# What could micro-propulsion do?

I have been following the NASA developments on the EM Drive with great interest, and so I was very interested to see that Paul March (a NASA Eagleworks engineer) today stated that he is measuring a force of over 100µN for 80W of electrical power.

I was curious to put this into context. A quick google reveals that you can purchase solar panels for spacecraft that produce 300W/kg.

Let’s say that you can scale up the solar panels such that the mass of the engine is negligible compared to the solar panel mass. (A big assumption)

Taking his Paul March’s figure of 100µN/80W at face value, 300W per kg solar panels would produce a force per kg of = 0.0004 Newtons/kg, so an acceleration of:

$a = F/m = 0.0004 m / s^2$

This sounds extremely small, but we can accelerate constantly over long periods of time.

After 1 year of this acceleration, we would have a delta-velocity, dV, of:

$dV = at = 12 km/s$  (43,000 km/hour)

A delta-V of 12km/s isn’t shabby at all. Here’s a delta-V map of our solar system:

DeltaV map of our solar system

So from Low Earth orbit, a dV of 12km/s gets us to the moon in very low orbit, and coming back again! $(2.44+0.68+0.14+0.68+1.73 = 5.67)$

(of course you can’t actually land and take off since the force is less than that of gravity at the surface, but you could do a very low fly-by).

So we could have a shuttle going to the moon and back every year. Not sure if that’s useful, but I’m sure someone could think of a use!

What else could we do with it?

From that delta-V map, we could get to a low orbit around Venus in about 6 months $(2.44+0.68+0.09 + 0.28+0.36+2.94 = 6.79 km/s)$.

We could get to pretty much any planet in up to a year. (e.g. low orbit Neptune is $2.44+0.68+0.09+0.39+2.7+0.99+0.69+0.27+0.35+6.75 = 15.35 km/s$)

Imagine what we could do with a spacecraft capable of visiting the furthest planets and returning in just over a couple years.

And that’s all without any sort of gravitational slingshots!

## Effect of moving away from the Sun

A reader pointed out that the solar energy collected would of course decrease as you moved away from the Sun.

We can do a quick calculation – consider a round trip to Mars and back.

Distance from Sun to Mars: 1.38 au

$(1/1.38)^2 = 52%$

So the power generated by the solar panels would be reduced by 50%. That means, roughly, on average you’d get 75% of the acceleration that I mentioned, and so would increase the round trip from lower earth orbit to lower Mars orbit from 0.95 years to 1.3 years.

(I’ve obviously linearised where I shouldn’t have, the effect being an underestimation of the time, but you can get a crude back-of-the-envelope feeling for this.)

## Solar Panels or Nuclear Power?

Nasa’s “Safe Affordable Fission Engine” can generate 100kW with 500Kg mass.  So 200W/kg.  Less than the 300W/kg that the solar panels can produce near earth, and on par with the average for a trip to Mars.  For a trip much further than the Mars then it starts to make sense.

## Batteries?

Batteries store around 250Wh/kg. So 1kg can generate 250 watts for one hour… which is what a solar panel can do continuously.

## Theoretical maximums

If it could be scaled to produce 1g acceleration, a human could explore this galaxy and even a neighbouring galaxy within their lifetime: http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

The problem however is that at high velocities, space dust becomes more and more dangerous. At even high velocities, the very bumps in the spacetime from gravity waves becomes dangerous.

# Balancing on a stick

Consider a pole of length $h$, with a motor and flywheel attached to one end.  The motor can spin up the flywheel.  The other end is on a non-slip floor.

