Hi Chris,
On Aug 11, 2006, at 2:31 PM, Esposito, Christopher wrote:
>
> Hello, All-
> I have the coordinates of 2 points Lat1, Long1 (which represent
> a sensor) and Lat2, Long2 (which represent a target. The sensor has a
> field of view of 30 degrees, and we currently assume that the
> target is
> centered in the sensors' field. What I would like to do is
> construct and
> draw a triangle (an OMPoly) with the sensor location at the apex of
> the
> triangle and the target location at the center of the base, so that a
> line from sensor to target bisects the triangle.
>
> 1) If this triangle was in Euclidean geometry, then the distance
> between
> the two points and the known angle (15 degrees) between the hypotenuse
> and the bisection line from sensor to target, plus a bit of
> trigonometry, would allow me to calculate the distance between the
> target location at the midpoint of the base and the other 2 corners of
> the triangle at either end of the base.
>
> 2) The spherical_distance method in the GreatCircle class seems to
> me to
> be the right method for calculating the distance between the two
> points
> on a sphere (right?). This method says it returns a float - what
> units
> (miles, meters, etc.) is this number in?
Yes, spherical_distance will give you that information. The answer
is in radians, you can convert it to the units you want using the
com.bbn.openmap.proj.Length class - i.e.,
com.bbn.openmap.proj.Length.METER.fromRadians(answer);
You can also use the com.bbn.openmap.geo.Geo class for this. The
GreatCircle class is based on a sphere, the Geo class is based on an
ellipsoid.
>
> 3) if I use the method from (1) and the distance from (2) to calculate
> the distance from the target to the other 2 corners that define the
> base
> of the triangle, what OpenMap method call will give me the coordinates
> of these points given their distance from a known point? I've looked
> through a few projection-related classes, but my knowledge of
> projections and spherical geometry is limited enough I'm not sure I
> would recognize the answer if I saw it.
If you just want to draw the 'triangle', I'd follow David Ward's
suggestion of using an OMArc. Your triangle isn't really be a
triangle on a sphere, after all. The OMArc will take care of the
curved edges for you.
If you need to know the points for some other calculations, you can
use the GreatCircle.spherical_between method, which will give you the
point at a known distance and azimuth from another point. Again,
Geo.offset(distance, azimuth) will give you the answer on an ellipsoid.
Hope this helps,
Don
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Don Dietrick, dietrick_at_bbn.com
BBN Technologies, Cambridge, MA
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
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Received on Tue Aug 15 2006 - 16:29:50 EDT