I am struggling calculating the correct torque needed to rotate a windmill object at the required speed. Suppose I have a windmill:

     <- o ->

with a motor at the centre, and the four blades are 2.4m long, each weighing about 10kg. I'd like to calculate the torque needed to rotate the windmill at around 5 rpm.

I'd be fine calculating the torque needed if there were just one blade, but my intuition tells me that since everything is balanced and symmetric, ANY non-zero torque would be sufficient. However, this answer seems "too easy".

I'm not an engineer, just a meagre pure mathematician, so any pointers or tips (or even answers!) would be greatly appreciated.

  • 1
    $\begingroup$ "ANY non-zero torque would be sufficient" - the only forces will be air resistance, friction in the bearings, etc, and those forces will be small at only 5 RPM. A more relevant question might be "how quickly do you want to accelerate the rotor from rest to 5 RPM". That will depend on the moment of inertia of the rotor about the axis and the acceleration required. $\endgroup$
    – alephzero
    Aug 15, 2016 at 16:23
  • $\begingroup$ As stated by @alephzero, what we need is the acceleration desired to reach from zero to 5 RPM. Or, equivalently, the time necessary to accelerate to the desired speed. $\endgroup$
    – Wasabi
    Aug 15, 2016 at 16:38

1 Answer 1


Torque, $\tau$, is required for rotational acceleration, $\tau = I\alpha$, and for viscous losses, $\tau = b\omega$. Here I'm using $\phi$ for angular position, $\omega$ for angular velocity, and $\alpha$ for angular acceleration. $I$ is the moment of inertia as "seen" by whatever you care about, and $b$ is damping constant, again as seen by whatever you care about.

For the windmill blades, they will have some moment of inertia about their centers of mass. You can use the parallel axis theorem to "convert" a moment of inertia from about the center of mass to about some other useful point in space - this would be the center of the hub.

It's important to note here that the hub is likely to have its own drum to which the blades attach - the blades don't all meet in the exact middle like a bunch of orange slices, which means that the center of the hub (about which the blades rotate) is probably a phantom or projected point in space. Use that distance for the parallel axis theorem.

So, once you have the moment of inertia for everything, then you need the viscous damping coefficient. This can be very difficult to measure, depending on the size of your windmill, but you can try to determine it by starting the windmill at a known speed and then removing the actuating device and timing how long it takes for the speed to decay.

Net torque is given by:

$$ \tau_{\mbox{net}} = \tau_{\mbox{inputs}} - \tau_{\mbox{outputs}} \\ $$

The inputs are the things that act on the windmill, the outputs are the things that are acted on by the windmill. Generally, the only thing acting on the windmill would be the torque applied by the wind, but this is a complex relationship that has a lot to do with wind speed, air quality, and blade shape. I would imagine you already have some modeled relationship, so we can just leave that as $\tau_{\mbox{wind}}$. Everything else described mathematically:

$$ \tau_{\mbox{net}} = \tau_{\mbox{wind}} - \tau_{\mbox{generator}} - \tau_{\mbox{acceleration}} - \tau_{\mbox{viscous}} \\ $$ $$ \tau_{\mbox{net}} = f(\mbox{wind}) - \tau_{\mbox{generator}} - I\alpha - b\omega \\ $$

The generator torque can be approximated by examining the power output of the generator (I'm assuming this is, in fact, a power generating windmill and not a grain windmill or something else). If the output power of the generator is $P = IV$, where $P$ is power in Watts, $V$ is voltage (in volts) and $I$ is current in amps, then you could approximate the generator torque via:

$$ P_{\mbox{out}} = \eta P_{\mbox{in}} $$

Where the output power is the electrical side, the input power is the mechanical side, and eta ($\eta$) is a constant denoting the efficiency of the power conversion process. I would probably put this around 0.8 (80%) if it were my simulation, but you could do anything from 0.7 to 0.95 and probably make a justifiable case. In the end, though, you want your simulations to be as accurate as possible, so if you're using an off-the-shelf (OTS) generator, ask for the efficiency of the device.

Another way of stating the above conservation of power formula is:

$$ P_{\mbox{out}} = \eta P_{\mbox{in}} \\ IV = \eta \tau_{\mbox{generator}}\omega \\ $$

From the mechanical power formula, $P=\tau\omega$. You can reformulate this as:

$$ \tau_{\mbox{generator}} = \frac{IV}{\eta\omega} \\ $$

So, finally,

$$ \tau_{\mbox{net}} = f(\mbox{wind}) - \frac{IV}{\eta\omega} - I\alpha - b\omega \\ $$

This net torque gets applied to the windmill shaft, which causes everything attached to accelerate or decelerate (again, from $\tau = I\alpha$). Integrate the acceleration for velocity and again for position.

If you're trying to control the speed, the only two things you really have control over are your wind inputs (by varying blade pitch) and your generator output (by varying output voltage). If you're connected to a "sufficiently large" electrical "body" (like the national power grid), then a very slight change in output voltage should make dramatic differences on how "loaded" or "unloaded" the generator is. For a small setup, though, you might want to consider adding bank of power resistors to ensure there is always sufficient "demand" for your electrical power, otherwise the automated algorithm may over-voltage everything attached in an attempt to increase generator output power.

  • 1
    $\begingroup$ Oh, one final comment about the "as seen by the thing you care about" proviso - You said the windmill is "connected to the motor" - if this is the case, then no problem. If the windmill is connected to a gearbox that is connected to the motor, then you need to take that into account. Speed is transformed by the gearbox ratio, but inertia is changed by the square of the gearbox ratio. Rotor inertia of a motor/generator is usually very significant. $\endgroup$
    – Chuck
    Aug 15, 2016 at 18:05

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