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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?




Incorrect tape length




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      
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 ±(2mm + 2ppm). The instrument operator is able to consistently set up the TSI with a centering error of ±0.004. What is the maximum reflector centering error? Express to 0.01 ft.


Determine total allowable error


MSA constant and proportion

Rearrange ESum equation to solve for reflector centering

Reflector centering error = ±0.03 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.


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)


Solve for e

One end may be up to 0.14 ft higher or lower than the other.


Return to: Self Study: Distances