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	<title>Comments on: World file calculator</title>
	<atom:link href="http://egb13.net/2009/03/worldfile-calculator/feed/" rel="self" type="application/rss+xml" />
	<link>http://egb13.net/2009/03/worldfile-calculator/</link>
	<description>Time-lapse, weather, maps, and more</description>
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		<title>By: John Reed</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-12988</link>
		<dc:creator>John Reed</dc:creator>
		<pubDate>Thu, 02 Feb 2012 06:12:48 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-12988</guid>
		<description>WONDERFUL Post.thanks for share..more wait ..</description>
		<content:encoded><![CDATA[<p>WONDERFUL Post.thanks for share..more wait ..</p>
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		<title>By: RMFRunyan</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-2391</link>
		<dc:creator>RMFRunyan</dc:creator>
		<pubDate>Wed, 20 Oct 2010 22:30:04 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-2391</guid>
		<description>I&#039;ve figured out what I am doing wrong.  The lats should be 1/4th the polar circ, not half.

Thanks for a very nice reference.  Bookmarked.

rmfr</description>
		<content:encoded><![CDATA[<p>I&#8217;ve figured out what I am doing wrong.  The lats should be 1/4th the polar circ, not half.</p>
<p>Thanks for a very nice reference.  Bookmarked.</p>
<p>rmfr</p>
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	<item>
		<title>By: RMFRunyan</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-2381</link>
		<dc:creator>RMFRunyan</dc:creator>
		<pubDate>Wed, 20 Oct 2010 20:30:39 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-2381</guid>
		<description>OK.  You&#039;re right.  I&#039;m just always used to listing South and West as negatives even if listed South and West.

However, line 4 still seems to be twice what it should.
7143.334479058464
0.00000
0.00000
-14272.22052182727
-25716004.124610472
25689996.939289085

Shouldn&#039;t it be: -7136.110260913635?

BTW: the 7201pxW by 3601pxH is Equirectangular Projection.  Thus, the height covers only 180 degrees, but the width will cover 360 degrees.  Does the JS script assume height also covers 360 degrees?

Thanks for the prompt reply.

rmfr</description>
		<content:encoded><![CDATA[<p>OK.  You&#8217;re right.  I&#8217;m just always used to listing South and West as negatives even if listed South and West.</p>
<p>However, line 4 still seems to be twice what it should.<br />
7143.334479058464<br />
0.00000<br />
0.00000<br />
-14272.22052182727<br />
-25716004.124610472<br />
25689996.939289085</p>
<p>Shouldn&#8217;t it be: -7136.110260913635?</p>
<p>BTW: the 7201pxW by 3601pxH is Equirectangular Projection.  Thus, the height covers only 180 degrees, but the width will cover 360 degrees.  Does the JS script assume height also covers 360 degrees?</p>
<p>Thanks for the prompt reply.</p>
<p>rmfr</p>
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	<item>
		<title>By: egb13</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-2380</link>
		<dc:creator>egb13</dc:creator>
		<pubDate>Wed, 20 Oct 2010 18:46:23 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-2380</guid>
		<description>Well, I see for one thing that the negatives are confounding it; &quot;West&quot; means negative East, so you should enter the absolute value instead of a negative, unless you make the direction East as well.  Likewise for N/S.  If I remove the negative signs the results are closer to what you expect:

7143.334479058464
0.00000
0.00000
-14272.22052182727
-25716004.124610472
25689996.939289086

The fourth value, delta-y, still doesn&#039;t agree.  Are you sure of your input and expected output?  E.g., is the image height correct?

It&#039;s a simple-minded bit of code that works for what I needed.  The code can be viewed at &lt;a href=&quot;http://egb13.net/jscripts/worldfile.js&quot; rel=&quot;nofollow&quot;&gt;http://egb13.net/jscripts/worldfile.js&lt;/a&gt;.  Any suggestions for improvements are appreciated.  One thing that would come in handy sometimes would be to let you enter the pixel size, and have it give you the resulting image dimensions, for when I want to crop a map image to a specific extent, e.g.</description>
		<content:encoded><![CDATA[<p>Well, I see for one thing that the negatives are confounding it; &#8220;West&#8221; means negative East, so you should enter the absolute value instead of a negative, unless you make the direction East as well.  Likewise for N/S.  If I remove the negative signs the results are closer to what you expect:</p>
<p>7143.334479058464<br />
0.00000<br />
0.00000<br />
-14272.22052182727<br />
-25716004.124610472<br />
25689996.939289086</p>
<p>The fourth value, delta-y, still doesn&#8217;t agree.  Are you sure of your input and expected output?  E.g., is the image height correct?</p>
<p>It&#8217;s a simple-minded bit of code that works for what I needed.  The code can be viewed at <a href="http://egb13.net/jscripts/worldfile.js" rel="nofollow">http://egb13.net/jscripts/worldfile.js</a>.  Any suggestions for improvements are appreciated.  One thing that would come in handy sometimes would be to let you enter the pixel size, and have it give you the resulting image dimensions, for when I want to crop a map image to a specific extent, e.g.</p>
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	<item>
		<title>By: RMFRunyan</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-2378</link>
		<dc:creator>RMFRunyan</dc:creator>
		<pubDate>Wed, 20 Oct 2010 18:03:43 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-2378</guid>
		<description>This program does not truly work.  When I enter the below data:
N: 25697133.04955
W: -25719575.79185
S: -25697133.04955
E: 25719575.79185
width: 7201
height: 3601

