274 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



[Skpt 



Fig. 3. 



posed to be drawn from L lo S, it is evident that the point E falls within 

 the triangle LPS, the angles of which are found by problem (3) to be as 

 follows : — 



ylSPL = 112' or 27" = a; 



ZPLS = 15 21 32 ^ h; 



Z PSL = 52 37 01 = c; we have also 



Z PES = 112 30 00 = »« = 10 points, or W.S.M'. 



/ PEL = 78 45 00 = n = 7 points, or N. by E., and 



Z LPE = X. Then by the general rule we have, 



log. cosec. a ■=10-0339082 



log. sin. c =9 9001454 



log. cosec. m =10 0313847 



log. sin. (m + n) =9 2902357 



log. cosec. (6— n) =10.0486213 



Reject 50 and 49'30G2953= 1-3062953 = log. of 0-2024395 ; from which 

 substract — -5002568, the natural cotangent of (6— n), and we have, 

 0-7033003 for the natural cot. of a-, .•. ^■=54'' 52' 40"=L P E. The dis- 

 tances L E, S E, C E, can be found by Plane Trigonometry, and are 

 53-11906; 17-1334; and 1391746 miles respectively. John Topham 

 makes them 53-04; 17-36 ; and 14333 respectively. 



II. Being at a town in Kent I observed three objects on the other side 

 of the river Medway, a castle (C), a windmill (W), and a spire (S), whose 

 distances from one another are known; from the castle, (the nearest object 

 seen,) to the spire, (CS) is 10 furlongs ; from the castle to the windmill 

 (W C), 23 furlongs ; and from the windmill to the spire (W S), 25 furlongs. 



I took with a theodolite, the angular 

 distance between the castle and the 

 spire (CT S) and found it to be 28° 

 34', and between the castle and the 

 windmill (C T W) 57° 45' ; what dis- 

 tance did I stand from each of these 

 objects ? From having the three sides 

 of C S W we can find the three angles 

 by problem 3 ; hence we have. 



Being calculated independant of 

 each other, and making 180° affords 

 a proof. 



Fig. 4. 



} 



00; and instead of (4 — n) take 



negative. Putting a-=zTCW; and 

 proceeding according to the general rule, we find the sum of the five loga- 

 rithmic quantities to be 501382785, or 0-1382785 which correspond to the 

 natural number 1-374923, which is negative. From — 1-3749236 take the 

 thenat. cot. (52° 08' 3S") = 0-7772423 and it leaves — 0-5976737 :— this 

 nat. cot. corresponds to 59° 08' 04" ; . • . .t = 120° 51' 56". Then (59° 08' 

 04") — (S7° 45' 00")= 1° 23' 04 '= Z T W C. To find the remaining parts 

 fall under the head of common-place Trigonometrical calculations, 

 T W = 23-3439 furlongs, T = 0-6570643, and C S= 10-57215. 



III. In a garrison there are three remarkable objects. A, B, C, the dis- 

 tance of which from one another are known to be as follows : — B A = 213, 

 B C = 424, and A C = 262 yards. lam desirous of knowing my position 

 and distance when standing at a station A, with respect to the three points 

 A, B, C ; at A T observed A to be the nearest object, and the angles 

 BS A=13° 30', A S C = 29° 50'. This example was originally propesed 

 in Dr. Hutton's Conic Sections, then reproposed in the Lady's Diary for 

 17B7, where two ditferent solutions are given the following year ; but none 

 of the different solutions to any of the cases of this problem are as practical 

 as following the general rule here given. A S will be found = 429-6814 • 

 C 8 = 524-2365 ; and B 8 = 605-7122 yards. 



