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Posted: 4/28/2006 7:47:58 PM EDT
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I bought a 100 pack of small sillouette targets. They are like the full size targets I think we've all seen, but the sillouette is only about 19 inches tall by about 9 or so inches wide. Made by the National Target Corporation, with an official NRA stamp on it. It also says B-29 which I'm assuming is the actual target 'model'. I'm pretty sure the full size target we shoot at department ranges is a B-27. What I'm wondering is at what ranges will it simulate a man size target at XXX yards. For example, If I put this target at 100 yards, will it represent the relative size of a human at 300 yards? Or would it be 400 yards, or whatever. What if I put it 50 yards, or 200? I'm thinking that if it's a half sized target, at 100 yards it would appear as a 200 yard full size target, and if I put it at 200 yards it will appear as a 400 yard target. Am I right about this? I know there should be a pretty simple mathmatical formula, but I have no idea what it would be. I apologize if I have done a poor job of articulating my question - it's been a long day and my brain is about shot right now.  If anyone can help I would greatly appreciate it. Thanks!-K |
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Here we go, this is one way to do it, if you shoot at a range. Measure the distance from your head to your shoulder, measure the overall width of your head at eye level. Measure your chin to your belly button, measure your torso across your nipples to the outside of your arms. Now that you have that dimension take cardboard and make torso/head targets cut them out and place them on stakes. Take them to the range and place them at the various distances you want to check. Use a Mil Dot reticle set of binoculars or a rifle scope to transfer each distance on a 25 meter target (this will require 2 people) You will then have the dimensions to use OR you can get a US Army M-16A1 25 Meter Target (used for zeroing and Alternate Course of Fire) and get the dimensions from that. You are really better off shooting the targets you made above at the proper distance so you can learn wind judgement and target aquisition IMHO |
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Yes, the scale seems to hold true. At 900", half the target would be 4.5" Thus, making a 30-60-90 triangle. The Tangent of the angle would equal the opposite leg from the angle over the adjacent leg, or 4.5" / 900", or 0.005 By similar triangles, this will equal any other opposite over adjacent if the angles dont change. Thus 0.005= x (our new width) / 3600" (100 yards) and x= 18" This is what we know from the targets intended use, posted at 25M to simulate 100M I used yards as a quick and dirty subsitution for Meters. So, our new triangle has the same opposite leg, 4.5" but an adjacent leg of 3600", then the Tangent of the angle is 0.00125. Agian by similar triangles, 0.00125 = 18" ( our found "actual width" from part 1) / y ( the unknown range that the percieved width equals standard target) . We get and answer of 14400" or 1200' or 400 yards. The 1 to 4 ratio seem to hold. |
Well alrighty then Professor Trig. Thanks!  So if it's a 1:4 ratio, then if I put it at 100 yards, then it will appear as a full size target would at 400 yards. If I post it 50 yards, then it will appear as a full size target at 200 yards. Right? -K |
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