Sec.4.7 - Planes and birds: Minimizing energy

Applied Project 2 in Sec.4.7, Calculus by Stewart
Chinese version: 飛機和鳥：能量最小化

Small birds like finches alternate between flapping their wings and keeping them folded while gliding. In this project we analyze this phenomenon and try to determine how frequently a bird should flap its wings. Some of the principles are the same as for fixed-wing aircraft and so we begin by considering how required power and energy depend on the speed of airplanes.

Question 1: The power needed to propel an airplane forward at velocity​ $v$ is

$$ P = Av^3 + \frac{BL^2}{v} $$

where $A$ and $B$ are positive constants specific to the particular aircraft and $L$ is the lift, the upward force supporting the weight of the plane. Find the speed that minimizes the required power.

Answer:

$$ v_P = \sqrt[4]{\frac{BL^2}{3A}} $$

Question 2: The speed found in Problem 1 minimizes power but a faster speed might use less fuel. The energy needed to propel the airplane a unit distance is​ $E = P/v$ . At what speed is energy minimized?

Answer:

$$ v_E = \sqrt[4]{\frac{BL^2}{A}} $$

Question 3: Hows much faster is the speed for minimum energy than the speed for minimum power?

Answer:

$$ \frac{v_E}{v_P} = \sqrt[4]{3} $$

Question 4: In applying the equation of Problem 1 to bird flight we split the term $Av^3$ ​into two parts:​ $A_b v^3$ ​for the bird’s body and​ $A_w v^3$ ​for its wings. Let​ $x$ be the fraction of flying time spent in flapping mode. If $m$ is the bird’s mass and all the lift occurs during flapping, then the lift is $mg/x$ and so the power needed during flapping is

$$ P_f = (A_b+A_w)v^3 + \frac{B(mg/x)^2}{v} $$

The power while wings are folded is $P_o = A_b v^3$. Find the average power over an entire flight cycle.

Answer:

$$ P_a = x P_f + (1-x) P_o = A_b v^3 + A_w v^3 x + \frac{B m^2 g^2}{xv} $$

Question 5: For what value of​ $x$ is the average power a minimum? What can you conclude if the bird flies slowly? What can you conclude if the bird flies faster and faster?

Answer:

$$ x = \frac{mg}{v^2}\sqrt{\frac{B}{A_w}} $$

It can be seen that $x$ is inverse proportional to $v^2$, so,

• 鳥飛的慢,拍翅時間佔總體飛行時間上升
• 鳥飛的快,滑翔時間佔總體飛行時間上升

Question 6: The average energy over a cycle is $E_a = P_a/v$. What value of $x$​ minimizes $E_a$?

Answer:

$$ x = \frac{mg}{v^2}\sqrt{\frac{B}{A_w}} $$