2

The drive-side crank of a bicycle must be more efficient than the non-drive side, right?

Not only does the power not have to go through the BB, but the drive side is usually also attached to the chainring out closer to the end, especially on high-end cranks like this one:

enter image description here

Is there any data on the size of the difference in efficiency? I'm not looking for data on this specific crankset, anything general is fine.

Improve this question
10
  • 1
    Makes you wonder if a left handed (therefore left-footed) person would benefit from a mirrored bike, one with the transmission on the left.
    – Criggie ♦
    Aug 7, 2016 at 7:31
  • If you're convinced there's a difference here then you'll love Z-shaped cranks.
    – Daniel R Hicks
    Aug 7, 2016 at 12:11
  • @DanielRHicks The science that tells us Z-cranks are inefficient is the same science that tells us that the drive side is more efficient than the non-drive side. Therefore, by throwing science out (as I've seen you do multiple times in the past), you are making yourself the one who loves Z-shaped cranks.
    – BSO rider
    Aug 7, 2016 at 13:47
  • Exactly how is the drive side "more efficient". Where is the energy disappearing on the "less efficient" side?
    – Daniel R Hicks
    Aug 7, 2016 at 13:55
  • 1
    On the non-drive side, energy is being used on BB flex, and more importantly, on crank flex (this lost energy ends up generating heat, if you were wondering). That flex is virtually non-existent on the drive side, due to the crank being attached to the chainrings all the way out to the end, as you can see in the pic.
    – BSO rider
    Aug 7, 2016 at 14:04

4 Answers 4

Reset to default
6

There will be windup (flex in the crank under force). Both the drive side (spiders and chain ring) and the non-drive side (shaft, spiders and chain ring) will experience windup during power transfer. The amount of wind up is however very small (and depends on the magnitude of the force applied), but will be slightly higher for the non-drive side as more components are involved.

That said, once the small amount of windup has occurred (at the start of the power stroke), both sides will transfer force with equal efficiency for the rest of the pedal stroke. Because the windup is so small, overall efficiency between sides will be essential identical.

As an aside frame flex will be a much bigger component, convince your self by locking your brakes and apply force to either crank and see what flexes. I find I have better power transfer with the right amount of frame flex as it allows me to get into a rhythm that I can best describe as like riding a wave. Some term this planing. What ever you call it, a bike and components that are too stiff will feel "dead" for lack of better words.

Improve this answer
2
  • Nice. I have noticed my newer carbon / alu frame favours a lower cadence than my 80's steel frame. At about 90 it sings, whereas the steel frame favours a cadence of 100 or more.
    – andy256
    Aug 7, 2016 at 6:10
  • @andy256 with 172.5mm cranks, a cadence of 100+ and power at 250+ watts the turn over feels almost magical on my steel bike. I was test riding a carbon bike last week and it felt dead in comparison, I will have to test ride again, but at a lower cadence as per your observation!!
    – Rider_X
    Aug 7, 2016 at 6:19
4

Just an observation:

With spider based power meters, such as SRM or Quarq, you can test the torque applied to the spider when a weight is hung from the pedal spindle of either crank arm. Torque can be measured down to small fractions of a Nm (which at typical cadences is equivalent to about 1/3rd of a watt).

There's not been any systematic asymmetry in the torque measured at the spider for the many dozens of meters I have validated calibration for.

Improve this answer
1
  • Check the point about windup from Rider_X's answer.
    – ojs
    Aug 7, 2016 at 16:36
3

Uh, you realize, don't you, that there is a shaft that connects the left and right sides, and that shaft, on any halfway decent bike, is incredibly stiff. There may be some very slight difference in efficiency due to the flexing of the shaft, but it would only be measurable in laboratory conditions, with a mechanical "rider".

And the fact that the drive-side shaft is usually longer means that the pedal position on that side will be subject to more flex, causing the leg on that side to "flail" about slightly more. But, again, the difference would only be measurable with laboratory instruments.

All in all, any measurable effect would be orders of magnitude less than effects due to asymmetries in the rider's physiology.

Improve this answer
5
  • The Ergomo power meter and the Rotor InPower meter are instrumented BBs that measure the twist in the BB, so BB twist is both measurable and exploitable. BB-based power meters only measure the power supplied by the left leg, and cannot measure power supplied by the right leg.
    – R. Chung
    Aug 7, 2016 at 21:52
  • @R.Chung - If you mean that these meters utilize strain gauges on the cranks axles, the amount of deformation they're measuring is microscopic.
    – Daniel R Hicks
    Aug 8, 2016 at 1:37
  • I agree that the forces are small but they are absolutely, undeniably, present and can be measured. I don't know whether the InPower uses strain gages (I suspect so but haven't seen one to know for sure) but the Ergomo didn't use strain gages; it used an optical sensor behind a "comb" to measure torsion. Similarly, the Power Tap rear hub also uses a torque tube and strain gages to detect torsion between the freehub and the hub shell.
    – R. Chung
    Aug 8, 2016 at 5:39
  • Yep, and they all measure microscopic amounts of deformation.
    – Daniel R Hicks
    Aug 8, 2016 at 11:52
  • Exactly. But they're actual, real-world, on-the-bike, all-weather instruments, not laboratory instruments. Most people can't distinguish the difference between 20 mph and 21 mph without a speedometer but that doesn't mean that the difference doesn't exist or can only be measured in the laboratory, or with a "mechanical rider." Most people can't distinguish the difference between 10 miles or 10.1 miles without an odometer or some other way to measure distance but that doesn't mean the difference doesn't exist, or can only be measured in a laboratory.
    – R. Chung
    Aug 9, 2016 at 12:30
2

Yes, the power from the drive side has to "go through" the bottom bracket. The leverage you're applying to create tension on the chain is pivoting around that bracket, creating a load on its bearings where energy is dissipated, even though the linkage between crank arm, chain wheel and chain "bypass" the bracket. Energy flows to and is dissipated by every place in the moving machine where there is friction.

If the drive side crank suddenly became separated from the spindle, you wouldn't be able to pedal; it's part of the machine.

If anything, the load inside the bottom bracket might be somewhat better distributed when the opposite crank is driving. That is just a hunch.

Improve this answer
2
  • I agree that bearing load is another factor (I didn't touch in my answer for the sake of brevity). I agree that non-drive side loads likely result in more even bearing loads (anecdote: I find drive side BB bearings always go first, despite being more protected from the elements). That said, I suspect the amount of realized friction and power loss is quite small, otherwise the moment BB went external manufacturers would have started spec'ing larger bearings on the drive side. Talk about an avenue for differential marketing.
    – Rider_X
    Aug 7, 2016 at 16:46
  • Drive side bearing is always under load from chain tension. I believe that is why it fails first.
    – ojs
    Aug 7, 2016 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.