Visibility distances have always been a pain in the crotch for me as a referee.
But I came upon this formula over at RPG.net today. Apparently I'm the last one on Earth to have heard of it, but just in case I'm not, I'll pass it along:
D=sqrt(1.5*H)
D is the visibility distance in unobstructed miles
H is the height (in feet) being viewed from
So, roughly, a 6' tall man can see 3 miles in every direction if the terrain is flat and featureless (plains or ocean?).
Very helpful to me at the moment, and maybe helpful to all you sandboxers out there.
So if there's a tower 50' high, is it simple enough to just say it can be seen from 8.66 miles away? If there's an escarpment 100' high, 12 miles or so away? Or is there a different formula for that?
But I came upon this formula over at RPG.net today. Apparently I'm the last one on Earth to have heard of it, but just in case I'm not, I'll pass it along:
D=sqrt(1.5*H)
D is the visibility distance in unobstructed miles
H is the height (in feet) being viewed from
So, roughly, a 6' tall man can see 3 miles in every direction if the terrain is flat and featureless (plains or ocean?).
Very helpful to me at the moment, and maybe helpful to all you sandboxers out there.
So if there's a tower 50' high, is it simple enough to just say it can be seen from 8.66 miles away? If there's an escarpment 100' high, 12 miles or so away? Or is there a different formula for that?
As I recall Judges Guild's Ready Ref Sheets has a half-page or so devoted to rules for visibility, including a random chance that something like a ridge or tall tree interrupts the field of vision.
ReplyDeleteNo you got it right.
ReplyDelete@Jeff aside from being overly generous with the distances page 45 of the Ready Ref rules has keen sighting rules. The basic gist is that if you want to see details there is a % chance of succeeding. Gross details use line of sight rules.
Good to know for the next time I run a hex crawl. Makes the 6-mile per hex of Classic D&D maps work pretty well for mapping purposes, with various terrain features allowing mapping of adjoining hexes.
ReplyDeleteI would only add that the curvature of an earth-sized spheroid occurs at roughly 9 mi., so that two 6' individuals would lose sight of each other over the horizon at or about 9 mi.
ReplyDelete--If the object in question is taller then it can still be seen beyond that distance.
Og,Gorilla King of Earth, demands more updates on Weird Fantasy Role Playing!
ReplyDeleteIf for some reason you want to fiddle with this formula, the distance goes as the square root of the radius of curvature, R, of the planet.
ReplyDeleteD = sqrt(h*R)
i.e. bigger planets have longer distances of visibility.
Don't forget to account for the height of the observer AND the height of the item being observed. (I learned something useful from being in the Navy!)
ReplyDeleteDrat, forgot a two above!
ReplyDeleteD = sqrt(2*h*R)
Just sum the two distances from the above formula to correct for observer height:
D = sqrt(2*h*R)*(1 + sqrt(h_o/h))
where h_o is the height of the observer.
So a 6' adventurer could see a 50' tower from 11.66 miles away (as opposed to 8.66) whereas a Fire Giant (12' according to the 1e Monster Manual 1) could see that tower from 12.9 miles away.