Monday, July 20, 2009

effect of mass reduction on climbing speed

A frequent question in cycling speed-power calculations is: "If I save a certain percent of weight, then what is the effect on my speed?". The same ananalysis applies to other mass-proportional power terms: the effect of grade and/or rolling resistance changes on speed. This has tangible significance when considering how much it's worth to invest in some über-light weight-weenie widget, or when considering the trade-off between aerodynamic versus mass in wheels or frames.

Consider a power-speed model which, simplified and neglecting wind speed and acceleration, is expressed as:

p = fm m s + fw

where fm is the mass-proportional coefficient and fw is the wind resistance coefficient. In the standard linearized model, fm = g (grade + Crr) / α, where g is the acceleration of gravity, grade is the road grade, Crr is the coefficient of rolling resistance, and α is the drivetrain efficiency (characterizing the drivetrain as a power-proportional loss term may be a poor approximation, but it's the one which is typically made). The wind-proportional term is ½ ρ CDA / α, where ρ is the mass density of air (typically around 1.1 kg/m³), CD is the coefficient of drag (typically around 0.7), and A is the effective cross-sectional area (typically around 0.5 m² for a racing position). The details of these coefficients aren't terribly important for this analysis, however. All that's important is the ratio of fm and fw.

We can then calculate the derivative of the dependence of mass m on subsequent speed s:

(m / s) d s / d m =

(m / s) (∂ p / ∂ m) / (∂ p / ∂ s)


∂ p / ∂ m = fm s,

∂ p / ∂ s = fm m + 3 fw.


(m / s) ∂ s / ∂ m =

(m / s) fm s / [fm m + 3 fw s²] =

fm m / (fm m + 3 fw s²)

which can be expressed as:

1 / (1 + 3 r)

where r is the ratio of wind resistance power to mass-proportional power. If you know instead the fraction of total power (neglecting drivetrain losses) due to wind resistance, then substitute r = f / (1 - f), where f is this fraction:

(1 - f) / (1 + 2 f)

If f is sufficiently small, this can be slightly simplied to second order as:

1 / (1 + 3 f + 3 f²)

So there it is. If you're riding such that, for example, 10% of the power is from wind resistance, and I reduce mass (or equivalently grade + Crr) by for example 1%, then the effect on speed will be 1% / [1 + 3 × 0.1 / (1 - 0.1)] = 0.75%. This shows that the frequent assumption that a 1% reduction in weight will increase speed by 1% is a significant overestimate of benefit: wind resistance, even if it's a relatively small fraction of total power, is still marginally important when balancing weight and wind resistance on climbs.

So how much is it worth to save a second up Old La Honda? If I run these numbers for my stats and assuming I'm within one second of my 17-minute goal, that second up Old La Honda requires 86.3 grams saved. A regression I did around a year ago on SRAM component prices (SRAM is a good choice, as they have the same design of each of their groups other than weight savings) is weight reduction came at a cost of $3.35/gram. I tend to stick with this number when contemplating other upgrade choices, as well. So it would appear for me saving a second up Old La Honda is worth at least $290. If not, I'd be using Rival on my race bike.

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