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Idly wonderings whilst following the Mrs on a little road pootle yesterday.
Say, fr'instance, you're in the big ring at the front and the big ring at the back, and the chain is therefore at quite an angle (lets say x) to the chainring and again at the same angle to the sprocket.
Am I right in thinking (dredging up A Level Physics from *ahem* years ago) that the resultant force driving the sprocket forwards is
force applied at chain ring * (cos x)^2 ?
(ignoring friction blah blah)
If not, why not?
Whatever modifier you apply at the sprocket, you will need to apply the same (but t'other way round) at the chainring. The chain is at x° to the chainring [b]and[/b] the sprocket.
I think the two will cancel out.
I know it's at the angle to the chain ring and the sprocket, in fact that's what I said...
I don't think they'll cancel, I think they'll be compounded:
Chain ring force is straight forwards, resulting in a lesser force on the chain cos of the angle. That lesser force on the chain then acts at an angle on the sprocket, resulting in a lesser force again turning the sprocket.
I think.
There is a force tending to pull the sprocket outwards and the chainring inwards (if you are on big/big) but that force is opposed fully by the lockring of the cassette and the spider of the chainring. So it won't make a difference.
Imagine if the chain could flex without friction in the horizontal plane as well as in the vertical. You could put idler sprockets in front of the rear sprocket and behind the chainring. Then you would not expect your pedaling force to be diminished because it would roll over the sprockets nicely. The angle doesn't matter.
It's only friction that reduces the efficiency as the chainline increases.
Cheers Molly, makes sense.