(7.) In running a mean line through a country, I arrived at the bank of 

 » river, B ; and having booked a station at O, I find by subtraction that 



O B is 25-36 chains : causing a flagstaff to be plaied at O, and another at 

 B, I crossed the river in a boat, and set up a third at A, in a right line 

 with B and O. Proceeding along the line to O', a distance of 2536 

 chains, a fourth flagstaff is set up ; then continuing the line to a convenient 

 point T, a theodolite is set up and the point in each staff as O, B, A, O', 



which is cut by the same hori- 



^ zontal plane is noted. Now 



at C any station where the 



points O, B, A, O', can be 



seen, I take the angles a, b, c 



and find them So be 12° 05'' 



28° 25', and 17° 23' respect.' 



ively ; required the breadth of 



the river A B ? Let B A = j; 



the angle A O' C = x, then will zB A C-c + x, and A B C= 



180°— 6-c-.r = <;-x;also(i=180°-4-c=134°12': BOC = 180°-a- 



4-c-x, putting e=180°-u-6-c = 122° 07'. In this problem we 



shall compare the ratios round two points O and O'. 



First round O, O B ; O A = n In + ij; 



O A: O C = sin. (0 + 6) : 



O C : O B = sin. {tl-x) ; 



Second round O', O' A- -O' B = k ; n -|- 1/ ; 



0'B;0'C = sin', (4+.;); 



O' C : O' A = sin. (c + .r) ; sin. c. 



^Ve obtain directly from compounding both the analogies and expunging 



the common factors. 



; sin. 

 ; sin. , 



{c-l-x); 



: sin. ('i — x) ; 



sin. (d — t) 



/siu. a sin. 

 ^\/ sin. c siu. 



(6 + C -) 

 {b + a) 



Let the right hand member of thi> 



sin. (c + x) 



equation for the sake of brevity be called K. But, sin. (d — x) 



sin. {b + (c + x)} 



(180°-6-c-i-) = sin. (4 + t + ar). .• 

 . • . sin. 4 cot. (c -f a-) + cos. 6 ^ K. 



sin. (c + x) 



= K. 



cot. (c + x) = ^ cot, 4. 



sin. 4 



X = cot. 



'{(n 



sin. a sin. (4 -)- c) 



sin. c sin. (4 4- a) sin 



jt^J-cot. 4j- 



From this equation the rule is deduced. We need scarcely remark, 

 that before finding any of the distances is necessary to find the unknown 

 angle x. 



Rdle. — Add into one sum the log. sin. of a, and the log. sin, of {b -\- c) ; 

 then add together the lug. sin. of c, the log. sin. of {b -j- a), and twice the 

 log. sin. of b: subtract the latter sum from the former and divide the re- 

 mainder bij 2, the neutral number coj responding to the <iuotient made less by 

 the nat. cot. of 4, will give the nat. cot. of (c -)- x), and consequcntii/ x be- 

 comes known. 



Log. sin. 12° 05' (a) = 9-3208400 

 log. siu. 45 48 (4-f c)= 9 8554654 



Reject 20 



F'rom 



Take 



19-1763050 

 1-1763050 

 2-6428665 



Log. sin. 17° 23' (c) = 9-4753271 



log. sin, 40 30 (a + 4) = 9-8125444 



2 log. sin. 28 25 (4) = 193549950 



Reject 40 38-6428663 



2-0428665 



This divided by 2)0-5334385 = 0'26G7192 which correspond to the nat_ 

 number 1-8480730, this is to be made less by the nat. cot. of 28° 25' (4) = 

 1-8181761, which is greater than 1.8480730 by 0001031 ; . • . this is the 

 negative nat. cot., and corresponds to 21" ; hence, (c -{■ x) — 90° 0' 21" ; 

 but c=17° 23', . • . ^=92^ 37 21''. The remainder of the calculations 

 are common place. B A = 43-88967; A C = 81-009 chains. Before 

 proposing the next problem, which is rather complicated, we wish to re- 

 mark, that it is given more to illustrate the properties of the foregoing 

 aialogies than for any practical utility. 



(8.) In superintending an extensive survey, I gave directions to as as- 

 sistant vrhu chained one of the base lines, A B, to set up flagstaffs equally 



