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Swaging Bullets: Core length by diameter CalculationsIt pains me to see people struggle to get answers to simple questions. I see several common questions associated with bullet swaging. One of them is some form of this: "How long do I need my core to be if I want it to weigh "X" grains?"
The formulas for obtaining the core length are not difficult to deal with, and I suspect that most folks 'get there' PDQ. Nonetheless, I thought I'd offer the following "condensation" for those that would like the 'thing' condensed into the most succinct, and hopefully useful form. Skip to the PS if you don't care about "how he got there". Cores are cylinders. The formula for the volume of a cylinder is: V = pi * r2 * h Where: "V" = the Volume, pi = 3.14, r2 = the radius of the core multiplied times itself, and h= the height or length of the core - the value we are looking for. Since it's "h" we are looking for, we rearrange the equation to: h = V / (pi * r2). In words: the height of the cylinder (our core) equals the volume of the core divided by pi times the radius of the core times itself. Another value is needed - the density of lead - to complete this "condensation". (I'll call that "d".) The theoretical density of pure lead is 11.34 GRAMS per cubic CENTIMETER. Since most folks want their core dimensions in inches and weight in grains, the density of lead in GRAINS PER CUBIC INCH is 2,867.35 One can get the mass (weight, here on earth) of some solid if one knows its density. The formula is: m = V * d Where: "m" = the weight we want the core to be, "d" = 2,867.35 grains per cubic inch. "V" = the volume of the core just like in the above equation, AND THAT IS IMPORTANT because it allows us to combine the equations into ONE, Since m = V * d, V = m/d. Therefore, we can substitute "m/d" into the FIRST equation (the volume of a cylinder) which becomes: h = (m/d)/(pi * r2) In words: The height of the core equals the weight of the core divided by the density of lead divided by pi times the radius of the core times itself. While that might not look "simple", stick with me a minute more and I think it will be. There are some constants and 'knowns' that are going to simplify matters. We 'know' what we want our core to weigh, so we 'know' "m". We 'know' what the value of pi is. We 'know' what the diameter of our core is, therefore we know what its radius is. We 'know' the density of lead. So... Here's an example: I swage 8mm bullets. The cores I use are 0.250" in diameter, making their radius 0.125". That value times itself is 0.125 * 0.125 = 0.015625 inches. That's r2 FOREVER for a cylinder that is 0.250" in diameter. Therefore, calculated ONCE forever for my cores, that value NEVER changes. My jackets weigh ~43 grains. I want a final weight of ~125 grains. Therefore, I want the core to weigh 125 grains minus 43 grains = 82 grains. That's my "m" for core for a 125-grain bullet. The density of lead in grains per cubic inch is 2,867.346. Substituting into the equation what we know, we get: h = (82/2867.346)/(3.14*0.015625), Therefore, our core length should be 0.58 inches. If I want to change the weight of the bullet, the ONLY number I have to change in the equation is "m", in this case the 82. The other numbers are CONSTANT - FOREVER for my bullets with 0.250" cores. "Doing the algebra", the density of lead, pi, and the radius of the core times itself can all be combined into ONE NUMBER: 2867.346 * 3.14 * 0.015625 = 140.75. THEREFORE, for MY 8mm bullet cores, all I have to do to find the length of a core for a specific weight finished bullet is divide the core weight I want by 140.75. For example, let's say I want a finished bullet weight of 150 grains. The core would need to weigh 150 - 43 = 107 grains. 107 divided by 140.75 = 0.76. Therefore, the core length necessary for a 150-grain bullet would be 0.76 inches long. Let me take one more moment and show the difference for a different core diameter (caliber). Let's say the core material for a .224 bullet is 0.125" in diameter. (1/8th inch lead wire.) The radius is 0.0625 inches. That value times itself is 0.00390 square inches. Let's say the jackets weigh 20 grains and I want the finished bullet to weight 55 grains. The core then needs to weigh 35 grains. Once again, 2867.346 * 3.14 * 0.00390 equals our "CONSTANT" for the .224 bullet with a 0.125 core. That number is 36.41. Therefore, dividing 35 grains (the weight we want our core to be) by 36.41, we get a core length of 0.96 inches. Before anyone howls too loudly, I don't swage .22 caliber bullets, so the above weight of the jacket, (20 grains) and the core diameter (0.125") are just WAGS on my part to illustrate how simple it is to go from one core diameter to another. I HOPE this has been helpful/useful. That was the intent. Paul PS In an attempt to reduce the number of operations to the minimum, you don't even have to find the radius. To get THE Number, just square the diameter and multiply by 2252.009. So the steps reduce to three: 1) Square the diameter of core material, (the more precise the measure of the diameter of the core material, the better the estimate of length will be), 2) Multiply by 2252.01, SAVE RESULT 3) Divide desired core weight by the result of #2. The answer to that division is the length of the core for the weight you want it to be. Therefore, all you really need to determine the length of a LEAD core of a given bullet weight is the diameter of the core and the weight you want it to be. (If you don't have a "square" key on your four-function calculator, you can divide the DIAMETER by 16. The result is the same as squaring the RADIUS.) Paul |

Re: Swaging Bullets: Core length by diameter CalculationsI decided that if I wanted to truly HELP, I would just "do the math" and provide the results in a table here. I found a guy on Ebay that is selling lead wire for bullet swaging. The smallest diameter he sells is 0.180" and the largest is 0.475". I'll run a table from 0.180 to 0.500. I can't get all of them "snipped" into one table so there are 10 separate ones. Here ya go:http://i122.photobucket.com/albums/o...pst537empg.jpg
http://i122.photobucket.com/albums/o...psvzacf9wg.jpg http://i122.photobucket.com/albums/o...psvrsrm0aq.jpg http://i122.photobucket.com/albums/o...pstadwx5sv.jpg http://i122.photobucket.com/albums/o...psjidncumi.jpg http://i122.photobucket.com/albums/o...ps7j6k5fcp.jpg http://i122.photobucket.com/albums/o...psovbncqma.jpg http://i122.photobucket.com/albums/o...psjz111brg.jpg http://i122.photobucket.com/albums/o...psbmrmf6ob.jpg http://i122.photobucket.com/albums/o...psdeepjv0y.jpg So, all 'you' have to do is take the number from the column labeled "THE Number" that corresponds to the diameter of your core and divide the weight that you want your core to be by that number. The result will be the length in inches of your core. Personally, since I swage my cores, I would add maybe "a little" (say, 0.05"), to that resulting LENGTH figure (NOT to THE Number), for "sqeeze out". (Note that at a wire diameter of 0.250", THE Number is 140.75 - the same as obtained in the OP.) Paul PS - For those that might be interested, I would appreciate it if you would please check this against what you have determined for your actual core length values for a given weight, and report in this thread if the numbers jibe. PPS - I should note that a difference/error of 0.001" in the TRUE diameter of your core WIRE and what you put in the "equation" for finding THE Number will result in an error of 2.252 units in THE Number. PPPS - I don't like the name "THE Number". Let's call it the "core length constant" or CLC. At least in this thread. Actually that could "work". As it has been described in this thread, it is unique for pure lead. Were one to use a different metal or alloy, a subscript could be added. For example: Using clip on wheel weights, the abbreviation could be CLCcoww. For lead, CLCpb. Paul |

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