1840.] 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



r> 



ON THE TESTING OF SURVEYS BY CALCULATING THE 



LINES OF CONSTRUCTION. 



Bv S. Hughes, C.E. 



In transferring to paper the measured lines of a large survey, it is 

 always considered by the surveyor a nratter of great satisfaction if the 

 lines prove or fit in to each other as it is called. 



That tlie meaning of this term may be understood by those who are 

 not conversant with the practice of surveying, suppose three lines 

 have been measured in the form of a triangle, A, B, C, and a fourth 

 line B, D has been measured from one of the angular jioints to D in 

 the opposite side. It is evident that the three sides of the triangle 

 being given, the length of B D is determined, and ought on the ground 

 to measure neitlier more nor less than the distance in a direct line 

 from B to D. 



Now, if on laying down the above diagram on paper it be found 

 that the distance between B and D either exceeds, or is less than that 

 measured on the groimd, the presumption is that an error has been 

 committed, and the work shouUl forthwith be examined in order to 

 discover it. B D is called a proof line, and the above example is 

 given to illustrate the nature of these lines. 



The object of this paper is to investigate a fe\y simple formulEe for 

 determining the lengths of proof lines by calculation, in order to save 

 the trouble of laying down at an inconvenient time the main lines of 

 extensive surveys, and to guard against the danger of error in laying 

 down the lines on paper. 



Pros. 1st. — Let a, b, c, be the three given sides of a triangle, it is 

 required to determine the perpendicular A B from the vertex to the 

 opposite side c, and also the segments into which the side is divided 



by such perpendicular. Put .r = one of the segments, and we have 

 a-^—x'=:b-^—(c—,vy or a-'—x'^=b^—c-—x--\- 2 c .r add .»■', and a-=r 

 b''—c-+ 2 c .1-; subtract^ 6» — c' and a^ — 6^4-c^= 2 c x: Divide by 



a'—b''+c2 'a^—b' , c 

 2 c and ~ =.v or — • -f-=.i-the greater segment. 



Now the difference of two squares is equal to the product of the 

 sum and difference of their roots. Let s and d be the sum and differ- 



ence of the two sides a and 6, then - 



2 



2c 



=: X the greater or less 



segment, according as the positive or negative sign is used in the 

 fornnda. The perpendicular A B of course will bo y'a- — .r-. From 



the nature of similar triangles it is also=:— where x is the lesser 







xb 

 segment, and ^ " — vvhere x is the greater segment. 



Suppose an obtuse angled triangle, then a' — (c+.f)^=62— .r' or a' — 



c' — x' — 2 c .r=6-' — ,1-2. Add ,i-' and a' — c-' — 2 c x^^b 

 and a- — c" = 6- + 2 c r. Subtract 5- and divide by 2 c, 



- =:,r, or substituting as before the sum and difference 

 ^i c 



-'. Add 2 c*- 



, a- — 6- 

 then ^; 



of (I and b we 



l);ive =,r, and the perpendicular here will be ^b' — x^. 



2c 2 



Api'lication I. — Given the three sides a 6 c of an acute emgled 



JS. 



triangle, also B D, and consequently D C the segments of the b.ise c, 

 required the length of the proof line A D. 



Put BD=rf the perpendicular AP as found by the preceding pro- 

 blem=^, and, the segment B P also found by the prublem=s, then 



Case II.— Let the triangle be obtuse as ABC, thou retaunng the 

 same letters as above A^p'+(d — s)-— A D. 



Case III.— In the triangle ABC, the three sides are given, also the 

 distances B A', B D' required the length of the proof line A'D'. 



B— p D 1' 



Through the point A draw AD parallel to a A'D', then B A' : BA : : 

 BD' : BD and AD may be found as shewn in case I. Then we have 

 BA : BA' : : AD : A'D' the length required. 



Or suppose the two sides B A, and B C arp given also B A', B D' and 

 A' D' and the length of the proof line AC be required. Through A' 

 and A draw A'P', and A P perpendicular to B C and find the length of 

 A' P' by the problem. T hen B A' •. B A : : A' P' : AP find also the 

 length BP, and then VAP^+(BC— B P)==AC. 



Corollary. By means of the formula in tliis case maybe determined 

 also any proof line measured on the opposite side of the base line to 

 that on which the triangle has been constructed. 



Thus let A B C be tlie triangle of which the sides are given, and of 



which one of tlieni A C has been Continued to D, and its extremity 

 connected by the line D E, with another of the sides B C also pro- 

 duced to E. Draw A F and G D perpendicular to B E, and find the 

 length of A F by the pro blem, the n AC : CD : : AF : D G. Th e 

 distance CG will then be= j/CD'-DG-. And ED=: vGD'+(BE- BG)' 



