(I'm Australian, so I tend to use metric measures. I did learn imperial measures as a child, but I don't use them much these days.)
Korbl wrote:It's for a D&D game, basically it's Yggdrasil.
I have no clue what an arc second is (I stopped at chemistry in High School, where I did poorly). Can you give me an idea of that distance in feet or miles or at least metric distances? Curvature and atmospheric distance aren't super important.
Ok. I asked about curvature because I guessed that this might be on a flat world, or at least on a planet with a different radius to the Earth. If it's not on a flat world then at large distances some (or all) of the tree will be below the horizon, hence not visible.
An arc second is a unit of angular measure. There are 60 seconds to a minute, and 60 minutes to a degree; the terms arc-second and arc-minute are used to distinguish these units from the units of time.
A line 2 units long at a distance of 412 529.612 units subtends an angle of 1 arc-second, so it will be visible under ideal viewing conditions, according to that Wiki link. So on a flat world the 8km high tree will be visible at a distance of over 1.6 million kilometres (ignoring atmospheric distortion). However, to get any kind of detail, we'd need to use the figure of 5 minutes of arc mentioned earlier in the article, giving a distance of a shade over 5,500 km. On Earth, that's a little over half the distance from a pole to the equator, so horizon issues will definitely be significant.
At a height of 8km, the horizon is roughly 320 km away, so the 8km high tree will be below the horizon for observers more than 320 km distant from it.
FWIW, the formula I used for relating height, distance and angular size is
(height / 2 ) / distance = tan(angle / 2)