32 CHAIN SURVEYING 



SOLUTION. Here AD = 206.1 and ED = 35.1; therefore. 

 substituting in the preceding formula, AB= A/206.1 2 +35.1* 

 = 209.1 ft. 



Survey of a Closed Field. If a closed field is to be surveyed 

 without the aid of an angle-measuring instrument, the area 

 is divided into triangles by means of diagonals, which are meas- 

 ured on the ground. The area of each triangle may then be 

 determined by the formula 



A=^s(s-a)(s-b)(s-c), 

 in which a, b, and c represent the three sides and s represents 



a+b+c 

 half of their sum, or - . 



When obstacles make it impossible to measure directly the 

 diagonals of a field, as, for instance, the diagonal BE, Fig. 6, 



a tie line FG parallel to BE 

 I! is run and measured. Then, 



GFXAB 



BB IF- 



To run the line FG, pro- 

 duce BA and select any 

 convenient point F and 

 \ --- -/* measure AF. Then pro- 

 duce EA and locate G 

 X from the relation 



Fie. 6 AG = ^^ E - 



AB 



EXAMPLE. In Fig. 6, let the lengths of the sides be as fol- 

 lows: AB = 32Q ft., BC = 217 ft., CD=196 ft., DE = 285 ft., 

 and EA =304 ft. It is required to calculate the length of the 

 diagonal BE by means of a tie-line. 



SOLUTION. Let the line BA be prolonged 100 ft. beyond A ; 

 that is, make AF= 100 ft. Then, AG must be equal to 

 AFXAE 100X304 



Let the length of GF, as found by measurement, be 125 ft. 



. GFXAB 125X320 

 Then . ,__ --- J55 



