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Aviation History
1919
1919 - 0501.PDF
Tangent. Cotangent. .000 90 .087 n.430 85 .176 5-671 80 .268 3.732 75 •364 2.747 70 .466 2.145 65 •577 1-732 60 .700 1.428 55 .839 1.192 50 1.000 1.000 45 •m APRIL 17, 1919 for all angles for use in calculations, from which the following short table has been extracted :— Angle. Sine. Cosine, o .000 1.000 5 .087 .996 i» .174 .985 15 .259 .966 20 .342 .940 35 .423 .906 3» .500 .866 35 -574 819 40 .643 .766 45 -7<>7 -7°7 Cosine. Sine. Cotangent. Tangent. Angle. The foregoing ratios are for angles of 900 or less. For angles between 90* and i8o\ the ratio of the supplement is used ; thus if A is an angle greater than 90°, Sin.A = sin(i8o° -A). 1 The following must be borne in mind when using the ratios of the supplements : for all angles of 90° or less, all the ratios have positive values, but for angles between 900 and 1800, the skie is positive, and the cosine, tangent and cotangent are all negative. Example:— Cos 8o° = .174. Cos ioo° = cos i8o° —ioo°) = —cos 8o° = —.174. In aay right-angled triangle, the use of these ratios permits any side or angle to be calculated when only two sides or one side and an angle are known; thus in Fig. 1, given A 6o°, b 40 miles, to find a, c, and B. B is the complement of A, hence it is 900 —60* = 300. b 40 c = — *m cos 6o° = .5 (from tables). Therefore c = 80 miles. Also— b 40 — «= — — cot. 6o° = .577 (from tables). Therefore a = 69.4 miles (nearly). In the majority of problems in 'plane navigation, the triangles which have to be solved do not contain a right- angle, and it is then necessary to know either one side and two angles, or two sides and one angle before the remaining sides and angles can be calculated. In order to solve any 'plane triangle ABC (Fig. 2), either of the following rules may be used according to the data available :— Rule 1 Sine Law: a b sin A sin C tan \ (A -B) tan i(A+B) tanj (B-C) tan f(B+C) tan I (A -C) tani(A+C) 12 13 sin B Rule 2.—Tangent Law, a -b a+b b-c or r— b+c a—c a+c Example :— The desired track is ioo° true, wind from 3000 true at 20 m.p.h., air speed 100 m.p.h., what is the true course ? Sketch roughly the diagram, Fig. 3, in which [NCT = 100° (NCW = 3000 It follows that psrcw,= 1200 and (TCW, == 200 CV represents the velocity of the wind, and VT the air speed and course. The line CT, is parallel to VT, hence the angle T,CT equals the angle CTV. and the course is therefore jNCT - |CTV This is a case for the use of the Sine Law, and oy sub stitution in Eq. 12, we get:— 100 sin 200 20 sin CTV sin CTV = 20 sin 20 u 20 x .342 100 = .0684. From the table it will be found that ICTV is less than 5°, being actually 30 55', say 40, and the course will therefore be ioo° - 40 = 960 true. With a little practice, this method will be found speedier and much more accurate than the graphical method. if P <»sag§^>l Another Example An aircraft flies on a course 230* true at an air speed of 90 m.p.h. In one hour the pilot finds that he has travelled 60 miles along a track 1800 true, what is the direction and velocity of the wind ? Sketch roughly the diagram 4, where CT is the track and represents 60 miles. CD is the course, representing also the air speed of 90 m.p.h., therefore DT represents the direction and velocity of the wind. W,C is parallel to DT, therefore WjCD = |CDT, and the direction of the wind will be given by [NCD (2300) + jCDT I |NCT = 1800, (NCD = 2300 . . pCT = 230° - i8o» = 500. The sine law cannot be applied in this case, because only one angle is known, and the side opposite to that angle is unknown. It is, therefore, necessary to use the tangent law, Eq. 13, which is applied thus :— 90 - 60 _ tan i(]CTD - CDT) 90 + 60 ™ tairy(CTD~+7:DTy hence Since jCTD + jCDT + (CTJD + £(130°) = 30 = 150 " tan J(CTD - jCDT) = from tables, jflCTD- |CDT) = and J(]CTD + |CpT) = subtract |CDT = DCT = i«o*. and jDCT => 5o°, then CDT = 1300 65* tan 1(;CTD - |CDT) tan 65s 30 x 2.145 23 150 13', say 230 65° = .429 42" The bearing of the wind is therefore 2300 + 42° = 272* true. Another angle of the triangle has now been found, and the sine law may be applied in order to calculate the velocity of the wind. 60 DT sin 42" DT = 69 m.p.h. (nearly). sm 50" 60 x .760 .669 This second example requires a slightly greater time for its solution than by the graphical method, but is more accurate. The two examples given above are representative of the practical problems which will arise in connection with course- setting during a journey. The usual method of solution by means of the course and distance indicator is sufficiently accurate for all practical purposes, but knowledge and practice of the mathematical method will be beneficial. Spherical Navigation The shortest distance between two points on the earth's surface lies along an arc of a great circle, and the length of the arc is measured in degrees, each degree being equal in length to 60 nautical miles. Three arcs of different great circles may meet to form a spherical triangle, as in 5, where a is a portion of a meridian, b is a portion of the equator, and c is the arc of the great circle joining A and B. As all meridians cross the equator at right-angles, the angle C will be 900, but in a spherical triangle, the sum of the three angles is always greater than 180° by an amount unknown until the triangle is solved. When a spherical triangle is right-angled, it may be solved when any two parts are known in addition to the right-angle, by applying Napier's rules. Napier's Rule 14 Any right-angled spherical triangle may be regarded as comprised of five parts in addition to the right-angle, i.e. Fig. 5 is regarded as comprising :— a, b (900 - A), (900 - e), (900 - B). If these are written in a circle, as in Fig. 6, any part taken as middle part will have two others adjacent, and two opposite, thus:— Take (900 - c) as Mid. Part, then (900 - B), and (900 - A), are adjacents, and a and b are opposites. Then f tangents of adjacents multiplied together sin Middle Part .. • or ^cosines of opposites multiplied together. Example:— a = 300, b = 500, to find c. 50I
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