Homework 4 - Answers
1 – Describe how the total station employs different wavelengths to measure a distance.
The EDM sends out varying wavelengths in order to measure each significant digit of a distance.
The phase shift is measured by phase shift/360 * wavelength = portion of length. In other words,
the EDM sends out a 1, 10, 100, etc wavelength and then calculates a fraction of each amount for
the total distance.
2 – What is the prism constant? Why is it important to know that value when using a new prism?
The prism constant compensates for the difference in the way light travels in glass and air.
Without it, an error would accumulate. Therefore it is initially necessary to enter the value into
the total station.
3–
Horizontal control is extended by establishing a traverse and vertical is extended by a
level loop.
4 – What are the purposes, as mentioned in the lecture, of establishing a traverse?
The purpose of establishing a traverse is to have a control point. By using a closed traverse, a
check can be performed on the measured angles and distances.
5- Explain the difference between closed and open traverse. Comment on the advisability of using
open traverses.
The difference between the two is that a closed traverse starts and ends on points with known
location and an open traverse starts with a known point, but ends on a point with unknown
location. An open traverse is usually not used since the error in location measurements cannot
be computed. However, if an open traverse is used, measurements should be taken repeatedly.
6- What should be the sum of the interior angles for a closed polygon traverse that has a) 6sides
b) 8 sides and c) 12 sides?
Sum of interior angles = (n – 2) * 180
a) (6 – 2) * 180 = 720
b) (8 – 2) * 180 = 1080
c) (12 – 2)*180 = 1800
7- The interior angles in a five sided closed polygon traverse were measured as A = 13910”11”,
B = 12617’43”, C = 9428’30”, D = 7104’59” and E = 10858’31”. Compute the angular
misclosure. For what order and class is this survey?
(5 –2) * 180 = 540
Sum angles = 53959’54”
Misclosure = 540 - 53959’54” = 0000’06”
C = KN: K = constant; N = # angles; C = misclosure
0000’06” = K5 K = 000”2.683”
 2.7”
2.7” gives second order class I
8- In adjusting measured traverse angles, why aren’t adjustments made in proportion to angle
sizes?
adjustments aren’t made in proportion to angles sized because the errors expected in measured
angles bear no relation to angle size. For example, a small angle is as likely to contain an error as
a large angle.
9- Compute and tabulate for the following closed-polygon traverse:
(a) bearings
(b) departures and latitudes
(c) linear misclosure
(d) relative precision
(e) adjust the traverse using the bowditch rule assuming the coordinates of A are 5000.00E and
5000.00N, determine the coordinates of all other points.
line
bearing
AB
BC
CD
DE
EF
FA
N32'21'E
N9'36'W
N79'13'W
S46'43'W
S47'21'E
N83'24'E
length(ft)
701.82
1009.91
1109.91
1320.51
1447.66
785.74
measured
angle
A = 128'58'
B = 138'04'
C = 110'24'
D = 125'57'
E = 85'57'
F = 130'46'
6375.12 Sum =720'6'
Adjustment
-1’
-1’
-1’
-1’
-1’
-1’
adjjusted
angle
128'57'
138'03'
110'23'
125'56
85'56'
130'45'
Sum = 720
Expected Angle Sum = (6-2) * 180 = 720
Misclosure = 720 - 7206’ = -6’
Correction -6’/6 angles = -1’
adjust each angle by –1’
Departure = L sin azimuth
Latitude = L cos azimuth
line
departures (ft)
latitudes (ft)
AB
BC
CD
DE
EF
FA
375.54
-168.42
-1089.89
-961.29
1064.76
780.53
592.89
995.77
207.58
-905.25
-980.82
90.31
Sum = 1.23
Sum = 0.38
Linear misclosure = ((1.23)2 + (0.38) 2) = 1.29 ft
Relative Precision = 1.29 ft / 6375.12 ft = 2.02 * 10-4 = 1/5000
Area = 2,818,400 sq. ft.
= 64.701 acres