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Aviation History
1920
1920 - 1219.PDF
NOVEMBER 25, 1920 In order that two airscrews may be strictly geometricallysimilar they ought not to vary in their form under load It is perfectly certain that the above model and the full-sizerotating wing will not deform to the same extent when rotat- ing since they differ in construction, but even if there is adifference it will be compensated for by the improvement due to the known superiority of full-size results over thosededuced from model tests. In spite of the great width of the blades of this airscrew the figures obtained from it very closelyconform to the theoretical formulae to Col. Renard. In order to interpret the results we must apply the two funda-mental relations :—• Thrust in Kg. F = a«a D4 H.P. T = /8«8D5. The results of the last column of the table of tests of September 5th, 1918, given for the model tested-— •«, = 1,169 r.p.m. =T' = 3-47H.P. and FJ = 18 Kg. The diameter and the maximum speed of rotation of the rotating wing of the full-size machine are D = 7 metres, n = 160 r.p.m. If we designate by T the power necessary at the shaft of the full-size screw and by F the corresponding lifting thrust we shall deduce from the results of the model test the following value for lift and the power : T ~ /N WD\ 5 / 160 \»/ 7 \s Tl(^)(r) = 3'47 (—r- I—— I =115 H.P. \iibo / \i.o5/ F = TK°8 ?n°Zmal ^efficient of resistance (flat plane).F = Thrust (left) in Kg. . ' P = Total weight of helicopter. n — Revolutions per second of airscrews Taking the case of a machine with two screws each 7 metres SSSl^ f°Ur WadeS ^ ^ gr°Undth * f P = 2F = 2OW2D«. If the force F becomes superior to P, the helicopter leavesthe ground. * The acceleration on leaving the ground is :— ^7 = 2F -P-KSV. The term KSVa is negligible at the start "has already been shown that the interpretation of theresults of Table No. 2 gave for D = 7 metres and n «= 160 r.p.m. F = 700 Kg. or 2F = 1,400 Kg. Weight of the Helicopter.—It may be taken that the weightof a helicopter, with 250 h.p. engines, will weigh in flving order 1,000 Kg. = P. 3 6 At the rate of 6 Kg. per square metre, the weight of thescrews (of 40 square metres) will be p = 6 X 40 = 240 Kg Experience has shown that the total weight P is aboutP ^ The actual helicopter which is in view uses two turningwings geometrically similar to that of the model tested, and •each of these wings will then be capable of a thrust of 700 k.for a power of 115 h.p. The supreme importance of this test lies in the fact that the actual helicopter as built only weighs1,200 kg. and consequently each of the wings will only have to support 600 kg. In addition, the useful power available per wing is 130 h.p.,and finally the fixed incidence given to the blades of the model was 10 degs. The excess of lifting force therefore is approxi-mately 200 kg. for two wings together. Stability of the Helicopter.—With reference to the centre of• gravity of the helicopter we have to take account of three conditions of stability :— (1) Longitudinal stability, which must be obtained around a horizontal axis through the centre of gravity and parallel to the plane of symmetry. (2) Lateral stability; stability around the axis parallel to the span and passing through the centre of gravity. (3) Rotational stability round a vertical axis through the centre of gravity. In order to maintain its equilibrium in all positions a heli- -copter must possess rudder, elevator and special arrange- ments capable of producing effects similar to those due to the ailerons of an aeroplane. This question of stability is the most important that one met with in the helicopter, and various Speed of Chmb at Start.—- y = 1,400 . ,. , 400 x 9-81 from which y = —7^— = 3 80 metres per sec. 1,000 = 400 Kg. 7^ The general equation for the vertical flight of a helicopteris of the form :— r P d\T P Where t\ M To -95)efficiency of the transmission gear (the density of air at the altitude Z Engine power at sea level. Speed of Horizontal TranslationIt will be seen that if 2 F is the total resultant thrust along the inclined axis OY' of the airscrews that the total horizontalcomponent (OA for the element ds) will be for the whole machine :— For 2F = 1,400 Kg. 7=io°. 2OA = 1,400 xo-i74 = 250 Kg. If the resistance to horizontal flight is the equivalent of 2 sq. metres of normal surface the speed of translation will be V = Vo- 250 X 3-6 = 140 km.p.h. Y RESULTANT THRUST ALONG AXIS VERTICAL COMPONENT - O8 X 2 The lower the resistance to translation at movement the less is the inclination of the airscrew axis necessary to give any required horizontal speed. Gliding Descent of a Helicopter with Stopped Motor.—This is the most vital point in the problem of the helicopter, for the adversaries of this type of heavier- than-air machines make it their essential argument that a safe descent under these conditions seems to them to be impossible. I wish to show in what follows that the vertical gliding descent of the helicopter with unclutched screws is actually possible, and to conclude that there should be no delay in solving this problem. w OF FL/QHT FIG. 5 •solutions, several of which have already given valuable results, but into the details of which it is not now permissible to enter, are now being tested. Equations of sustentation and practical coefficients Calling T = H.P. supplied.p = weight of airscrews. 11 = efficiency of transmission gear.q = " qualite " of airscrews. S, = Surface of normal plane equivalent to the resistance to vertical ascent. •*••• A descent with stopped motor can occur in two manners. First, airscrews declutched, turning round their axis as windmills, with a suitable incidence on the blades so that their direction of rotation may not be replaced. Second with airscrews stopped in first position as soon as the engine stops. Fixed Screws.—I shall have little to say upon this second solution of the problem, in which I have very little confidence. It is equivalent to turning the helicopter into a very bad glider which will descend in a path very much as an aeroplane does, and this thanks to the use of appropriate gearing. The total surface of the glider in this case will always be less 1221
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