Blackey Cole Posted August 3, 2017 Share Posted August 3, 2017 I have several lasers from several sources And a few targets for them. Problem is they are roughly 1.5" in size if I recall correctly. I'm wonder at what distance to set these so that it is equivalent of a 16' target at 15 yards.then would also need a conversion of a foot at same distance for spacing targets I guess I could calculate that myself if I had too.buti tried so math I thought I remembered from algebra but I couldn't done it c orrectlu as it wasalmst touching the barrel. Link to comment Share on other sites More sharing options...
Doc Shapiro Posted August 3, 2017 Share Posted August 3, 2017 Just off the top of my head, 5 feet should be close. Might have to bring them in even more. Link to comment Share on other sites More sharing options...
Cheyenne Ranger, 48747L Posted August 3, 2017 Share Posted August 3, 2017 did it using cross products 1.5" = 16ft X is distance to target for 1.5" target X 15 yds = sign should be in middle but can't get it to do that on computer convert everything to inches You can say this as, " 1.5 is to X as 192 is to 540" 1.5 = 192 X 540 now we're looking at oranges vs. oranges cross multiply 192X = 1.5 x 540 capital X is unknown, don't get lower case x (multiply operation confused) multiply right side together 192X = 810 divide both sides by 192 192X = 810 192 192 results with X = 4.2 inches as we converted everything to inches a couple of steps ago if you what to check plug in 4.2 to equation where everything is converted to inches 1.5 = 192 4.2 540 divide numerator (top) by denominator (bottom) or other way around just so long as you do the same on both sides: left side gives you .357 right side = .355 these numbers are arbitrary it is just a check; if both come out same (or close in our case as we rounded to nearest 1/1000) then 4.2 inches is correct which means 1.5" targets need to be 4.2" from barrel of your gun Link to comment Share on other sites More sharing options...
Cheyenne Ranger, 48747L Posted August 3, 2017 Share Posted August 3, 2017 of course question I have is the target really 16' (feet) in size???? Link to comment Share on other sites More sharing options...
Cheyenne Ranger, 48747L Posted August 3, 2017 Share Posted August 3, 2017 if is really 16" (inches) then the math works out to 50.6" or rounding it, to 4' 3" (4 ft 3 inches) which shows us Doc did a great job of estimating Link to comment Share on other sites More sharing options...
Shooting Bull Posted August 3, 2017 Share Posted August 3, 2017 7 hours ago, Blackey Cole said: a 16' target And Creeker brags about big targets. He's got nothing on you. Link to comment Share on other sites More sharing options...
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