If the stick is at angle $\theta$, and the total mass of motor $M_m$ plus mass of flywheel, $M_f$ is $M_m + M_f = M_t$, the torque due to gravity, $\tau_g$ is:

$\tau_g = M_t gh \cdot \sin(\theta)$

For a solid and hollow flywheel, the moment of inertia is:

$I_s = \dfrac{M_f r^2}{2}$ and $I_h = M_f r^2$

If the motor applies torque $\tau_m$, then the angular acceleration of the flywheel is:

$\alpha = \dfrac{d\omega(t)}{dt} = \dfrac{\tau_m}{I}$

And this will require a power at time t, $P(t)$ of:

$P(t) = \tau_m \cdot \omega(t) = \tau_m \cdot \int{\dfrac{\tau_m}{I}}dt$

If we start the flywheel from rest, and keep the torque constant:

$P(t) = \dfrac{\tau_m^2}{I}\cdot t$

(In units, we’ve got on the left: $J/s = Nm/s$ and on the right $(Nm)^2 (kg^{-1} m^{-2}) s = Nm (kg m^2 s^{-2}) (kg^{-1} m^{-2}) s = Nm/s$, thus the units match and check our work)

So the power required to keep a given torque increases linearly with time.

Notice that $I \propto M_f$ and the torque required to counter gravity is $\tau_m \propto M_t$, so if the mass is dominated by the flywheel then $P \propto M_t$.

So we want to minimize the mass of the flywheel (while keeping the assumption that the flywheel mass dominates the total mass) while maximizing the moment of inertia.

That means that we want to use a hollow flywheel, since this has twice the moment of inertial for a given mass.

Combining the equation $P(t)$ with the equation for $\tau_m$, we get

$P(t) = \dfrac{\tau_m^2}{I}\cdot t = \dfrac{(M_t gh \cdot \sin(\theta))^2}{I}\cdot t = \dfrac{M_t^2 g^2h^2 \cdot \sin^2(\theta)}{I}\cdot t$

And substituting in the equation we had for $I_h = M_f r^2$:

$P(t) = \dfrac{M_t^2 g^2h^2 \cdot \sin^2(\theta)}{M_f r^2}\cdot t$

If we want to minimize the power, then we want to minimize $M_t^2/M^f = (M_m + M_f)^2/M_f$. Differentiating gives $(M_f^2 - M_m^2)/M_f^2$ which has a zero at $M_f = M_m$.

So the minimum power required is when $M_f = M_m$ !

## Three flywheels

If we consider the case of having two or three flywheels on the box, perpendicular to each other to give us more control, then we can consider each flywheel individually, and consider $M_m$ to include the mass of the other flywheels. For minimum power usage, this would mean that we’d want to minimize $(M_m + 3M_f)^2/M_f$ which is minimum at $M_f = M_m/3$.

So if we have 1kg of motors, payload, etc, then we’d want the flywheels to be 1/3 kg each.

Since $P \propto 1/r^2$ we want to maximize the radius of the flywheel.

## Example

A random example EDF motor has weight 148 grams (which we’ll round up to 0.2kg) and produces 450 Watts of power. We’ll consider this on a stick of length 1 meter and with a hollow flywheel.

If we begin at an angle of 10 degrees (0.17 radians) and desire a torque sufficient to balance against gravity:

$\tau_g = (M_f + M_m) gh \cdot \sin(\theta) = (M_f + M_m) \cdot (1m * 10m/s^2) \cdot 0.17 = (M_f + M_m) \cdot (1.7m^2/s^2)$

So:

$P(t) = \dfrac{\tau_m^2}{I}\cdot t = \dfrac{(M_f + M_m)^2 \cdot (2.89 m^4/s^4)}{M_f r^2} \cdot t$

Let’s make the radius 0.1m – i.e. 10% of the length of the stick. We’ll consider the case of using a single flywheel and minimizing the power usage.

So this gives a power of $P(t) = (2M_f)^2/M_f \cdot (0.0289 m^2/s^4) \cdot t = 4 M_f \cdot (0.0289 m^2/s^4) \cdot t = M_f \cdot (0.1156 m^2/s^4) \cdot t$

Using the EDF motor mass of 0.2kg:

$P(t) = (0.023 kg m^2/s^4) \cdot t = (0.023 J/s^2) \cdot t$

The motor can provide a maximum of 450 Watts, meaning that it could maintain a 10 degree angle for 450/0.023 s = 19565 seconds = 5.4 hours.