Problem 1 -- Propel by lift or drag?

1. Let the axis point from the nose of the boat to its tail. The velocity of the boat relative to water would be LaTeX: (-V,0,0) . To produce a drag in the forward (negative x) direction, the oar must move backward relative to the water (positive x direction). Denote the speed of the oar relative to the water is $LaTeX: V_{o-w}$ , then the drag, which must equal to the thrust T, is

$LaTeX: T = \frac12 \rho V_{o-w}^2 \,A\,C_D$

Because LaTeX: C_D=1 , we can calculate the speed of the oar relative to the water:

$LaTeX: V_{o-w} = \sqrt{\frac{2T}{\rho A}}$

The speed of the oar, relative to the boat, is then

$LaTeX: V_{o-b} = V + V_{o-w} = V + \sqrt{\frac{2T}{\rho A}}$

The mechanical power it takes for a person on the boat to operate the oar is the force it takes, T, times the speed of the oar relative to the boat,

$LaTeX: P_{in} = T\,V_{o-b} = T\,\left( V + \sqrt{\frac{2T}{\rho A}}\right)$

The useful power is the thrust times the speed of the boat $LaTeX: P_{useful} = T\,V$ . The efficiency is thus

$LaTeX: P_{useful} / P_{in} = \frac{V}{V + \sqrt{\frac{2T}{\rho A}}}$

2. To produce a lift in the forward (negative x) direction, the oar must move orthogonal to the x axis. Let the y axis be the direction the oar moves relative to the water, then the relative velocity can be denoted as $LaTeX: (0, V_{o-w}, 0)$ . The relative velocity of the oar relative to the boat is then $LaTeX: (V, V_{o-w}, 0)$ . Unlike in the previous question, in which all the forces and velocities are aligned with the x-axis, this problem involves forces and velocities as vectors. To push the oar against a lift force of (-T, 0, 0), the person operating the oar must push it with a force of (T,0,0). The input power, the vector dot product of input force to the oar and the relative velocity of oar to the boat, is thus

$LaTeX: (V, V_{o-w}, 0) \cdot (T, 0, 0) = T\,V$

Because this is the same as the useful power, the efficiency is 1.

Problem 2 -- Arctic Tern

1. The tube of air is equal to the cross sectional area of $LaTeX: \frac32 b^2$ times its lifetime travel distance of $\frac{3}{2} b^{2}$ 2.4 million kilometers, which is 2.3 billion cubic meters. With an air density of 1.225 kg per cubic meters, the total air weighs 2.8 billion kg, or 2.8 million metric tons, same as about 54 Titanic ships.

2. At 10 m/s, the tube of air moved per second is $LaTeX: \frac32 b^2$ times 10m, which is 9.6 cubic meters, weighing 11.8 kg. To sustain the bird's weight of 1N, which requires an upward momentum of 1kg m/s per second, the 11.8 kg of air needs to move downward at a speed of 1/11.8 = 0.085 m/s.

3. The total mass of 2.8 billion kg (Q1) moving at a speed of 0.085 m/s (Q2) contains a total kinetic energy of 10 million Jouls, roughly equal to the chemical energy released by burning 0.29 litters of jet-A.

4. The skin friction drag is $LaTeX: \frac12 \rho V^2 S\,C_f = 0.018N$ . Assuming the bird is streamlined, this is approximately the total form drag. This drag equals the momentum flux of a viscous wake of cross sectional area LaTeX: A_w moving at the same speed V as the bird. The mass of the wake formed per second is $LaTeX: \rho A_w V$ , and the momentum of the wake formed per second is $LaTeX: 0.018 N = \rho A_w\, V^2$ . We can then calculate that LaTeX: A_w = 15E-5 square meters, or 1.5 square centimeters.

5. Multiplying Q4's answer by $\frac{3}{2} b^{2}$ 2.4 million kilometers, we get 260 thousand cubic meters of air, which weighs 0.44 million kg, a tiny fraction of Q1's answer.

6. Multiplying Q5's answer by $LaTeX: \frac12 V^2$ , we get 22 million Jouls, or 0.62 litters of jet-A.