Grid points - note circles shown are 1 degree in radius (i.e. I've spent a few days looking for a better method but without any luck.įig 2. This is repeated for every remaining point in the four quadrants. The dot product of the rotation matrix using the same rotation axis but rotation angle of opposite sign to that given above with vector D' gives the coordinates of vector D, which can be converted to the latitude and longitude for point D. The dot product of the rotation matrix and each of vectors A, B and C allows vector D' (the equivalent of vector D in the xy-plane) to be found from their components, The rotation matrix requires a rotation axis and rotation angle that I found here: To find D, the approach I've used is to rotate the plane on which A/B/C/D lie to align it with the xy-plane. For the upper left quadrant, the rectangle (sometimes a square) would have corners clockwise from the upper-left D, B, A, and C.įirst compute vectors AB and AC then a unit vector normal to the triangle BAC can be found from their cross-product, i.e. This is repeated outwards in steps of 1 degree for one half of the central row then the other half using a unit vector in the opposite direction to N.įill in the remaining points in each quadrant:Įach point in the grid forms a rectangle lying on the same plane as three other points in the grid - let A be the point at the grid centre, B is a point 'above' or 'below' A in the same column and C is a point on the central row, and the unknown point D is in the same row as B and same column as C. The vector M can then be used as an axis to rotate N by (90-x) degrees^, where x is the offset (in this case 1 deg) from the central column. The cross product of the unit vector N with a unit vector in the direction of A gives another normal vector (M). The two points can be expressed in Cartesian coordinates, and so the unit vector (N) normal to the great circle that contains A and B is found using the cross product, i.e. Consider two points at the given central longitude let point A be the grid centre at 0 degrees longitude and 0 degrees latitude, and point B at 0 degrees longitude and 1 degree north. Since the row of points through A follow a great circle that intersects the meridian at a right angle, we need to find the plane that this second great circle lies on. if the current point's latitude is less than -90 or greater than +90).įind the points along the `central' row (through the grid centre): following the meridian running through the centre point) and wrap the latitude and longitude if a pole is crossed (i.e. All calculations assume a sphere of radius 1.įind the points along the `central' column: Note: a 10-by-10 degree grid would consist of 11-by-11 points each separated from the nearest four points in the adjacent rows and columns by 1 degree. longitude=0, latitude=0 let's call this point A) involves three stages: The procedure I've followed so far to generate the n-by-n degree grid centred at a given point (e.g. Does anyone know of a method that can determine the required grid points to greater precision? However, for a 20-by-20 degree grid the angular separation of the corner point to the central column is ~4.5 arcseconds smaller than the ideal 10 degree separation. The updated method described below finally generates a regular grid. Each point in the second plot (showing the result of my latest attempt) is plotted with a 1 degree-radius circle surrounding it, which ideally should coincide with the surrounding four points. The previous approach I followed using vectors only found great circles perpendicular to the central column (like the coloured grids shown in the first plot) and so the points were not evenly spaced, especially at the edges. I have little experience with spherical trigonometry (aside from knowing about the Haversine formula and the spherical law of cosines) but after some reading I'm fairly certain that each cell will require four unique great circle arcs. The grid should be oriented with the central column of points along the same meridian as the given point. I've been trying to generate an evenly spaced grid centred at a given point on a sphere, such that the angular separation between any neighbouring pair of points is the same (e.g., 1 degree).
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