In Prof. Langley’s lifetime, we had many discussions regarding the width and shape of aeroplanes. The Professor had made many experiments with very small and narrow planes, and was extremely anxious to obtain some data regarding the effect that would be produced by making the planes of greater width. He admitted that by putting some two or three aeroplanes tandem, and all at the same angle, the front aeroplane a ([Fig. 57]), would lift a great deal more than b, and that c, would lift still less. He suggested the arrangement shown at a′, b′, c′, in which b′ is set at such an angle as to give as much additional acceleration to the air as it had received in the first instance by passing under a′, and that c′, should also increase the acceleration to the same extent. With this arrangement, the lifting effect of the three aeroplanes ought to be the same, but I did not agree with this theory. It seemed to me that it would only be true if it dealt with the volume of air represented between j, and k, and that he did not take into consideration the mass of air between k, and l, that had to be dealt with, and which would certainly have some effect in buoying up the stream of air, j, k. Prof. Langley admitted the truth of this, and said that nothing but experiment would demonstrate what the real facts were. But it was a matter which I had to deal with. I did not like the arrangement a′, b′, c′, as the angle was so sharp, especially at c′, that a very large screw thrust would be necessary. I therefore made a compromise on this system which is shown at a′′, b′′, c′′. In this case a′′, has an inclination of 1 in 10, b′′ an inclination of 1 in 6, and c′′ an inclination of 1 in 5. It will be seen that this form, which is shown as one aeroplane at a′′′, b′′′, c′′′, is a very good shape. It is laid out by first drawing the line c, d, dropping the perpendicular equal to one-tenth of the distance between c and d, and then drawing a straight line from c, through e, to f, where another perpendicular is dropped, and half the distance between d and e laid off, and another straight line drawn from e, through g, to h, and the perpendicular h, i, laid off the same as f, g. We then have four points, and by drawing a curve through these, we obtain the shape of the aeroplane shown above, which is an exceedingly good one. This shape, however, is only suitable for velocities, up to 40 miles per hour; at higher velocities, the curvature would be correspondingly reduced.
Fig. 57.—Diagram showing the evolution of a wide aeroplane.
THE ACTION OF AEROPLANES AND THE POWER REQUIRED EXPRESSED IN THE SIMPLEST TERMS.
In designing aeroplanes for flying machines, we should not lose sight of the fact that area alone is not sufficient. Our planes must have a certain length of entering edge—that is, the length of the front edge must bear a certain relation to the load lifted. An aeroplane 10 feet square will not lift half as much for the energy consumed as one 2 feet wide and 50 feet long; therefore, we must have our planes as long as possible from port to starboard. At all speeds of 40 miles per hour or less, there should be at least 1 foot of entering edge for every 4 lbs. carried. However, at higher speeds, the length may be reduced as the square of the speed increases. An aeroplane 1 foot square will not lift one-tenth as much as one that is 1 foot wide and 10 feet long. This is because the air slips off at the ends, but this can be prevented by a thin flange, or à la Hargrave’s kites. An aeroplane 2 feet wide and 100 feet long placed at an angle of 1 in 10, and driven edgewise through the air at a velocity of 40 miles per hour, will lift 2·5 lbs. per square foot. But as we find a plane 100 feet in length too long to deal with, we may cut it into two or more pieces and place them one above the other—superposed. This enables us to reduce the width of our machine without reducing its lifting effect; we still have 100 feet of entering edge, we still have 200 feet of lifting surface, and we know that each foot will lift 2·5 lbs. at the speed we propose to travel. 200 × 2·5 = 500; therefore our total lifting effect is 500 lbs., and the screw thrust required to push our aeroplane through the air is one-tenth of this, because the angle above the horizontal is 1 in 10. We, therefore, divide what Prof. Langley has so aptly called the “lift” by 10; 50010 = 50. It will be understood that the vertical component is the lift, and the horizontal component the drift, the expression “drift” also being a term first applied by Prof. Langley. Our proposed speed is 40 miles per hour, or 3,520 feet in a minute of time. If we multiply the drift in pounds by the number of feet travelled in a minute of time, and divide the product thus obtained by 33,000, we ascertain the H.P. required—
50 × 3,52033,000 = 5·33.
