Problem (1)

A TSI is set up 5.41' over point Q whose elevation is 745.39'. The following readings are taken:

Point Horizontal
angle
Vertical
angle
Slope
distance
101 0°00'00" 4°24'10" 145.25'
102 39°56'15" -6°00'25" 89.32'
103 79°05'45" -2°58'30" 118.95'

The reflector height in each case was 5.64'. The instrument and reflector heights were measured with the same cloth tape.

What are the horizontal distances to and elevations of the observed points? Compute values to 0.01'.

Solution

Each vertical angle must be converted to a zenith angle.

Point 101

Because the curvature and refraction is so small, it can be ignored for the other horizontal distances.

Because the curvature and refraction is so small, it can be ignored for the other vertical distances.

Point 102

Point 103

 

Problem (2)

A TSI is set up over the SW corner of a property. The TSI height is 5.73' above the corner, measured with a fiberglass tape. The elevation of the top nut on a fire hydrant (FH) across the street is 827.43'.

From the TSI, the following readings are taken

Point Reflector
Height1
Slope
distance
 Horizontal
angle
Zenith
angle
FH 5.00 84.39' 0°00'00" 95°30'45"
NW 6.25 194.48' 44°51'55" 87°05'35"
NE 6.25 240.68' 80°56'30" 85°48'05"
SE 7.50 142.02' 134°51'35" 92°23'50"

1Measured with the same fiberglass tape.

What are the elevations of the four property corners? Compute to 0.01' to minimize rounding error.

Solution

We saw in the previous problem that over short distances, curvature and refraction are too small to affect elevations.

Point SW

Determine the elevation of the TSI at point SW from the fire hydrant, then the elevation of point SW from the TSI.

Point NW

Point NE

Point SE

 

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