## Introduction

While doing some reading on lighthouses, I needed a formula to compute the distance to the horizon as a function of height. This formula would give me an idea of how far away a lighthouse could be seen. As I usually do, I started with the Wikipedia and it listed two interesting equations:

• Eq. 1: $\displaystyle s\left[ \text{km} \right]=3.57\cdot \sqrt{h\left[ m \right]}$

where s is the distance along the Earth’s surface in kilometers and h is the height of the observer in meters. This formula assumes that light moves strictly in straight lines. You will also see it in the form $s=\sqrt{2 \cdot h \cdot r_E}$, where $r_E$ is the radius of the Earth.

• Eq. 2: $\displaystyle s\left[ \text{km} \right]=3.86\cdot \sqrt{h\left[ m \right]}$

This formula assumes that light is refracted and travels along a circular arc.

Allowing for refraction means that light will curve around the geometrical horizon (Eq. 1) because means that we will see objects that are just beyond the will compute a horizon distance that is about 7% further than Eq. 1. Figure 1 illustrates this point.

Figure 1: Illustration Showing that Refraction Increases the Perceived Distance to the Horizon.

While I frequently see refraction modeled using Eq. 2, I have not seen anyone go into the details of Eq. 2 — I am sure this analysis exists — I just have not seen it. So I will dive into the details of Eq. 2 in this post. I have seen the Eq.1 derived in a number of places, so I will not repeat its derivation here.

The material presented in this blog is similar to that presented on mirages. I have included some mirage information in Appendix B.

## Background

### Basic Idea

Light bends because of density differences in the atmosphere. These density differences are caused by (1) altitude and (2) temperature. Of course, temperature and pressure vary with time. To get some idea of the effect of refraction on the distance to the horizon, we need to assume a “typical” atmosphere. We can derive Eq. 2 using the typical atmosphere assumption, but we need to understand that those conditions are often not present.

## Analysis

Here is my problem solving approach:

• Show that the air’s index of refraction is a simple function of density.
• Develop an expression for the rate of change of air’s index of refraction.
• Develop an expression for the radius of curvature of a refracted light beam.
• Develop an expression for the arc length along the Earth’s surface that a refracted light beam will traverse.
• Simplify the expression by assuming typical atmospheric conditions and common units.

### Air’s Refractivity as a function of Density

Because air’s index of refraction is very close to 1, physicists usually talk in terms of refractivity. The refractivity of air is defined as $\displaystyle {{N}_{Air}}\triangleq \left( n-1 \right)\cdot {{10}^{6}}$, where n is the index of refraction. One common formula for the refractivity of dry air is given by Equation 3 (Source).

 Eq. 3 $\displaystyle {{N}_{Air}}=\left( 776.2+\frac{4.36\cdot {{10}^{-8}}}{{{\lambda }^{2}}} \right)\cdot \frac{P}{T}$

where

• P is air pressure [kPa].
• T is the air temperature [K].
• $\lambda$ is the wavelength of light [cm].

Using the ideal gas law, we can show that temperature and pressure are closely related to density as shown in Equation 4.

 Eq. 4 $\displaystyle P\cdot V=n_m\cdot R\cdot T\Rightarrow \frac{P}{T}=\frac{m}{M{{W}_{Air}}}\cdot \frac{R}{V}=\frac{m}{V}\cdot \frac{R}{M{{W}_{Air}}}=\rho \cdot \frac{R}{M{{W}_{Air}}}$

where

• V is the volume of air.
• m is the mass of air.
• nm is the number of moles of air = $\frac{m}{MW_{Air}}$, where $MW_{Air}$ is the molecular weight of air.
• R is the universal gas constant.
• $\rho$ is the air’s density.

Equation 4 states that the ratio of P/T is proportional to $\rho$ (R and MWAir are constants). This means we can restate Equation 3 in the simpler form shown in Equation 5.

 Eq. 5 $\displaystyle {{N}_{Air}}=a\left( \lambda \right)\cdot \rho$

where

• $a\left( \lambda \right)$ is a function of the wavelength of light.

The index of refraction is a function of wavelength, as is shown in Figure 2. Note how the shorter wavelengths have a larger index of refraction than the longer wavelengths.

Figure 2: Air’s Index of Refraction Versus Wavelength.

For this post, I will assume that the wavelength of light is fixed and that $a\left( \lambda \right)$ is a constant.

### Air’s Index of Refraction Rate of Change

For light to refract in the atmosphere, it must encounter air with an index of refraction that varies. The air’s index of refraction is a function of its density, which varies with altitude and temperature. We can express the variation of the atmosphere’s index of refraction using Equation 6.

 Eq. 6 $\displaystyle \frac{dn}{dz}=-\frac{\left( n-1 \right)}{T}\cdot \left( \frac{\rho \cdot g\cdot T}{P}-\gamma \right)$

where

• n is the air index of refraction.
• z is the altitude.
• g is the acceleration due to gravity.
• $\gamma$ is the atmospheric lapse rate, which is defined as the rate of temperature decrease with height (i.e. $\gamma =-\frac{dT}{dz}$).

Deriving this expression requires manipulation of the ideal gas law. The derivation is straightforward but tedious. I have included it in Appendix A.

Equation 6 has a couple of interesting aspects:

• $\frac{dn}{dz}=0$ when $\gamma = \frac{\rho \cdot g\cdot T}{P}$

This value of $\gamma$ is known as the autoconvective lapse rate. This rare condition means that light does not refract. It occurs over surfaces that are easily heated, like open fields and deserts. This condition requires that the rate of temperature decrease with altitude to cause an air density increase that exactly cancels the air density decrease with altitude.

• Usually, $\frac{dn}{dz} < 0$

The change of index with altitude is normally negative because the air density increase with altitude associated with decreasing temperature is not enough to cancel the air density decrease with altitude. Since light bends towards regions of higher density, light projected straight from a lighthouse will refract toward the Earth’s surface under normal circumstances (as shown in Figure 1).

We can compute the autoconvective lapse rate as shown in Figure 3. The value I obtain here can be confirmed at this web page.

Figure 3: Calculation of the Autoconvective Lapse Rate.

### Radius of Curvature for the Refracted Light Beam

Assuming a typical atmosphere, we can model the path of a refracted beam of light in the atmosphere as an arc on a circle. Figure 4 shows a derivation for the radius of curvature for a refracted beam of light. The radius of curvature will be constant (i.e. a circle) when $\frac{dn}{dz}$ is a constant.

Figure 4: Derivation of the Radius of Curvature Expression.

In the case of a lighthouse beam, $\theta =90{}^\circ$. This means we can write the refraction radius as $\displaystyle {{r}_{R}}=-\frac{n}{\frac{dn}{dz}}$.

### Arc Length Traversed By A Refracted Light Beam

We are going to model the refracted light beam as moving along the arc of a circle of radius larger than that of the Earth. Figure 5 illustrates this situation.

Figure 5: Illustration of the Refracted Light Beam Moving Along the Arc of a Circle.

Given the geometry shown in Figure 4, we can derive an expression for the arc length upon the Earth’s surface that the light beam traverses as shown in Figure 6.

Figure 6: Derivation of Formula for the Distance Traveled Along the Earth’s Surface By Refracted Light Beam.

Equation 7 shows the key result from the derivation in Figure 6.

 Eq. 7 $\displaystyle {{s}_{R}}\left( h,{{r}_{E}},{{r}_{R}} \right)=\sqrt{\frac{2\cdot {{r}_{E}}\cdot h}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}}$

where

• rE is the radius of the Earth.
• rR is the radius of the refraction circle.
• h is the height of the lighthouse.

Notice how Eq.7 can be viewed as Eq. 1 with an enlarged Earth radius of ${{{r}'}_{E}}=\frac{{{r}_{E}}}{1-\frac{{{r}_{E}}}{{{r}_{R}}}}$. I see refraction often modeled by engineers who simply use Eq. 1 with an enlarged Earth radius. The radius used depends on the wavelength of the photons under consideration.

### Simplified Arc Length Expression

Figure 7 shows how I obtained Equation 2 from Equation 7.

Figure 7: Equation 2 from Equation 7.

## Conclusion

While the derivation was a bit long, I was able to derive Equation 2 from first principles. The derivation shows that the final result is sensitive to the choice of lapse rate, which varies throughout the day. I should note that the use of a refraction radius that is a multiple of the Earth’s radius is often applied to other types of electromagnetic signals. For example, radar systems frequently use a “4/3 Earth radius” for refraction problems (Source). The refraction radius for radar is different than for optical signals because the index of refraction in the radio band is different from that of optical signals.

## Appendix A: Derivation of Rate of Change of Atmospheric Index of Refraction with Lapse Rate.

Figure A: Derivation of Rate of Change of Atmospheric Index of Refraction with Lapse Rate.

## Appendix B: Excellent Discussion of Index of Refraction, Lapse Rate, and Mirages.

I really like the way this author discusses mirages (Source).

Figure B: Excellent Description on the Formation of Mirages.

I am an engineer who encounters interesting math and science problems almost every day. I am not talking about BIG math here. These are everyday problems where a little bit of math really goes a long way. I thought I would write some of them down and see if others also found them interesting.
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### 9 Responses to Distance to the Horizon Assuming Refraction

1. Scott Lundin says:

Thanks for this derivation. It’s very helpful for something I’m working on. At the bottom of figure 7 you restate equation 2, however the square root term is SQRT(h/m) instead of simply SQRT(h) as shown above. How/why does the (/m) term drop out?

• mathscinotes says:

It does look confusing. That is actually the units notation from within Mathcad. This notation is a bit strange, but it says that this equation works with h in units of meters. The nice thing about Mathcad and this unit notation is that if you input the number with units of feet, Mathcad will convert it to meters automatically.

What I am telling you is that the “/m” is a Mathcad unit thing and is not present in derivations as normally done by people.

mathscinotes

• Scott Lundin says:

Got it. Thanks.