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Air kinetic energy theory

Wednesday, May 10th, 2017 at 2:40 pm Written by:

My theory is that airspeed is related to kinetic energy. Like a person walking up and down a train, the airmass provides the reservoir of momentum and energy so that it all adds up.

The balance of kinetic energy when walking in a train has always bothered me. If the train is moving at a velocity v then your body’s kinetic energy is m*v^2/2 while seated in the train. If you walk in the towards the front of the train at walking speed w your kinetic energy will be m*(v+w)^2/2 and if you walk towards the back of the train your energy will be m*(v-w)^2/2. That means the difference in kinetic energy going forwards is m*v*w + m*w^2/2 and going back is -m*v*w + m*w^2/2.

The m*w^2/2 is your kinetic energy from walking on a stationary ground. But how do we account for the +-m*v*w?

Well, it turns out that if the train has mass M then, by the conservation of momentum, when you start walking up the train at velocity w the train must slow down by a factor of w*m/M where M is the mass of the train. This tiny tiny decrease in the train’s velocity releases kinetic energy of about w*m/M*M*v, which accounts for that factor. (I’m ignoring lower order terms that take account of the fact that your forward velocity is not actually v+w because you need to subtract off for the slightly slower train.)

Air is not the same as a train — an enormous solid object to which one can directly transmit kinetic energy and momentum — but we’ll treat it as such until I can think of it in a better way to handle it.

To illustrate the difference, here’s my top landing at Camlo Hill which transmitted a good deal of momentum via a dent in the ground.

dentcamlo

While in the air, speeding up and slowing down and in turns, the glider must somehow be shedding air in the opposite direction to its change in velocity. Air being 1.17kg/m^3 and the glider being about 120kgs, we could be sending a cube of air that’s 4.68metres on each side in an equal and opposite direction — about the same size as the wingspan.

Now, my attempt at producing data for a polar curve (rate of sink vs airspeed) during one particularly stable flight, resulted in this graph of airspeed and rate of descent:
dentpolarcurve

The dip is when I covered up the airspeed sensor to signal that I was attempting to make a measurement. Then I flew at progressively faster speeds for what seemed an age, but was actually no more than 3seconds a piece, and produced a set of data that was completely rubbish.

The blue line is the rate of descent, and I’d hoped it would be a stepped function, but instead it seems that every time I pulled the bar in a bit more the glider swooped slightly and didn’t steady out.

The two sensors at least agree here. Where the blue line is high I’m losing altitude the fastest, and these correspond with the steepest up curves in the green airspeed curve. So there’s hope.

But there was air was moving upwards there, which would confound the calculations.

Let’s look at a later flight when I flew straight down from a great height (after successfully getting up over the Blorenge) and encountered no thermals.

I was flying fast to get away from sinking air, though I could have just been sinking because I was flying fast (which is done by pulling your body forwards to bring the angle of the nose down).

dentdescend

With everything multiplied out by the mass of the glider, the potential energy consumed will be the altitude lost during this time period:

dentpoteng

The kinetic energy at any time is airspeed^2/2 and looks like:

dentkineng

Subtracting the kinetic energy from the potential energy should give an idea of the energy that has been consumed between 12:37 and time t on the graph.

denttoteng

This is obviously wrong because the energy should never go down. There is no way to recover the energy that has been transmitted from the glider to the surrounding air. It must either be a mistake, or a sensor error, or both.

But for now I’ll try the quick and dirty trick of smoothing the windspeed using a butterworth filter so it evens out.

denttotengsmooth

To get to the polar graph, I need the energy consumed at different air speeds:

k = pandas.DataFrame(data={"w":airspeed, "e":lostenergy.diff(5)}).dropna()
s = k.groupby(numpy.floor(k.w*2)).mean()  # group every 0.5m/s
plt.scatter(k.w, k.e)
plt.plot(s.w, s.e, color="r", linewidth=2)

denttotengscatter

And so, this is my first sighting of the glider’s polar curve, where the rate of energy lost is equal to the sink rate at a constant speed.

It’s unacceptably noisy, and I need to sort out the sensor response properties of the baro-altimeter and the windspeed. The barometer might be effected by the windspeed, and the air speed sensor is probably effected by the angle of the air flow. And I’m not accounting for the kinetic energy of descent either.

I would like this to be a clean curve if possible, like I’ve got with the thermal temperature entropies so that the signal is not swamped by the noise.

What I do like about the potential and kinetic energy difference is the angle of the glider doesn’t matter, whether it’s in a turn or flying straight and level.

A banked glider sinks faster because some of the force of the lift being perpendicular to the wing is directed horizontally outwards to make the turn, meaning there is less available in the vertical dimension to counteract its weight, so that it will begin to fall.

Logically, this will always mean that a sudden sharp turn will always be followed by a fall after an resultant acceleration downwards. It’s always going to be a series of rolls and swoops. It’s not just me and my bad steering.

But that fall will have to turn into kinetic energy. I will need an airspeed meter that doesn’t care about the exact direction. A sonic or a hotwire meter will do.

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