Wind Speed, vT
m/s

Newton
Pulling
Force,
F

, cD =

### Drag Machine: Drag Power Calculator

Although most wind turbines and windmills - new and old - are based on the lift principle rather than drag, it is worthwhile to explore the physics of drag-based wind machines. There are many similarities in the physics of the two kinds of machines.
Let us test the simplest possible wind machine built on the principle of drag: The animation shows a car without rolling resistance equipped with a sail with an area of exactly 1 square metre facing the wind. The car is pulling a wire running over a pulley with no friction, and a mass attached to it. The work done (energy produced) by this machine is the lifting of the weight to the left.
The power produced, (i.e. the energy production per second) is measured in W (Watts). You can see how much power the machine produces next to the wind speed. We disregard the mass of the car, sail and wire. It does not matter for the final result anyway, it just means that we disregard the initial acceleration needed to start the machine.

How to Use the Calculator
You can easily change the values in the input boxes to the left. You can move them up or down by one unit by clicking the arrows, or you may enter a number. If you enter a number, press the tab key or click outside the box to tell the programme to calculate the result. Remember to use decimal points, and not commas when entering decimal fractions in the boxes. The calculator will not accept unreasonable quantities.
You can read the input and output data for each experiment in the Test Results to the right. Each variable is explained in the section How does the Calculator Work? at the bottom of this page.
You can sort the test results by any column by clicking on the variable name at the top of the column. You can delete a row by clicking on the row number (in the first column). You can plot any two variables against one another by selecting the variables from the popup menus at the top of the list and then clicking the Plot button. Remember to sort your data by the variable on the horizontal (x-) axis before you plot. Finally, you can delete all of the test results by clicking the Delete button.

Find the Power Output, P, and the Ground Speed, vP, for a Constant Load
Try a reasonably low load, F, e.g. 2 Newton, which could represent the rolling and friction resistance in the machine. (The figure is just about the same as the rolling resistance plus the bearing friction of a good racing bicycle). How does the power output, P and the ground speed, vP of the machine vary with the wind speed?

and the Optimal Operating Speed Ratio, λ
Whether you are a workman, a horse or a racing bicyclist, there is some optimal load you can carry to maximise your work (or energy) output. Bicyclists use the gear mechanism to achieve exactly the load, which maximises their pedalling output. It may be an advantage to pedal twice as fast with half the load, in order to deliver maximum power.
How much power, P, can this machine deliver at different wind speeds? You can see for yourself, that it depends on the load, F.
If we start with a zero force pulling the string, obviously the car will be rolling at the same speed as the wind, doing no useful work at all. It is also clear that if we pull with too large a force, the car will simply not move forward, and do no useful work either. The machine can of course do its best with an ideal load somewhere in between.
Your job is now to find out the optimal loading of this machine, i.e. the force that maximises the power output for a given wind speed. Go ahead, you simply increase the force, F, (the weight to the left ) until you get the maximum power, P, from the machine. When you have found the result for the wind speed vT = 6 m/s, what is the speed the car will be travelling with, vP, i.e. what is the optimal ground speed?
Next, try another wind speed, vT, say 9 m/s. Which ground speed, vP, is the car travelling with, in order to deliver maximum power output? Try a third time with another wind speed, VT, and see if you can find the general result for the optimal speed for the car.

Is there an optimal ratio, λ (lambda), between the ground speed of the device and the wind speed? i.e.

λ = ground speed / wind speed = vP / vT

What is the Maximum Efficiency of the Machine?
How large a share of the power in the wind can the machine capture, i.e. what is the maximum power coefficient, cP, for this drag-based wind machine using

1. a flat plate perpendicular to the wind?
2. a semisphere open towards the wind (parachute shape)?

Make sure you do these experiments before you go to the next page and compare your results with ours.

Weight and Mass
*) For some people it is confusing that physicists distinguish between weight and mass. Mass is measured in kg (kilogrammes). Weight (force) is measured in N (Newton). The mass is important to determine e.g. how difficult or easy it is to accelerate a body. Weight is actually the force pulling objects towards the ground due to gravity. An object with a mass of 1 kg is pulled towards the earth with a force of 9.82 N. The factor 9.82 m/s2 is gravity acceleration, and varies slightly in different areas of the globe.

How does the Calculator Work?
We need the following variables:

vT = tailwind velocity in m/s
vP = propulsion velocity (ground speed) of the vehicle in m/s
λ = vP / vT = the operating speed ratio (lambda)
A = area of the sail in m2
= the density of air, 1.225 kg/m3 (at 15° C, and standard atmospheric pressure at sea level)
F = the force applied to put a load on the machine measured in N (Newton)
FD= the aerodynamic drag force on the sail measured in N
cD = the drag coefficient of the sail measured in N/m2 (Newton per square metre frontal area)
P = the mechanical power produced by the machine
cP = the power coefficient, i.e. the share of the power of the wind, which the machine is able to convert into mechanical power

As we explained on the previous page about aerodynamic drag:

FD = cD 0.5 A (vT - vP)2

If the drag force on the machine is larger than the pulling force of the weight, the vehicle will be accelerating. If the drag force is smaller than the pulling force force, it will be decelerating. When the two are equal, i.e.

F = FD

we have a constant speed, vP. Thus, to find vP, we have to solve the equation

0 = cD 0.5 A (vT - vP)2 - F

for its only unknown, vP. We can solve this second order polynomial in vP for each wind speed, vT (entered in the box to the top left) and for each load, F applied to the vehicle (entered in the box to the bottom left).
To find the power produced by the machine, P, we simply multiply F by vP. The power coefficient, cP, is the share of the power of the wind, which our machine is able to capture. As we explained on the page about the power of the wind, the power passing though one square metre area perpendicular to the wind is

P = 0.5 (vT - vP)3,

hence the power coefficient, cP, is

cP = (F vP) / P

The drag coefficients used for the hollow sphere, 1.42 and the flat plate, 1.1 are from White (1999) - see the bibliography. The drag coefficients are not independent of the size of the object, as mentioned on the previous page about the Reynolds Number.

Solving a Second Order Polynomial Equation
Remember, it is fairly easy to find the solutions of a second order polynomial equation: One of the roots of
A x2 + B x + C = 0 is:

x = 0.5 (- B - [B2 - 4 AC]0.5 )

Actually, there are two roots, because you should first add and then subtract the square root of the discriminant, i.e. the expression in the brackets. In our case, we only want the smallest of the two roots, since obviously the car cannot be moving faster than the wind.

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