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681

answers:

3

Hi, using a Latitude and Longitude value (Point A), I am trying to calculate another Point B, X meters away bearing 0 radians from point A. Then display the point B Latitude and Longitude values.

Example (Pseudo code):

PointA_Lat = x.xxxx;
PointA_Lng = x.xxxx;
Distance = 3; //Meters
bearing = 0; //radians

new_PointB = PointA-Distance;

I was able to calculate the distance between two Points but what I want to find is the second point knowing the distance and bearing.

Preferably in PHP or Javascript.

Thank you

A: 

dx = sin(bearing)
dy = cos(bearing)
x = center.x + dist*dx;
y = center.y + dist*dy;

Chris H
This would work on a plane if you swap sin() and cos()
stacker
centerX=41.88592 (Latitude) and centerY=-87.62788 (Longitude)My distance is 500meters. The above calculation does not give me the coordinates 500meters right of my original location. The distance is in the wrong format so im trying to figure that out next.
Pawel
true, it depends on the convention for the bearing. if 0 is "to the right" and x+ is "to the right" then swap sin and cos... the convention wasn't given in the question so I made something up at random.
Chris H
dx = cos(0) = 1, dy = sin(0) = 0, x = 41 + 500*1, y = -87 + 500*0
Chris H
If you want to express dx and dy in degrees longitude and latitude, there's a cos(latitude) correction factor involved.
Jim Lewis
+1  A: 

It might help if you knew that 3600 seconds of arc is 1 degree (lat. or long.), that there are 1852 meters in a nautical mile, and a nautical mile is 1 second of arc. Of course you're depending on the distances being relatively short, otherwise you'd have to use spherical trigonometry.

Arthur Kalliokoski
Spherical trig is a good point to raise. If the distances are anything larger than a few miles they will diverge from the globe and be up in the air. If this matters or not depends on your use-case, precision and goals. Perhaps he's not working on a globe but some flat space.
Karl
Karl and akallio, thanks for the input. Im actually working on a flat surface and the distance will be about 500 meters.
Pawel
so just to confirm using the conversion above: 500meters = 0.2699784 sec of arc?
Pawel
@akallio: A nautical mile is one arcmin, not arcsec. And that is only true along a great circle -- one nautical mile east or west does notcorrespond to 1/60 degree of longitude except at the equator.
Jim Lewis
+3  A: 

It seems you are measuring distance (R) in meters, and bearing (theta) counterclockwise from due east. And for your purposes (hundereds of meters), plane geometry should be accurate enough. In that case,

dx = R*cos(theta) ; theta measured counterclockwise from due east
dy = R*sin(theta) ; dx, dy same units as R

If theta is measured clockwise from due north (for example, compass bearings), the calculation for dx and dy is slightly different:

dx = R*sin(theta)  ; theta measured clockwise from due north
dy = R*cos(theta)  ; dx, dy same units as R

In either case, the change in degrees longitude and latitude is:

delta-longitude = dx/(111320*cos(latitude))  ; dx, dy in meters
delta-latitude = dy/110540                   ; result in degrees long/lat

The difference between the constants 110540 and 111320 is due to the earth's oblateness (polar and equatorial circumferences are different).

Here's a worked example, using the parameters from a later question of yours:

Given a start location at longitude -87.62788 degrees, latitude 41.88592 degrees, find the coordinates of the point 500 meters northwest from the start location.

If we're measuring angles counterclockwise from due east, "northwest" corresponds to theta=135 degrees. R is 500 meters.

dx = R*cos(theta) 
   = 500 * cos(135 deg) 
   = -353.55 meters

dy = R*sin(theta) 
   = 500 * sin(135 deg) 
   = +353.55 meters

delta_longitude = dx/(111320*cos(latitude)) 
                = -353.55/(111320*cos(41.88592 deg))
                = -.004266 deg (approx -15.36 arcsec)

delta-latitude = dy/110540
               = 353.55/110540
               =  .003198 deg (approx 11.51 arcsec)

Final longitude = start_longitude + delta_longitude
                = -87.62788 - .004266
                = -87.632146

Final latitude = start_latitude + delta_latitude
               = 41.88592 + .003198
               = 41.889118
Jim Lewis
Thank you Jim for the edit example! I really appreciate it. I fully understand the concept now.
Pawel