I get these results:
0
0.00000
0.00000
0
25719575.79185
25697133.04955

However, should the results actually have been:
7143.334479
0.00000
0.00000
-7136.110261
-25716004.12461
25693564.99442

Just wondering...

rmfr</description>
		<content:encoded><![CDATA[<p>This program does not truly work.  When I enter the below data:<br />
N: 25697133.04955<br />
W: -25719575.79185<br />
S: -25697133.04955<br />
E: 25719575.79185<br />
width: 7201<br />
height: 3601</p>
<p>I get these results:<br />
0<br />
0.00000<br />
0.00000<br />
0<br />
25719575.79185<br />
25697133.04955</p>
<p>However, should the results actually have been:<br />
7143.334479<br />
0.00000<br />
0.00000<br />
-7136.110261<br />
-25716004.12461<br />
25693564.99442</p>
<p>Just wondering&#8230;</p>
<p>rmfr</p>
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	<item>
		<title>By: egb13</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-56</link>
		<dc:creator>egb13</dc:creator>
		<pubDate>Thu, 15 Apr 2010 15:28:42 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-56</guid>
		<description>Is this program available for download?  The JavaScript code for it can be found at http://egb13.net/jscripts/worldfile.js.  It&#039;s just a couple lines of arithmetic.</description>
		<content:encoded><![CDATA[<p>Is this program available for download?  The JavaScript code for it can be found at <a href="http://egb13.net/jscripts/worldfile.js" rel="nofollow">http://egb13.net/jscripts/worldfile.js</a>.  It&#8217;s just a couple lines of arithmetic.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Donald</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-52</link>
		<dc:creator>Donald</dc:creator>
		<pubDate>Fri, 26 Mar 2010 15:14:59 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-52</guid>
		<description>Is this program available for download?</description>
		<content:encoded><![CDATA[<p>Is this program available for download?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: egb13</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-23</link>
		<dc:creator>egb13</dc:creator>
		<pubDate>Thu, 26 Nov 2009 19:48:47 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-23</guid>
		<description>Replying to myself...  my response to NickMcW is for Google Maps, but he asked about Google Earth.  It&#039;s a more interesting question, but not one I have an answer to.  Yet.</description>
		<content:encoded><![CDATA[<p>Replying to myself&#8230;  my response to NickMcW is for Google Maps, but he asked about Google Earth.  It&#8217;s a more interesting question, but not one I have an answer to.  Yet.</p>
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	<item>
		<title>By: egb13</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-22</link>
		<dc:creator>egb13</dc:creator>
		<pubDate>Thu, 26 Nov 2009 18:41:02 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-22</guid>
		<description>If in Google Maps you click on &quot;Link&quot; you get a choice two strings to copy and paste.  The first one, for email or IM, is the bare URL of the displayed map; if you past it in the browser&#039;s address box, you&#039;ll get the map you see displayed.  That Maps URL looks like this:

    http://maps.google.com/?ie=UTF8&amp;ll=36.779042,-103.971863&amp;spn=0.370679,0.801315&amp;z=11

The &quot;ll=&quot; parameter is the latitude and longitude of the center of the map.  The &quot;spn=&quot; parameter must stand for &quot;span&quot;, because it appears to be the height and width of the displayed map in degrees.  Test this by trying to find the coordinates of the upper-left corner: add half the vertical span to the latitude and subtract half the horizontal span from the longitude.  (Moving northwest increases latitude and decreases longitude.)  Applying that formula to the URL above gives this new URL:

    http://maps.google.com/?ie=UTF8&amp;ll=36.964381,-104.372520&amp;spn=0.370679,0.801315&amp;z=11

That indeed looks like the upper-left corner of the original map, and what was at the center (an extinct cinder-cone volcano that you can drive to the top of and hike through the crater) is now at the lower-right corner.

Since you&#039;re already given the degree span, you don&#039;t really need this calculator; divide the vertical and horizontal span values given in the link URL by the pixel dimensions of the image and you have the 1st and 4th parameters for the worldfile.

