Shot dripping arithmetic, part 2
This is from the last thread.
quote:Baubun says:
Sorry, your math stinks," Or one hole 0.021785" in diameter (equals the area of 15 holes of 0.0225" diameter)." Your answer is smaller than any one of the 15 holes.
You're right. My mistake, and this is what I get from transcribing numbers back and forth from a crappy web-based calculator without double-checking them. [:(]
As a simple ball park answer, lets consider 16x instead of 15x. To get one hole with 16x the area of a smaller hole, the larger hole would need to be exactly 4x the diameter of the smaller hole. For a smaller 0.0225" hole, that would mean one larger hole just under around 0.09" in diameter. For a 15X hole, it would need to be a little bit less in diameter than the 16x hole.
For the exact math, lets try this again:
-One hole 0.0225 inches in diameter is 0.01125 inches in radius. It has an area of (Pi x 0.0001265) square inches.
-Fifteen holes of that area have a combined area of (15 x Pi x 0.0001265) square inches, or Pi x 0.0018975 square inches.
-If you had ONE hole of Pi x 0.0018975 square inches AREA, its RADIUS would be the square root of 0.0018975, or 0.04356 inches.
-The DIAMETER of said hole would be approximately 0.08712 inches. . .which is what TSR1965 said in the last thread on this, and close to my approximation above.
Again, the problem here is that the rate of flow of a liquid through a hole is *NOT* exactly proportionate to the area of the hole because of friction and turbulence effects. It might be pretty close with certain frictionless "superfluids" like liquid helium, but molten lead alloy isn't going to qualify.
Also, with respect to shot making in general, you don't really NEED to drip the shot through one big hole, then a number of smaller ones. . .just the smaller ones should do.
EG:
http://www.littletonshotmaker.com/
http://www.svartkrutt.net/articles/vis.php?id=8
quote:Baubun says:
Sorry, your math stinks," Or one hole 0.021785" in diameter (equals the area of 15 holes of 0.0225" diameter)." Your answer is smaller than any one of the 15 holes.
You're right. My mistake, and this is what I get from transcribing numbers back and forth from a crappy web-based calculator without double-checking them. [:(]
As a simple ball park answer, lets consider 16x instead of 15x. To get one hole with 16x the area of a smaller hole, the larger hole would need to be exactly 4x the diameter of the smaller hole. For a smaller 0.0225" hole, that would mean one larger hole just under around 0.09" in diameter. For a 15X hole, it would need to be a little bit less in diameter than the 16x hole.
For the exact math, lets try this again:
-One hole 0.0225 inches in diameter is 0.01125 inches in radius. It has an area of (Pi x 0.0001265) square inches.
-Fifteen holes of that area have a combined area of (15 x Pi x 0.0001265) square inches, or Pi x 0.0018975 square inches.
-If you had ONE hole of Pi x 0.0018975 square inches AREA, its RADIUS would be the square root of 0.0018975, or 0.04356 inches.
-The DIAMETER of said hole would be approximately 0.08712 inches. . .which is what TSR1965 said in the last thread on this, and close to my approximation above.
Again, the problem here is that the rate of flow of a liquid through a hole is *NOT* exactly proportionate to the area of the hole because of friction and turbulence effects. It might be pretty close with certain frictionless "superfluids" like liquid helium, but molten lead alloy isn't going to qualify.
Also, with respect to shot making in general, you don't really NEED to drip the shot through one big hole, then a number of smaller ones. . .just the smaller ones should do.
EG:
http://www.littletonshotmaker.com/
http://www.svartkrutt.net/articles/vis.php?id=8
0
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That's much better !!!! I deal with simple math like that all day long,
and I NEVER make misstakes....[:)]0
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