# Refraction and Height of Light Rays above Surface

By analysing the data from the video Angular Size Change & Curvature - Clarifying an FE Misunderstanding by The Maine Surveyor we can calculate the refraction coefficients k for each light ray separately from the following data (all lengths and heights in meters):

Distance Height Refracted Angle a Curvature Drop c EyeLevel Drop of Target y Geom.Drop Angle of Target δ Refraction Angle ρ Refraction Coefficient k 0 23.42 14,447 24.38 0°02'47" 16.45 15.49 0°03'41" 0°00'54" 0.23 37,642 56.08 0°05'09" 111.20 78.54 0°07'10" 0°02'01" 0.20

As we can see the refraction k of the light rays to the 2 targets is not exactly the same. How can this be and why is refraction to the farther target less than that of the nearer target?

Refraction depends on the atmospheric conditions along a light ray. These are not constant, especially over very long distances like in this example. So it should not be surprising that the mean refraction coefficients for the 2 light rays are slightly different.

Is this a problem? It depends on what you want to measure and how accurately.

If you only want to get an estimation of the size of the earth, you can assume a standard refraction coefficient of say k = 0.17 and use the following equation to calculate the radius of the earth from the measurements:

(1)
where'
 $R$ ' =' 'radius of the earth from the measured data $d$ ' =' 'distance to target $k$ ' =' '0.17 = assumed standard refraction coefficient $\alpha$ ' =' 'measured apparent drop angle from eye level to target top in radian $h_\mathrm{T}$ ' =' 'target height from surface to top $h_\mathrm{O}$ ' =' 'observer height

In this case we get R = 6843 km for the nearer target Halfway Rock and R = 6603 km for the farther target Seguin Island. This is better than 8% accuracy. If we did measure refraction at the observer location to be about k = 0.2 we would get R = 6596 km or 3.5% too much for the nearer target Halfway Rock and R = 6365 km or 0.1% too less for the farther target Seguin Island. Not bad. We can certainly conclude the earth is a sphere with a radius of 6000..7000 km.

If you want to accuratly measure the elevation of a target, you can't do that without measuring the refraction along the line of sight very accurately. This is not practicable over long distances, so such measurements are not done this way and not over long distances.

Another question is, how can we explain the different refraction coefficients for the 2 targets. At first we could argue, that the light ray to the farther target passes nearer the ground and if we assume that nearer the ground refraction is stronger than higher up, we would expect that refraction for the farther target is stronger. But if we plot the height of the light rays above the surface depending on the distance, we can see that the light ray to the farther target in its last 1/3 gets way higher than the the light ray to the nearer target and in the first 2/3 it is not much nearer to the surface. So if we assume that refraction gets less with altitude, it is possible to get less refraction for the farther target in this case. But this is only a quantitative conclusion. We don't really know the conditions along this very long observation.

The following image shows the height of each light ray over the surface as a function of distance.

2020-12-01 Height of Light Rays above Surface.zip  for the GeoGebra file

## Equations

(2)
where'
 $c$ ' =' 'curvature drop from the surface tangent $d$ ' =' 'distance from the observer $R$ ' =' '6371 km
(3)
where'
 $y$ ' =' 'geometric drop of target top from eye level $h_\mathrm{O}$ ' =' 'observer height $h_\mathrm{T}$ ' =' 'target top height $c$ ' =' 'curvature drop from the surface tangent, see (3)
(4)
where'
 $\delta$ ' =' 'geometric from angle from eye level to the target top in radian $y$ ' =' 'geometric drop of target top from eye level, see (4) $d$ ' =' 'distance from the observer
(5)
where'
 $\rho$ ' =' 'refraction angle in radian, apparent lift due to refraction $\delta$ ' =' 'geometric from angle from eye level to the target top in radian $\alpha$ ' =' 'measured angle from eye level to target top in radian
(6)
where'
 $k$ ' =' 'refraction coefficient $\rho$ ' =' 'refraction angle in radian, see (6) $R$ ' =' '6371 km $d$ ' =' 'distance from the observer

Equation (1) is derived from the equations above.

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