XPost: alt.windows7.general, alt.comp.os.windows-10   
   From: nospam@needed.invalid   
      
   On Sat, 1/24/2026 9:16 AM, J. P. Gilliver wrote:   
   > On 2026/1/24 8:28:44, R.Wieser wrote:   
   >> John,   
   >>   
   >>> There are three stages to this   
   >> ...   
   >>> 1. The mathematical conversion from latitude (and longitude) to the   
   >>> position on the plane map *relative to its centre*   
   >>   
   >> Nope.  There is no rule that that center must be taken as the origin.   Take   
   >> the formule Paul provided for instance.  It returns a result in the range of   
   >> 0 to 1 .   
   >   
   > It makes it easier to understand what's going on if done in these three   
   > stages. And assuming we're talking of the Web Mercator image as shown at   
   > https://upload.wikimedia.org/wikipedia/commons/e/ec/Web_maps_M   
   rcator_projection_SW.jpg,   
   > 0 degrees latitude and longitude _is_ at its centre.   
   >   
   >>   
   >>> 2. The _scaling_ required, depending on the size of the map image.   
   >>> Which again will depend on the size of the image (and may be different   
   >>> for X and Y).   
   >>   
   >> Worse : Mercator maps go from +85 to -85 latitude.  That must be part of the   
   >   
   > (Or 85.051129.)   
   >   
   >> formule (I've found several maps where the south-pole is cut off, and a bit   
   >> of the north too. iow, useless without further information).   
   >   
   > No, just because the map is cut off at those latitudes, that figure does   
   > NOT have to appear in the formula. The poles _have_ to be cut off,   
   > otherwise the map would be infinitely tall, and very distorted at the poles.   
   >   
   >>   
   >> But again no.   The result of the formule *is* the scaling (in your and my   
   >> usage ranging from +1 to -1, in Pauls formule case, from 0 to +1).  You just   
   >> apply it on whatever size Mercator-style map you have handy.   
   >   
   > The above image, according to my browser, is 2068 by 2060 pixels.   
   >   
   >>   
   >>> I think _most_ of those contributing to this discussion know that,   
   >>> but have not been making it very clear which bits of their formula(e)   
   >>> do what.   
   >>   
   >> I do not need to know what all the parts of a car do, as long as I can drive   
   >> it.  The same goes for these two formules.   Latitude goes in, something I   
   >> can apply comes out.   
   >>   
   > True, if that really is all you want. As a scientist/engineer/just   
   > enquiring mind, I don't like to blindly use a formula without knowing   
   > what it does - or perhaps _why_.   
   >   
   > Yes, you can drive a car without knowing what each bit does. But knowing   
   > at least some of them will improve your longevity (wear and tear on the   
   > mechanisms), fuel economy, performance ... as you drive.   
   >   
   >>   
   >> Though the whole problem isn't that nobody understood what you said there,   
   >> but that nobody was willing to compare the (intermediate) results I posted   
   >> with what they got themselves, allowing me to locate where I made my   
   >> mistake(s?).   
   >   
   > That is indeed one of many problems.   
   >   
   > It would be good to see where the following points come out on e. g. the   
   > above image, using any formula (longitude given first):   
   > 0, 0   
   > +/- 180, 85   
   > +/-180, -85   
   > and some known place, such as London or New York.   
   >   
   >>   
   >> Not when I asked for it in my first post, and not when I rather explicitily   
   >> asked for it a few days back. :-(   
   >>   
   >> Regards,   
   >> Rudy Wieser   
   >>   
   >>   
   > I guess if you just want a formula that works, and _aren't_ bothered   
   > about the three steps - the mathematical conversion from angle to linear   
   > dimension, the scaling, and the offset - then we're very different   
   > minds. Which is of course fine; if we were all the same, it'd be a   
   > boring world.   
   >   
      
   As an engineer, I run a range of values through the equation, to get   
   a feel for both the rate-of-change of my (broken-or-working) function   
   as well as the absolute displacement. When we design logic blocks that   
   cannot be changed, in silicon, we waste a whole month on a test bench to   
   test all boundaries for mis-behavior. If trig functions were involved,   
   that just magnifies the amount of work, because then there are lots   
   of ways to break a thing.   
      
   The code I got, has a "clamp" function that sets any value outside 85.051129   
   back to exactly 85.051129 so that the function cannot produce any   
   infinities by accident.   
      
   #include    
   #include    
      
   /* gcc -lm -o merc.exe merc.c */   
      
   int main(void) {   
       double lat, lon;   
       const double PI = 3.14159265358979323846;   
      
       while (1) {   
           printf("\nEnter Latitude in degrees: ");   
           if (scanf("%lf", &lat) != 1) break;   
      
           printf("Enter Longitude in degrees: ");   
           if (scanf("%lf", &lon) != 1) break;   
      
           /* Clamp latitude to Mercator limits (approximately 85.05113 degrees)   
   */   
           if (lat > 85.05113) lat = 85.05113;   
           if (lat < -85.05113) lat = -85.05113;   
      
           /* Convert longitude to X in [0,1] */   
           double x = (lon + 180.0) / 360.0;   
      
           /* Convert latitude to Y in [0,1] using Mercator projection */   
           /* the C log function is an ln() below, the C log10 function would be   
   a log() */   
           double rad = lat * PI / 180.0;   
           double y = 0.5 - (log(tan(PI/4.0 + rad/2.0)) / (2.0 * PI));   
      
           printf("Your screen data point is at X=%f  Y=%f\n", x, y);   
       }   
      
       return 0;   
   }   
      
   My conclusion from examining the pattern, is   
      
      (0,0) is on the upper left of the unit square   
      and Y is going down the page. That makes -80 near the   
      bottom.   
      
   Enter Latitude in degrees: -80   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.887741   
      
   Enter Latitude in degrees: -50   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.660855   
      
   Enter Latitude in degrees: -30   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.587425   
      
   Enter Latitude in degrees: 0   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.500000  <=== (0,0) is at (0.5,0.5)   
      
   Enter Latitude in degrees: 30   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.412575   
      
   Enter Latitude in degrees: 50   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.339145   
      
   Enter Latitude in degrees: 80   
   Enter Longitude in degrees: 0   
   Your screen data point is at X=0.500000  Y=0.112259   
      
      Paul   
      
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