drivetrain losses: simple model

My model for drivetrain losses, which I'm sure is unoriginal as it's so obvious, is to model the power loss as each chain link completes a circuit:

Bicycle drivetrain from Wikipedia.

1. Start at the bottom of the crank. The chain link is at a fixed tension determined by the derailleur spring. It moves backward, not bending, until it reaches the bottom derailleur pulley. This motion dissipates negligible energy. Note there is some bending here due to a nonideal chain line. But I'll believe Spicer's assertion, which I found surprising, that chainline is relatively unimportant.
2. The chain link bends as it contacts the bottom deraileur pulley. It does so under low tension. The amount of bending is independent of the gear selection. This dissipates some energy.
3. The chain link moves along with the pulley. This dissipates negligible energy.
4. The chain link unbends after losing contact with the bottom pulley. This dissipates energy. Still the tension is low, set by the derailleur springs.
5. The chain link moves to the upper pulley: no dissipation.
6. Still under low tension, it bends at the upper pulley. Energy is dissipated.
7. The chain link moves along with the upper pulley. No energy is dissipated.
8. The chain unbends as it exits the upper pulley. Energy is dissipated. Tension is still low, determined by the derailleur springs.
9. The chain now moves to the rear cog. No energy is dissipated.
10. The chain bends as it contacts the rear cog. Tension is low. The chain dissipates energy as it bends. The amount of this bending is inversely proportional to the number of rear cog teeth.
11. The chain link moves along with the rear cog. Tension increases as it moves. The chain link is doing work against the rear cog. Work done requires is proportional to a tension difference. It may well be, especially for a slightly stretched chain and a relatively less worn cog, that the majority of this tension increase occurs near the end of the period where the link contacts the cog.
12. The chain unbends as it exits the cog. The amount of this bending is inversely proportional to the number of rear cog teeth. Here the chain link is under maximum tension.
13. The chain link moves to the front chainring under high tension. No dissipation.
14. The chain bends as it contacts the front chainring. The amount of this bending is inversely proportional to the number of chainring teeth, which is typically larger than the number of cog teeth.
15. The chain link moves around with the chainring. Tension decreases as the chain moves, with the chainring doing work on the chain.
16. The chain exits the chainring, unbending and therefore dissipating energy. The bending is inversely proportional to the chainring size. Tension is fixed, determined by the rear derailleur spring.

That's it for chain dissipation. I'll also consider dissipating in the pulleys, which spin under fixed load determined by the derailleur spring. This power is proportional to the rate at which the pulley spins, which equals the ratio of the rate chain links move to the number of teeth on each pulley.

I'll assume when a chain bends, its frictional losses are proportional to the tension plus a constant. Even a chain under no tension experiences some friction, due to the built-in stresses within the chain structure. Further, the loss is proportional to the amount of bending, which is inversely proportional to the number of teeth on the associated cog, chainring, or pulley. Note it doesn't matter for how long the chain is in contact with the cog, ring, or pulley. All that matters is the amount the chain bends to make contact, and subsequently unbends to leave contact.

So the model I'll assume for a bending (or unbending, which is equivalent) event is:

ΔE_bend = K (T + T₀) / N,

where T is the extra tension on the chain (0 when the chain is moving from the chainring ot the cog, positive when moving from the cog to the chainring), T₀ is the combination of the tension of the chain on the return path and the built-in stress leading to friction of an unloaded chain, N is the number of teeth in the cog or chainring or pulley (assuming these are round), and K is a proportionality constant.

As an aside, if the chainring is not round (such as a Rotor ring, then what matters is how much the average link bends, which equals the average curvature of the ring. That will be slightly more than the curvature of a round ring with the same number of teeth. This suggests an exccentric ring (like a Rotor) will have a bit more resistance than a round ring. Additionally, the chain tension and speed become a function fo time, which complicates things further. But that's an aside. I'll keep the analysis to round rings.

Total energy dissipated in one circuit is the sum of bending and unbending events. There's four such events at the pulleys, all at low tension. Then there's one each for the cog and chainring at low tension, and one each for the cog and chainring at high tension. Summing these:

ΔE_rev = K [T (1/N_f + 1/N_r) + 2 T₀ (1/N_f + 1/N_r + 2/N_d)],

where ΔE_rev is the energy dissipated by a link in a revolution, N_f is the number of chainring teeth, N_r is the number of cog teeth, and N_d is the number of teeth in each pulley. Note the tensioned part of the chain is assumed to have a total tension T + T₀, so it appears in both terms.

To calculate the power dissipated, I need to multiply these energies, which are per link, by the rate at which links complete revolutions. This rate is simply number of teeth in the chainring multipled by the rate at which the chainring is turned:

R = Nf × C,

where R is the rate at which chain links complete the circuit and C is the cadence.

So in total:

P_c = C K [T (1 + N_f/N_r) + 2 T₀ (1 + N_f/N_r + 2 N_f/N_d)],

where P_c is the power dissipated by chain bending.

Further simplification is possible by solving for T:

T = P / chain speed

   = P / ( N_f × C × L ),

where L is the chain half-pitch.

So:

P_c = K [(P / L) (1 / N_f + 1 / N_r) + 2 C T₀ (1 + N_f/N_r + 2 N_f/N_d)],

where L is the chain half-pitch and C is the cadence.

I then need to add in power dissipated by the derailleur pulleys:

P_d = K_d R / N_d
    = K_d C N_f / N_d.

These can be combined to yield the total power dissipated in the drivetrain:

P_dt =
P_c + P_d =
(K / L) (1 / N_f + 1 / N_r) P +
2 K C T₀ (1 + N_f/N_r + 2 N_f/N_d) +
K_d C N_f / N_d.

So this leaves us with three terms:

1. A power-proportional term, which depends on the front and rear chainring teeth. This term is larger for a compact crank, where both N_f and N_r are proportionately smaller.
2. A power-indendent term, which depends only on ratios of teeth. This term is smaller for a compact crank, where N_f/N_r may be the same but N_f/N_d is smaller.
3. The pulley friction, which is less for the compact crank, since the pulley spins more slowly.

I've seen it argued the compact should be more efficient, as the chain is under more tension. But chain tension is just one part of the story: chain speed and the amount of bending are other components. At the pulleys, which are assumed to have the same number of teeth, the compact drivetrain results in slower chain speed and therefore the pulleys spin slower and chain links are bending less often there, both reducing the dissipation fo the compact relative to the conventional drivetrain. But then there's the contribution of T₀ at the chainrings and cogs. Here the chain speed is an advantage to the compact, but that advantage is cancelled by the proportionally increased amount of link bending. So for the T₀ component at the cogs and chainrings, it's a wash. Finally there's the T-proportional components at the cogs and chainrings. Here there's the additional factor that the tension is higher in the compact set-up. So in this component the compact set-up dissipates more. So whether the compact is more or less efficient depends on the details. There's no simple answer, at least with this model.

I'll compare with Spicer data next time.