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Distance to the horizon

Submitted by johng on 16 March, 2012 - 17:29

Have you ever wondered how far can you see to the horizon from an elevated position ? Using simple trigonometry the distance to the horizon along the Earth's surface can be easily determined. Two cases are considered: (i) light travels in a straight line and (ii) light travels along a curved path as a result of atmospheric refraction. In both cases it will be assumed that the Earth is a sphere resulting in a circular cross section when examining the problem in two dimensions.

Simple case

Consider the case of an observer standing on an elevated position such as a lighthouse viewing platform. In figure 1 below there is a clear and unobstructed view of the sea to the right of the lighthouse. The furthest distance visible from the observer to the horizon (which is represented as the quantity s) can be uniquely determined by the beam of light shown as a dashed line in Figure 1. The straight beam of light meets the horizon on the Earth's surface out to sea at a tangent. Given the height of the light house and the radius of the Earth the problem is reduced to a case of calculating the arc length of s shown Figure 1.

 

 

 
Figure 1

Equations

The geometry can be examined in more detail in Figure 2. In the triangle PHO, the following points are defined:

  • P - the peak or viewing platform
  • H - the point marking the horizon. This is the furthest point visible from P on the Earth's surface
  • O - the center of the earth

In addition the following distances are defined:

  • s is the distance along the arc length of the Earth's surface from base of the observer at P
  • h is the height of the observer above sea level
  • r is the radius of the Earth
  • $\theta$ - the angle at the center of the Earth, defined by the arc length s
Figure 2

A tangent to a circle touches a circle at only one point which is at right angles to the radius at the point of contact. PH and OH form a 90 degree angle in figure 2. Therefore PHO is a right angle triangle with the following quantities:

  • PO = r + h
  • OH = r

Using elementary trigonometry in triangle PHO, the cosine of the angle $\theta$ is expressed as:

$$\cos(\theta) = \frac{r}{r+h} \qquad_{...(1)}$$

The arc length is easily calculated as

$$s = r \theta \qquad_{...(2)}$$ where $\theta$ is measured in radians. Substituting the expression for $\theta$ from equation (1) into (2), the following is obtained:

$$s = r \arccos \left( \frac{r}{r+h} \right)\qquad_{...(3)}$$

Factoring out $r$ from the numerator and denominator, the arc length along the Earth's surface can be expressed as the ratio $h/r$.

$$ s = r \arccos \left[ \frac{1}{1+\frac{h}{r}} \right] \qquad_{...(4)}$$

Examining the limits of equation (4) is a useful check to see if the expected results are obtained in two extreme cases.

  • At h=0, s=0. An observer lying on the ground will not able to see along any distance along the horizon
  • At an infinite height above the Earth's surface ($h \to \infty$), we would expect to see the full face of the Earth. In equation (4), the expression in arccos approaches zero. $\arccos(0) = \pi / 2$. Therefore s approaches $\pi r/2$ This is equivalent to the quarter circumference of a circle with the radius of r. Taking into account the symmetry (see figure 2) the total arc length on either side is $\pi r$

For practical applications equation (4) is cumbersome. The expression for $\theta$ in equation (1) can be simplified if one assumes that the height of the observer above sea level is a small fraction of the radius of the Earth, i.e. h/r is a small quantity. Indeed this condition is true for most terrestrial situations, such calculating the distance to the horizon from a mountain, lighthouse or ship.

In mathematics any continuous function can be represented as a series of polynomials known as a Taylor expansion. In the case of $\cos(\theta)$ the Taylor series is:

$$\cos(\theta) = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + ...$$

The valid assumption that h/r is small also implies that $\theta$ will be a small angle. As a result the first two terms are sufficient to approximate $\cos(\theta)$. Rewriting equation (1):

$$1 - \frac{\theta^2}{2} = \frac{r}{r+h} = \frac{1}{1+\frac{h}{r}}$$

Multiplying the numerator and denominator of the right side of the equation by $1 - \frac{h}{r}$ the following expression is obtained

