Visualising Spherical Grids

I have found a much faster way to display spherical grids based on IDL object graphics. This uses OGL hardware support. I chose the warmest month ever recorded (March 2016) to show  results. First here is the grid rotated about the ‘y-axis’ which shows coverage from pole to pole. The intense gridding visible are the US stations in GHCNV3 and to a lesser extent also in Europe. Australia and Africa are also clearly visible.

The next animation shows the temperature anomalies calculated on the same grid, as described in the last post. I have to use the original aspect ratio this time otherwise the animated gif washes out the blue colour. Yes it is still abnormally cold in Antarctica.

The temperature scale for anomalies is ± 10C (blue to red). For comparison here is one of the coldest months in the last decade : January 2008. There was exceptional cold conditions over Siberia and the global average was about the same as the normalisation period 1961-1990.

Animations are all a bit of a gimmick, but I just can’t resist them. However I will  try to make some better quantitative visualisation for a given month.

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6 Responses to Visualising Spherical Grids

  1. Hans Erren says:

    You can also use polar stereographic projections

    • Clive Best says:

      No I don’t think you can because triangles span the poles. So they have vertices from say Lon = +10 Lat = 80 to Lon = -200 Lat =78 . Map projection software wraps it half way round the world. You have to use 3D Cartesian coordinates.

  2. Hans Erren says:

    Then clip at the equator and show two hemispheres. It’s a transformation of spherical coordinates to polar coordinates, you can chose from a plethora of azimuthal polar projections, My preference would be lambert equal area.
    https://en.wikipedia.org/wiki/Lambert_azimuthal_equal-area_projection

    • Nick Stokes says:

      With a projection you can show a hemisphere, with some distortion.Clive’s rotating sphere shows all that without distortion. But more to the point, as Clive says, you can’t usefully triangulate a projection. As I said on another thread, a useful thing about a sphere is that it has no boundaries. Triangulating is a homogeneous exercise, and is in fact just a convex hull. As soon as you cut, you have to figure how to connect across he cut.

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