home bbs files messages ]

Forums before death by AOL, social media and spammers... "We can't have nice things"

   alt.comp.os.windows-xp      Actually wasn't too bad for a M$-OS      17,273 messages   

[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]

   Message 17,156 of 17,273   
   R.Wieser to All   
   Re: Worldmap mercator projection - Latit   
   24 Jan 26 18:10:40   
   
   XPost: alt.windows7.general, alt.comp.os.windows-10   
   From: address@is.invalid   
      
   John,   
      
   > It makes it easier to understand what's going on if done in these   
   > three stages.   
      
   It is, and its why I tried to convert Pauls CoPilot formule into one which   
   would returne a +1 ... -1 result (and made a mistake I only much later   
   recognised).   
      
   > 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,   
      
   I started with the image just above it (as mentioned in my initial post).   
      
   And thats another problem (which I ignored for the time being) : which of   
   those two where either formule for ?   Can you tell ?   I definitily can't.   
      
   > 0 degrees latitude and longitude _is_ at its centre.   
      
   For /that/ image, yes.   
      
   I don't think you noticed, but it and the one above it have a border around   
   the map-image itself, throwing the precision off a bit.  And imagine a map   
   with un-equally sized borders.  And yes, I encountered those too. Then there   
   are the maps I mentioned earlier, which could be in a Mercator projection,   
   but not in a +85 to -85 latitude range.   
      
   > 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.   
      
   You mis-understood : I've seen maps where the top of the image is at about   
   +75 latitude (just above Russia), and the bottom at about -60 latitude (just   
   below south america).  Meaning that *none* of the south-pole ice is visible   
   on the map, and the ice-plate at +45 longitude is pretty-much cut in half.   
      
   iow, the equator of such an image is /definitily not/ at half the height of   
   the image.   
      
   I've also found a few maps in which 0 degrees longitude was *not* in the   
   horizontal center of the image (its left started at about -170 degrees   
   longitude, just between the russian peninsula and canada).   
      
   >> 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_.   
      
   Bothering about /how/ something works is only a good use of time and energy   
   *after* you make sure /that/ it works.   
      
   > 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.   
      
   Sigh.  I already gave the latitude of Washington DC.  Multiple times even.   
   In my initial post I already provided the result and mentioned that it   
   pointed somewhere into canada.   
      
   But it doesn't really matter which value is plugged into the formula (as   
   long as its a valid one ofcourse).  The only thing that I needed was someone   
   who would repeat the calculations I provided and compare the result with the   
   one I also gave.   
      
   I can give you the results for +85 and -85 latitude - which will ofcourse   
   mirror each other - but those values will be of zero use to anyone - other   
   than to be able to tell me that those results are wrong.  And thats   
   something I already know.   
      
   Do you still want them, now you know that they would be of no real use to   
   you ?   
      
   One thing bothers me though : why would I need to provide results for +180   
   and -180 longitudes too ?  The longitude isn't any part of the formule.   
      
   Regards,   
   Rudy Wieser   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]


(c) 1994,  bbs@darkrealms.ca