$$1 - \frac{\theta^2}{2} = \frac{1 - \frac{h}{r}}{1-\frac{h^2}{r^2}}$$

Taking into account $h / r$ << 1, then $\frac{h^2}{r^2} \approx 0$, therefore

$$\frac{\theta^2}{2} \approx \frac{h}{r}$$

$$\theta \approx \sqrt{\frac{2h}{r}}$$

Substituting $\theta$ into equation (2) we get:

$$s \approx r\sqrt{\frac{2h}{r}} = \sqrt{2hr} \qquad_{.....(5)}$$

The resulting equation will work provided the height (h), the radius of the Earth (r) and the distance to the horizon (s) are all in the same units of length. A conversion of units is required to produce a more practical result that will work in both metric and imperial units of measurement.

Equation (5) can be rewritten where the height is specified in meters and the arc length is expressed in km. Refer to Metric and Imperial Units in the Appendix:

$$s_{km} = 3.57 \sqrt{h_{m}}   \qquad_{.....(6)}$$

The imperial unit equivalent is:

$$s_{miles} = 1.23 \sqrt{h_{feet}}   \qquad_{.....(7)}$$

The resulting equations are in close agreement with other sources: $3.57 \sqrt{h}$ for metric units and $1.22 \sqrt{h}$ for imperial units - (see reference (2))

Refraction

In the simple case light travels in a straight line. However light traveling through a medium such as the Earth's atmosphere will deviate from a straight line due to atmospheric refraction. The amount of refraction will depend on a number of factors such as temperature, air pressure and density. In the case of the Earth's atmosphere, the refractive index of air (a measure of how light is refracted) will be a function of height above the Earth's surface. One would expect the effect of refraction would diminish at greater heights as the atmosphere thins out into empty space.

To treat the case of refraction accurately requires a detailed model of the atmosphere which goes beyond the scope of the article. However a very good approximation of the light path is an arc of a large circle with a radius much larger than that of the Earth's radius. As a result of the curved light path shown in red (in figure 3), the distance to the horizon (sf) in the presence of an atmosphere is greater that the simple case (s) from a straight beam of light shown as the dashed line in the same figure.
Figure 3

 

The complete geometric configuration is shown in figure 4. The cross section of the Earth is represented as the grey circle. The key points to note are:

  • P - the peak or viewing platform
  • B - the base of the viewing platform
  • X - the point marking the horizon as defined by the refracted beam of light. This is the furthest point visible from P on the Earth's surface
  • O - the center of the earth
Figure 4

The red circular arc between the points P and X represents the refracted light path. Q is the center of the red circle. Both circles meet at the point X. Therefore a perpendicular line projected from a tangential line at point X will pass through both O and Q.

Triangle POQ is formed with clearly defined quantities to determine the angle POX, denoted as $\alpha$. Once $\alpha$ is determined the arc length BX along the Earth surface is computed, corresponding to the distance to the horizon for a refracted light beam.

In triangle POQ the following quantities defined:

  • PQ = R, the radius of the refracted beam of light
  • PO = h + r
  • QO = R - r

Using the cosine rule:

$$R^2 = (R-r)^2 + (r+h)^2 - 2.(R-r).(r+h)\cos(\pi-\alpha) \qquad_{.....(6)}$$

Rearranging equation (6) and using the identity $\cos(\pi-\alpha) = - \cos(\alpha)$:

$$\cos(\alpha) = \frac{2Rr-2r^2-2hr-h^2}{2(R-r)(r+h)} \qquad_{.....(7)}$$

Therefore:

$$s_f = r . \arccos \left[ \frac{2Rr-2r^2-2hr-h^2}{2(R-r)(r+h)} \right] \qquad_{.....(8)}$$

