# Energy Maneuverability Theory Applied to WW1 Fighters

I am trying to test the Fokker D.VII, Spad S.XIII, Sopwith Camel, Nieuport 28, and the Se5a using the model in this paper:

It's called Combat Performance Advantage and is derived from EM theory. My question: How are they reaching these Ps values, and how can I do the same using the data I have? They have an example calculation on page 14/15, and before that provide some details on the model.

I have the aircraft weight, fuel weight, zero lift drag coefficient, wingspan (reference area), maximum lift coefficient, aspect ratio, maximum instantaneous and sustained turn rates (at 4km altitude), and a few other things. I was able to estimate the Oswald efficiency factor and a couple of other metrics. I'm still not clear on how to combine these to get the power lost during a turn in the attached paper, although I was able to calculate t Alpha and t Theta.

• EM theory may have more to do with retaining speed and the ability to go vertical. WW1 planes did not have the power to do much of that. Instead, they flew tight turns as near to stall as possible to try to gain advantage. They also were very draggy compared to more modern aircraft. May 22, 2023 at 19:06

$$P_s=\frac{(T-D)V}{W}$$

Which is written for a jet aircraft (in terms of thrust). If we write it for a prop...

$$P_s=\frac{P_a-P_r}{W}$$ where $$P_a$$ is power available and $$P_r$$ is power required.

Power available is the propeller efficiency times the available shaft power.

$$P_a=\eta_p \, P_\mathrm{shaft}$$

For WWI aircraft (fixed pitch props), this calculation is more complex than you might think to do right. In particular, the difference in props may be the difference between two aircraft's capabilities at a certain point in the flight envelope.

$$P_r=D \, V$$

$$D = C_D \, q \, S$$

Here we'll assume a simple parabolic drag polar. There really should at least also be a term that is linear with lift. However if you don't have detailed drag polars for the aircraft, that won't matter.

$$C_D = C_{D,0} + K \, {C_L}^2$$

$$C_L=\frac{L}{q\,S}$$

And here is where turning flight comes in...

$$L=n \, W \, \cos(\theta)$$

Where $$n$$ is the load factor -- if the aircraft is pulling two gees, then $$n=2$$.

The $$\cos(\theta)$$ here is often ignored (particularly for relatively low performance aircraft). It introduces two complications -- 1) the solution becomes iterative as $$P_s$$ depends on $$\theta$$ while it also determines $$\theta$$. If you need, some quick iteration will sort this out. 2) it depends on pilot intent -- are you putting the excess power into climb, or into horizontal acceleration, or both. You can even dive with positive $$P_s$$ to really pick up speed.

In particular, when most interested in maximum sustained turn rate, the definition of that point is $$\theta=0$$, so the problem is avoided.

So, to answer what I believe is your question... The additional induced drag from pulling gees during a turn increases the power required and thereby decreases the Excess Power.

I believe this is what you mean by the "power lost during a turn".