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Question about wimd meters....
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<blockquote data-quote="Guest" data-source="post: 116897"><p>[ QUOTE ]</p><p><strong>Why can't I take a reading 90 degrees to my rifle and use that number? Isn't this the wind that my bullet will see? Why do i care what the full wind is unless that happens to be at 90 degrees?edge. </strong></p><p></p><p>[/ QUOTE ]</p><p></p><p>I think you mean Zero degrees - you want the *** component of the wind. The question is, will keeping the wind meter parallel with my *** give me the *** component (ie, the COS(alpha) of true wind speed). It depends on the geometry of the wind meter. On a unit like my </p><p>-</p><p> <img src="http://lib.store.yahoo.net/lib/giftngadget/meteoss.jpg" alt="" class="fr-fic fr-dii fr-draggable " style="" /> </p><p> <a href="http://www.ambientweather.com/skmewime.html" target="_blank"><strong>Skywatch Meteos</strong></a> that's Omnidirectional, you obviously can't get the *** component. I'm guessing most wind meters have too much turbulence and would read load in comparison to the cos(alpha).</p><p></p><p>I think I'll add the -</p><p> <img src="http://lib.store.yahoo.net/lib/giftngadget/KestrelL.jpg" alt="" class="fr-fic fr-dii fr-draggable " style="" /></p><p>-</p><p> <a href="http://www.ambientweather.com/ke25nivipowe.html" target="_blank"><strong> Kestrel 2500NV Night Vision </strong></a> to my wind bag collection. I'll be adding the <a href="http://www.ambientweather.com/windtunnel.html" target="_blank"><strong>Speedtech Wind Tunnel</strong></a> </p><p></p><p>-</p><p> <img src="http://us.st11.yimg.com/us.st.yimg.com/I/yhst-37697109791737_1904_181070502.jpg" alt="" class="fr-fic fr-dii fr-draggable " style="" /> </p><p>-- </p><p>( Meets NCAA Track and Field requirements and provides precise wind velocity in a specific direction for precision measurements. ) to test this conjecture.</p><p></p><p>BTW, the *** wind component at 45 degrees is not 1/2, but sqrt(2)/2 or aprox .707</p></blockquote><p></p>
[QUOTE="Guest, post: 116897"] [ QUOTE ] [b]Why can't I take a reading 90 degrees to my rifle and use that number? Isn't this the wind that my bullet will see? Why do i care what the full wind is unless that happens to be at 90 degrees?edge. [/b] [/ QUOTE ] I think you mean Zero degrees - you want the *** component of the wind. The question is, will keeping the wind meter parallel with my *** give me the *** component (ie, the COS(alpha) of true wind speed). It depends on the geometry of the wind meter. On a unit like my - [img]http://lib.store.yahoo.net/lib/giftngadget/meteoss.jpg[/img] [url="http://www.ambientweather.com/skmewime.html"][b]Skywatch Meteos[/b][/url] that's Omnidirectional, you obviously can't get the *** component. I'm guessing most wind meters have too much turbulence and would read load in comparison to the cos(alpha). I think I'll add the - [img]http://lib.store.yahoo.net/lib/giftngadget/KestrelL.jpg[/img] - [url="http://www.ambientweather.com/ke25nivipowe.html"][b] Kestrel 2500NV Night Vision [/b][/url] to my wind bag collection. I'll be adding the [url="http://www.ambientweather.com/windtunnel.html"][b]Speedtech Wind Tunnel[/b][/url] - [img]http://us.st11.yimg.com/us.st.yimg.com/I/yhst-37697109791737_1904_181070502.jpg[/img] -- ( Meets NCAA Track and Field requirements and provides precise wind velocity in a specific direction for precision measurements. ) to test this conjecture. BTW, the *** wind component at 45 degrees is not 1/2, but sqrt(2)/2 or aprox .707 [/QUOTE]
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