After considerable manipulation and taking into account the valid approximations that  $h / r$ << 1, $\frac{h^2}{r^2} \approx 0$  and $\alpha$ is a small angle so that $\cos(\alpha) \approx 1 - \frac{\alpha^2}{2}$ it can be shown that:

$$\alpha \approx \sqrt{\frac{2h}{r(1-r/R)}}     \qquad_{.....(9)} $$

A detailed analysis on obtaining equation (9) can be found in the Refraction section of the Appendix. The circular arc length BX can be easily computed one the angle $\alpha$ is determined:

$$s_f = r \alpha = \sqrt{\frac{2hr}{1-r/R}} \qquad_{.....(10)}$$

The result is equivalent to the simple case where light travels in a straight line for a "fictitious" planet with a radius of $r/(1-r/R)$. A quick examination of the limit of equation (10) for $R \to \infty$ corresponds to a straight line path where $s_f \to s = \sqrt{2hr}$

The radius of the light beam is dependent on atmospheric conditions such as temperature, altitude and air pressure. A detailed analysis beyond the scope of this article would be required to determine the value of R. A common value used for correcting surveyor's data for over a century is r/R=0.13 (i.e. R/r=7.69) - ref (1). Another estimate for R/r is 7 based on a simple method by Young - ref (2).

Taking the first estimate of r/R = 0.13 equation (10) becomes:

$$\begin{align} s_f &= \sqrt{2.2989 hr} \\ &= 1.07 \sqrt{2hr} \qquad_{.....(11)} \\ &= 1.07 s\end{align}$$

The result indicates that one can see 7% further as a result of atmospheric refraction compared to the simple case where no atmosphere is taken into consideration. As in the simple case a conversion of units is required to produce a useful formula for both metric and imperial units. Refer to Metric and Imperial Units (Appendix) for the conversion:

$$s_{f (km)} = 3.83 \sqrt{h_m} \qquad_{.....(12)}$$

The imperial unit equivalent is:

$$s_{f (miles)} = 1.33\sqrt{h_{feet}} \qquad_{.....(13)}$$

Using a similar comparison as in the simple case, the resulting equations are in close agreement with other sources: $3.86 \sqrt{h}$ for metric units and $1.32 \sqrt{h}$ for imperial units - (see reference (2)).

Calculations

The resulting equations from the analysis of the simple and refracted light cases above lead to the generation of a number of practical tables and graphs. This allows us to get a quantitative picture of what distances to expect for various structures or viewing platforms above the Earth's surface.

Tabulation

Equation (12) can now be used to generate real data of a number of landmarks and viewing platforms including moving aircraft and orbiting structures.

Structure or viewing platform height (m)

distance to
horizon (km)
The Cape Byron Lghthouse (Byron Bay, NSW, Australia) 118 42
Eureka Tower (Melbourne, Australia) 297 66
Petronas Twin Towers (Kuala Lumpur, Malaysia) 458 82
Burj Khalifa (Dubai, United Arab Emirates) 829 110
Mount Kilimanjaro (Tanzania) 5895 294
Mount Everest (Nepal) 8848 360
cruising altitude of a 747 12000 420
International Space Station 350000 2266
Table 1

Table 1 assumes the horizon out to sea is visible without any obstructions, which may not be the case for some structures listed, e.g. Mount Everest. The table simply illustrates the expected distance to the horizon at the given height above sea level.

The second table below compares the results of a number of equations derived from the theory:

  • determine the distance to the horizon for both metric and imperial units
  • compare equation (12) with the exact formulae (eq 8 and 3) and the simple approximation (eq 5)

Distance to horizon
miles

 Height above sea level (feet/metres) Distance to horizon km Exact value
(refracted light - eq 8)		 Simple case
approximation (eq 5)		 Simple case
no approximations (eq 3)
4.2 10 12.1 12.1 11.3 11.3
5.9 20 17.1 17.1 16.0 16.0
7.3 30 21.0 21.0 19.6 19.6
8.4 40 24.2 24.2 22.6 22.6
9.4 50 27.1 27.1 25.2 25.3
10.3 60 29.7 29.7 27.7 27.7
11.1 70 32.0 32.0 29.9 29.9
11.9 80 34.3 34.2 31.9 31.9
12.6 90 36.3 36.3 33.9 33.9
13.3 100 38.3 38.3 35.7 35.7
21.0 250 60.6 60.5 56.4 56.5
29.7 500 85.6 85.6 79.8 79.9
42.1 1000 121 121 113 113
66.5 2500 192 191 179 179
94.0 5000 271 271 252 253
133 10000 383 383 357 357
297 50000 856 854 798 796
421 100000 1211 1204 1129 1122
Table 2

