<div dir="ltr">Or better yet: <a href="http://www.movable-type.co.uk/scripts/latlong.html">http://www.movable-type.co.uk/scripts/latlong.html</a><div><br></div><div style>Look for: <b>Destination point given distance and bearing from start point</b></div>
</div><div class="gmail_extra"><br><br><div class="gmail_quote">On Fri, May 31, 2013 at 3:17 PM, Wally Atkins <span dir="ltr"><<a href="mailto:wallyatkins@gmail.com" target="_blank">wallyatkins@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Maybe this would do the trick for what you need?<div><br></div><div><span style="font-family:Arial,Helvetica,sans-serif;font-size:13px">function destination(lat, lng, dist, heading) {</span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px">
<span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> lat *= Math.PI / 180;</span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px"><span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> lng *= Math.PI / 180;</span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px">
<span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> heading *= Math.PI / 180;</span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px"><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px">
<span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> var lat2 = Math.asin(Math.sin(lat) * Math.cos(dist) + </span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px"><span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> Math.cos(lat) * Math.sin(dist) * Math.cos(heading));</span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px">
<span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> return [180 / Math.PI * lat2, 180 / Math.PI * (lng + Math.atan2(Math.sin(heading) * Math.sin(dist) * Math.cos(lat2), Math.cos(dist) - Math.sin(lat) * Math.sin(lat2)))];</span><br style="font-family:Arial,Helvetica,sans-serif;font-size:13px">
<span style="font-family:Arial,Helvetica,sans-serif;font-size:13px"> }</span><br></div><div><br></div><div>From: <a href="https://groups.google.com/forum/?fromgroups#!topic/google-earth-browser-plugin/s1A_kizmolU" target="_blank">https://groups.google.com/forum/?fromgroups#!topic/google-earth-browser-plugin/s1A_kizmolU</a></div>
</div><div class="HOEnZb"><div class="h5"><div class="gmail_extra"><br><br><div class="gmail_quote">On Fri, May 31, 2013 at 2:02 PM, <a href="mailto:mail.roliveira@gmail.com" target="_blank">mail.roliveira@gmail.com</a> <span dir="ltr"><<a href="mailto:mail.roliveira@gmail.com" target="_blank">mail.roliveira@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Thanks, I will have a look.<div><br></div><div><br></div><div>The move(x, y) method of OpenLayers.Geometry.Point may be helpful. But I don't understand the x, y arguments... What are the units of those values?</div>
<div><br></div><div>this could help if I move a clone of pointA.<div><br><br><br><div>Sent from Alto - <a href="http://altomail.com" target="_blank">altomail.com</a></div><br><br><br></div><hr style="border:0;min-height:1px;color:#999;background-color:#999;width:100%;margin:0 0 9px 0;padding:0">
<b>From: </b><a href="mailto:wallyatkins@gmail.com" target="_blank">wallyatkins@gmail.com</a><<a href="mailto:wallyatkins@gmail.com" target="_blank">wallyatkins@gmail.com</a>><br><b>To: </b><a href="mailto:mail.roliveira@gmail.com" target="_blank">mail.roliveira@gmail.com</a><<a href="mailto:mail.roliveira@gmail.com" target="_blank">mail.roliveira@gmail.com</a>><br>
<b>cc: </b><a href="mailto:openlayers-users@lists.osgeo.org" target="_blank">openlayers-users@lists.osgeo.org</a><<a href="mailto:openlayers-users@lists.osgeo.org" target="_blank">openlayers-users@lists.osgeo.org</a>><br>
<b>Sent: </b>Friday, May 31, 2013<br><b>Subject: </b>Re: [OpenLayers-Users] Draw a line knowing its length in meters e it's angle<div><div><br><br>Maybe look into the JSTS?<div><br></div><div><span></span><a href="https://github.com/bjornharrtell/jsts" target="_blank">https://github.com/bjornharrtell/jsts</a><br>
<br>On Friday, May 31, 2013, <a href="mailto:mail.roliveira@gmail.com" target="_blank">mail.roliveira@gmail.com</a> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">How do I draw a line, starting from a known point, and knowing only it's length in meters and it's angle?<div>
<br>
</div><div>Suppose I have a point at:</div><div><br></div><div>PointA:</div><div><div><span>lat: </span>-22.29417905166, lon: 19.022827148438</div><div>(EPSG:4326)</div><div><br></div><div>Now I want to draw a line 500 meters long with an angle of 15 degrees (0 degrees is an horizontal line).</div>
<div><br></div><div>OpenLayers.Geometry.LineString allows to draw only a point from two points...</div><div><br></div><div>Any ideas?</div><div><br><br><div>Sent from Alto - <a href="http://altomail.com" target="_blank">altomail.com</a></div>
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