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I have seen the explanation on "How bicycle gear works?" but this seem to be using a standard round chain ring. I would like to know if there is any difference when an oval chain ring is used. As I can clearly see, one can have an oval chain ring on a fixed/single gear bike. There fore there is no slaking or tightening of the chain. So I would say that there is no difference in gear ratio during the complete rotation of the oval-chain-ring. So the torque will remain the same.

Is there an answer to my problem? and if there is could you proved any equations

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The principal benefit of elliptical chainrings is to reduce the resisting torque when the pedals are oriented vertically (one foot higher than the other) and increase the generated torque when the pedals are oriented horizontally (both feet at the same height). This is because a person can produce much more torque in the horizontal case by shifting weight to the forward foot.

Elliptical chainrings are not as particularly common because they are more complicated to produce and complicate the design of chain tensioners.

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  • $\begingroup$ I know what they are designed for. I was hoping for a more in-depth answer. $\endgroup$ – Jeff Shepherd Mar 12 '17 at 21:12
  • $\begingroup$ I do not see why there should be difference in torque. after all, the amount of chain use is equal during the complete rotation of the drive chain. (IE the chain will not become slack or tight during the rotation of the driver ring.) So would that, not mean the rotation of the driven ring be uniform? If this is the case then the toque should be the same. $\endgroup$ – Jeff Shepherd Mar 12 '17 at 21:37
  • $\begingroup$ The chain will become slack or tight during the rotation. That's why elliptical chainrings complicate the design of tensioners. $\endgroup$ – Schlusstein Mar 13 '17 at 3:11
  • $\begingroup$ I am sorry but I feel that your comment is untrue. These oval chain can be used on a fixed/single gear bike. So they can not experience any change in tension due to the rotation of the driver ring. If your comment was true then you would have problems with the chain falling off or snapping. $\endgroup$ – Jeff Shepherd Mar 14 '17 at 7:04
  • $\begingroup$ Can any one provide a mathematical answer to my question? $\endgroup$ – Jeff Shepherd Mar 14 '17 at 7:06
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If you consider the chain ring as delivering torque about a point where the chain to the rear sprocket and front chain ring is tangential (the top portion of the chain is always under tension when delivering power) you'll see that torque is not constant. In the pedal cycle, when the bulge of the chain ring is tangent to the chain the virtual lever will produce less force for a given torque. When the dip is tangential you'll produce more force (torque = force * distance).

Conversely you could express this as requiring more torque (to drive the rear wheel) when the bulge is tangential and requiring less torque when the dip is tangential.

This is used to compensate for the lack of power you can produce at the 12/6 pedal position. In effect the chain ring is actually normalizing your power output throughout the revolution of the pedals. The slackness of the lower chain is taken up by the pulleys and springs in the rear mech.

In a fixed gear bike I'd suggest there may simply be a little extra slack in the bottom portion of the chain and/or an additional chain tensioner.

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