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Aviation History
1917
1917 - 0083.PDF
JANWARY 25. UNITS EMPLOYED IN RECORDING THE RESULTS OF WIND TUNNEL EXPERIMENTS. IN spite of all that has been written on wind tunnel experiments, the nature of the units employed inpresenting the results of these tests still remains a mystery to many of those who are daily taking amore direct interest in matters aviatic. This is increasingly evident from the number of enquiries we receive week by week upon the subject. A briefdescription and explanation of the various units may therefore be helpful in giving a better under-standing of the precise nature of these " coefficients " that are constantly referred to in all matters dealing with the investigation of aeroplanes and their parts. The experiments with a dropping apparatus carriedout by Mons. G. Eiffel at the tower bearing his name have definitely proved that the resistance offered by A body passing through the air, or—what amountsto the same thing—by a stationary body placed in a current of air, can be expressed by the formulaR = KSV2, where S is the area of the surface—(the area is taken as the actual area of a plate, either inclined or at right angles to the air current, and, inthe case of a rounded body, as the area of the cross- section)—V the velocity of the air current, and K a-coefficient, depending on the form of the body, independent of V and increasing slightly with S. The unit resistance K is the resistance in kilogrammesoffered by a solid body of i square metre area moving at a velocity of i metre per second in air of normaldensity. The normal density of air at 150 and a barometric pressure of 760 millimetres is 1,220. In Eiffel's books the experimental results are all reduced to a velocity of 10 metres per second. If,for a certain body, the value of K is stated to be 0.05, the resistance of that body at a velocity of 1 metre per second is: R =0.05 x 1 x I2 = 0.05 kilo-grammes per square metre. At a velocity of 10 metres per second the resistance per square metre isof course, 0.05 x 1 x 10- = 5 kilogrammes. If the velocity is expressed in kilometres per hour,the co-efficient K must, if the formula is to " hold good," be altered accordingly. Thus, 10 metres per second = 10 x 60 x 60 = 36,000 metres per hour= 36 kilometres per hour. Therefore, if V is 36 kilometres per hour, the coefficient K must obviously be multiplied by (-^) = 0.077. In the previous example, if the velocity is expressed as 36 kilometresper hour instead of as 10 metres per second, the formula becomes : R — 0.077 KSV* = 0.077 *0.05 xi x 362 >=» 4.9716 kilogrammes per square metre. The slight discrepancy (from 5 to 4.9716)is, of course, due to the fact that only three decimal places were taken in the number 0.077. If it is desired to employ this formula with Englishunits (pounds per square foot and miles per hour), K must be multiplied by a certain number represent- ing the difference between English and metric units.As one kilogramme ==2.2 lbs., 1 square metre = 10.76 sq. ft., and 1 kilometre per hour = 0.62 mileper hour, this number is obviously—as Eiffel's results are reduced to a velocity of 10 metres per second or 36 kilometres per hour—0.2 (—- ) — 0.0408. For v 22.3'purposes of conversion the formula then becomes : R = 0.0408 xKSV2, where K is the coefficient givenin Eiffel's works, S the area in sq. ft., and V the Telocity in miles per hour Thus, if Eiffel gives the coefficient for a certain plate or body as 0.05, the resistance per square foot of that body at a speed of 50 miles per hour is : R — 0.0408 x 0.05 x 1 x 50-* =5-i lbs. As it is sometimes inconvenient to employ the original formula, owing to the care necessary in choosing the right units, it is frequently replaced by another which has the advantage that it is inde- pendent of the units employed. This formula is8 written by Continental authorities as R — z- SV*, in which 8 is the specific gravity of air, g the acceleration due to gravity, and z a coefficient depending only on the shape of the body. S and V, as before, represent the area and the velocity respectively. With Eiffel's definition of K,- — ~^~gj ™g —0.125. Therefore z = 8K. The quantity 2 in this equation, which, as will be shown presently, is generally indicated by the letter C in this country, is termed an " absolute " quantity, since it is independent of the units em- ployed, so long, of course, as these are dynamically self-consistent. In this country the same formula is usually written in a slightly different way—that is to say, the notation is different. Thus, in the annual reports of the Advisory Committee for Aeronautics, pressures and forces are recorded by tabulating the quantity C in the equations :— P «= Cpv2 F - CApv°~ where P and F are the pressures and forces in what-ever units of mass, length and time are employed, A is the area, p is the density of the air and v is the velocity. If it is desired to find the pressures in lbs.per square foot when v is in feet per second, the value of p is 0.00238. For example, if the " absolute "lift coefficient of a wing section is given in the N.P.L. reports as being 0.3, the lift per square foot of thatsection at, say, 100 ft. per second is found by sub-* stituting in the formula P •= Cpv2, which gives:P =-0.3 x 0.00238 x ioo2 "=7.14 lbs./sq. ft. If it is desired to find the lift of the same wing section,and the speed is given in miles per hour instead of in feet/seconds, the value of p becomes 0.0051, and the lift per square foot at the same velocity asbefore, only expressed in miles per hour (1.46 ft./sec. = 1 m.p.h., therefore 100 ft./sec. —68.4 m.p.h.) is:P =0.3 x 0.0051 x 68.4* =7.14 lbs. per sq. ft. When the loading is in kilogrammes per square metre and the velocity in metres per second, the value ofP is 0.125. This corresponds, it will be seen, with the value of £ or 0.125 f°r ~ m *ne French equation o R = z- SV2, so that we have here proof of the simi- larity of the two formulae. It will be seen that z = C, _•• = p = 0.125, when the loading is in kilo- 6grammes per square metre and the velocity in metres per second, - 0.0051 when the loading is in lbs./ square feet and the velocity in miles per hour, —0.00238 when the loading is in lbs. per square foot and the velocity in feet/second. It thereforefollows that to convert Eiffel's coefficient K into the " absolute " coefficient C all that is necessary is to $ F 2
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