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- From: jack@cs.com (Jack Hudler)
- Newsgroups: sci.astro
- Subject: Re: Distance of horizon
- Message-ID: <1992Nov20.131437.8385@cs.com>
- Date: 20 Nov 92 13:14:37 GMT
- References: <lglhj3INNb0c@appserv.Eng.Sun.COM> <1992Nov19.021430.13833@sfu.ca> <75439@hydra.gatech.EDU>
- Organization: Computer Support Corporation, Dallas,Texas
- Lines: 28
-
- In article <75439@hydra.gatech.EDU> collins@emperor.gatech.edu (Tom Collins) writes:
- >In article <1992Nov19.021430.13833@sfu.ca> palmer@sfu.ca (Leigh Palmer) writes:
- >>In article <lglhj3INNb0c@appserv.Eng.Sun.COM> fiddler@concertina.Eng.Sun.COM
- >>(steve hix) writes:
- >>>Anyone have handy a function for figuring the distance of the
- >>>horizon from a viewer based on the viewer's height from the
- >>>surface?
- >> -1
- >>Try d = R arccos ( 1 + h/R )
- >>
- >> d = horizon distance
- >> h = height above MSL (assuming horizon is at sea level)
- >> R = radius of Earth
- >
- > 2 2 2
- >since R + d = (R+h) (Pythagorean theorem)
- >
- > 2 2
- >d = sqrt( (R+h) - R )
- >
- >This says that the horizon for a six-foot tall person is about
- >3 miles away.
-
- But these equations would give the 'mean' horizion, shouldn't you include
- refraction to give the 'apparent' horizion?
- --
- Jack Computer Support Corporation Dallas,Texas
- Hudler Internet: jack@cs.com
-