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Aviation History
1929
1929 - 1127.PDF
;,UT 30, 1929 39 THE AIRCRAFT ENGINEER SUPPLEMENT TO FLIGHT (e) Parasite Drag ^Outside Slipstream :—Similarly, parasite V2drag outside slipstream = DB 2 = da x where d2 = parasite drag outside slipstream at 100 ft./sec. V = forward speed in ft./sec. /) Drag due to Wheel Friction :—In the case of a normal aerodrome, a coefficient of friction p of 0-05 should be used, but for aircraft operating from a deck, y. = 0-04 may be taken. Then if W = gross weight of machine, 'Drag due to friction = D? = (W — total lift) x [i. ^The total lift is given by (lift in slipstream + lift outside slipstream) found previously, the vertical component of the thrust being neglected, as it is usually very small when the aircraft is in the assumed attitude. We are now in a position to find the total drag by adding the above items, i.e., Total drag D = Dw, + Dw2 + Dw3 + DBX + DB2+DF. The total drag values are now subtracted from the thrusts at appropriate speeds, giving values of accelerating force F and a curve of accelerating force F against air speed V plotted. This completes the first part of the investigation, dealing with the aircraft in the " tail up " attitude. The next step is to determine the minimum speed at which the aircraft will take off with the tail down and the wings at their maximum angle of attack. The angle of attack with the skid on the ground is found from the G.A. of the machine, and the total lift calculated at this attitude for a few speeds near the expected take-off speed. The total lift consists of— (1) Lift in slipstream. (2) lift outside slipstream. (3) Vertical component of thrust. (1) and (2) need not be explained, as the method employed is identical with that given previously under headings (a) and (b), the only variation being in the angle of attack. As the angle of the thrust line is much greater than in the " tail up " attitude, the vertical thrust component is now too large to neglect, and we obtain for item (3):— Lift due to vertical thrust component=thrust X sin(a3—oc0)° where a3 is the angle of attack of the wings in the '' tail down " attitude. On the assumption of no acceleration, the drag is unneces- sary for this stage, and need not be calculated. The total lift L is obtained by adding items (1), (2) and (3), and is then plotted against air speed V. Then the speed at which the total lift is equal to the weight W is the minimum take-off speed Vs. We can now proceed to evaluate the run required against any head wind or deck speed. As previously explained, the run is divided into two distinct parts :— (A) Bun with Tail Up :—Let the constant wind or deck speed = Vo ft./sec. Then, if at any instant the speed of the aircraft relative to still air is V ft./sec. its speed relative to the deck will be (V - Vo) ft./sec. Now, starting from rest relative to the deck or ground (actual air speed V,,), take constant increments of velocity, say 5 m.p.h. to slightly above the minimum take-off speed. For each of these velocity increments there is a constant increment of momentum, given by Increment of momentum = W- X increment of velocity g (velocity in ft./sec..). Now from the curve of accelerating force find the mean value of F over each velocity increment, and hence the time ' occupied in each velocity increase. „, increment of momentum since Time t = -.—-.—2 accelerating force. Then the multiplication of t by the mean velocities (relative to the deck or ground) will give the run required for each velocity increment, and a progressive addition of these runs will enable a curve of run against true air speed to be plotted. A tabular method should be employed for the above calculation, and a typical example, which should make the explanation much clearer, is given in Pig. 2. The weight of the aircraft was 7,040 lbs., and the deck speed was 16 knots. Velocity increments of 8 knots were taken, giving 7 040Increment of momentum = ~- X (8 X 1 • 69) = 6ZZ 2,960 lbs./eec. The following columns are employed :— (1) Velocity relative to still air = V. (2) Velocity relative to deck = (V — Vo) = (V— 16 knote) in this case. (3) Mean velocity VM relative to still air. (4) Mean velocity V], relative to deck. (5) Accelerating force F at each value of VM- increment of momentum. (6) Time t sees. = = -. -, 1 ' accelerating force. (7) Increment of run = t x mean velocity Vi>. (8) Total run by progressive addition. Column (8) is plotted against column (1), and the value of the total run found when V is equal to the minimum take-off speed. In the example the minimum take-off speed was 43-5 knots, giving a total run of 142 ft. against a head wind or deck speed of 16 knots. FIG.2. CO VELOCITY RELATIVE TO STILL AIR V KNOTS 16 £4 32 1 1 40 4-8 56 (2) VELOCITY RELATIVE TO OECK (V-I61KN0TS 0 8 16 24 32 40 (3) MEAN VELOCITY RELATIVE TO STILL AIR VM KNOTS 20 28 36 44 52 OS)MEAN VELOCITY RELATIVE. TO DECK v 0 KNOTS 4 .12 20 28 36 (9MEAN ACCELERAT- ING PORCE OVER EACH INCREMENT 1882 1808 1680 1521 1547 C6>TIME. 1- SECS. FOR EA.CH INCREMENT 1-572 1-640 1-764 1-945 2-200 C7) RUN FOR EACH VELOCITY NCREMENT FEET 10-6 33-3 59-6 92-2 134 (8)TOTAL RUN IN FEET 0 10-6 43-9 103-5 195-7 329-7 This process is carried out with different values of head wind, or deck speed, and a curve of total run against deck speed plotted. The total run will, obviously, become zero when the deck speed equals the minimum take-off speed. This curve,^of course, does not give the complete run necessary to take off, but only the length of run required to accelerate to'the minimum take-off speed, with the tail up. We'must now find the run while'dropping the tail. (B) Run while Dropping Tail:—As stated in the original assumptions, the aircraft during this operation runs for 500 10 20 30 AIR SPEED M.PH 446c
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