You know that in logarithms, instead of multiplyingyou just add logarithms; also, instead of dividing onenumber by another, you just subtract the logarithm ofthe second from the logarithm of the first. Note that thelogarithm of 1 is 0.000000. Therefore, the aboveequation can be expressed as follows:(log sin 1 + log sin 3 + log sin 5 + log sin7) - (log sin 2+ log sin 4 + log sin 6 + log sin 8) = 0Suppose, now, that after the second figure adjust-ment, the values of the angles shown in figure 15-28 areas follows:A table of logarithmic functions shows the log sinesof these angles to be as follows:By subtracting the two sums, you get the following:9.243442–10-9.243395–100.000047Therefore, the difference in the sums of the log sines is0.000047. Since there are eight angles, this means theaverage difference for each angle is 0.0000059.The next question is how to convert this log sinedifference per angle into terms of angular measurementTo do this, you first determine, by reference to the tableof log functions, the average difference in log sine, persecond of arc, for the eight angles involved. This isdetermined from the D values given in the table. Foreach of the angles shown in figure 15-28, the D value isas follows:The average difference in log sine per 1 second ofarc, then, is 20.01/8, or 2.5. The average difference inlog sine is 5.9; therefore, the average adjustment foreach angle is 5.9 +2.5, or about 2 seconds. The sum of the log sines of angles 2, 4, 6, and 8 isS greater than thatof angles 1, 3, 5, and 7. There for, you add 2 secondseach to angles 1, 3, 5, and 7 and subtract 2 seconds eachfrom angles 2, 4, 6, and 8.CHECKING FOR PRECISIONEarly in this chapter the fact was stated that theprecision of a triangulation survey may be classifiedaccording to (1) the average triangle closure and (2) thediscrepancy between the measured length of a base lineand its length as computed through the system from anadjacent base line.Average Triangle ClosureThe check for average triangle closure is made afterthe station adjustment. Suppose that, for thequadrilateral shown in figure 15-28, the values of theas follows:15-37

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