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Aviation History
1920
1920 - 0559.PDF
MAY 20, 1920 MODEL AEROPLANES NOTE."— All communications should be addressed to the Model Editor. Designing a Model Aeroplane - THE lift of an aerofoil varies directly as the angle and asthe square of the velocity ; that is to say, double the angle and you double the lift; double the speed and you get four• times the lift; halve the speed and the lift is reduced to one-fourth. Now at 40 miles an hour the weight borneby a square foot of surface is about 2.25 lbs. At 20 miles an hour, the lift per square foot will therefore be 0.56 lb.,and at 10 miles per hour only 0.14 lb., or 2.24 ozs., will be carried—this when the angle of incidence is about 1 in 8,or 70 approximately, which is about the angle used in most of the well-known machines. In models, I have found that such an angle is the greatestpracticable—much finer angles being generally best; and as, with models, the difficulties of providing large surfacesare not so great as with the full-sized machines, angles of 1 in 10 to 1 in 15 will be found most suitable. There is' another reason why finer angles can be used in models. Suppose we make a scale replica of, say, a Sopwith machine,we reduce the size of the machine by linear measurements, the area of surfaces as the square and the weight as the cubeof the original. The model should therefore have better chances of success than the original, because the weight inproportion to surface is not so great. True, the speed is not so great—but the speed is not reduced by scale ; a modelone-twenty-fourth full size will fly easily at from one-fourth to one-eighth of the speed of the original; so it will be seenthat less inclination and less power is required in proportion to size. Many of my own experimental models have flownat about 15 miles an hour in still air, and the lift has been 4.4 ozs. per sq. ft. of surface. Other models having finerangles of inclination have flown at higher speeds, but the pressure on the planes has been the same. In designing models, it is not usually tiecessary to go deeply into mathematical calculation, because it is so very easy to vary the factors governing the success of the machine. The source of power (generally india-rubber) can be increased or reduced in quantity ; angles of the aerofoils and the areas of same can be altered. But suppose we have a motor other than rubber, of known power and weight—which power and weight cannot be varied—and we wish to fit that motor to drive an aeroplane The matter is then somewhat different, and we shall have to find out what is the proper area of surface to fit to the machine to support the weight. Suppose our motor is a piece of clockwork weighing, we will say, 4 ozs., we find by a formula which I shall give later that the clockwork is capable of giving at the propeller a thrust of 3.4 ozs. ; we shall find presently, in addition to the . thrust, the velocity which the motor is capable of giving to the machine. Now, as I have already stated, the theoretical ratio of lift to the drift of ah inclined plane or aerofoil varies directly as the angle of inclination. Thus, if an aerofoil is placed at an angle of 1 in 10, the power required to drive it will be 1 oz. for every 10 ozs. carried—i.e., if the model aerofoil weighs 10 ozs., a pressure of 1 oz. will be sufficient to push and sustain it. If the angle be 1 in 15, then a thrust of 1 oz. will sustain 15 ozs.-—and so on. Now a single aerofoil is not a complete machine, and there _ are, therefore, resistances to be overcome other than that of ';> purely dynamic kind. There is that of the head resistance of the whole framework—which is pretty considerable ; iand that of skin friction, which (whatever may be argued ." regarding full-size machines) may be considered negligible 1 in well-made models, especially those having polished wooden planes. It will be seen, therefore, that if we take say one in ten as our angle, the thrust required rises from one-tenth of the weight to one-fifth—or even as much as one-fourth. Designers usually adopt a standard for thrust of one-third or one-fourth ; but on large aeroplanes there is the resistance offered by the area of the pilot's body, petrol tanks, etc. For models, since the speeds will be low, we may safely take the thrust at one-fifth of the weigh:. Having, then, a thrust available of 3-4 ozs., we may drive and support with this a machine having a total weight- including the clockwork—of 3.4 X 5 = X7O ozs. We now require to know the velocity which the motor is capable ofgiving to the machine. This we shall arrive at presently, along with the thrust; but for the present we will assume itto be 12 miles an hour. Having then before us both the weight and the velocity, we can by the following find the pianoarea required :— Equation, A = kof ys gin ^ Where W = weight carried «= 17.0 ozs., or 1.065 lbs.,kor •= 0.005 for cambered surfaces, and 0.004 for flat planes. V =s velocity in miles per hour = 12. Sin 03 X angle of plane to the horizontal path of flight = 1.10th. 1.065 We have therefore ^To.Oo5 X «• X I/To = 7-39 ^ iU area. This area may represent the whole surface, includingelevators and tail, if one be fitted, or it may be taken as the area of the main surfaces only.Before we can arrange how the area of surface arrived at in the foregoing is to be disposed, we have of course todecide whether the main aerofoils are to be at the leading end of the machine with the elevator tail behiud, or at therear with the elevator in front. I have given examples of both these types.For the purposes of the theoretical machine we are now dealing with, we will assume that we are going to put theelevator in front and a pair of super-posed foils at the rear- Now, I have found that it may be taken as a very good rulethat the area ratio of front planes to rear should be about 1 to 6. That being so, we take our total area—which wefound was 7.39 sq. ft.—and divide it by seven, i.e., the sum of 6 and 1. We thus obtain -*— — 1.06 sq. ft., approximately.. This is to be the area of the front plane. Now, subtracting:this from 7.39 gives us 6.33 sq. ft., as the area of the main planes, or 3.165 ft. for each~£erofoil. If the framework carrying the surfaces is made justsufficiently large and substantial to do its work properly,, and is not needlessly heavy, the centre of gravity of thewhole model will be found to lie on a transverse line some little distance in front of the leading edge of the main surfaces :that is, of course, before the motor is put on. The weight of the motor and propeller will probably shift the centre ofgravity to a point a little aft of the leading edges. To complete the design, we must calculate the weights ofthe various parts of the model. We have the weight of the motor, 4 ozs. ; to this we may add, say, I oz. for the propellerand its attachments, bearings, etc. This 4 ozs. + I oz, must be deducted from the total weight of 17 ozs., leaving12 ozs. as the weight of the aerofoils and framework. From this again, we take, say, 2 ozs. to allow for fabric, brads,.wire, etc., leaving 10 ozs. for woodwork. The best way to arrange that this weight shall not be exceeded will beto sketch out the design, full size if possible, on paper, and measure up the cubic contents of all the parts. Then bymeans of a table of weights of different materials, which may be found in engineering works and pocket-books, the weightof the model may be found. If the weight thus arrived at exceeds the 10 ozs. allowed, the sizes of the parts must bereduced. For instance, suppose the whole framework and the frame of the aerofoils are to be made of birch, we findfrom a table that a cubic foot of that wood weighs 45 lbs. 720 . or 720 ozs.; that is to say, j-^ = 0.4 oz. per cubic inch. Now suppose we find by measuring up the quantity repre-sented in our sketched design that there are, say, 30 cubic ins., we know that we must reduce the sizes of some parts, because 10 ozs. —j—— = 25 cubic ins. It may be thought that this is a long, tedious job, but itis both quicker and easier than making the framework foils, etc., and weighing them afterwards, only to find thatconsiderable alteration is necessary. [ (To be continued.) 559
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