied by the chain at that distance from the rear axle, imparting a torque of ##T_{sprocket}##=##F_{chain}\cdot R_{sprocket}## on the wheel.
    • The radius of the wheel is ##R_{wheel}##. The torque ##T_{sprocket}## translates into a force of ##F_{ground} = \frac{T_{sprocket}}{R_{wheel}}## at the ground, pushing the bicycle forward.
    • Expanding ##F_{ground}## by substituting all the previous elements, we get the formula for the force on the ground as ##F_{ground}=\frac{F_{pedal}\cdot R_{crank}\cdot R_{sprocket}}{R_{chainring}\cdot R_{wheel}}##
    So you can make it easier - and slower - to pedal by lengthening the cranks, using a smaller chainring, using a larger rear sprocket or using a smaller rear wheel.

    All of these devices are used, separately and in combination, from time to time. For instance, on track (velodrome) bikes, cranks are typically shorter (165mm vs 170mm or 175mm on a road bike), because on the track cyclists have only one gear and so can sometimes have to pedal at cadences above 160rpm when at top speed. Shorter cranks makes that easier (less distance for the foot to travel) at the expense of requiring slightly greater force on the pedals.
     
  4. Nov 12, 2017 at 5:47 PM #3

    sophiecentaur

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    I agree, broadly, with Andrew Kirk's post, above but I have to point out that no machine is 100% efficient and that the equations quoted in the above post are theoretical.
    The 'effectiveness' of a machine does not just depend on lever lengths, numbers of teeth or pulley diameters. Bicycles are pretty good machines on the whole but efficiency has to be considered.
    I was taught in, perhaps, old fashioned terms and there were two measures involved:
    Velocity Ratio (VR) is the ratio of the distance moved by the Input Force / distance moved by the Output Force. This is just down to the geometry of the situation and it's the first calculation you do (as above).
    Mechanical Advantage (MA) is the ratio of the Output Force / Input Force. It's what you actually get from the machine.
    The Efficiency E is given by MA/VR and it's always less than ONE. A new modern bike is pretty efficient as machines go but still there are losses.
    Most cycle gearing (Chainwheel / rear sprocket combinations) will have some 'overlap' in which different combinations of sprocket will give the same 'gearing ratio' but cyclists are usually aware that one setting can be better than another with the same nominal ratio. (It is generally better to avoid using the smallest sprocket when possible because of the tight angles of the chain as it goes round). Also, as Andrew says, the crank length is relevant. It is also relevant to the cyclist's muscles and leg dimensions and will affect the actually efficiency of cycling (miles per meat pie).
     
  5. Nov 13, 2017 at 2:33 AM #4
    Thank you very much for the detailed answer. Mathematically, this seems all clear to me.

    I guess what I am having trouble understanding are the implications of this part: ##F_{chain}=\frac{T_{crank}}{R_{chainring}}##

    Because this implies that ##F_{chain}## decreases as ##{R_{chainring}}## increases, I would expect pedaling to become easier (less Force) as we switch to a larger front chainring. However, experience tells me the opposite is true. What am I missing in my understanding of the physics?

    Thanks.

    EDIT: I think I might have understood. What makes pedaling "heavier" isn't ##F_{chain}## per se - what makes pedaling heavier is friction from the ground. So if there is less force applied to the chain, there will be more net force acting on the bike to prevent it from acceelrating and hence why it feels harder (ie. more Force required) to accelerate. Does this sound right?
     
    Last edited: Nov 13, 2017 at 3:05 AM
  6. Nov 13, 2017 at 3:23 AM #5

    A.T.

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    Think of the pedal crank as a lever. Larger chain ring means that the load (chain force) has a larger lever arm, so you need to apply more pedal force.
     
  7. Nov 13, 2017 at 4:46 AM #6

    sophiecentaur

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    I think your problem may be with the 'cause vs effect' thing. That equation is just based on the fact that the torque caused by the foot force is the same value as the torque on the chainwheel. The mathematical re-arrangement ignores which is the cause and which is the effect. I doubt that you would have the same problem is you were just dealing with a simple lever and the situation is actually just the same - as A.T. points out.
     
  8. Nov 13, 2017 at 5:34 AM #7
    I might be getting confused on the terminology (I'm assuming "lever arm" is the same as "moment arm"), but my understanding was that a larger lever arm should require less pedal force, as opposed to more?

    Thank you. Still not sure I understand. Regardless of how the equation was derived, isn't it implying that, all else being equal, increasing the ring size decreases the force? And if that is true, why does a reduction in force lead to "harder" pedaling? (assuming my "EDIT:" explanation under post #4 is incorrect?)
     
