3I( 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



[October. 



only to be contimietl at a future day, whilst the river Stour carried 

 througli Rainsgate H.irbour would form a third great shingle trap in 

 Pegwell Bay, and scour that highly valuable harbour eli'ectivelv. 



PRACTICAL PROBLEMS 

 IMPORTANT IN PLANE TRIGONOMETRICAL SURVEYING^ 

 By Professor Oliver Byrne, Mathematician. 

 Tlie common phrase, "Things that are really useful are always 

 simple," IS far from being generally true; people do not like to give 

 themselves the trouble to understand a subject that may appear a 

 little compound, in fact, that which appears difficult, whatever may 

 be its usefulness or excellence, often falls into disuse, especially if one 

 of those simple, clumsy substitutes be convenient. The truth of the 

 foregoing observations will be readily admitted by the mathematician, 

 and a more striking illustration conld not be given than that afforded 

 by the class of problems which is here arranged under the title 

 " Plane Trigotwmttrical Siirrtying" and will be found of great use to 

 the practical surveyor, if only in the way of tests; the principles upon 

 which they are based are simple, and may be explained as follows:— 

 Proposition I. — If any number of lines A, B, C, D, &c., be drawn, 

 the ratio compounded of the ratios of A : B, B : C, C : D, &c., 

 continued in order to A, is a ratio of equality: or which is thesame 

 thing, when each becomes an antecedent and a consequent, taken in the 

 above-mentioned order, the continued product of the antecedents is 

 equal to the continued product of the consequents. 



Proposition II.— If triangles be formed by j'dning in succession 

 _. , the extremities of the lines of any contour and 



any point O, fig. 1, the continued product of the 

 sines of the angles opposite the antecedents is 

 equal to the continued product of the sines of 

 the angles opposite the consequents taken as in 

 the first proposition. 



O A : O B : : sin O B A : sin O A B 

 O B ; O C : 



o c : o D : 

 o D : o E : 

 o E : o A : 



Because by the first proposition the product of the first antecedents 

 is equal to the product of the first consequents ; therefore, the pro- 

 duct of the second antecedents must be equal to the product of the 

 second consequents. Or, sin O B A X sin O C B X sin O D C X sin 

 O E D X sin O A E is equal to sin O A B X sin O B C X sin O C D 

 X sin O D E X sin O E A, which are the sines of the angles subtended 

 by the lines drawn from the point O, alternately taken. If from the 

 angular points of any rectilineal contour right lines be drawn to any 

 point O, a ratio of equality will exist if, instead of the ratio of any 

 two consecutive lines, the ratio of the sines of their opposite angles 

 be substituted ; or if, instead of the ratio of the sines of the angles, 

 the ratio of the lines be substituted, the ratio of equality will also 

 exist. It is often convenient in practice to deal with the angles and 

 at other times with the lines; with the latter especially when two 

 points or more fall in a line drawn to the point O, and with the former 

 when more than two lines are drawn to the point O, from points given 

 in the same right line. 



Examples. 

 (1.) The distance between two stations A and B, fig. 2, is known (26105 

 feet) ; C and O are stations in the same plane 

 with AandB; the known angles are — A B = 

 a = 23° 29', B A = 4^35' 27', C B = c = 

 19° 9', C A = d=yo° al'. Required tlie re- 

 maining angles and the distances A, OB, 

 C, AC, and B C. Let 20105 = m, angle 

 O B C = x; then will angle A C = 180-a- 

 i — c — d—x = e—x, by making e = 180 — a— 4 

 _c — (i=66° 4'. 'Then, by the last propo'ition, 

 OA : OB = sini : sin o,'0 B : OC = sinc : 

 8ina-, OC : A=sin (e— r) : sin rf. .-.sin* 



sin O C B : sin O B C 



sin O D C : sin O C D 



sin O E D : sin ODE 



sin O A E : sin O E A 



sin e sin (?—«•) = sin a sin x sin d, by prop. 2. 



But sin (e— jr) = sin e cos a-— cos e sin x. 

 cos e sin j-) = sin a sin d sin x. 



Hence sin 4 sin c sin e cos x — sin b sin c cos e sin 



By dividing by sin x, and transposing, &c., we have 



sin a sin d 



cot^' = cote + -: — ; — : 



sin sin c sin e 



sin li sin c (sin e cos x~ 

 = sin a sin d sin x. 



