1
CVEN303 Engineering Measurement, Fall 2009
Homework Assignment # 5 KEY
Due
: One week
after the assignment is handed out or posted on CENotes (for Monday-
Wednesday lecture sections, due date is Wednesday 11/11).
Note:
Homework assignments are due to beginning
of class (see syllabus for HW
policy).
GPS and Total Station Problems
(
Read sections 8.13 and 11.6 in the book and lectures
13 and 14
)
1. The latitude of a certain survey (Survey Fondue I) in the Central Zone of the Texas
State Plane Coordinate System is 30º16’26.3582”.
The average elevation of Survey
Fondue I is 1,267.54 ft.
a)
(5 points)
Using Appendix B for the Central Zone from the Lambert Conformal
Conic Projection Tables, what is the Scale Factor (K
scale
).
From Appendix B, Texas Central Zone, Lambert Conformal Conic Projection
Tables:
Latitude
Scale Factor
30º15’00”
0.99996708
30º16’00”
0.99996334
30º17’00”
0.99995969
(If needed, interpolate linearly between values.)
Solution:
30º16’00”<30º16’26.3582”<30º17’00”
Difference between 30º16’00” and 30º17’00” is 60”
Difference between 0.99996334 and 0.99995969 is 0.00000365
Between 30º16’00” and 30º17’00”, Scale Factor decreases by 0.00000006083 per second
(0.00000365/60 = 0.00000006083).
At Latitude 30º16’26.3582”, Scale Factor = 0.99996334 – (0.00000006083 * 26.3582) =
0.99996174
b)
(5 points)
What is the elevation factor (K
elev
)?
Solution:
Elev. Factor = 20,902,000/(20,902,000+1,267.54) = 20,902,000/20,903,267.54 =
0.99993937
c)
(5 points)
What is the combined factor?
Solution:
Combined factor = Scale Factor * Elevation Factor = 0.99996174 * 0.99993937
= 0.99990110 = Combined Factor

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