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Lens for my blinking Christmas Lights
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<blockquote data-quote="WayneF" data-source="post: 344116" data-attributes="member: 12496"><p>If staying in your yard means from say 30 feet, then it would take a 12 mm DX lens to cover 60 feet width. That certainly is wide angle, and before, I just meant that 100 feet would not be. At this extreme, you would encounter some lens distortion (field curvature, straight lines on house would show a a little curve, which some software could correct, for example, Adobe Raw).</p><p></p><p>Here is a calculator that will show the numbers:</p><p></p><p><a href="http://www.tawbaware.com/maxlyons/calc.htm" target="_blank">http://www.tawbaware.com/maxlyons/calc.htm</a></p><p></p><p></p><p>4th calculator down "Dimensional field of view". Use 1.5 for DX "Focal length multiplier". You sort of have to enter numbers trial and error to hit your goal.</p><p></p><p>And that is showing DX 3:2 format, and not HDTV 16:9 format, which my notion is, HD will be a little less tall. 16:9 is not far from 2:1, but closer to 60:33.75 feet will fit in the format.</p></blockquote><p></p>
[QUOTE="WayneF, post: 344116, member: 12496"] If staying in your yard means from say 30 feet, then it would take a 12 mm DX lens to cover 60 feet width. That certainly is wide angle, and before, I just meant that 100 feet would not be. At this extreme, you would encounter some lens distortion (field curvature, straight lines on house would show a a little curve, which some software could correct, for example, Adobe Raw). Here is a calculator that will show the numbers: [URL]http://www.tawbaware.com/maxlyons/calc.htm[/URL] 4th calculator down "Dimensional field of view". Use 1.5 for DX "Focal length multiplier". You sort of have to enter numbers trial and error to hit your goal. And that is showing DX 3:2 format, and not HDTV 16:9 format, which my notion is, HD will be a little less tall. 16:9 is not far from 2:1, but closer to 60:33.75 feet will fit in the format. [/QUOTE]
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Lens for my blinking Christmas Lights
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