Maximum draw distance

[BigWalleye]

Originally Posted by Sailor Steve

The question then becomes "Okay, you can see ten miles. Is that the horizon, meaning the waterline of the target ship? Is it the top of the superstructure? The top of the masts? Or the top of the smoke coming from its funnels?

A simple and fairly accurate approximation (good to within 6%) is distance to the horizon in nautical miles is equal to the square root of the observer's height in feet. So for 16 feet (5 meters, an good number for the eye level of the lookouts on a VII), the distance to the horizon is 4 nm. For the distance to see something over the horizon, the distance is the sum of the square root of the observer's height and the square root of the object's height. A typical merchant ship has its highest point about 25 m above the waterline, or about 81 feet. That's 9+4=13, so the lookout could see the top of the ship's mast at 13 nm. Putting that back into metric units, 13 nm is 24 km. That's to see the extreme tip of the highest mast. At 20 km, the lookout could see a point 15 m above the waterline, say the top of the ship's superstructure. So the 20 km horizon seems like a good practical limit. It certainly is not the distance at which a lookout could see smoke, though.