1. trigonometrical heighting??

as part of a measurement scheme to monitor subsidence a target at P2 was mounted on the face of an embankment and observed from a theodolite at P1.
Height of Collimation of theodolite at P1
zenith angle P1-P2 95d30m00s
Coefficient of refraction (K) -0.28
R=6375km

From the above compute the elevation of the target at P2

2. Originally Posted by jane??
as part of a measurement scheme to monitor subsidence a target at P2 was mounted on the face of an embankment and observed from a theodolite at P1.
Height of Collimation of theodolite at P1
zenith angle P1-P2 95d30m00s
Coefficient of refraction (K) -0.28
R=6375km

From the above compute the elevation of the target at P2

[1] The slope distance from P1 to P2.

[2] The elevation of P1.

You are probably aware that the coefficent of curvature and the coefficient of refraction have opposite signs.

Curvature & Refraction become important when the distance from P1 to P2 is larger than 300 metres.

3. this is all that was givin i also left out the value for the height of colimation of the theodolite at P1 is 100m

4. Originally Posted by jane??
this is all that was givin i also left out the value for the height of colimation of the theodolite at P1 is 100m
The slope (or line of sight) distance from P1 to P2 is required to
compute the Curvature & Refraction correction for elevation determination.

The
coefficient of refraction = -0.28
doesn't fit the normally acceptable range using miles or kilometers.

The polar radius of the earth is approx: 6356.9 kilometers = 3950.0 miles
The equatorial radis of earth is approx: 6378.4 kilometers = 3963.4 miles

The R given as 6375km is within that range.
The (C)orrection (vertical adjustment) required to compute the difference
from a true horizontal line to a level line, at a (D)istance on the earth
C = R - sqrt( R^2 - D^2)
C = correction in units given
R = Radius of the earth
D = Horizontal distance from observer

If using kilometers:
C = 6375 - sqrt( 6375^2 - 1^2 ) = 0.00007843
and multiply by 1000 to get the correction in meters
The curvature correction from a true horizontal line to a level line
at 1 kilometer is 0.07843 meters

If using miles:
C = 3961.241 - sqrt( 3961.241 - 1^2 ) = 0.00012622
and multiply by 5280 to get the correction in feet
The curvature correction from a true horizontal line to a level line
at 1 mile is 0.66644 feet

The refraction correction (in earth's atmosphere) is approximately 1/7 the curvature correction.
If using kilometers the coeffiecient for refraction would be close to
0.07843/7 = 0.0112 per kilometer squared
If using miles it would be 0.0952 per mile squared.
The given coefficient does not fit either.

"THE REFRACTION CORRECTION IS -0.28"
The the horizontal (or slope) distance could be computed.

Pedantic Note:
The Earth curvature formula works uniformly well for
a horizontal distance out to 32 kilometers (approx 20 miles). Distances greater than
HOWEVER:
The correction for refraction will vary (very noticeably) with air density. As the
air temperature and barometric pressure change, so will the coefficient of refraction.