A Worked Example - Small Pigeons

Figure 1: Mechanical energy consumption curve (in bold) for a carrier pigeon of mass 31 grams flying at about 400 m ASL. Location of absolute minimum power consumption is illustrated by the dashed line; $v_{am}$ = 9.07 m/s. Location of the maximum range speed, Vmr, located by use of a tangent line (small dashes), is given by the solid lines. For this bird, $v_{mr}$ = 11.9 m/s. Power required is 2.67 J/sec and 3.04 J/sec for $v_{am}$ and $v_{mr}$, respectively.

Let's work this out for a small common pigeon (Columba livia) which will migrate at 400 meters of altitude. Here are measurements for the pigeon:

Air Density (400 m)	$\rho = 1.19$	kilograms/cubic meter
Total Mass	$m=.31$	kilograms
Drag Coefficient	$C_{Db}=.366$	unit free
Body Area	$A_b=3.73 \times 10^{-3}$	meters squared
Wing Disc Area	$A_{wd}=.283$	meters squared
Fat Fraction	$.24$	unit free

We can calculate coefficients for the power curve:

$$C_{par} = \frac12 \rho \ A_b \ C_{Db} = .5 \times 1.19 \times (3.73 \times 10^{-3} ) \times .366 \doteq 8.12\times 10^{-4} ,$$

and

$$C_{ind} = K_i \ \frac{m^2 g^2}{2 \ \rho \ A_{wd}}= 1.2 \frac{.31^2 \times 9.81^2}{2 \times 1.19 \times .283 } \doteq 16.48.$$

The equation for mechanical power requirements is then

$$P = 1.1 \left[ \frac{16.48}{v} + 8.12\times 10^{-4} v^3 \right].$$

Energy required for a flight of distance $D$ is

$$E_{req} = D \frac{P(v)}{v} = 1.1 D \left[ \frac{16.48}{v^2} + 8.12\times 10^{-4} v^2 \right].$$

Taking a derivative to determine where the minimum $E_{req}$ occurs,

$$\frac{d}{dv}E_{req} = 1.1 D \left[ -2 \frac{16.48}{v^3} + 2 ... ... 8.12\times 10^{-4} v \right] \stackrel{\mbox{\tiny set}}{=}0.$$

This can be solved for the maximum range speed, $v_{mr}$,

$$v_{mr}^4 = \frac{16.48}{8.12\times 10^{-4}} \quad \mbox{whic... ...r} = \sqrt[4]{\frac{16.48}{8.12\times 10^{-4}} } \doteq 11.93.$$

The rate of energy consumption at this optimum speed is

$$P_{mr} = P (v_{mr}=11.93) \doteq 3.04 \frac{\mbox{Joules}}{\mbox{sec}}$$

See Figure 1 for a graphical depiction of this speed, the power curve, and the overall minimum power speed.

On the other hand, the total energy available to the pigeon depends on its stored fat. Since the fat fraction for this pigeon is $F=.24$, the total mass of the fat available is $(m-m_e)$, which satisfy

$$F = \frac{m-m_e}{m_e} \longrightarrow me \ F = m-m_e \longrightarrow m=(1+F) m_e \longrightarrow m_e = \frac{m}{1+F}.$$

Thus $$m_e = \frac{.031}{1.24} = .025$$ and therefore the mass of fat is $m_{fat} = m-m_e = .031-.025 = .006$, or 6 grams of fat. Then the energy available is

$$E_{avail} = \eta \ e \ m_{fat} = .23 \times (3.9 \times 10^7) \times .006 \doteq 5.38 \times 10^4 \mbox{ Joules}.$$

Therefore the time the pigeon can stay airborne at a speed of $v_mr$ is

$$\mbox{Time } = \frac{E_{avail}}{P_{mr}} = \frac{5.38 \times ... ...mbox{Joules}}{\mbox{sec}}} \doteq 1.77\times 10^4 \mbox{ sec},$$

and the maximum distance the bird can travel is

$$D = v_{mr} \times \mbox{ Time }\doteq 11.93 \frac{\mbox{mete... ...mes 10^4 \mbox{ sec}) \doteq 21.11 \times 10^4 \mbox{ meters},$$

or $D = 211.1$ kilometers (about 132 miles).