In the sport of indoor rowing (and probably others) I know that to double your speed you must increase your power output by a factor of eight.

I am wondering why this is the case. The indoor rower works by measuring data from a fan e.g. speed of fan, deceleration, acceleration etc.

Normally with objects moving through air there are (I believe) two factors to consider when speed is doubled. One is that you will hit the air twice as fast and the other is that you will hit twice as much air. Hence the power required increases by a factor of four.

What is the additional factor I'm forgetting to consider in the case of indoor rowing?

  • 1
    $\begingroup$ Two things. First, power goes by $v \cdot F_{drag}$. Second, recall that the thing you're trying to model is a boat, so in addition to Plank (simple momentum transfer, $F_{drag} \propto v$) and Prantl (viscous friction; $\propto v^2$) drags there is a wave-making term ($\propto v^3$) and eventually turbulent drag (still higher powers). The problem is complicated. $\endgroup$ Jan 26, 2016 at 15:15

1 Answer 1


Hitting the air twice as fast takes four times as much energy ($E = {1\over 2} m v^2$). So there's an extra factor of two.

To expand a little; you're fighting drag ($F_d$) in a rowing machine and drag increases as the square of velocity. ($F_d \propto v^2)$.

Now, $Energy = force \times distance$ and $Power = {energy \over time}$ so

$Power = force \times {distance \over time} = force \times velocity$

Since $F \propto v^2$, it follows that $P \propto v^3$.


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