# Rainy Lake Experiment: Equations

This page contains the derivation of most equations used by the Rainy Lake Experiment.

## Atmospheric Refraction

See Deriving Equations for Atmospheric Refraction for how terrestrial atmospheric refraction is calculated from basic physical laws and how the equations and constants used in refraction equations are derived.

See Method to calculate the Radius of the Earth from 3 Points for how the GNSS Data Viewer calculates the radius from 3 selected GPS Vectors.

## Coordinate Transformations

See WGS84 Coordinate System for how vectors can be transformed between ECEF Cartesian coordinates and Geodetic Ellipsoidal coordinates.

## Drop of Surface

The triangle c, b, R is a right angle triangle. We can use trigonometry to express c:

 (1)

The angle φ can be calculated from the distance d:

 (2)

The line c extended to the horizontal tangent line is also R. So x is the difference Rc:

 (3)
(4)

exact

(5)

good approximation if dR

where'
 $x$ ' =' 'vertical drop from tangent to surface $R$ ' =' 'radius of the earth $d$ ' =' 'distance along surface

Approximation: We can use the Taylor series of the cosine function up to the quadratic term cos(φ) ≈ 1φ2/2 to get an approximation for small angles:

 (6)

If we use miles for d and 8 inches/miles2 for 1/2R we get the famous approximation "8 inches per miles squared".

## Drop of Horizon from Eye Level

Drop x is h + c. The blue dash line is a symmetry line. So we can find the height h on the right side too, building the hypotenuse of a right angle triangle with angle φ. With trigonometry we find c = h · cos(φ). So we have:

 (7)

We have another right angle triangle with angle φ: R+h, b, R. In this triangle is R = (R+h) · cos(φ). So we can solve for cos(φ):

 (8)

and replace the cosine of (7):

 (9)
(10)

exact

(11)

good approximation if hR

where'
 $x$ ' =' 'drop of the horizon from eye level $h$ ' =' 'observer eye level height $R$ ' =' 'radius of the earth

Approximation: for hR we can approximate R + hR:

 (12)

## Drop Angle of Horizon from Eye Level

We can find the drop angle φ also at the not visible corner of the right angle triangle R+h, b, R. The cosine of this triangle is:

 (13)

Solving for φ yields:

(14)

exact

(15)

good approximation if hR

where'
 $\varphi$ ' =' 'drop angle from eye level in radian $h$ ' =' 'observer eye level height $R$ ' =' 'radius of the earth

Approximation: The angle φ can also be expressed by an asin function if we use the third side b of the triangle, which is b = √ (R + h)2R2 . So φ = asin( b / (R+h) ). The asin function can be approximated to asin(φ) ≈ φ for small angles φ and R + hR for hR, so we can make multiple simplifications without loosing much accuracy:

 (16) (17)

## Drop of Target from Eye Level

The angle φ = d / R is one of the angles of the right angle triangle R+hO, R+hT+x, b. The cosine of this angle is the adjacent R+hO divided by the hypotenuse R+hT+x:

 (18)

We can solve for the denominator, which contains our unknown x:

 (19)

And rearranging for x yields:

(20)
where'
 $x$ ' =' 'drop of the target from eye level, negative for targets above eye level $d$ ' =' 'distance to target along surface (line of sight or along surface is approx the same) $h_\mathrm{O}$ ' =' 'observer eye level height $h_\mathrm{T}$ ' =' 'target height $R$ ' =' 'radius of the earth

Note, for small angles φ, i.e. if hOR and hTR, then yx. y is the adjecent of a similar right angle triangle with the hypotenuse x. So we can get the exact value for y by multiplying (20) with cos(d/R):

(21)
where'
 $y$ ' =' 'drop of the target from eye level, negative for targets above eye level $d$ ' =' 'distance to target along surface (line of sight or along surface is approx the same) $h_\mathrm{O}$ ' =' 'observer eye level height $h_\mathrm{T}$ ' =' 'target height $R$ ' =' 'radius of the earth

## Angular Size

(22)

exact

(23)

good approximation if sd

where'
 $\theta$ ' =' 'angular size in radian $s$ ' =' 'size of the object in the distance measured perpendicular to line of sight $d$ ' =' 'line of sight distance

Approximation: For small angle φ we can approximate atan(φ) ≈ φ. The angle φ is small if sd. If this is the case we get:

 (24)

## Refraction Angle and Lift

The magenta line is the direct line of sight from the observer to a target. The orange arc is the corresponding refracted light ray with radius of curvature r. The refraction angle is ρ.

We have a right angle triangle r, d/2, b with an angle ρ. We can use trigonometry to calculate the sine of ρ:

 (25)

The curvature 1/r of the light ray is defined by the refraction coefficient:

 (26)

So we get:

 (27)

For small angles ρ, that is if dR, we can approximate sin(ρ) ≈ ρ, and using the equation for Angular Size we can calculate the apparent lift due to refraction l to get:

(28)
(29)
where'
 $\rho$ ' =' 'refraction angle $l$ ' =' 'apparent lift of target due to refraction with respect to a fixed reference line like eye level of the observer $k$ ' =' 'refraction coefficient $k/R$ ' =' '1/r = curvature of light ray $r$ ' =' 'radius of curvature of light ray $d$ ' =' 'distance of target from observer along the surface $R$ ' =' 'radius of the earth

## Line of Sight Distance to Horizon

We can simply apply Pythagoras to the right angle triangle R+h, d, R to get our unknown:

(30)

exact

(31)

good approximation if hR

where'
 $d$ ' =' 'line of sight distance from observer eye to horizon $h$ ' =' 'height of observer eye level $R$ ' =' 'radius of the earth

Approximation: We can approximate (R+h)2R22·R·h if hR:

 (32)

The expression 2·R·h is much greater than h2 if hR. So we can neglegt h2.

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