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.
Fig. 58.—In a recently published mathematical treatise on Aerodynamics, an illustration is shown, representing the path that the air takes on encountering a rapidly moving curved aeroplane. It will be observed that the air appears to be attracted upwards before the aeroplane reaches it, exactly as iron filings would be attracted by a magnet, and that the air over the top of the aeroplane is thrown off at a tangent, producing a strong eddying effect at the top and rear. Just why the air rises up before the aeroplane reaches it is not plain, and as nothing could be further from the facts, mathematical formulas founded on such a mistaken hypothesis can be of but little value to the serious experimenter on flying machines.
Fig. 59.—An illustration from another scientific publication also on the Dynamics of Flight. It will be observed that the air in striking the underneath side of the aeroplane is divided into two streams, a portion of it flowing backwards and over the top of the edge of the aeroplane where it becomes compressed. An eddy is formed on the back and top of the aeroplane, and the air immediately aft the aeroplane is neither rising nor falling. Just how these mathematicians reason out that the air in striking the front of the aeroplane would jump backwards and climb up over the top and leading edge against the wind pressure is not clear.
Fig. 60.—This shows another illustration from the same mathematical work, and represents the direction which the air is supposed to take on striking a flat aeroplane. With this, the air is also divided, a portion moving forward and over the top of the aeroplane where it is compressed, leaving a large eddy in the rear, and, as the dotted lines at the back of the aeroplane are horizontal, it appears that the air is not forced downwards by its passage. Here, again, formula founded on such hypothesis is misleading in the extreme.
Fig. 61.—This shows the shape and the practical angle of an aeroplane. This angle is 1 in 10, and it will be observed that the air follows both the upper and the lower surface, and that it leaves the plane in a direction which is the resultant of the top and bottom angle.
Fig. 62.—This shows an aeroplane of great thickness, placed at the highest angle that will ever be used—1 in 4—and even with this the air follows the upper and lower surfaces. No eddies are formed, and the direction that the air takes after leaving the aeroplane is the resultant of the top and bottom angles.
Fig. 63.—Section of a screw blade having a rib on the back. The resistance caused by this rib is erroneously supposed to be skin friction.
Fig. 64.—Shows a flat aeroplane placed at an angle of 45°, an angle which will never be used in practical flight, but at this angle the momentum of the approaching air and the energy necessary to give it an acceleration sufficiently great to make it follow the back of the aeroplane are equal, and at this point, the wind may either follow the surface or not. Sometimes it does and sometimes it does not. See [experiments with screws].
Fig. 65.—The aeroplane here shown is a mathematical paradox. This aeroplane lifts, no matter in which direction it is driven. It encounters air which is stationary and leaves it with a downward trend; therefore it must lift. However, if we remove the section b, and only subject a to the blast, as shown at [Fig. 66], no lifting effect is produced. On the contrary, the air has a tendency to press a, downwards. The path which the air takes is clearly shown; this is most important, as it shows that the shape of the top side is a factor which has to be considered. All the lifting effect in this case is produced by the top side.
Fig. 66.
Fig. 67.—In this drawing a represents an aeroplane, or a bird’s wing. Suppose that the wind is blowing in the direction of the arrows; the real path of the bird as relates to the air is from i to j,—that is, the bird is falling as relates to the air although moving on the line c, d, against the wind. In some cases, a bird is able to travel along the line g, h, instead of in a horizontal direction, thus rising and apparently flying into the teeth of the wind at the same time.