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Aviation History
1917
1917 - 0839.PDF
/ AUGUST 16, 1917. It is interesting to note that, putting v — «V as before, (19) can be written in the form Tp (or y) - ?XT („ J (2O) Which should be compared with (18). It will be noted that the value v •= V is a critical velocity m that, at this value, there is discontinuity in the law of pressure variation for the front and back of the plane, and for the total resistance. This discontinuity occurs, according to accepted views, at a velocity equal to 1 • 34 times the velocity of sound. There is, however, no geometrical dis- continuity in the pressure-velocity curves. With the alternative view suggested by the writer, a similar discontinuity is found, but at a velocity v = V identical with the velocity of sound. To obtain ex- pressions corresponding to those found above, we useJ- oi a2 sin <t>. cos B. dd-in place of a2 sin2 <p. cos 6. d$, the method of procedure being the same in each case as before. Thus Bin 4, cos e.dO = V ( v>-"')\«-«*« lav J Vv the general integral to be evaluated between the propel limits. We will here give only the results. Case i, v<V.—Pressure on the front of the element is given by * - pV2( — + — n + TTH2 + ---n* - ••„«* + terms involving higher\z 3 a 15 48 powers of »). (21) pV2When n — o, p — -— . For low values of v, the increase of pressure is proportional to the wind velocity, approximately. When v — V, n — 1, and the pressure is given by 16,.. .2p =- • 6^ times the static pressure, a value to which (21) approximates as n increases. For the back of the element i«2 - f5«'-&c- When n-o,p- .^J and when „ 1. P (22) 0, a value to which (22) approximates as « increases. The total resistance is given by 8 8s Case 2, v >V.—The pressure on the back of the element is zero, as before. The pressure on the front, which is equal also to the total resistance, is given by ~2PV2(n2 f 2 + _ va - V4 we see that p — 2pv2 when V — o, and can only be proportional to v* for all values of v when the second and third terms are negligible, i.e., when V»- 0, or t; =• oc. It is interesting to note that, with both views of kinetic theory, the expressions for the pressures on the back and front of the element involve the first, second and third powers of v or n up to the value n — 1, whilst the total resistance nvolves the first and third powers only. For values of n greater than unity the expression giving both pressure and resistance omits the odd powers depending only on the second power in (19), and depending chiefly on the second power in (24). The above results are shown in the graph (Fig. 2), which gives the pressure-velocity curves—(a) for the front of the element; (b) for the back of the element ; (c) for the total resistance—the full-line curves corresponding with expres- sions (14), (17), (18) and (19), and the. broken-line curves with (21), (22), (23) and (24). As a further development of more particular interest to aeronautics, we may apply the foregoing methods to deter- mine how, for a given wind velocity v, the pressures on the front and back of an element of area vary when the inclina- tion of the relative wind varies. It will, be seen from Fig. 3 (in which the lettering is similar to that of Fig. 1 to facilitate comparison) that the effect of compounding the relative velocity v of the wind with the molecular velocity V is to displace the centre C of the sphere of reference a distance CK ( — v) from the element K along a line inclined to the plane at an angle EKN (- \p, say), equal to the inclination of the relative wind. Thus, for any stream such as GC (— V) the resultant velocity is given by GK (=•«). If from G we draw a perpendicular GS to the plane, meeting the plane at S, and join SK, the angle GKS (•=• B, say) is the resultant inclination of the stream to the plane. By l/elocity reasoning similar to that already given, we see that we must find the mean value of aasina(8 for all points such as G uni- formly distributed over the surface MHEN, and apply this mean value to gas of effective density corresponding to the same surface. We cannot, however, take a zone GH as our element of surface, because 0 has not the same value for all M Fig. 3. points in this zone. Now, a? sina 18 = GS2, and if we draw CW perpendicular to the plane MN, join GW, SW, we see that, putting ax for GW. and <pl for the angle GWS, «,*sin2<p, - GSa-aasinB/3. We therefore have to find the mean value of ata sin2 ^>j for all points such as G uniformly distributed over the surface MHEN. This is the problem 839
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