What would be the power requirements and performance of a Scroggs-type delta-wing aircraft with a 250 cm span?

Question

Roy Scroggs, a tailor, patented in 1917, US1250033, and in 1932 US1848578, a Delta Wing plan form machine 'Flying Dart', flight tested

with an OX-5 engine, they say anything would fit an OX-5 powered airplane, as in Besler Steam Airplane, it had same Leading Edge swept, 60º, as Handley Page HP 115 of 1961, that tested the Concorde wing (

) The page of an study from the French ONERA about vortex over a Delta plan wing is an experiment conducted with the type of pointed wing in the description by Jones, and in the Langley tests on A Lippisch DM-1, but it's not sure if flow will be identical on a truncated point Delta, as used in first Saab 'Draken'.??

I wonder what the power requirements and performance would be of an Ultralight Airplane with a wing span of 250 cm, maximum width allowed on the road, and a total weight less than 299 kg. 115 kg empty weight is the class limit for Ultralight

Answer 1

First, thank you for unearthing this unorthodox design!

However, it shows that paper airplanes don't scale well. Nevertheless, it follows the same laws of physics, so the same equations should apply for a quick performance evaluation. What have we got?

1. A flat plate airfoil. Maximum lift coefficient will be between 0.7 and 0.8, so minimum speed at a given wing loading will be quite high.
2. A sweptback delta wing of low aspect ratio, so the lift curve slope will be low. For the same lift coefficient it needs much more angle of attack than a regular airplane. Since upper side flow is separated, airfoil drag will grow with the square of the angle of attack, and the low lift curve slope ensures that drag will be needlessly high.
3. A big fuselage with a lot of wetted surface area.
4. A massive separation area at the back which will produce a lot of drag without any benefit, except for improved directional stability.

Let's start with induced drag. The low aspect ratio ensures a nearly elliptic lift distribution, and lifting 300 kg with 2.5 m of wing will create $$D_i = \frac{2\cdot(m\cdot g)^2}{\rho\cdot v^2\cdot\pi\cdot b^2} = \frac{180000\cdot g^2}{1.225\cdot v^2\cdot3.14159\cdot 6.25} = \left(\frac{g}{v}\right)^2\cdot 7483\;N$$ At 20 m/s flight speed this will equal 3,600 N of drag and at 40 m/s flight speed a quarter of that (900 N). Multiply by speed to get the power needed to overcome just induced drag: 36 kW at 20 m/s and 18 kW at 40 m/s.

But what speed is achievable? We need to guess the wing area in order to find the stall speed. I assume 4 m chord and 1 m span at the leading edge, so the wing area is 7 m². Using a maximum lift coefficient of 0.75, this means you cannot fly slower than 30.24 m/s. Whoa! That is 58.8 kts or just below the 61 kts limit for the old FAR 23.

Let's add some safety margin and continue with 40 m/s flight speed, which corresponds to a lift coefficient of 0.43. Now to the angle of attack - this requires to know the lift curve slope. As a slender body it will be slightly below $\pi\cdot\frac{AR}{2}$, and with an aspect ratio $AR$ of below 1 this is maybe 2.2 per radian or 0.04 per degree. At $c_L$ = 0.43 this is 11°. Multiply the weight with the sine of 11° and you get the cost for flying a fully separated top wing: 506 N. Dragging this through the air at 40 m/s needs another 20.2 kW.

Now to the wetted surface: The vertical surfaces add maybe another 4 - 5 m² to the 14 m² of the wing. The Reynolds number is 11$\cdot 10^6$, so the friction drag coefficient is as low as 0.003 if the surface is smooth. Since the upper wing shows separated flow, I will continue with 15 m² surface area. At 40 m/s this equates to 44.1 N of drag, needing 1.76 kw of power.

The fixed and unfaired landing gear will be another source of drag; here I simply double the friction drag in order to account for it, too. I expect that a detailed drag analysis will not change the result substantially.

Your idea of filling the separation area with a ducted fan is probably costing even more performance. Please consider that now you have a high-speed stream of air flowing through a duct running along the length of the fuselage. Better to narrow the fuselage in the rear half so it ends in a vertical rudder.

If we now assume that you avoid any boattail drag this way, flying straight and level at 40 m/s will need a total of 42 kW. Add 50% for manoeuvring and climbing (those extra 21 kW translate to a climb rate of more than 6 m/s - looks OK) and a propeller efficiency of maybe 75% (look at this answer - that could be as low as 60% in reality!), and you end up with an engine requirement of 84 kW or 112.6 HP. It looks like you need a Rotax 914 or similar engine, the mass of which will eat up one third of your budget.