1844.] 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



295 



ORIGINAL PRACTICAL SOLUTION OF AN 

 IMPORTANT PROBLEM. 



By Oliver Byrnb, Mathematician. 



Given the angles of elevation of any distant object, taken at any 

 three stations in the same horizontal plane, and the distances between 

 the stations, or data sufficient to determine them; to find the height of 

 the object, and its distance from any of the stations. 



Let A, B, C, be the three stations in the same horizontal piano, T 



any distant object, T O the 

 perpendicular let fall from T 

 to the horizontal plane passing 

 through A, B, C; the distances 

 A B, B C, C A, and the angles 

 of elevation T A O, T B O, 

 TCO, are given. As AO, 

 BO, CO, are in the same 

 horizontal plane each is per* 

 pendicular to T O. 



The right angled triangles 

 O A T, O B T, O C T, having 

 a common perpendicular, their 

 bases AO, BO, CO, are in 

 the ratio of the cotangents of the angles of elevation at A, B, C, 

 lespectively. Consequently, in the figure AOCB, vre have given 

 the lengths of the lines A B, B C, CA, which we shall call a, 6, c, 

 respectively, and the ratios of O A to O B to O C, which we shall call 

 m : ti : p. It is evident, if we find any of the lines O A, O B, O C, 

 we can then readily find the altitude O T, or any other line in the 

 figure ; for this purpose we shall first give the following lemma: — 



If a straight line AB, be divided in a given ratio in the point C, 

 and AB be produced to S, so that, C S : CAasCB : AC— CB, 

 then if a circle be described with S as a a centre and S C as radius : 

 lines drawn from A and B to any point P in the circumference of this 

 circle will be in the ratio of AC : CB, i.e., 



A c : c B : : A P : B P. 



This lemma is proved by that celebrated mathematician Thomas 

 Simpson nearly as follows: — let AC = »», CB=:«, BS=;a:; wis 

 supposed to be greater than n. 



n + .r ■.)«■.:?!; m — n 

 .'.n ■{■ X : m -\- n -\- z '. '. n : m (Q) 

 .' .n ■\- X '. n '. : m -\- 11 -\- X : m 

 .'.n -\- X : X : : m -\- n ■{- x '. n -\- X, 



thatis,sp : BS : : as : sp. 



/. because the angle ASP is common to the two triangles APS, 

 EPS, they are similar ; consequently, 



B P : A P : : n -\- x : m -\- h -\- x, 



but by (Q) 



)i + .r : )» + « + * '. '. n '. m 



.-.BP : AP : : H : ?n; 



, • , JiP \ BP '. '. m '. n which was to be demonstrated. 



S C or SP = K + ^ = the radius of the circle, is evidently equal 



■ ""■; andBSor;!r= «= . 



Now to return to the original horizontal figure A B CO. 



Suppose B to be less than either O A or O C, or suppose n^ not 



greater than m, ot p: at B the above construction is to be made: 

 A B, orals to be divided in the point D, so that AD ". DB : ', m : n; 



. . A D = JfL^, and D B = 

 m -\- n 



B C or 6 is to be divided in thi 



CE 



b p 



, and E B : 



in -\- n 



point E, so that, CE 



n b 



E B : : ^ ; «; 



Produce AB to G, so thati 



n -\- p' M 4- P 



according to the foregoing lemma, if a circle be described from G as a 

 centre and G D as a distance, all lines as A O, B O, drawn from A and 

 B, to any point O in the circumference may be in the ratio of m : n. 

 Again produce C B to F, so that if a circle be described with F as a 

 centre, and F E as a distance, all lines drawn from B and C, to any 

 point O, in the circumference, will be in the ratio of n *. p. 



a in n a 



Therefore, D G = O G = 



BG = 



FE:= FO 



It p b 



{m + n)(in — n) 



(m + n){m — n) 

 np b 



P- — n- {P + n){p — n) 



FB = 



n^b 



n- b 



P- — n' (J> + n) ( p — n) 



The four lines FO, FB, GO,.GB, are readily found, as the ex- 

 pressions for them are suited to logarithmic calculation. Then as the 

 three sides of the triangle A B C are given, the angle A B C := F B C, 

 is readily found ; next, in the triangle B F G, there are given FB, 

 B G, and their contained angle F B G, therefore F G and the remain- 

 ing angles B F G, B G F, may be found ; also, in the triangle O F G 

 the three sides are given, consequently the three angles can be deter- 

 mined. The ditference of the angles O F G and B F G, is the angle 

 O F B. Now in the triangle B F O we have given O F, F B, and the 

 contained angle O F B, hence O B becomes known. When O B is 

 found, the height of the object T, and its distance from any of the 

 stations are easily obtained. The whole of this detail of execution 

 may be briefly expressed thus : — The four sides F O, F B, GO, G B, 

 of two triangles on the same base F G, being known to find O B, the 

 distance of their vertices. The following rule gives the method of 

 calculation. 



Rule. 



The station at which the angle of elevation is the greatest, is the 

 vertex of one of the triangles, and the foot of the perpendicular height 

 of the object is the vertex of the other triangle, having the same base 

 as the former, all the sides of which are unknown but mav be calculated 

 thus : — add the natural cotangent of the angle of greatest elevation to 

 the natural cotangents of each of the other elevations, and then sub- 

 tract it from them : find the sub. logs, of these sums and differences. 

 Then find the radii of the circles that determine the point at the foot 

 of the perpendicular height of the object, or two sides of one of the 

 triangles: add together the log. of either of the sides which meet at 

 the station where the greatest angle of elevation is observed, the log. 

 cotangents of the angles observed at the extremities of that side, and 

 the sub. logs, of the sum and difference of the same natural cotangents ; 

 this sum will be the log. of the radius on that side produced. The 

 part producd is found by adding together twice the log. cot. of the 

 greatest angle of elevation, the log. of the side produced, and the sub. 

 logs, of the sum and difference of the natural cotangents of the angles 

 of elevation taken at the extremities of that side. Thus we may de- 

 termine the parts of the sides produced at the station where the 

 greatest angle of elevation is observed, and as the angle contained by 

 them is the angle contained by the horizontal lines meeting at that 

 station which can be readily found, as the three lines joining the sta- 

 tions are given, the common base is readily determined. Hence 

 having all the sides of the two triangles on this common base the dis- 

 tance of their vertices may be found by the rules of plane trigonome- 

 try, and therefore the altitude of the distant object and its distance 

 from any of the stations. 



Examples. 



I. At three stations, A, B, C, in the same plane, whose distances 

 AB, BC, CA, are 462, 429, and 495 feet respectively, the angles 

 of elevation of an object standing perpendicularly over O, are 

 36° 22' 07"'93, 48° 45' 50"-53 and 39° 0' 26"-91. Required the per- 

 pendicular height of the object and the distance from station B to O. 



25* 