To determine the distance to the horizon in metric units:

  • Look up the height above sea level in the central column of the table (e.g. 100 meters)
  • Read the value to the right in km. In the example highlighted in red, the distance to the horizon for an observer 100 m above the Earth's surface is 38.3km

To determine the distance to the horizon in imperial units:

  • Look up the height above sea level in the central column of the table (e.g. 50 feet)
  • Read the value to the left in miles. In the example highlighted in blue, the distance to the horizon for an observer 50 feet above the Earth's surface is 9.4 miles

Note: The table above is a combination of two tables. The lookup table for the imperial units consists of columns 1 and 2. The lookup table for the metric units consists of columns 2 and 3, i.e. 4.2 miles does not equal 12.1 km

The last three columns of the table are to compare the metric values with (i) the exact value as determined from equation 8, (ii) the simple case approximation from equation (5) and (iii) the exact value for the simple case as determined from equation (3). In all cases it can be seen that the values differ no more than 7% up to a height of 100 km above sea level. The result given in equation (11) is sufficient for most practical purposes.

Log-Log Graph

The tabulated values can be represented as a log-log graph. It can be easily shown using elementary high school maths that is one takes the logarithm of both sides of equations (5) and (6) a graph of the relationship between the $s$ and $h$ will result in a straight line with a slope of 0.5.

One advantage of using a log graph is a large range of values can be represented in increasing or decreasing orders of magnitude along the horizontal axis. However there is a loss of accuracy due to the limitation of the number of grid lines in the log-log graph.

For consistency and clarity the red line corresponds to metric units and the blue line for imperial. Two graphs have been combined into one (as is the case of the table above).

Examples:

  • At 1000 meters above sea-level the distance to the horizon is approximately 120 km (using the red line to read the value)
  • At 10 feet above sea-level the distance to the horizon is about 4 miles (using the blue line to read the value)

In both cases the examples are consistent with the tabulated values.

Comparison of formulas

In the simple case, the relationship between the exact formula (equation 3) and the approximation (equation 6) as shown in Table 1 can be better visualised with a log-log graph. It can be clearly seen in figure 5 that both formulas are in close agreement out to a height of 1000 km (106m)above sea level.
Figure 5

The exact calculation (marked as the solid black line) clearly shows the asymptotic limit where s approaches $\pi r /2 \approx $ 10018 km (as already discussed in the limits to equation 3). The cyan coloured straight line depicts equation 6. Based on the comparisons in table 2 and figure 5, we can be confident that $3.57\sqrt{h}$ is a good approximation for determining the distance to the horizon out to 1000km above the Earth's surface.

The case for refracted light as shown in Figure 6 also depicts a similar comparison. Both the exact formula (equation 8) and the approximation (equation 12) are in close agreement out to 1000 km above sea level.
Figure 6

In the case of refracted light in Figure 6 the dark cyan line depicts the approximation $3.83\sqrt{h}$ and the solid black line is the exact calculation (equation 8). The grey curve denotes the fraction (num/den) of the arccos argument of equation 8 which is defined in the range -1 to 1. The results are summarised below:

  • The exact calculation (eq 8) and the approximation (eq 12) are in precise agreement out to $h \approx 1000$ km.
  • The solid black line (eq 8) and the grey curve are not defined for values greater than a height of 108m (100,000 km) above sea level

Let A denote the argument of arccos of equation 8:

$$ A = \frac{num}{den} = \frac{2Rr-2r^2-2hr-h^2}{2(R-r)(r+h)} \qquad_{.....(14)}$$

