Drivetrain losses: friction-based tension dependence

After reviewing data from Spicer, from Kyle and Berto, from d'Herripon and Verhoeven, all I get is a headache. So to heck with experimental data. It's always a mistake to let experimental data get in the way of a good theory. So I'm going to forge ahead, and pull the numbers I like from the experiments, and resort to a simple theoretical model for the effect of chain tension on drivetrain loss.

I'll start by trusting the pulley friction measurement data I used before, based on experimental data, sure, but experiments on a simpler system (a pulley on bearings) than on a more complex system (a full drivetrain):

Kd = 94 mJ/rev for Shimano Dura-Ace
Kd = 2.4 mJ/rev for CeramicSpeed pulleys

Data measured by Mark Kelly:

Kd = 37 mJ/rev (standard bearings)
Kd = 6.4 mJ/rev (high quality steel bearings)

Then based on a pin diameter of 5/32 inches = 3.97 mm, and using a coefficient of friction for steel-on-steel with a smooth oxide coating = 0.27, with a tooth pitch L = 0.5 inches = 1.27 mm, I calculate:

K / L = π (5/32 inches) (0.27) / (1/2 inch) = 0.265.

That leaves T0. From Spicer's data, I got

2 (K / L) T0 = 50.65 N.

I can guess at the slack-side chain tension from measuring the oscillation frequency, but that wouldn't include the plate friction, so I'll stick with this number.

This gives me:

T0 = 95.6 N.

If I campare these results to Spicer's reported measurements, it's not too terrible, if you allow for the possibility Spicer had an offset in his efficiency measurements:


So what does this all give me?

Suppose I'm riding at 70 rpm up a hill at 290 watts. I have a choice of using a 36/18 or a 50/25. Which gear has the lower drivetrain loss? Following measurement-based Spicer's recommendation I'll assume cross-chaining has little effect. I also neglect differences in derailleur tension between the two gears. Additionally, I'm not using no freakin' Dura-Ace pulleys: instead I have high-quality SRAM pulleys with ceramic bearings (harumph!) and 12 teeth.

Well, here's the model, slightly rewritten:
Ploss = (K / L) P (1 / Nf + 1 / Nr) + C T0 K (1 + Nf/Nr + 2 Nf/Nd) + Kd C Nf / Nd

where

K/L = 0.265,
T0 K = 0.322 J.
Kd = 2.4 mJ (negligible)

Losses are 3.54% of total power.

So then I switch to the 50/25. I then get losses are 3.27% of total power.

So in this case, the smaller gear costs me 0.27% of total power. Assuming 12% of my power goes into wind resistance, the time saved in the climb from 0.27% less power = 0.27% / (1 + 2f) where f = 0.12, which equals 0.22% or 2.2 seconds out of 17 minutes. So that's worth thinking about.

Now what if I get my Red derailleur upgraded to a 15 tooth lower pulley? I'm back in the 36/18. Well, now losses are 3.45% of total power. I save 0.09% of total power. This increases my speed 0.073% = 0.7 seconds out of 17 minutes.

So at least with this model, that 15 tooth lower pulley doesn't do too much for me. Although every second counts... And just knowing I've got a blinged-out rear derailleur is probably good for a solid 10 seconds!

What about Dura-Ace pulleys? Back to the original derailleur, 36/18, I get 3.57% power loss. This costs me 0.03% of my power, and I get to the top 0.25 seconds slower. The difference between Mark's "high quality steel bearings" and the ceramic bearings is too small to even discuss.

So the end result is the compact gearing does seem to hurt a bit under this sort of climbing effort. And the pulleys, which sit on the low-tension side of the chain run, don't make much difference as long as they're clean.

Comments

Ron George said…
Thanks Dan for linking to the drivetrian losses study. I'm building a mini baja car to race powered by 10hp engine with a simply chain driven reduction gearbox , maybe I'll start looking at efficiency losses. ;0

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