Looking for a decent triple crankset



artemidorus said:
I argue that the perpendicular vector does no work, as the sprocket is not accelerated perpendicularly, and is consequently not a power loss, except in so far as it increases hub friction.
We must not be communicating here. What I'm saying is that the vector component perpendicular to the rear axle is the only part turning the cogs; the component parallel to the rear axle only serves to pull the axle sideways.

Take an extreme example with the chain were running at a 45* angle to the plane of the cogs (also at a 45* angle to the axle). If you were producing 100 lbs of pull on the chain, only the tension x cos 45 deg, or 70.7 lbs would be doing work to turn the cogs. Another 70.7 lbs of force would be pulling the axle to the left, which of course is doing no work; only trying to flex your rear triangle.
 
dhk2 said:
We must not be communicating here. What I'm saying is that the vector component perpendicular to the rear axle is the only part turning the cogs; the component parallel to the rear axle only serves to pull the axle sideways.

Take an extreme example with the chain were running at a 45* angle to the plane of the cogs (also at a 45* angle to the axle). If you were producing 100 lbs of pull on the chain, only the tension x cos 45 deg, or 70.7 lbs would be doing work to turn the cogs. Another 70.7 lbs of force would be pulling the axle to the left, which of course is doing no work; only trying to flex your rear triangle.

Yep, terminology problem. My fault - if you substitute "parallel" (to axle) for perpendicular in my last post, you'll get what I meant to write in response to your first post.

Correct me if I'm wrong, but if you apply a crankset torque resulting in 100lbs of chain tension with perfect chainline, and then apply the same torque with the chain crossing at x degrees (but the same lever arms at both ends), won't you get a chain tension >100lbs due to the bracing of the sideways vector between two laterally immovable points? The perpendicular to axle vector would remain 100lbs. My point is that the parallel to axle tension vector does no work and is consequently not an inefficiency.
 
artemidorus said:
My point is that the parallel to axle tension vector does no work and is consequently not an inefficiency.
By conservation of energy, if it's lost, then that's inefficiency.
 
artemidorus said:
If no parallel-to-axle work is done, then no energy is wasted.
But it has. You have assumed a perfectly rigid system which the rear triangle isn't.
 
sogood said:
But it has. You have assumed a perfectly rigid system which the rear triangle isn't.
Deformation of the rear triangle under chain tension occurs with or without cross chaining and is a separate cause of inefficiency. In addition, such deformation is elastic and may not cause energy loss. How much does the rear triangle heat up during hard riding?
 
artemidorus said:
Deformation of the rear triangle under chain tension occurs with or without cross chaining and is a separate cause of inefficiency. In addition, such deformation is elastic and may not cause energy loss. How much does the rear triangle heat up during hard riding?
Rear triangle deformation occurs to different degree depending on the degree of cross chaining. So with extreme small-small chain line, the forces acting on the deformation is increased.

Yes, it's elastic deformation, but unfortunately it's not a perfect system, so again energy is dissipated and lost. How much does it heat up? That depends on how hard you ride. Looks like small increments of energy loss is still being summated... :p
 
artemidorus said:
Yep, terminology problem. My fault - if you substitute "parallel" (to axle) for perpendicular in my last post, you'll get what I meant to write in response to your first post.

Correct me if I'm wrong, but if you apply a crankset torque resulting in 100lbs of chain tension with perfect chainline, and then apply the same torque with the chain crossing at x degrees (but the same lever arms at both ends), won't you get a chain tension >100lbs due to the bracing of the sideways vector between two laterally immovable points? The perpendicular to axle vector would remain 100lbs. My point is that the parallel to axle tension vector does no work and is consequently not an inefficiency.
We're on the same page now with the terms. And, must agree with you that the chain tension would increase due to a cross-chain angle, cancelling out the power-loss effect from the angle. I was thinking only about the rear half of the problem; of course the same vectors would apply to the BB axle to increase the chain tension.

I was also having a problem with where the energy (work) was going in my "half-analysis". Sure there are still some small friction losses with bending of the chain, but that's not going to be measured in watts.