$\arccos(x)$ is defined for values of x between -1 and 1. Therefore the right hand side of equation 14 will also have a range between -1 and 1. One can solve for h where A=-1, corresponding to $\alpha = 180$o. A quadratic equation is solved to get $h=2(R-r)$. Taking the ratio of r/R=0.13 (i.e. R/r=7.69) - ref (1), the maximum possible value of h in equation 8 and 14 is:

$$h_{max} = 13.38r = 8.533 \times 10^7 m$$
Figure 7

Figure 7 illustrates the behavior of refracted light for various scenarios. An interesting result of the analysis of the limits of equation 8 is that at $h=h_{max}$ and $\alpha=180o$ one can see a point on at the opposite side of the Earth. In reality this effect is not observed taking into account: (i) the atmosphere only extends out to a small fraction of the radius of the Earth and (ii) the approximation that refracted light will follow an arc of a circle is not likely to be valid for an extended atmosphere.

Distance between objects above horizon

Imagine the following scenario. A ship out to see is heading towards the coastline as shown in figure 8. There is a critical light path where the lighthouse will first become visible to an observer on the ship. What is the distance between the lighthouse and ship ? The problem is simply a case of applying the "distance to the horizon" problem for both the lighthouse and the ship using the results obtained in the previous sections. The scenario can apply to any pair of elevated objects and not just a lighthouse and a ship.
Figure 8

The simplest case is to assume there is no atmospheric refraction where light travels in a straight line. Using equation 6 one can calculate s1 and s2 and ad the distances to obtain the total distance between the lighthouse and the ship:

$$\begin{align} s_1 &= 3.57 \sqrt{h_1} \\ s_2 &= 3.57 \sqrt{h_2} \end{align}$$

therefore

$$d = s_1 + s_2 = 3.57 \sqrt{h_1 + h_2} \qquad_{.....(14)}$$

The case for atmospheric refraction is shown in figure 9. Even though the refracted beam of light in red follows an arc of a circle, the same principle applies as for the simple case above. Using equation 12 once can calculate s1 and s2 for refracted light:

$$\begin{align} s_1 &= 3.83 \sqrt{h_1} \\ s_2 &= 3.83 \sqrt{h_2} \end{align}$$

therefore

$$d_f = 3.83 \sqrt{h_1 + h_2} \qquad_{.....(15)}$$
Figure 9

Comparing figures 8 and 9 it can be seen that distance between the lighthouse and the ship is increased as a result of atmospheric refraction. A simple comparison between equations 14 and 15 indicates that the increase is 7%, which is consistent with the analysis for distance to the horizon problem for refracted light .

A table can be generated to show the critical distance between two objects as function of h1 and h2 in the Earth's atmosphere using equation 15.

Critical visible distance between two objects above the horizon (km), s1 + s2
h2 (m) h1(m)
  0 5 10 25 50 75 100 250 500 1000
0 0.0 8.6 12.1 19.1 27.1 33.2 38.3 60.6 85.6 121.1
5 8.6 12.1 14.8 21.0 28.4 34.3 39.2 61.2 86.1 121.4
10 12.1 14.8 17.1 22.7 29.7 35.3 40.2 61.8 86.5 121.7
15 14.8 17.1 19.1 24.2 30.9 36.3 41.1 62.3 86.9 122.0
20 17.1 19.1 21.0 25.7 32.0 37.3 42.0 62.9 87.3 122.3
25 19.1 21.0 22.7 27.1 33.2 38.3 42.8 63.5 87.8 122.6
30 21.0 22.7 24.2 28.4 34.3 39.2 43.7 64.1 88.2 122.9
35 22.7 24.2 25.7 29.7 35.3 40.2 44.5 64.7 88.6 123.2
40 24.2 25.7 27.1 30.9 36.3 41.1 45.3 65.2 89.0 123.5
45 25.7 27.1 28.4 32.0 37.3 42.0 46.1 65.8 89.4 123.8
50 27.1 28.4 29.7 33.2 38.3 42.8 46.9 66.3 89.8 124.1
Table 2

The table is asymetrical where $h_1$ spans a larger range than $h_2$ to be consistent with $h_1$ denoting the height a tower, lighthouse or mountain and $h_2$ a significantly shorter object or structure such as a ship.

