On a clear day, and at normal cruising altitude, how far ahead can an aircraft pilot see?
|
2.5 million Light years - Andromeda Galaxy? Although it may be further without atmospheric pollution at 40,000 ft?
|
>> 2.5 million Light years - Andromeda Galaxy?
Night before last, a clear night, despite several attempts didn't see a single meteor in the promised Perseid shower. And those are in atmosphere so no distance at all.
|
mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
In short, it is square root of (2 * radius of earth * height from ground)
Last edited by: movilogo on Wed 14 Aug 13 at 13:30
|
On a clear day at 40K ft I have seen a ground feature which I could recognise and identify on a map at about 180 miles. Clear Middle East air, not this polluted European muck.
To a good approximation, the relevant equation is
h*2R = d^2,
where h is the height, R is the radius of the Earth, and d is the distance to the horizon.
(This is actually a property of chords and tangents of a circle in the approximation that h << R.)
So d = sqrt (2 R h).
It is most convenient to express this in miles. Therefore, for h = 20,000 ft, this gives:
d = sqrt (2*3693*20,000/5280) miles = 167.3 miles.
Since d is proportional to h^(1/2), for 35,000 ft height d is
167.3 x (7/4)^(1/2) = 221.3 miles.
Note that sqrt (20,000) = 141.4, while the distance to the horizon for h = 20,000 ft = 167.3 miles.
So the distance d to the horizon in miles is (167.3/141.4) sqrt (h in ft.), so that
d (in miles) = 1.183 sqrt (h in ft.).
That means in practice that if you simply find the square root of h in feet, you only have to increase that number by approximately 20% (1/5th) to have an estimate of the distance to the horizon in miles that is accurate to almost a percent.
|
Where's Number_Cruncher now we don't need him?
|
I've never really looked any further than the Times crossword puzzle.
|
How do you manage that with your feet up, a fag in one hand and a coffee in the other?
edit. Substitute fag for trolley dolly, depending on your preferences.
|
A fag could be a trolley chap, be careful out there!
|
>> On a clear day ............
This is getting interesting.
So if there was a wind turbine with a tip height of 754 feet situated 20 miles out to sea from the shore, what height of the turbine would be visible from the shore and what height would be hidden beyond the horizon, assuming that the eyes of the viewer are 66 inches from the ground?
|
none of it. Sea haze, mist or fog will block its view.
|
Dunno, but a quick search threw up a photo of the turbines off Skegness taken from Hunstanton:
m.flickr.com/#/photos/davidswilson/4112965274/
Almost all of each structure visible at 15 miles away.
|
On a level/plane surface I would expect the bottom 10ft to be unseen, at 15 miles, due to Earth curvature
|