. • . cot ^ = cot e + sin a cosec b cosec c sin d cosec e. Therefore to find 

 the value of x we have the following 



Practical Rule. — Add together the logs of the sine of a, of the cosecant 

 of b, of the cosecant ofc, oftlie sine of d, and of the cosecant of e ; the na- 

 tural number corresponding to this sum, rejecting 50 in the index, added to 

 the natural cotangent of e, will give the natural cotangent of x. 

 sin a =sin 23^29' .. .. Iog= 9 C004090 

 cosec i = cosec 35° 27' .... log= 10-2365778 

 cosec c = cosec 19° 9' .... log= 10-4810700 

 sind =sin 35° 51' .. .. log= 9-7676494 

 cosec e= 66° 4' .... log= 10-0390452 



501277514 

 The natural number corresponding to -1277514 of a logarithm = 1-3419970 

 to which if nat cot of e =0-4438352 



1-7858322 



be added we have 

 = the natural cotangent of x = 29° 14' 50" 

 .-.6-^ = 30° 49' 10" 

 When x is obtained, the other linear and angular distances may be easily 

 determined hy the rules of plane trigonometry. OA = 17676-01 feet; 0C=^ 

 18087-388 feet ; A = 2881093 feet ; B C = 27082-2 feet. 



(2.) From the first of four stations, figs. 3, 4,5, in the same plane, the an- 

 Fig. 3. Fig. 4. Fig. 5. 



gular and linear distances of the second and third stations are known, as welt 

 as the angular distances of those three stations observed from the fourth. 

 Determine the remaining Hnear and angular distances of those stations. In 

 any way this problem may be taken,— the first station is the best for the 

 point O, round which, as before, when we have particularized the distances, 

 we shall compare the ratios. 



The linear distances from 1 station. 1, 2, = 3755 ffiet = n ; 1, 3, = 4000 

 feet = m. Angular distances at 4 station. 2 4 = 44° 33' = a; 0, 4, 3,= 

 55° 17' = 4. And the angular distance at station I, in fig. 3, or Z 2 1 3 = 

 180° 0'; in fig. i, 12 1 3 = 23-2° 16'; in fig. 5, /2 1 3 = 127° 44'. It 

 is required to find the linear distances of 2, 4 ; 4, 3 ; and 4, 1 ; as well as 

 the angular distances 12 4 and 13 4. Supposing A, li, and C, to be at the 

 stations 2, 4 and 3. 



Put the angle OC B=.r, then will A B = 360 — a— 4 — c— j-=d— a:, 

 makingd=3G0-a-4-c = 80'' 10'(fis. 3); =27° 54' (fig. 4) ; =132° 26' 

 (fig. 5). 

 Then, as usual, A : B=sin a ; sin {d—x) 

 OB : C = sin J? : sin (J) 

 C : A = m : n 

 . • . m sin a sin x=n sin b sin {d—x) ; but sin (rf— .r) = sin d cos:r— cosrf 

 sin .r. 



. • . m sin a sin x= n sin b (sin d cos x— cos d sin x) 

 Dividing by sin x we have, 



>n sin a = n sin b (sin d cot a-— cos d) 

 m sin a n sin 4 cos d 



. • . cot X = — : — I — '• — J -^ 1 — '■ — J 



n sin b sin a n sin b sin a 



cot a- = cot rf+ — sin a cosec b cosec d. 

 It 



Which expression in words gives the following — 



Rule. — Jdd together the sub log of n, the log of m, the log sine of a, lotf 



cosecant of b, and the log cosecant of d ; the natural number corresponding 



to this sum, when a proper allowance is tnade in the index, added to the na- 



tural cotangent of d will give the natural cotangent of x. 



n =3755 .. .. sub log = 6-4253201 



m =4600 .. ..log = 3-6626578 



sin a ■= sin 44° 33' .. .. log sin = 9-8460471 



cosec A = cosec 55° 17' .. .. log cosec = 100851396 

 cosec d = cosec 132^ 26' .. .. log cosec = 10-1319066 



40 1512510 

 Rejecting 40 in the index we have •1512412, to which log the natural 

 number 1-41658 corresponds. The natural cotangent of d = nat tan 42° 

 26' = -9141929, which is negative. 



. ■ . from 1-4165S00 take -9141929, the natural cot of .r = -5023871. 

 . • . J- = 0:'.° 19' 32-, fig. 5 ; = 13° 38' 9", fig. 4. 

 When these two cases are understood there can be no difiiculty in finding 

 X in fie. 3, as it is a mere repetition of the last case with the exception of 

 taking-rf = 80° 10' 



