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Drag is exponentially related to speed
Not quite. Drag is proportional to the square of velocity, so power required to overcome drag is proportional to the cube of velocity. Ptown's broad point stands, small absolute or relative changes in velocity make larger differences to the time taken to cover a distance if the initial velocity is small, because time taken is inversely proportional to velocity.
Small absolute change (ΔV=1mph):
15mph to 16mph saves 6m15s over 25 miles
25mph to 26mph saves 2m18s over 25 milesSmall relative change (ΔV=5%):
15mph to 15.75mph saves 4m45s
25mph to 26.25mpg sames 2m51sEven after allowing that slow riders expend a smaller proportion of total power overcoming drag, because other resistance forces are proportional to the first power of velocity, time savings for the same incremental reduction in CdA can be greater for slower riders.
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gbj_tester
skinny
Force is proportional to velocity squared. So as required increases the force needed increases exponential.
Force (drag) = Density * coefficient drag * area * (Velocity^2)/2
So at low speeds (15mph) drag isn’t that high. At if you were to reduce coefficent drag, the proportional increase in speed is quite small, at overall low power production.
However if you’re travelling 25mph then due to the relationship the drag saved correlates to quite a reduction in power needed to over come the drag.
Margins aside (not a factor in my decision).