Some Quick Hexcrawl Horizon Visibility Calculations

How far away can an object be before it dips below the horizon and is no longer visible?

For simplicity, treat the planet like a smooth perfect sphere, and ignore the atmosphere and any other obstacles.

Table of Examples

Assume that your eyes are 1.8 meters above the surface of an earth-sized sphere. From how far away can you see the tops of various objects?

Object	Object Height (m)	Visable Distance (m)	Vis. Dist. (miles)
The Ground Itself	0	4,800	3.0
Garden Gnome	0.5	7,300	4.5
Another Person	1.8	9600	6.0
A Male Giraffe	5.2	13,000	8.0
An Aspen Tree	16	19,000	12
Statue of Liberty (with base)	93	39,000	24
Great Pyramid of Giza	140	47,000	29
Statue of Unity (with base)	240	60,000	37
Burj Khalifa	830	110,000	67
Mount Everest	8800	340,000	210
Severe Thunderstorm	18000	480,000	300

The practical takeaways for hexcrawling are that if you're standing in the middle of a flat twelve-mile hex, then: 

• You can just barely see the top of someone's head if they're standing at the edge of the hex. (But there's probably a shrubbery in the way.)
• You can see the top of a mature tree standing in the middle of the neighboring hex. (If you use a telescope and there are no other trees in the way.)
• You can see the top of a giant robot from two hexes away. (Very important to know.)
• You can see storms and mountains from quite some distance. (Haze will be more important than horizon by that point. I don't know whether you'd actually be able to recognize a storm at 300 miles (25 hexes), but here are a couple anecdotes of people noticing thunderstorms from more than 120 miles (10 hexes) away under the right conditions.)

I've seen similar numbers before, but wanted to recalculate everything in a tidy little list for my own use. 

Calculator widget below

Calculator

Parameters

Eye Height (in meters):

Height of Object (in meters):

Radius of Planet (in meters):

Maximum distance at which object is visible:

Arc distance along ground from feet to base: meters. ( miles)
Straight-line distance from eyes to tip of object: meters. ( miles)

Equations

Straight-line Distance

Assume the earth is a perfect sphere with radius $R_⊕$.

If your eyes are at height $h$ above the surface of the sphere, then the distance to the horizon is:

\[d = \sqrt{ {(R_⊕+H)}^2 - R_⊕^2}\]

Now suppose there’s an object of height $H$ which is so far away that you can just barely see the top of it over the horizon.

The distance from the horizon to the top of that object is

\[D = \sqrt{ {(R_⊕+H)}^2 - R_⊕^2 }\]

And the distance from your eyes to the top of that object is

\[distance = d + D = \sqrt{ {(R_⊕+h)}^2 - R_⊕^2} + \sqrt{ {(R_⊕+H)}^2 - R_⊕^2 }\]

Distance Along the Ground

Unless you’re looking at something really tall, the arc-length distance along the ground will be practically the same as the straight-line distance.

But it you want to be exact, the formula is

\[arclength = (\theta_1 + \theta_2) \cdot R_⊕\]

Where the angles are

\[\theta_1 = \arccos \left(\frac{R_⊕}{R_⊕+h}\right)\] \[\theta_2 = \arccos \left(\frac{R_⊕}{R_⊕+H}\right)\]