Thanks for persisting and providing me with a refreshing BFO (blinding flash of the obvious) moment :)
 
dhk2 said:
...must agree with you that the chain tension would increase due to a cross-chain angle, cancelling out the power-loss effect from the angle.
Would the chain tension increase when one shifts across the cogs? Isn't that chain tension being balanced by the RD? Further, if there's a significant tension variations, then how would Polar's depend on chain tension for its system of power measurement?
 
sogood said:
Isn't that chain tension being balanced by the RD? Further, if there's a significant tension variations, then how would Polar's depend on chain tension for its system of power measurement?
The RD has no effect on the tension of the active (top) segment of the chain, only the part that is returning to the RD.
The effect of cross-chaining on chain tension would be small, as the angle of deviation is never large. Polar admit that their system is less accurate than a system directly measuring torque at rear hub or BB.
 
artemidorus said:
The RD has no effect on the tension of the active (top) segment of the chain, only the part that is returning to the RD.
I am not sure this is a true statement. The tension, certainly at resting state is clearly quite uniform top and bottom. And you'll have to agree that the RD plays a major part in this. Dynamically the proportion may be smaller due to drive from the chainring, but surely RD's effect is still there.
 
sogood said:
I am not sure this is a true statement. The tension, certainly at resting state is clearly quite uniform top and bottom. And you'll have to agree that the RD plays a major part in this. Dynamically the proportion may be smaller due to drive from the chainring, but surely RD's effect is still there.

The RD can have no effect on tension in the top chain run. If you simplify the model such that the cassette cog and the chainring are represented each by a vertical pin through the chain, you see that the chain tension in the top is only a function of the resultant force normal to the two pins. You can further show this by cutting a chain such that it is a length such that it will engage the top of the cassette and the chainring, with a little extra length added for good measure. Apply a torque to the system by pressing on a pedal. Tension is now on the chain with no input from the RD at all.....especially since there's no chain running through it.

You can see this even more easily just by realizing that for the RD to place tension on the top chain run, all of the teeth on either the chainring or cassette cog would have to move, i.e. be broken off, for this tension to be transmitted!
 
But isn't that a special case where there's a force being applied through the pedal/chainring?

Assuming there's no force being applied on the pedal, then what would happen if the lower chain gets pulled? Doesn't it get transmitted through to the top segment?

I have always understood the chain sytem as one that's similar to a pulley system, where the tension throughout all segments of the string/rope is considered to have equal tension. In bike's case, the friction in the freewheel, RD can breakdown this uniformity to an extent. Isn't this correct?
 
sogood said:
I have always understood the chain sytem as one that's similar to a pulley system, where the tension throughout all segments of the string/rope is considered to have equal tension. I
True when you are not pedalling. But I guess that we are not really interested in pedalling inefficiency when we are not pedalling.
If tension were equal throughout the chain, then no torque would be applied to the cassette.
 
artemidorus said:
True when you are not pedalling. But I guess that we are not really interested in pedalling inefficiency when we are not pedalling.
If tension were equal throughout the chain, then no torque would be applied to the cassette.
True, but surely there's still a component of the chain tension that's being exerted by the RD, just the proportional contribution is different b/n static and pedalling states.
 
sogood said:
True, but surely there's still a component of the chain tension that's being exerted by the RD, just the proportional contribution is different b/n static and pedalling states.
No, under load the tension of the active segment, being firmly fixed between the teeth of the rear sprocket and the teeth of chainring, is independent of the RD. The torque exerted by the idle segment on the front and rear sprockets would be so tiny as to be a completely ignorable term.
 
artemidorus said:
The torque exerted by the idle segment on the front and rear sprockets would be so tiny...
Although very small, you just chose to ignore it. So the answer should have been a yes. :p
 
matagi said:
The current guidelines put out by Campy and Shimano about derailleur capacity etc etc assume that you WILL be cross chaining at some stage.

But they also explicitly suggest you *not* cross-chain. First page of the Ultegra 6600 FD manual specifically warns of potential noise problems and advocates moving to an adjacent cog (i.e., straightening the chain line) as a solution.
 

Similar threads