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    Re: Azimuthal equidistant grid
    From: Gary LaPook
    Date: 2009 Oct 1, 10:43 -0700

    That's beautiful.
    On Sep 29, 6:23�pm,  wrote:
    > I thought of a modification of a known method of approximate graphic solution
    > of spherical triangles. It involves plotting points on a grid that resembles a
    > grid of parallels and meridians.
    > Similar method was mentioned by Gary Lapook inhttp://www.fer3.com/arc/m2.aspx?i=108398&y=200905
    > The referenced ARG1 device uses equatorial aspect of stereographic projection.
    > I made similar grid, but for equatorial aspect of azimuthal equidistant
    > projection. It has the benefit of having angular distance from center of grid
    > to any other point be simply linear distance. If one chooses scale 1 degree=1mm
    > (which I did), the angular distance can be measured with a regular mm rule.
    > Also, by freak coincidence, grid of 18 cm diameter has just the right size and
    > fits on a sheet of paper comfortably.
    > As an example, I will explain how navigational triangle can be solved using this
    > method. (see picture)
    > Let's have both GP and AP in northern hemisphere. AP is 52N, GP is 25N, LHA is
    > 61 degree.
    > In the beginning the center of the grid represents the north pole, outside
    > circle is equator and right half of horizontal line -- is the meridian of AP.
    > Let's mark the AP at on the meridian 90-52=38 mm from pole (or 52mm from
    > equator). Then mark GP on the line 61 degrees down from meridian and 65mm from
    > pole (or 25 mm from equator). These are points on a celestial sphere and 
    they are projected on the grid. Now we rotate the sphere along the meridian 
    until AP moves to the center of the projection. Both N and AP move to the 
    left 28mm.
    > AP ends up in AP', in the center of the grid, N moves to N'.
    > Now GP also move 38 degrees along one of the "horizontal" curved lines that
    > look like parallels. GP ends up in GP'. Alternatively, GP may be projected on
    > the horizontal line along one of the "vertical" curved lines, then moved along
    > horizontal line 38 mm to the left and then returned to its original "latitude"
    > along another "meridian" line. The transition along the horizontal line may be
    > peformed with a dividers or a rule.
    > Now with AP in the center we can measure distance from AP to GP with a rule (or
    > select it with a dividers and then transfer this distance to a horizontal or
    > vertical line, which are both subdivided into 1 degree=1mm intervals). The
    > azimuth or may be also read directly as angle beween N and GP along the limb.
    > Different triangles can be solved in a similar way. The idea is to construct 
    the triangle by bringing points to the center using suitable combination of 
    rotations along the vertical axis and along the center and constructing other 
    points that are in relation with the central point using the fact that 
    distances and angles from the central point can be measured directly.
    > I noticed that different printers handle thin lines differently and I could 
    not figure out the thicknesses of lines that look good on all printers.
    > That is why i include original postscript file. The line thicknesses can be adjusted in the
    > /LW [0.1 0.3 0.5] def
    > These are points (=1/72")
    > Andrew Nikitin
    > �azeq.png
    > 4KViewDownload
    > �grid_azeq.pdf
    > 97KViewDownload
    > �grid_azeq.zip
    > 54KViewDownload
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