Jeremy McCreary
Well-Known Member
- Region
- USA
- City
- Carlsbad, CA
@Kilowatt and @dodgeman — You're both right. Air resistance and the power lost to it depend on ground speed in different ways.
In still air, ground speed Vg and air speed are the same. The air resistance Ra encountered at Vg in this scenario is a net aerodynamic force given by
Ra = k Vg²,
where Ra is in Newtons, and k is a constant for a given bike and rider in given clothing and riding position. Note that Vg is only squared here.
Ra directly opposes forward motion. The power Pa lost in moving forward against it at speed Vg is
Pa = Ra Vg = k Vg³,
where Pa is in watts, and Vg is now cubed.
Sometimes you'll get farther with a cycling physics problem using a resistance analysis. In other cases, a power analysis is more useful. A lot depends on the empirical data available.
In discussions like these, it's absolutely essential to keep the distinction between resistance and power front and center.
Things are simpler when dealing with relative resistance and power. Let Rt be the total resistance encountered at Vg, and Pt, the total power lost to Rt at Vg. The relative air resistance is just Ra / Rt — the fraction of total resistance due to air resistance.
The relative aerodynamic power loss is then
Pa / Pt = (Ra Vg) / (Rt Vg) = Ra / Rt
In other words, the relative air resistance at Vg equals the relative aerodynamic power loss at Vg. When Ra becomes 50% of total, so does Pa.
In still air, ground speed Vg and air speed are the same. The air resistance Ra encountered at Vg in this scenario is a net aerodynamic force given by
Ra = k Vg²,
where Ra is in Newtons, and k is a constant for a given bike and rider in given clothing and riding position. Note that Vg is only squared here.
Ra directly opposes forward motion. The power Pa lost in moving forward against it at speed Vg is
Pa = Ra Vg = k Vg³,
where Pa is in watts, and Vg is now cubed.
Sometimes you'll get farther with a cycling physics problem using a resistance analysis. In other cases, a power analysis is more useful. A lot depends on the empirical data available.
In discussions like these, it's absolutely essential to keep the distinction between resistance and power front and center.
Things are simpler when dealing with relative resistance and power. Let Rt be the total resistance encountered at Vg, and Pt, the total power lost to Rt at Vg. The relative air resistance is just Ra / Rt — the fraction of total resistance due to air resistance.
The relative aerodynamic power loss is then
Pa / Pt = (Ra Vg) / (Rt Vg) = Ra / Rt
In other words, the relative air resistance at Vg equals the relative aerodynamic power loss at Vg. When Ra becomes 50% of total, so does Pa.
Last edited:
