# What is the “dropoff” to the ground from the eye line of an observer straight across a curved globe earth?

Please note that we assume the observer's eye line is exactly at sea level (0 inches) and we are assuming a perfect spherical earth with no atmospheric effects. The idea here is an alternative approach to evaluating the curvature of the earth since "distance to horizon" appears to be already covered eg. http://www.wikihow.com/Calculate-the-Distance-to-the-Horizon

Diagram of Problem

PLEASE VIEW DIAGRAM HERE - TL;DR SOLVE FOR D KNOWING R AND X

Is this correct?

$$D = \sin{(90-\tan^-1{(\frac{X}{R})})} * (\sqrt{R^2+X^2}-R)$$

Long Version:

An observer stands at point $P^0$ at $0$ inches of elevation and looks in a direct straight line over a distance of $X$ miles to point $P^1$.

The ground curves away from the observer's eye line, the eye line being tangential from point $P^0$ to point $P^1$ (and beyond) over a spherical earth.

Hence a right-angle triangle is formed between $P^0$, $P^1$ and the centre of the earth ($C^0$)with radius $R$, acute angle $a$, and hypotenuse $R+H$.

Clearly the "concentric height" $H$ ie. from $P^1$ to the ground (point $P^2$) in this right-angle triangle is: $$H = \sqrt{R^2+X^2}-R$$

As the distance $X$ inreases the "concentric height" $H$ increases in accordance to the curvature of the earth. This is known colloquially as "8 inches per miles squared", for example, approximately: 1 mile gives 8 inches, 2 miles gives 32 inches, 3 miles gives 72 inches and 10 miles gives 800 inches.

As the observer is looking at 0 inches of elevation in a straight line any objects in their eye line higher than sea level will obviously be visible but will eventually curve away until it is impossible to see whether the observer uses eyesight, optical zoom or laser techniques.

An issue arises when we want to calculate the "dropoff" straight "down" to the ground perpendicular to the eye line of the observer. That is, a "dropoff" $D$ which is the opposite face of a right-angle triangle between $P^1$, a point along the straight eye line (a distance less than $X$, point $P^3$) and the ground point $P^2$.

How do you calculate this "dropoff"? Here is my method, could there be a mistake in it?

As you can see from the diagram, we can establish the following.

The "concentric height" $H$ is: $$H = \sqrt{R^2+X^2}-R$$

The acute angle (arc originating from the centre of the earth) $a$ is: $$a = \tan^-1{(\frac{X}{R})}$$

The smaller right-angle triangle relevant to $D$ within the larger right-angle triangle is formed between $P^3$, $P^1$ and $P^2$ with acute angle b where: $$b = 90 -a$$

The dropoff $D$ can then be defined as: $$D = \sin{(b)}* H$$

Hence the final formula of: $$D = \sin{(90-\tan^-1{(\frac{X}{R})})} * (\sqrt{R^2+X^2}-R)$$

"Dropoff" Example Calculations - Corrected Aug 29 2016

Eg. Find dropoff in inches, for 10 miles "eyeline distance" $X$ per diagram above, with an earth of radius 3959 miles, enter into Wolfram Alpha online: (sin(pi/2-arctan(10/3959))) * (sqrt(3959^2+10^2)-3959) * 63360. Colloquial(A) is "8 inches times miles squared". Colloquial(B) is "2/3 feet times miles squared".

X Miles, D Dropoff (Inches), D Dropoff (Feet), Colloquial(A)(Inches), Colloquial(B)(Feet)
1, 8.00, 0.664, 8, 0.667
2, 32.0, 2.66, 32, 2.67
3, 72.0, 5.98, 72, 6.00
5, 200, 16.6, 200, 16.7
10, 800, 66.4, 800, 66.7
20, 3200, 266, 3200, 267
30, 7202, 598, 7200, 600
40, 12802, 1063, 12800, 1067
50, 20002, 1660, 20000, 1667
100, 79982, 6639, 80000, 6667
1000, 7638400, 633987, 8000000, 666666
2000, 26947774, 2236665, 32000000, 2666666

Further Research

• A formula should be derived to factor in the height of the observer from sea level
• A formula should be derived for the distance $X^B$ which is the arc length (ground distance) from the observer. (Edit: one of the answers below has provided this).
• A formula should be derived to factor in the angle of observation.
• Empirical tests should be conducted using optical zoom (300mm and greater magnification) and highly-focused lasers all at different points on the earth at different dates and times.
• The radius of approx. 4000 miles used is for observations longitudinally North to South at 0 degrees Longitude, or for observations latitudinally East to West at 0 degrees Latitude. This radius would be smaller for observations across a sphere at different points, eg. observations latitudinally East to West at 40 degrees Latitude cf. "Circumference of the earth at different latitudes" formula.
• There is a bit of inconsistency in defining the "dropoff" as both "straight down to the ground" and "perpendicular to the eye line of the observer". While the discrepancy will be negligible for practical measurement, the line "perpendicular to the eye line of the observer" would not pass through the center of the earth as the line "straight down to the ground" would. – hardmath Feb 20 '16 at 9:34
• Please, do upload that diagram - this could be stated with just a few lines of text if you had a drawing to accompany it. – Bobson Dugnutt Feb 20 '16 at 9:45
• @hardmath I mean "straight down to the ground in a perpendicular fashion" (it doesn't pass through the centre of the earth, that is the "concentric height"). – SaltySub2 Feb 20 '16 at 9:51
• @Lovsovs Uploading now. Glad to see such quick interest in this question. – SaltySub2 Feb 20 '16 at 9:51

D can actually be expressed as :

$$D = R-\frac{R^2}{\sqrt{R^2+X^2}}$$

There's no need for sin or tan or other trigonometric functions. They're basically two similar triangles with 'D' corresponding to H in the same proportion as H corresponds to H+R. So you can just multiply R by H over (H+R) to get D :

how to get D

• Took some time but marking this as the accepted answer as results are identical to my less-simplified equation. – SaltySub2 May 12 '19 at 13:37

The equation for the drop-off is simply $R - R\cos\left(\dfrac cR\right)$, where $R$ is the radius of the earth and $c$ is the distance along the ground.

For a full treatment, check out https://chizzlewit.wordpress.com/2015/05/13/working-with-the-curvaure-of-a-spherical-earth/

• Hi @james-m, thanks for your response, that link is very useful and certainly makes certain calculations easier than what I'm doing!. One thing though, the $c$ that refers to is the distance along ground which is arc length not straight eyeline distance from the observer... My "eyeline distance" $X$ is not defined as $m$ in that link per se (compare my diagram in my question). I'm hoping for verification of my original question as posed (the question is phrased for a specific purpose we can't go into here). Cheers. – SaltySub2 Aug 28 '16 at 17:45

I did manage to find one error, which had been throwing me off because your calculations were not yielding the right answer. Wolfram Alpha defaults to Radians for trig functions. If you change sin(90 - tan⁻¹(X/R)) to sin(π/2 - tan⁻²(X/R)), your answers will come out correct.

To add, I believe the main issue with this method, is that because your laser will have refracted over the distance measured, this will introduce errors into your measurement. Laser range finding will have to take into account refraction, which means knowing temperature and atmospheric pressure of the air that the laser is passing through.

• Hi @user362649, thanks for your comments. Using a scientific calculator it appears that indeed WolframAlpha is assuming radians and thus my results in the "Dropoff Example Calculations" are wrong. I have corrected accordingly. Regarding laser experimentation indeed we will have to look at the physics aspect of that to determine refraction/ bending/ etc. perhaps on physics.stackexchange separately. – SaltySub2 Aug 28 '16 at 17:47

First problem, you're measuring distance as P_0 to P_1, but P_1 is an imaginary point above the surface of the earth. Distance should be measured as an arc length along the surface of the earth.

Second problem, you've defined the point "below" $P_1$, which you label $P_2$, as based on the direction of down relative to $P_2$, but then define the vertical drop as a measurement parallel to down at $P_1$. To my understanding of the mentality of the people involved, you have found a way to measure the vertical drop at the point $P_2$.

I would say, redefine your input distance as the arc length $P_1$ to $P_2$, which I will call $S$. Now $$a = S / 180 * R * pi$$.

You can then construct the same triangle you're using now using the Law of Sines to get $R+H$ and $X$ and then the equation you're currently using to arrive at $D$.

• Welcome to the site! OP is asking to find the distance $D$ as shown in the diagram, not the arc length. And what is the point of instructing OP to find $D$ by using "the equation you're currently using to arrive at $D$"? If this is not what you meant, please edit your answer so as to make it more clear what you are suggesting. Thank you. – Bobson Dugnutt Feb 20 '16 at 14:42
• @Ryvaken Thanks for the response. I think I understand conceptually what you mean in terms of using the arc length, chord, etc. and using ground distance not eye line distance etc. I still need to verify though if my equation for finding $D$ is correct. I'm trying to determine the "dropoff" ie. how far "below" $P^0$ is $P^2$. – SaltySub2 Feb 20 '16 at 15:03
• @Ryvaken For dropoff $D$ for $X$ (let's call it $X^A$) ie. the direct eyeline distance, versus let's say $X^B$ ie. the ground distance over the curved globe earth (arc length), I have found it is quite tiny. As you suggest we can derive a formula for $D$ based on a given $X^B$ not $X^A$. But for now, I have found: $X^B = \tan^-1{(\frac{X^A}{R})} * \frac{\pi}{180} * R$. For example, $X^A$ of 10 miles leads to $X^B$ of 9.99998 miles, $X^A$ of 100 miles leads to $X^B$ of 99.979 miles, $X^A$ of 100 miles leads to $X^B$ of 99.979 miles, $X^A$ of 1000 miles leads to $X^B$ of 979.512 miles. – SaltySub2 Feb 21 '16 at 4:20