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Aviation History
1910
1910 - 0978.PDF
THE PROBLEM OF THE HELICOPTER. THE conclusions deduced in the following article, which treats of the theory of static thrust of screws, may be summed up as follows :— For a given load : the screw of greatest diameter will give the greatest efficiency. For a given load with a given diameter : the steepest effective blade angle will give the greatest efficiency. For given revolutions : the blade angle of least resistance will give the greatest efficiency. By efficiency is meant ratio of thrust to power in the shaft. THE helicopter, or direct lifting machine, unquestionably has a very great fascination for a large number of people whose ideas on the power required for dynamic support by this means seem to vary so much that it should not be without general interest to investigate the iheory of the problem. There are two broad aspects of the helicopter problem, one associated with ascent, the other with propulsion, but in this article we purpose confining our atteni ion solely to the former. There are also two subdivisions of this aspect, one associated with the mere sustentaiion of the machine stationary in the air, the other dealing with the actual rise of the machine vertically through the air. Again, we purpose dealing only with the former case. Air, in common with every other form of matter, derives its capacity for acting as an abutment to a force from its inertia. The inertia of air is exceedingly small, consequently an enormous volume of this fluid has to be used fur the support of even small loads. When air is at rest it is incapable of supporting anything of greater density than itself, but when its molecules are in a state of accele ration, resulting from their inertia being overcome by force, air can support solid bodies by reaction. If a volume of air is accelerated downwards the reaction is upwards. This is the basis on which the principle of proposed helicopter flying machines is founded ; one or more screws on vertical spindles being employed to maintain the downward acceleration of the air that is essential to the sustentation of the load. When air or any other fluid is maintained in downward acceleration throughout a certain area located in a fixed position in space, a stream of uniform velocity comes into existence below the plane of operation ; thus there will be a steady downward draught from a helicopter screw. The measurement of the velocity of this draught, or slip stream as it is technically called, is also a measurement of the acceleration, the dimensions being in feet per second per second. It is assumed, in what follows, that the acceleration is uniform throughout the area, and that the area in question is the disc area of the helicopter screw, as defined by a circle passing through the tips of the blades. From what has already been said, it may be deduced that when the area is large the velocity of the slip stream requisite to support a given load will be less than when the area is small. Air, in common with all other bodies in motion, possesses energy, and since the energy in the slip stream is obviously wasted, it is eminently desirable that this amount should be as little as possible. Energy in a stream of air varies directly as the area and proportionately to the cube of the velocity, consequently it is essential to keep the velocity as low as possible, which means that the larger the area for a given load the less the power required to sustain it. Area is proportional to the square of the diameter, hence it is economical to employ few screws of large diameter rather than many sort ws of small diameter, the total area being the same in both cases. The problem of economically sustaining a load in the air by means of a helicopter thus resolves itself into the use of the largest lifting screw that is practicable. On the other hand, the circumstances may be such that the problem is to lift the load at any price, in which case the solution is to use the largest permissible engine. Such increase in the size of the engine necessarily saciifices efficiency in the ratio of thrust to power, but, owing to the fact that the engine represents only a fraction of the total weight, the increase in the total lift is greater, up to a point, than the increase in the total load. The engine cannot be indefinitely increased in size, because sooner or later its proportion of the total load will be such that any increase in the weight of the engine will no longer be adequately compensated for by the increment in the thrust. Thus, for example, if we double the power of an engine we only obtain I-6 times the thrust from the same screw, or, expressing this in another way, we may say that if the original engine weighs 1*5 times the net load (i.e., total load less weight o' engine) then the to'al lift will increase in greater ratio than the total load until an engine of twice the weight and twice the power of the original engine has been substituted. After this condition has been reached, any further increase in the power of the engine will be accompanied by a greater increment in weight than increment in thrust, consequently the machine will no longer be self- supporting. It is important to remember that the example given above is a particular case, and that the ratios therein do not apply for different condition?, thus, for example, if we quadruple the original engine we shall only obtain 2'5 times the original thrust, and not (2 by I "6) times. These are some of the more important mathematical considerations governing the fundamental problem of the helicopter, and likewise, of course, the static thrust of a propeller. Their application may, perhaps, be best seen in the following examples, which, for con venience, are worked out by the aid of the summarised list of formula; herewith :— FORMULAE FOR HELICOPTERS AND THE STATIC THRUST OF SCREWS. (From "Flight Manual," F 102.) Symbols :— T as thrust (lbs.); V as velocity of slip stream (ft./sec.); D = T diam. screw (ft.) ; P = horse-power in slip stream ; E = ^ ; w = weight of engine (Ibs./h.p.); e = efficiency of screw, and P transmission = — . Horse-power [It is assumed that V is uniform over the disc area •78540'', and that the mass per cub. ft. of air £-= "1 g 420-£ D'V2. v= v/535T\D= J535T, p= D»V« . 535 ' D V 587,000' T? I.IOO V If P is increased to KP with the same D, then T is increased to (VK)aT. If T is increased to KT with the same D, then P is increased to (VKf? = K>'5P. Example :— 1. Required a thrust of 1,000 lbs. from a 15-ft. screw. ? P. V = ^535 x 1.000 _' 730 15 15 1,100 _ 0,.fi n.r /P -. P = '»<*X> = 44-5. 22"6 3 ; ? net load supported. T = 48-6 ft./sec. E = *J£~= 22-6 lbs./P. 48*6 In the above let e = "5 and w Horse-power of engine = — 89. 44"5 •5 Weight of engine = 89 x 3 = 267 lbs. Net load = (1,000 - 267) = 733 lbs. 3. In the above let e = '25 and w = 5 ; ? net load supported. Horse-power of engine = -—- = 178. Weight of engine = (178 x 5) = 890 lbs. Net load — (1,000 - 890) = 110 lbs. 4. A thrust of 400 lbs. is given by a 50-h.p. engine operating a 10-ft. screw. ? e. V = ^535 * 4QO = v/iT^ = 460 = 46 ft/sec E = = 24 lbs./P. . = 33 %• 45? = 167. 24 1,100 W e = P- =, l617 HP 50 It will have been observed in connection with the foregoing problems that no mention whatever has been made of the kind of screw to be used, neither as regards its pitch, number of blades, or revolutions. This omission has been made on purpose in order to emphasise a fact that is too often overlooked, namely, that the thrust of a screw is only limited by the cross-sectional area, and the velocity of its slip stream. It has been assumed in the previous examples that the velocity of slip is uniform over an area repre sented by the disc area of the propeller,' which is defined by a circle passing through the tips of the blades. If it is the designees object to obtain a certain static thrust from a given propeller, as is 976
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