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Rifles, Reloading, Optics, Equipment
Long Range Scopes and Other Optics
Picatinny Rails?
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<blockquote data-quote="Razor18" data-source="post: 858817" data-attributes="member: 11395"><p>Thanks, but my question was not about wiggle room, it was pure geometry of the MIL-STD as shown in every image on th net, but I probably couldn't explain it precise enough.</p><p></p><p>The way it's drawn it can't be machined, cause some of the dimensions seem to be "over-defined", meaning there are 3 specific dimensions where if 2 are made right, the third can't be the value specified.</p><p></p><p> If you look at the drawing I'm sure you can follow. The heart of the rail is in the middle of the rail profile, datum C, which is a rectangle .748 +/- 0.002 wide, and exactly 0.108 high. On both sides, the corners of this datum C are touching 45° lines, these lines intersect in a 0.835 +/- 0.005 width (rail width), without any chamfer.</p><p> </p><p> The vertical sides of datum C and the 2x45° lines form two little triangles on both edges of the rail. The base of those triangles is the side of datum C, which must be 0.108. Perpendicular to this vertical base, the "height" of those triangles (on the image it's actually horizontal, and isn't drawn) from the middle of the base to opposite corner of triangle is obviously half of the difference between datum C width and rail width (0.835 - 0.748) / 2 = 0.0435.</p><p> </p><p> Now in a symmethrical triangle, if the height (which is 0.0435 here) is not equal to half of the base length (0.108/2=0.054 here), then the angle of the triangle sides can not be 2x45°, it's geometrically impossible. So, back to the MIL-STD drawing, if you require datum C to have the drawn dimensions, AND require the side angles to be 2x45°, then rail width can not be 0.835. From those 3 conditions, only 2 can be met at a time.</p><p> </p><p> So which one of the required dimensions must not be met so strict in your opinion to still have a match between rail and any proper mount?</p></blockquote><p></p>
[QUOTE="Razor18, post: 858817, member: 11395"] Thanks, but my question was not about wiggle room, it was pure geometry of the MIL-STD as shown in every image on th net, but I probably couldn't explain it precise enough. The way it's drawn it can't be machined, cause some of the dimensions seem to be "over-defined", meaning there are 3 specific dimensions where if 2 are made right, the third can't be the value specified. If you look at the drawing I'm sure you can follow. The heart of the rail is in the middle of the rail profile, datum C, which is a rectangle .748 +/- 0.002 wide, and exactly 0.108 high. On both sides, the corners of this datum C are touching 45° lines, these lines intersect in a 0.835 +/- 0.005 width (rail width), without any chamfer. The vertical sides of datum C and the 2x45° lines form two little triangles on both edges of the rail. The base of those triangles is the side of datum C, which must be 0.108. Perpendicular to this vertical base, the "height" of those triangles (on the image it's actually horizontal, and isn't drawn) from the middle of the base to opposite corner of triangle is obviously half of the difference between datum C width and rail width (0.835 - 0.748) / 2 = 0.0435. Now in a symmethrical triangle, if the height (which is 0.0435 here) is not equal to half of the base length (0.108/2=0.054 here), then the angle of the triangle sides can not be 2x45°, it's geometrically impossible. So, back to the MIL-STD drawing, if you require datum C to have the drawn dimensions, AND require the side angles to be 2x45°, then rail width can not be 0.835. From those 3 conditions, only 2 can be met at a time. So which one of the required dimensions must not be met so strict in your opinion to still have a match between rail and any proper mount? [/QUOTE]
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