It therefore takes 5·33 H.P. to carry a load of 500 lbs. at a rate of 40 miles per hour, allowing nothing for screw slip or atmospheric resistance due to framework and wires. But we find we must lift more than 500 lbs., and as we do not wish to make our aeroplanes any longer, we add to their width in a fore and aft direction—that is, we place another similar aeroplane, also 2 feet wide, just aft of our first aeroplane. This will, of course, have to engage the air discharged from the first, and which is already moving downwards. It is, therefore, only too evident that if we place it at the same angle as our first one—viz., 1 in 10—it will not lift as much as the first aeroplane, and we find that if we wish to obtain a fairly good lifting effect, it must be placed at an angle of 1 in 6. Under these conditions, the screw thrust for this plane will be 1⁄6th part of the lift, or 8·88 H.P. against 5·33 H.P. with our first aeroplane. In order to avoid confusion, we will call our first plane a′′, our second plane b′′, and the third c′′, the same as in [Fig. 57]. Still we are not satisfied, we want more lift, we therefore add still another aeroplane as shown (c′′, [Fig. 57]). This one has to take the air which has already been set in motion by the two preceding planes a′′ and b′′, so in order to get a fair lifting effect, we have to place our third plane at the high angle of 1 in 5. At this angle, our thrust has to be 1⁄5th of the lifting effect, and the H.P. required is twice as much per pound carried as with the plane a′′, where the angle was 1 in 10; therefore, it will take 10·66 H.P. to carry 500 lbs. As there is no reason why we should have three aeroplanes placed tandem where one would answer the purpose much better, we convert the whole of them into one, as shown (a′′′, b′′′, c′′′, [Fig. 57]), and by making the top side smooth and uniform, we get the advantage of the lifting effect due to the air above the aeroplane as well as below it. The average H.P. is therefore 5·33 + 8·88 + 10·66 ÷ 3 = 8·29 H.P. for each plane, or 25 H.P. for the whole, which is at the rate of 60 lbs. to the H.P., all of which is used to overcome the resistance due to the weight and the inclination of the aeroplanes, and which is about half the total power required. We should allow as much more for loss in screw slip and atmospheric resistance due to the motor, the framework, and the wires of the machine. If, however, the screw is placed in the path of the greatest resistance, it will recover a portion of the energy imparted to the air. We shall, however, require a 50 H.P. motor, and thus have 30 lbs. to the H.P.
From the foregoing it will be seen that at a speed of 40 miles an hour, the weight per H.P. is not very great. If we wish to make a machine more efficient, we must resort to a multitude of very narrow superposed planes, or sustainers, as Mr. Philipps calls them, or we must increase the speed. If an aeroplane will lift 2·5 lbs. per square foot placed at an angle of 1 in 10, and driven at a velocity of 40 miles an hour, the same aeroplane will lift 1·25 lbs. if placed at an angle of 1 in 20, and as the lifting effect varies as the square of the velocity, the same plane will lift as much more at 60 miles per hour, as 60² is greater than 40²—that is, 2·81 lbs. per square foot instead of 1·25 lbs. At this high speed, providing that the width of the plane is not more than 3 feet, it need be only slightly curved and have a mean angle of 1 in 20.
An aeroplane 100 feet long and 3 feet wide would have 300 square feet of lifting surface, each of which would lift 2·81 lbs., making the total lifting effect 843 lbs. 843 ÷ 20 = 42·15, which is the screw thrust that would be necessary to propel such a plane through the air at a velocity of 60 miles per hour. 60 miles per hour is 5,280 feet in a minute, therefore the H.P. required is 42·15 × 5,280 ÷ 33,000 = 6·7 H.P. Dividing the total lifting effect 843 by 6·7, we have 843 ÷ 6·7 = 125·8, the lift per H.P. If we allow one-half for loss in friction, screw slip, etc., we shall be carrying a load of 843 lbs. with 13·4 H.P. It will, therefore, be seen that a velocity of 60 miles an hour is much more economical in power than the comparatively low velocity of 40 miles an hour; moreover, it permits of a considerable reduction in the size and weight of the machine, and this diminishes the atmospheric resistance.