In the case of $h_1=0$ or $h_2=0$ the results in the table reduce to the distance to the horizon for only one object, consistent with equation 12.

Conclusion

The distance to the horizon of an observer looking from an elevated position above the Earth's surface can be determined using simple trigonometry. A number of assumptions were made: (i) the Earth is a perfect sphere, (ii) light travels in a straight line and (iii) there is a clear and unobstructed view to the horizon. The resulting equation was found:

$$ s = r \arccos \left[ \frac{1}{1+\frac{h}{r}} \right]$$

where r is the radius of the Earth and h is the height above the Earth's surface. Further simplification was possible by considering the case where h is a small fraction of r which would be applicable in most practical situations.

$$ s = 3.57\sqrt{h} $$

A more complicated case was considered taking into account the bending of light as a result of the presence of the Earth's atmosphere. To a first order approximation it was assumed that a refracted beam of light follows an arc of a circle with a radius of R. The distance to the horizon was shown to be:

$$ s_f = r . \arccos \left[ \frac{2Rr-2r^2-2hr-h^2}{2(R-r)(r+h)} \right] $$

After considerable mathematical reduction and additional valid approximations it was shown that:

$$ s_f = 3.83\sqrt{h} $$

In both cases it was interesting to see that both approximate estimates for straight and refracted light are propotional to $\sqrt{h}$. The distance to the horizon is increased by 7% in comparison to the simple case.

The resulting equations were compared with a table and a numbers of graphs (Table 2, Figures 5 and 6) to show the deviation between the approximations and the exact formulae. In both cases it was shown that the approximations start to deviate only after 1000 km above the Earth's surface.

Finally, the results can be extended to determine the maximum visible distance between two objects looking along the horizon.

$$d_f = 3.83 \sqrt{h_1 + h_2}$$

In conclusion to determine how far you can see to the horizon in km (i) calculate the square root of your height in meters above sea level and (ii) multiply the result by 3.83. Alternatively for imperial units: (i) calculate the square root of your height (in feet) and (ii) multiply the result to 1.33 to obtain the distance to the horizon in miles.

 

References

  1. Dip of the Horizon: http://mintaka.sdsu.edu/GF/explain/atmos_refr/dip.html
  2. Distance to the Horizon: http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
  3. Horizon: http://en.wikipedia.org/wiki/Horizon

Related Articles

Appendix

Metric and Imperial Units

Equation (5), $s=\sqrt{2hr}$ is not useful from a practical point of view. The units of measurement for the height and radius can be converted for imperial and metric units. Two cases are considered:

  • Given the height in meters how far is it to the horizon in km ?
  • Given the height in feet how far is it to the horizon in miles ?

Metric Units:

Using units of km for all distances,

$$s_{km} = \sqrt{2h_{km}r_{km}}$$

rewriting the height in terms of meters,

$$h_{km} = h_{m} \times \frac{1}{1000}$$

... and taking the average radius of the Earth to be 6378 km, the distance to the horizon in km is:

$$\begin{align} s_{km} &= \sqrt{h_{m}} \times \sqrt{2/1000 \times 6378} \\ &= 3.57 \sqrt{h_{m}} \end{align}$$

Imperial Units:

Using a similar treatment for metric units, express all distances in miles:

$$s_{miles} = \sqrt{2h_{miles}r_{miles}}$$

using the relation between miles and feet,

$$h_{miles} = h_{feet} \times 0.3048 \times \frac{1}{1000} \times \frac{1}{1.609344}$$

... and taking the average radius of the Earth in miles to be $\frac{6378}{1.609344} \approx$ 3964 miles, the distance to the horizon is:

$$\begin{align}s_{miles} &= \sqrt{2} . \sqrt{h_{feet}} . \sqrt{0.3048 \times \frac{1}{1000} \times \frac{1}{1.609344}} \times \sqrt{\frac{6378}{1.609344}} \\ &=  \sqrt{2 \times 0.3048 \times \frac{1}{1000} \times \frac{1}{1.609344} \times \frac{6378}{1.609344}} \times \sqrt{h_{feet}} \\ s_{miles} &= 1.225 \sqrt{h_{feet}} \end{align}$$

