Example. We will take the case of an aeroplane carrying a personnel load of 340 pounds, oil and gasoline 370 pounds, and baggage amounting to 190 pounds, instruments 100 pounds. Total live load will be 1000 pounds. Taking the live load percentage as 0.30, the total load will be 1000/0.30 = 3333 pounds. If the low speed is 50 miles per hour, and the maximum Ky of the chosen wing is 0.003 at this speed, the area will be A = W/KyV² = 3333/0.003 x (50 x 50) = 444 square feet. Since this is a biplane with a correction factor of 0.85, the corrected area will be: 444/0.85 = 523 square feet. The unit loading, or weight per square foot will be: 3333/523 = 6.36 pounds. The corrected area includes the ailerons and the part of the lower wing occupied by the body.
Empirical Formula for Loading. After investigating a large number of practical biplanes, the author has developed an expression for determining the approximate unit loading. When this is found, the approximate area can be found by dividing the total weight by the unit loading. This gives an idea as to the area used in practice.
It was found that the unit loading increased with the velocity at nearly a uniform rate. This gave an average straight line formula that agreed very closely with 128 examples. If V = Maximum velocity in miles per hour, and w = weight per square foot, then the unit loading becomes:
w = 0.065V - 0.25 for the average case. For high speed scouts this gives a result that is a trifle low, the formula for a fast machine being more nearly w = 0.65V - 0.15, for speeds over 100 miles per hour.
A two-seat machine of average size weighs 2500 pounds, and has a maximum speed of 90 miles per hour. Find the approximate unit loading and area. The loading becomes: w = 0.065V - 0.25 = (0.065 x 90) - 0.25 = 5.6 pounds per square foot. The approximate area will be: 2500/5.60 = 446 square feet.
If the above machine had a speed of 110 miles per hour, the formula would be changed for the high-speed type machine, and the loading would become:
w = 0.065V - 0.15 = (0.065 x 110) - 0.15 = 7.00 pounds per square foot. The required area will be: 2500/7.0 = 372 square feet. When the unit load is also determined in this way it is a very simple matter to choose the wing section from Ky = w/V².
Area From Live Load and Speed. By a combination of empirical formula we can approximate the area directly. For the average size machine, w = 0.065V - 0.25. And the total weight W = U/0.32 where U is the useful or live load. Since A = W/w, then A = U/(0.65V - 0.25) x 0.32 = U/0.021V - 0.08.
Thus if an aeroplane travels at 90 miles per hour and has carried a useful load of 800 pounds (including gas and oil), the approximate area is: A = U/0.021.V = 0.08 = 800/(0.021 x 90) - 0.08 = 442 square feet. This assumes that the useful load is 0.32 of the total load and that the speed is less than 100 miles per hour.