The Physics of Rowing covers the theoretical derivation of the effects of a variety of factors on erg and water speed in rowing. In particular they derive a formula for the work rate (power) required to move the mass of the rower up and down the slide of a static erg:
$P = 4 m s^2 \left(\frac{R}{60}\right)^3$
This is useful for looking at the theoretical relationship between mass, body movement and rate, but it assumes energy is spent on the drive and recovery phases of each stroke. When rowing “efficiently” an athlete will try to minimise the amount that they are “pulling” with their legs at the finish of the drive phase to slow themselves down and instead slow their momentum by pulling on the erg handle. The effect of this is that they are transferring the kinetic energy developed by their body’s mass into the flywheel, so it will be accounted for by the erg’s measurement. To adjust for this, the above formula can be altered to take into account the the fraction of time spent on the recovery phase of the stroke.
The following is the full derivation of a similar formula that accounts for energy only being “spent” on the recovery phase of the stroke.
The time for a single stroke $t_s$ and the time span of the recovery $t_r$ are:
$t_s = \frac{60}{R}$
$t_r=(1-f)\frac{60}{R}$
Where $R$ is the stroke rate in strokes/min and $f$ is the proportion of the time of the stroke spent on the drive. If the speed of the recovery $v_r$ is assumed to be constant, and the distance the rower’s mass moves $s$ is known, then the kinetic energy developed by the rower’s mass on the recovery $E_r$ can be calculated:
$v_r = \frac{s}{t_r}= \frac{sR}{60(1-f)}$
$E_r = \frac{1}{2}mv_r^2 = \frac{1}{2} m \left(\frac{sR}{60(1-f)}\right)^2$
Work is done to accelerate the rower’s mass to $v_r$ and back to 0 each stroke, equivalent to this calculated kinetic energy each time. To get the power expended each stroke in the recovery $P_r$, the total work done must be divided by the time for the whole stroke.
$P_r = \frac{2 E_r}{t_s} $
$P_r = m \left(\frac{s}{1-f} \right)^2 \left(\frac{R}{60}\right)^3$