But I don&#039;t know what projection Google uses to display their maps.  Is it simple latitude x longitude (EPSG:4326)?  Or is it a UTM projection?  Or does it vary by the size of the area covered?  UTM is not good for maps that cover more than a certain distance horizontally.  EPSG:4326 is not good for lots of things.  That&#039;s what is used in the video in my Road Trip! posting, and even though that doesn&#039;t cover a terribly large area (a little over 100 mi horizontally) and the counties shown are noticeably flatter than they should be.  And a worldfile (at least with the 2nd &amp; 3rd parameters zero) is not good for &quot;curved&quot; projections.

Might this be the wrong approach, though?  If you haven&#039;t already investigated Google&#039;s Maps API you might want to.  I see lots of places around the internet making pretty good use of it.  It&#039;s been a very long time since I dabbled in it and am unable to offer any help with that, not to mention that I don&#039;t know what exactly your after anyway.</description>
		<content:encoded><![CDATA[<p>If in Google Maps you click on &#8220;Link&#8221; you get a choice two strings to copy and paste.  The first one, for email or IM, is the bare URL of the displayed map; if you past it in the browser&#8217;s address box, you&#8217;ll get the map you see displayed.  That Maps URL looks like this:</p>
<p>    <a href="http://maps.google.com/?ie=UTF8&#038;ll=36.779042,-103.971863&#038;spn=0.370679,0.801315&#038;z=11" rel="nofollow">http://maps.google.com/?ie=UTF8&#038;ll=36.779042,-103.971863&#038;spn=0.370679,0.801315&#038;z=11</a></p>
<p>The &#8220;ll=&#8221; parameter is the latitude and longitude of the center of the map.  The &#8220;spn=&#8221; parameter must stand for &#8220;span&#8221;, because it appears to be the height and width of the displayed map in degrees.  Test this by trying to find the coordinates of the upper-left corner: add half the vertical span to the latitude and subtract half the horizontal span from the longitude.  (Moving northwest increases latitude and decreases longitude.)  Applying that formula to the URL above gives this new URL:</p>
<p>    <a href="http://maps.google.com/?ie=UTF8&#038;ll=36.964381,-104.372520&#038;spn=0.370679,0.801315&#038;z=11" rel="nofollow">http://maps.google.com/?ie=UTF8&#038;ll=36.964381,-104.372520&#038;spn=0.370679,0.801315&#038;z=11</a></p>
<p>That indeed looks like the upper-left corner of the original map, and what was at the center (an extinct cinder-cone volcano that you can drive to the top of and hike through the crater) is now at the lower-right corner.</p>
<p>Since you&#8217;re already given the degree span, you don&#8217;t really need this calculator; divide the vertical and horizontal span values given in the link URL by the pixel dimensions of the image and you have the 1st and 4th parameters for the worldfile.</p>
<p>But I don&#8217;t know what projection Google uses to display their maps.  Is it simple latitude x longitude (EPSG:4326)?  Or is it a UTM projection?  Or does it vary by the size of the area covered?  UTM is not good for maps that cover more than a certain distance horizontally.  EPSG:4326 is not good for lots of things.  That&#8217;s what is used in the video in my Road Trip! posting, and even though that doesn&#8217;t cover a terribly large area (a little over 100 mi horizontally) and the counties shown are noticeably flatter than they should be.  And a worldfile (at least with the 2nd &#038; 3rd parameters zero) is not good for &#8220;curved&#8221; projections.</p>
<p>Might this be the wrong approach, though?  If you haven&#8217;t already investigated Google&#8217;s Maps API you might want to.  I see lots of places around the internet making pretty good use of it.  It&#8217;s been a very long time since I dabbled in it and am unable to offer any help with that, not to mention that I don&#8217;t know what exactly your after anyway.</p>
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	<item>
		<title>By: NickMcW</title>
		<link>http://egb13.net/2009/03/worldfile-calculator/comment-page-1/#comment-21</link>
		<dc:creator>NickMcW</dc:creator>
		<pubDate>Thu, 26 Nov 2009 16:53:47 +0000</pubDate>
		<guid isPermaLink="false">http://egb13.net/?p=59#comment-21</guid>
		<description>Hi - nice utility, thanks for making it available. It prompts a related question: does anyone know a way to programmatically obtain the corner coordinates of the current view extent in Google Earth? The reason for asking is in order automate the creation of world files - as done with this calculator - for google earth screen-saved images. (The image size, in pixels, will depend on the resolution option chosen in GE; the options available depend on whether the free or pro version is used.)</description>
		<content:encoded><![CDATA[<p>Hi &#8211; nice utility, thanks for making it available. It prompts a related question: does anyone know a way to programmatically obtain the corner coordinates of the current view extent in Google Earth? The reason for asking is in order automate the creation of world files &#8211; as done with this calculator &#8211; for google earth screen-saved images. (The image size, in pixels, will depend on the resolution option chosen in GE; the options available depend on whether the free or pro version is used.)</p>
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