Engineering ? Drag from Tensioner vs jackshaft


100 kW
Sep 6, 2012
Saint Louis MO
I know gear reductions through a jackshaft add losses to a system through drag but how does that compare to a single stage with a tensioner? I am trying to compare 2 similar designs.

One design uses a jack shaft to locate the final drive to the pivot point of the bike's suspension to eliminate chain growth during travel. The jackshaft has almost no reduction. Its 18tooth in and 16tooth out to the rear wheel.

The other design would be a single stage from motor to wheel with a 2 sprocket derailleur type tensioner similar to the LMX bikes.

If both designs have the same amount of sprockets and bearings so would they both have equal frictional and inertial losses? If the bearings are equal are the losses just a matter of how many teeth are contacting the chain? Does a jackshaft incur more losses because the chain is under more load on more teeth compared to the relatively small load on derailleur/tensioner teeth?
The loss difference is problem negligible compared to the regen you could get from the first setup. The first setup will have higher losses (slightly less if you use a little bigger sprockets) but it won't be very much. The friction loss in the chain occurs when a link is bent and that amount of friction depends on the tension the link is under, so in the tensioner this is very little while going around driven sprockets it's higher. And so in turn the amount of bend is related to the loss so smaller sprockets means more losses because the links are bending more. But either way chains are pretty efficient in this configuration. The real problem is maybe getting the chain to clear the lower chainstay on the first setup and making sure that suspension link will clear the motor.
I think the more important difference would be suspension squat under power, if you're applying that second design to the first bike. The first option seems like a better design even if it comes with a slight efficiency cost. There is a program called linkage that might help you compare the differences here.
Scianiac is correct that the two stage version allows regenerative braking and a sprung tensioner does not. There are mechanical, practical, and ride dynamics reasons not to use regen, but if you want to use it then the two stage reduction you pictured is good for that.

Regarding your question, bends and pulleys on the unloaded side of the chain don't add nearly as much friction as as the same features would on the load carrying side of the chain. Friction in a chain drive is a function of both the articulation angle of individual chain links and the tension on the articulating links.

For a simple two-sprocket chain run (no idlers), the first three rollers or so on either end of the tight load-carrying length of chain can be considered to be under tension, and the rest only under however much static tension is on the chain (preferably negligible). Likewise, the friction in an idler axle increases with the force exerted on the idler.
Thanks for the replies.

I had a gut feeling a lot of the losses were from the interplay between chain and loaded teeth rather than just the amount of sprockets and bearings.

My biggest objection to the jackshaft route is that there is not room to get much reduction done at that spot. Its only substantial purpose is moving the pivot point.

The regen and chain retention that the pivot/jackshaft plan offers will be nice but also catastrophic for the bike if something jams in the chain or sprocket like a stick or my pants!

On the other hand, the idea of a sacrificial spring tensioner as a mechanical fuse is kinda nice.
I'm not a jack shaft fanboy. More complex, more parts to purchase and potentially fail, more noise and vibration. Yes there are times when there's no other options... i get that. But simplicity and reliablility to me is very important, and ranks near the top. If I take a moderate hit on drive efficiency, I just add more batteries. JMO