The unreasonable relationship between speed and power
After physics class finished, I approached my teacher and asked him how much power is required for an “ideal” helicopter to hover.
He grabbed a piece of paper drew a helicopter, with an arrow pointing down and another point up and said, “Well, clearly the thrust has to be equal to weight of the helicopter.” He then added, “But the helicopter isn’t moving. That means there is no work being done and no power is required”. As he was saying these last words, his face transitioned from confusion to excitement, it was this simple but challenging problem that has awaken his the passion for the field.
Let me show you the equation that changed Yves life, it show that the larger the propeller, less energy is required to generate thrust. (read this great article by Rhett Allain for a detailed proof)
Imagine our propeller covers a circle of area A, after a certain time, the propeller has pushed down a cylinder of air of size A * h.
This will generate a reaction F pushing the propeller upwards, and making are helicopter hover.
We want to substitute the variables in the force with our auxiliary equations to arrive to something that depends on the radius of the propeller and the speed.
If you are like me, equations mean nothing until I plot them. Let’s try to answer, “How large does my propeller have to be, and what is the speed of wind required to lift a 1kg helicopter?”
import numpy as np
import matplotlib.pyplot as pltF = 9.8
r = np.arange(0.01, 1.0, 0.01)
v = np.sqrt(F / (0.5 * 1.225 * np.pi )) / r
plt.plot(r, v)
plt.xlabel('radius [m]')
plt.ylabel('air velocity [m/s]')
plt.grid(True)
plt.show()
The answer is that a propeller of 0.05 cm has to spin fast enough to move the air at 50 m/s and make a kilogram helicopter hover. We can instead choose to use a very large propeller with a radius of 0.5 meters, which will require air moving at 2.7 m/s.
Both options produce the exact same amount of lift. And you read up to here, congratulations! The exciting part is about to start. The amount of power needed is wildly different.
Power is defined as the work done by unit of time. In this case we are making work by accelerating the particles of air.
We again substitute using the auxiliary equations from before to arrive to:
We now want to plot the power required for every point in the curve of the last plot.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3DF = 9.8
r = np.arange(0.01, 1.0, 0.01)
v = np.sqrt(F / (0.5 * 1.225 * np.pi )) / r
p = 0.25 * 1.225 * np.pi * r**2 * v**3
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(r, v, p, 'r')
ax.plot(r, v, 'b', zdir='z', zs=0)
ax.set_xlabel('radius [m]')
ax.set_ylabel('air velocity [m/s]')
ax.set_zlabel('required power[watts]')
plt.grid(True)
plt.show()
Going back to our two examples, the required power for a 5 cm propeller is 300 watts, but for a 50cm propeller is 5 watts.
That is 60x more efficient. Crazy!
This means you can lift anything by having a propeller large enough and by using no power. Is this actually possible? The answer is that it kinda is:
