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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 17,156 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)    |
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