  9. Nov 13, 2017 at 5:39 AM #8

    andrewkirk

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    It's the ratio of (the distance from pivot to point of application of input force) to (the distance from pivot to point of application of output force) that determines the mechanical advantage of a lever. That ratio is the ratio of the length of the crank to the radius of the chainring. So the larger the crank, and the smaller the chainring, the bigger the mechanical advantage (which means a 'lower gear').
     
  10. Nov 13, 2017 at 5:42 AM #9
    Ahh, thank you. I think I see now what piece of information I was missing. I'll ponder this a little longer and come back if I have any more questions.

    Thanks everyone.
     
  11. Nov 13, 2017 at 6:00 AM #10

    sophiecentaur

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    This is the Cause and Effect thing at work. The 'Equation' tells you the relationship between the variables but not which way the casual chain happens to be working.
    No. It tells you the (ideal) Velocity Ratio. It tells you nothing about any friction or, often more importantly, any dead weight that could be operating. I don't think I'm being very picky here, on a Forum that tries to give the right message whenever possible.
    The distinction between VR and MA is very important in some circumstances. Imagine a system of wooden pulleys and sisal ropes. Just making it work with no load can require a lot of work (low MA for a start) when the VR could be 10.
    Question: why don't screws come out of the wood under load or even under their own weight?
    Question: why do people spend hundreds of pounds on a posh set of gears and chain when they can have the same velocity ratio for just a few quid?
     
  12. Nov 13, 2017 at 6:17 AM #11
    I'm somewhat confused on this. Could you tell me what the actual cause-and-effect chain is in our case?
     
  13. Nov 13, 2017 at 6:36 AM #12

    sophiecentaur

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    I may have got this wrong and you may well have sorted it out by now but it's possible to confuse "produces more force" with "needs more force". It could be that your interpretation of that equation could be the other way round. Thing is, the topic seems to be obvious to me and I can't actually appreciate where your difficulty lies. I realise that you 'know' a longer pedal arm makes things easier but the equation seems to be giving you a problem.
     
  14. Nov 13, 2017 at 6:42 AM #13

    sophiecentaur

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    I just re-read this. It 'decreases the force on the chain' for a given pedal force or it "increases the pedal force needed for the same force on the chain". i.e. you didn't declare which force you were discussing. :smile:
    You are going to be doing this in your sleep soon!
     
  15. Nov 13, 2017 at 6:56 AM #14
    I'm already not sure if I'm asleep or awake. :)

    Sorry, the confusion is a result of me cramming 1-2 years' worth of school physics in under a week at home (Khan Academy, PhysicsClassroom). I'm trying to relate quite a few new concepts to one another to get a better conceptual understanding of the whole. As a result my questions may seem neither here or there, but all answers are really helping me "complete the puzzle," so to say. I think I'm close to the "click" moment - or at least closer than I was yesterday.

    If possible, please let me know if this sounds right:

    The chain is on the smallest front sprocket. I'm applying force ##F_{pedal}## to the pedals. This transmits ##F_{chain}## to the chain. I now proceed to move the chain to a new front sprocket, which is double the size of the sprocket I've been using so far. Because the radius of the sprocket has doubled, ##F_{chain}## will decrease by half, assuming ##F_{pedal}## remains unchanged. As a result I now require twice as much ##F_{pedal}## as before to accelerate the bike at the same rate (assume in both cases we were accelerating from zero).

    Does this sound... right? :)
     
  16. Nov 13, 2017 at 6:58 AM #15

    sophiecentaur

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    Yes. For same output force, you need a bigger input force for a bigger chain wheel.
     
  17. Nov 13, 2017 at 7:00 AM #16

    jbriggs444

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    Focusing on cause and effect is often irrelevant. You have a relationship. You have knowns. You have unknowns. You can solve for the unknowns without ever bothering to decide which is "cause" and which is "effect".
     
  18. Nov 13, 2017 at 7:02 AM #17
    Thank you very much.
     
  19. Nov 13, 2017 at 7:09 AM #18

    A.T.

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    Yes.

    Make sure you understand levers, before learning about gears. Making the load arm greater, will require more effort (for the same load and effort arm):

    levers_classes_2.gif


    This is a great related puzzle:

     
  20. Nov 13, 2017 at 8:50 AM #19
    I thought this was a very good video that added to my understanding. Just thought I'd post it for anyone who might run into this thread in the future.

     
  21. Nov 14, 2017 at 3:22 PM #20

    CWatters

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    I think the key is to understand that humans feel most comfortable pedaling at specific combinations of torque and rpm. The purpose of gears is to try and keep the rider in this comfort zone despite changing loads. Essentially the perform a matching function.
     
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