28.5(110 - 75)] = 6590 ft
2
(612.2 m
2
). Hence, both equations yield the same result. The
second equation has a distinct advantage over the first because it has only positive terms.
DIFFERENTIAL LEVELING PROCEDURE
Complete the following level notes, and show an arithmetic check.
Point BS, ft(m) HI FSft(m) Elevation, ft (m)
BM42 2.076(0.63) ... 180.482(55.01)
TP 1 3.408(1.04) ... 8.723(2.66)
TP 2 1.987(0.61) ... 9.826(2.99)
TP3 2.538(0.77) ... 10.466(3.19)
TP4 2.754(0.84) ... 8.270(2.52)
BM43 ... ... 11.070(3.37)
Calculation Procedure:
- Obtain the elevation for each point
Differential leveling is used to ascertain the difference in elevation between two succes-
sive benchmarks by finding the elevations of several convenient intermediate points,
called turning points (TP). In Fig. 8, consider that the instrument is set up at Ll and C is
selected as a turning point. The rod reading AB represents the backsight (BS) OfBM 1 , and
rod reading CD represents the foresight (FS) OfTP 1. The elevation of BD represents the
height of instrument (HI). The instrument is then set up at L2, and rod readings CE and
FG are taken. Let a and b designate two successive turning points. Then
Elevationa + BSa = HI (8)
HI-FS^elevation^ (9)
Therefore,
Elevation BM 2 - elevation BM 1 = 2BS - SFS (10)
FIGURE 8. Differential leveling.
Line of sight