Refraction (detailed analysis)
Figure 5

Determining the distance to the horizon with refracted light (shown in red in Figure 5) is simply a case of determining the angle POX ($\alpha$). The main assumption is the refracted light beam (PX) is an arc of a circle with a radius of R with the following properties:

  • The circle with a radius of R touches the Earth's surface at the point X
  • A perpendicular line drawn at from a tangential line at the point X for both circles with radius r and R will pass through O and Q

The problem is reduced to defining the lengths of the sides in triangle POQ in Figure 5 and determining the angle POX ($\alpha$).

In triangle POQ, applying the cosine rule:

$$(PQ)^2 = (QO)^2 + (PO)^2 - 2.(QO).(PO)\cos(\angle POQ)$$

$$ R^2 = (R-r)^2 + (r+h)^2 - 2.(R-r).(r+h)\cos(\pi-\alpha) \qquad_{.....(1a)}$$

$$ \cos(\pi-\alpha) = \frac{(R-r)^2+(r+h)^2-R^2} {2(R-r)(r+h)} \qquad_{.....(2a)}$$

Using the identity $\cos(\pi-\alpha) = - \cos(\alpha)$, equation (2a) becomes:

$$ \begin{align} \cos(\alpha) &= \frac{R^2 - (R-r)^2-(r+h)^2} {2(R-r)(r+h)} \\ &= \frac{2Rr-2r^2-2hr-h^2}{2(R-r)(r+h)} \end{align}$$

Multiply the numerator and denominator by $(r-h)$:

$$ \begin{align} \cos(\alpha) &= \frac{[2Rr-2r^2-2hr-h^2](r-h)}{2(R-r)(r^2-h^2)} \\ &= \frac{r(1-h/r)[2Rr-2r^2-2hr-h^2]}{2r^2(R-r)(1-h^2/r^2)} \\ &= \frac{(1-h/r)[2Rr-2r^2-2hr-h^2]}{2r(R-r)(1-h^2/r^2)} \end{align}$$

In most cases, $h / r$ << 1, therefore $\frac{h^2}{r^2} \approx 0$, simplifying the expression further:

$$ \begin{align} \cos(\alpha) &= \frac{(1-h/r)[2Rr-2r^2-2hr-h^2]}{2r(R-r)} \\ &= \frac{1}{2r^2} \frac{(1-h/r)[2Rr-2r^2-2hr-h^2]}{(R/r-1)} \\ &= \frac{(1-h/r)}{2(R/r-1)} [2(R/r) - 2 - 2(h/r) - h^2/r^2] \\ &=\frac{(1-h/r)}{(R/r-1)}[R/r-h/r-1] \\ &=\frac{1}{(R/r-1)} [ R/r - h/r -1 - Rh/r^2 +h^2/r^2 + h/r ] \\  \cos(\alpha) &= \frac{[ R/r-1-Rh/r^2 ]} {(R/r-1)}\end{align}$$

Multiplying the numerator and denominator by $r/R$ gives:

$$ \begin{align} \cos(\alpha) &= \frac{[(1-r/R)-h/r]}{(1-r/R)} \\ &= 1 - \frac{h}{r}\frac{1}{(1-r/R)} \end{align}$$

For small values of $\alpha$, $\cos(\alpha) \approx 1 - \frac{\alpha^2}{2!}$:

$$ 1 - \frac{\alpha^2}{2} = 1 - \frac{h}{r}\frac{1}{(1-r/R)}$$

Therefore:

$$\alpha = \sqrt{\frac{2h}{r(1-r/R)}}$$

The arc length $s_f$ is:

$$s_f = r \alpha = \sqrt{\frac{2hr}{1-r/R}}$$

 

Tags: 
horizon
trigonometry
refraction
light
lighthouse