|
|
Back to UserFriendly Strip Comments Index
|
Geodesics (kind of).. | by Unicorn | 2003-04-24 00:51:10 |
| hhm.. |
by Unicorn |
2003-04-24 18:11:41 |
Figure lat1 and lat2 as you would with an Equator-centered circle. If a
calculated latitude happens to exceed 90 degrees, then this doesn't
really apply because then your circle encloses a pole and then you can't
inscribe the circle in any "square" (it's really more like a trapezoid
in the general case) of latitude and longitude lines. I'll assume here
that the circle does not in fact enclose a pole.
Now for lon1 and lon2. Draw a spherical triangle with one vertex C at
the center of the circle, a second vertex P at the North or Sout Pole,
and a third vertex X on your circle, with X so placed that PX is tangent
to the circle. Then triangle PXC is a RIGHT triangle with the right
angle at X. The arc measures of the hypoteneuse (PC) and the leg PX are
known from the construction, and the Spherical Law of Sines gives
immediately:
sin(angle CPX) =
[sin(angle PXC)] * sin(arc XC)/sin(arc PX)
where the term is brackets is 1. To decide between two supplementary
solutions, angle CPX is taken to be acute. From angle CPX we can get
each of the bounding longitudes, adding and subtracting that angle to
the given longitude at C. Be sure to make proper provisions for when
you straddle the 180-degree meridian |
|
[ Reply ] |
|
|
[Todays Cartoon Discussion]
[News Index]
|
|
