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Subject: Re: Q: how to calculate refraction at higher altitudes on land?
From: Craig (c.scott@XXX.XXX)
Date: Thu Feb 28 2002 - 15:40:42 EST
I was thinking, in using trigonometry to solve a right triangle, you know
the adjacent side from your GPS (Difference between your lat/lon position
and lat/lon of mountain peak. You measure the angle, can get the tangent,
and solve for the opposite side, which is the difference between height of
mountain peak and your height. Subtract length of oppposite side from
mountain peak height, doesn't that give you your height?
Tangent=opposite/adjacent, therefore opposite=tangent*adjacent.
-----Original Message-----
From: Navigation Mailing List
[mailto:NAVIGATION-L@XXX.XXX]On Behalf Of Dov Kruger
Sent: Thursday, February 28, 2002 14:27
To: NAVIGATION-L@XXX.XXX
Subject: Re: [NAV-L] Q: how to calculate refraction at higher altitudes
on land?
Craig wrote:
>If you can see the top of a nearby mountain, measure the angle above (or
>below) horizontal, use the topographic map for mountain top coordinates,
>waypoint mountain top and your position, obtain distance (easy with GPS),
>use trigonometry for difference in opposite side, which is difference
>between your height and mountain top, simple math, and voila!
>
This is an interesting idea, but you don't know your own altitude, and
you therefore don't know the difference in altitude.
You can use the current uncertain altitude as a starting point and
iterate perhaps.
If GPSs display the straight line distance between two points including
altitude, then I am unaware of it. I do not even know if I can enter the
altitude of a waypoint on my GPS.
I don't think this is quite so easy as Craig says ;-)
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