Wednesday, February 15, 2023

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: 

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
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

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

4 comments:

  1. Some Even Quicker Hexcrawl Horizon Visibility Calculations:

    There'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.

    ReplyDelete
    Replies
    1. Ha, true.

      The 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."

      Delete
    2. A perfectly valid question. 👍

      Delete
  2. Been 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