effect of wind resistance on climbing speed
To continue the previous analysis: what about the effect of wind resistance on climbing?
Well, the result should be easily derived from the last two analyses. But for completeness, starting again from the simple power-speed model, assuming zero wind:
p = fm m s + fw s³
we can then calculate the derivative of the dependence of wind resistance on subsequent speed s, where we're interested in fractional rather than absolute changes of each:
(fw / s) d s / d fw =
(fw / s) (∂ p / ∂ fw) / (∂ p / ∂ s)
where:
∂ p / ∂ fw = s³,
∂ p / ∂ s = fm m + 3 fw s².
So:
(fw / s) ∂ s / ∂ fw =
(fw / s) s³ / ( fm m + 3 fw s² ) =
fw s² / ( fm m + 3 fw s² )
which can be expressed as:
r / ( 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:
f / ( 1 + 2 f ).
So back to our previous example, if I reduce wind resistance by 1%, and if wind resistance is 10% of total power, then my speed is increased by 0.083%. For example, suppose I'm riding on Old La Honda and have the choice whether to draft a rider or ride at my own pace without the draft: what's the potential benefit from that drafting? Well we know that drafting can be worth 30% of wind power on the flats, implying the same reduction in wind drag force (since the comparison is done at a given bike speed). Suppose only a 10% reduction in wind drag force at the speeds involved in climbing (around half the speed on the flats). Then given wind drag is responsible for 10% of the total power to start with, the increase in speed from drafting would be 0.83% = 8.5 seconds out of 17 minutes. Using the previous analysis of the effect of mass change on speed, this is similar to a 1.1% reduction in mass, or for rider + bike + equipment = 64 kg, 710 grams saved. So drafting is a big deal, even on a climb like Old La Honda.
Well, the result should be easily derived from the last two analyses. But for completeness, starting again from the simple power-speed model, assuming zero wind:
p = fm m s + fw s³
we can then calculate the derivative of the dependence of wind resistance on subsequent speed s, where we're interested in fractional rather than absolute changes of each:
(fw / s) d s / d fw =
(fw / s) (∂ p / ∂ fw) / (∂ p / ∂ s)
where:
∂ p / ∂ fw = s³,
∂ p / ∂ s = fm m + 3 fw s².
So:
(fw / s) ∂ s / ∂ fw =
(fw / s) s³ / ( fm m + 3 fw s² ) =
fw s² / ( fm m + 3 fw s² )
which can be expressed as:
r / ( 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:
f / ( 1 + 2 f ).
So back to our previous example, if I reduce wind resistance by 1%, and if wind resistance is 10% of total power, then my speed is increased by 0.083%. For example, suppose I'm riding on Old La Honda and have the choice whether to draft a rider or ride at my own pace without the draft: what's the potential benefit from that drafting? Well we know that drafting can be worth 30% of wind power on the flats, implying the same reduction in wind drag force (since the comparison is done at a given bike speed). Suppose only a 10% reduction in wind drag force at the speeds involved in climbing (around half the speed on the flats). Then given wind drag is responsible for 10% of the total power to start with, the increase in speed from drafting would be 0.83% = 8.5 seconds out of 17 minutes. Using the previous analysis of the effect of mass change on speed, this is similar to a 1.1% reduction in mass, or for rider + bike + equipment = 64 kg, 710 grams saved. So drafting is a big deal, even on a climb like Old La Honda.
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