Rower center of mass

Here’s the result of some other home experiment. In the previous post, I wrote about the angular velocity of the scull, and in an earlier post I discussed the velocity of the rower’s center of mass with respect to the boat, and how this velocity affects boat speed. Only in emergency situations does the handle leave the rower’s hand, so in all practical situations there should be a relation between handle speed and the speed of the rower’s center of mass.

In the picture, I show a measurement on my own body. For various positions in the stroke,  I measure the distance of the hands, feet, shoulder, and slide position from the hands position at the catch. All values are in centimeters. On an internet discussion forum for crime writers (!), I found some weight percentages for the weight of different body parts: Head 9%, legs, 33%, arms 11%, trunk 46%. I used these weight percentages to calculate the center of mass (marked CM in the picture).

The center of mass velocity is roughly half the handle velocity. However, at the start of the stroke, both velocities are almost equal. I approximate the relationship between handle and center of mass position by, using relative coordinates, x_{\rm handle}=1 corresponding to the finish position:

x_{\rm CM} = x_{\rm CM, 0} + x_{\rm handle} - \frac{x_{\rm handle}^2}{2 }

This approximation is a little too fast at the start of the stroke and a little too slow at the end of the stroke. I expect this to be of minor influence. On the other hand, Kleshnev is discussing these differences here, and indeed I try to row without the “shoulder swing” myself, so perhaps I should improve the model at this point.

The position of the handle is given by

x_{\rm handle} = L_{\rm in} (\sin\phi_0) - \sin\phi)

and, thus:

v_{\rm handle} = \dot{\phi} L_{\rm in} \cos\phi

The crew velocity is given by:

v_c = v_{\rm handle} \left( 1 - x_{\rm handle}\right)