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|    alt.comp.os.windows-xp    |    Actually wasn't too bad for a M$-OS    |    17,273 messages    |
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|    Message 17,133 of 17,273    |
|    J. P. Gilliver to R.Wieser    |
|    Re: Worldmap mercator projection - Latit    |
|    15 Jan 26 17:25:26    |
      XPost: alt.windows7.general, alt.comp.os.windows-10       From: G6JPG@255soft.uk              On 2026/1/14 8:24:17, R.Wieser wrote:       > Hello all,       >        > I have a world map using the Mercator projection, and would like to plot        > some stuff on it using Latitude and Longitude.       >        > The problem is that I can't seem to get the formule I found to spit out the       > correct value for Y.       >        > -- The map :       >        > https://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Merc       tor_projection_Square.JPG/250px-Mercator_projection_Square.JPG       >        > -- The formula :       >        > Y = ln( tan(latitude) + sec(latitude) )       >        > where "ln(...)" is log(...) / log(10)       >        > ... at least, that is what I could google about it.              Rather than Googling the formula, my first thought was to go back to       first principles: as the name implies, a map projection can be thought       of as being created by shining a light through the earth (globe) onto a       sheet of paper - either flat and fixed to the globe at one point, or -       more commonly - a cylinder wrapped round the globe, touching at one       circle (often the equator), and then unrolled. Once this is realised,       basic geometry should make calculation of the co-ordinates fairly simple.              Unfortunately, to do this, one needs to know where the nominal light       source is. And - despite it being a _long_ article - I can't find a       statement of this in the Wikipedia article about the Mercator projection       - other that that - I _think_ - it _isn't_ the centre of the earth       (so-called "radial"), though is often thought to be (certainly Google's       AI thinks it is).              If anyone _can_ find out where the light source point is for the       Mercator projection, I'd love to know! (And it would answer Rudy's       question.)              (One other position for the light source I remember from last time I       looked into this - which was probably over 40 years ago! - is on the       opposite surface of the globe to the projection point (i. e. sort of       tracking round opposite the map "printing"); I don't think it's that,       though, as that would show the poles, though still distorted.              If it _is_ light-source-at-centre, then the Y co-ordinate would just be       the tangent of the latitude (scaled appropriately for the map size).       Even if it isn't, this _may_ be close enough - try a few places.              --        J. P. Gilliver. UMRA: 1960/<1985 MB++G()ALIS-Ch++(p)Ar++T+H+Sh0!:`)DNAf               It has been my experience that folks who have no vices have very few       virtues       -- Abraham Lincoln, quoted by Mark Lloyd in alt.windows7.general 2018-12-27              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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