## Friday, October 28, 2016

### Olympic Rowing

This past weekend I went to watch the world’s biggest rowing regatta: the Head of the Charles (held in Boston!). While I was there I saw 8 of the world’s best single scullers who all competed in the Olympics in single sculling boats this past summer, including Kim Brennan (Australian who won gold) and Dr. Gevvie Stone (American who won silver). For some context, single scullers row boats by themselves with two oars (shown below). Ever since we talked about torque in class I thought about how it applied the rowing stroke, also diagrammed below:

When rowing, you pulling on the handle moves your boat forward by rotation around a center of rotation, which is the location at which the blade enters the water. If you’ve ever rowed before it’s intuitive to think that the water is moving just as much as the boat, but in fact when the blade enters the water the boat itself is propelled forward but the water doesn’t move. Thus, if we think of the wrench example we talked about in class, the oar is the wrench and the bolt it’s rotating around is the point where the blade meets the water.
I decided to calculate the torque exerted by Gevvie Stone per stroke during her race at the Head of the Charles in her single. The course is 3 miles (4.8km) long, and she finished in a time of 19:15.6 min. I first converted into hours and calculated her average speed:
19 min x 60s  =  1,140 s + 15.6s = 1,155.6s
1 min

1,155.6 x 1 hr  = .321 hrs
3600s
4.8 km = 14.7 km/hr
.321 hr

To calculate torque, we need Fperpendicular and the radius. The force can be calculated by simply F = ma, where the mass of Gevvie and her boat combined are 85.3 kg.

F = (85.3kg)(Δv/Δt)

It takes approximately 1 second for the handle to travel 1.5m from the catch to the finish (see diagram above), thus the acceleration is 1.5 m/s2.

F = (85.3kg)(1.5m/s2) = 128.0 N

The length of the oar (approximately 285 cm) serves as the radius, so we can create a triangle from the aerial view:

The average length of a sculling oar is 2.85m, and the approximate horizontal length from the blade to the boat is 2.9m, thus we have the hypotenuse and adjacent side lengths, and can calculate the angle to be 11.5 degrees. Now we can calculate the torque:
T = Fcos x * r
T = (128.0)(cos11.5)(2.8)
T = 351.2 Nm

And since she has two oars (one in each hand) her total torque each stroke is 2(351.2) = 702.4 Nm

References