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 |
Arc distance along the ground is same as straight-line distance, up to the precision I used for these numbers. Once you get up to the Mount Everest row, the difference is on the order of one mile.
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
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
\[\theta_1 = \arccos \left(\frac{R_⊕}{R_⊕+h}\right)\] \[\theta_2 = \arccos \left(\frac{R_⊕}{R_⊕+H}\right)\]
Some Even Quicker Hexcrawl Horizon Visibility Calculations:
ReplyDeleteThere's a bush over there: that's your horizon.
There's a hillock in that direction: that's your horizon.
You're in a forest: there's no horizon.
There are mountains in the distance: yeah, you probably can see them unless it's really hazy.
You're in the mountains: your horizon is the top of them.
Calculating the horizon distance is fun and interesting, but let's not forget that in most places that aren't open sea, the horizon varies a lot, depending on terrain undulations and trees and stuff in the way even more than elevation.
Ha, true.
DeleteThe original specific question I was curious about was: "There's a giant magical stationary super-storm. From how far away on the open sea can it be seen."
A perfectly valid question. 👍
DeleteBeen loving all your blog posts! I have been working on a Crimson Skies campaign that has a lot of hex based mechanics. I also love your map making skills, excellent stuff all around!
ReplyDelete