Self Study Distance Solutions
Problem (1)
Calibration data
Tape is 99.995 ft long, fully supported, at 68°F, and under 18 pounds of pull.
Its cross-sectional area is 0.003 sq inches with a 3.0x107 psi modulus of elasticity, 6.5x10-6 ft/ft/°F thermal coefficient, and 0.0195 lb/ft weight.
Use data
A distance was measured in two 50 ft end support segments, two 100 ft end support segments, three fully supported 100 ft segments, and one 78.46 ft end support segment, all using a 25 pound pull. The temperature was 80°.
To the nearest 0.001 ft, what is the corrected distance?
Solution
Sketch
Incorrect tape length
Temperature
Pull
Sag
Corrected Distance
Corr'd dist = 678.462 ft
Problem (2)
Given the EDM measurements table, determine the horizontal and vertical distances to 0.01 ft. Account for curvature and refraction. Instrument and reflector heights were 5.69 ft and 6.00 ft, respectively.
| To point | Zenith Angle | Slope Dist (ft) | H (ft) | V (ft) |
| 101 | 79°05'20' | 702.56 | 689.86 | +132.98 |
| 102 | 88°35'30" | 1193.25 | 1192.89 | +29.30 |
| 103 | 92°58'45" | 856.39 | 864.22 | -44.99 |
Sample Comps: Point 101
Problem (3)
Given the stadia readings in the table, determine the horizontal and vertical distances. Express to reasonable levels of accuracy.
The instrument used had a stadia multiplier and constant of 100 and 0 ft, and 5.50 ft height.
| Readings (ft) | Zenith | half-intervals | |||||||
| To |
Top | Mid | Bot | Angle | T | B | i | HD | VD |
| R | 6.76 | 5.50 | 4.25 | 88°50'35" | 1.26 | 1.25 | 2.51 | 250.9 | +0.05 |
| S | 7.03 | 6.25 | 5.48 | 93°05'50" | 0.78 | 0.77 | 1.55 | 154.8 | -8.37 |
| T | 6.01 | 4.95 | 3.90 | 91°18'55" | 1.06 | 1.05 | 2.11 | 210.9 | -4.84 |
Sample Comp
Problem (4)
A distance of 1400.00 ft is to be laid out with a maximum distortion of 1/50,000. The TSI used has a MSA of ±(4 mm + 3 ppm). The instrument operator is able to consistently set up the TSI with a centering error of ±0.004 ft. What is the maximum reflector centering error? Express to 0.001 ft.
Solution
Determine total allowable error
MSA constant and proportion
Rearrange ESum equation to solve for reflector centering
Reflector centering error = ±0.022 ft
Problem (5)
How much higher or lower can one end of a suspended 100 ft tape be held and still meet a distortion of 1/10,000 or better? Ignore the effects of pull, sag, and temperature.
Solution
The horizontal distance for the non-level tape would be 100.00-1/10,000 = 99.9999 ft (carrying additional sig fig to minimize rounding errors)
Sketch:
Solve for e
One end may be up to 0.14 ft higher or lower than the other.
Problem (6)
To calibrate her pacing, a surveyor paced a known 176.5 ft ling base line four times: 93.5, 94.5, 95.0, 93.0. She then preoceded to pace the perimeter of a warehouse in order to determine its area. Her three pacing trips around the perimter were:
- 144.5, 110.0, 144.0, 111.5
- 145.5, 110.5, 143.5, 110.5
- 146.0, 109.5, 144.0, 110.5
What is the area of the warehouse to the nearest 10 sq ft?
Solution
Her pace conversion is (176.5 ft) / [(93.5+94.5+95.0+93.0)/4] = 176.5 ft / 94.225 p
Average length is (144.5+145.5+146.0+144.0+143.5+144.0) / 6 = 144.5833 p
Average width is (110.0+110.5+109.5+111.5+110.5+110.5) / 6 = 110.4167 p
Area = (144.5833 p x 176.5 ft / 94.225 p) x (110.4167 x 176.5 / 94.225) = 56,015.733 = 56,020 sq ft