Re: Worldmap mercator projection - Latitude to Y calculation. - last try

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_Mercator_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