Interesting stuff

Plugged some examples from recent rides into this formula.

First up was a flat stretch of ground where we had a two way average of 19mph. It was a long enough stretch and not much time passed between the runs so it is assumed this cancels out the wind.

Using a test case with flat ground (slope = 0) revealed something interesting to me about the equation. You can break it in two parts. The first part is all about energy to push through the wind. The second part looks like it all about fighting gravity, but in the case where there is no elevation change it still has a contribution, and that is loss through the tires. That friction loss is a bigger part of the overall picture than I would have guessed, given the small (.005) estimate for coefficient of rolling resistance.

In our first example speed was 19mph which is 8.5 meters/sec. That gives 183 Watts steady output. About 145 aero, 38 rolling.

For the second example, there's a big hill near my house with a tough portion of about a mile of 11% grade. In a recent ride I averaged 5.6 mph through it. This version of the calculation shows only 4 Watts aero, but 256 for climbing.

Taking the whole climb including a few easier spots the averages were 7.2 mph at 8.1%. Overall power = 254. More than I thought!

I guess that means if I really got down to it I could push faster on the flats if I were motivated. Let's see how fast that should be, starting with the power and solving for speed. Answer: 21.7 mph

If that holds up in reality I'll be satisfied with that for sure.

Useful references for this formula:

mph to m/s is approximated by multiplying mph by .4469
grade is simple rise/run (elevation/distance)
If you want to convert grade to degrees (not required for this formula) take the arctan of the grade %. One spot on the climb mentioned here is 17.4% grade, but that only works out to 9.8 degrees, which sure doesn't sound as hard as it really is.