Hw 5 solution

Hw 5 solution - CVEN303 Engineering Measurement, Fall 2009...

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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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2 2. (5 points) Two points (A and B) are part of Survey Fondue I. A GPS unit is used to determine the Texas State Plane Coordinates of these two points. The Texas State Plane Coordinates are then used to calculate the length of the line A to B with the following result: 1,567.82 ft. A total station was used to measure the distance between point A and point B, this surface distance was found to be 1,567.98 ft.
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Hw 5 solution - CVEN303 Engineering Measurement, Fall 2009...

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