Home Presses Intro Site Map P.R. Chemicals Info Prices Specials How To Bullets B.Makers Books Classified Topics Jackets Terms Training Software Products Contact us Copper Strip/Jacket Production The following is a chart of production output for a given material input (that is, how many bullets or jackets you can make with a certain number of pounds of copper strip, lead wire, etc.). Naturally, there are variables in the design of bullets that require certain general assumptions, so that you may actually make a few more or less from the same material depending on the specific bullet. But in general this is a reliable planning guide for matching jacket and core material purchases, and calculating your production costs. For a discussion of how to determine production speed, see Production Speed page. To find the current cost of copper, jackets, lead wire, and other supplies, see the Price List page. Material Jackets/50-lb spool Max.Wt./Jacket 5/8-in x .030-T CU30-625 50-lb coil 8,205 18.97 gr. 1-in x .030-T CU30-100 50-lb coil 5,128 50.43 gr. 1.25-in x .030-T CU30-125 50-lb coil 4,102 79.78 gr. 1.25-in x .050-T CU50-125 50-lb coil 2,461.5 132.97 gr. 1.4-in x .094-T CU94-140 50-lb coil 1,169 315.23 gr.

How to Calculate Jacket Quantity and Maximum Weight
Assumptions: You will use Corbin's deep-drawing grade ASTM C11000 copper to draw the jackets. The density of the copper is 0.325-lb/cu-in. The copper is sold in "pancake" or flat coils of 50-lb weight, in standard width/thickness dimensions as follows:
• 5/8-in x .030 for .22 and smaller calibers
• 1-in x .030 for handgun calibers, up to 1-in x .018 wall .308
• 1.25-in x .030 for .358 and smaller calibers in normal length, thin walls
• 1.25-in x .050 for long or heavy wall .358 and smaller, thin wall .458-.375
• 1.4-in x .093 for .50 BMG, long .475-.416 w/heavy walls, and similar

Calculation Methods

The product of jacket wall thickness, caliber, and length determines the volume of material in the finished jacket. This volume must be the same or less than the volume in the original disk from which the jacket is drawn. Therefore, any jacket can be made longer if it is thinner walled or smaller diameter, thicker walled if it is made shorter or smaller diameter, and larger caliber if it is made with thinner walls or shorter length.

The number of jackets per a given weight of copper strip depends on the width and thickness of the strip. Copper strip is sized to provide the least amount of scrap for each jacket drawn, and is therefore matched as well as standard inventory allows to the desired jacket parameters. The jacket material is used to punch a disk, which must allow for a small buffer space between the edge of the strip and the start of the disk. The assumption is that this buffer will be 0.015-inches on each side of the disk. Therefore, subtract 0.015 x 2 = .030 inches from the width of the strip, and this will be the maximum disk diameter. (The actual buffer may vary somewhat, but for calculation purposes .015-inches is reasonable.)

Therefore, a 0.625 wide strip (5/8-in) can make a disk of 0.625-.030 = .595 inches diameter (maximum). The number of these disks that can be cut from a 50-lb coil is the length of the strip in that coil, in inches, divided by the width of strip.

The number of inches of length of any weight of coiled strip is determined by dividing the volume by the strip width times thickness. The volume is determined by the weight divided by the density. (Density is weight divided by volume). The density of Corbins material is 0.325-lbs/cubic inch. Therefore, 50 lbs divided by 0.325 lb/cubic inch is 153.846 cubic inches. This is the volume of ANY 50-lb coil, regardless of the width and thickness, so long as it is the same density.

To determine the length in inches of any coil, calculate the square inches of cross sectional area for the strip, and divide this into the volume. The cross sectional area is the width times the thickness. Here are the cross sectional areas for standard Corbin copper coils:

 Width in. Thickness in. Cross-sectional Area sq-in. Length in./50 lb Jackets per coil Pounds Max.Jacket Wt. Grains 5/8 .030 0.01875 8,205 0.00271 18.97 1 .030 0.03 5,128 0.0072049 50.43 1.25 .030 0.0375 4,102.5 0.0113975 79.79 1.25 .050 0.0625 2,461.5 0.0189959 132.97 1.4 .094 0.1316 1,169 0.045034 315.23

To calculate the scrap weight for recovery (sale of scrap copper), simply subtract the area of each disk from the area of the square (strip width squared) from which it is punched, and subtract the disk area from the square area. Multiply this remaining area by thickness, to obtain volume. Multiply the volume times the density of copper for the weight. You can then call a local scrap metal recovery firm, and ask what the current purchase rate may be for pure copper, by the pound. Multiply this dollar value by the scrap weight, subtract from your material cost, and you have the net price of the jackets.

• S = Scrap Weight, pounds
• W = strip width, in inches
• T = strip thickness, inches
• D = Density, lb/cu-in (.325)
• Q = Quantity of jackets per 50 lbs

S = (W^2 - ((W - .03)/2)^2 * Pi) * T * D * Q

To break this down in simple terms...
• Calculate area of the disk = width of strip minus .03, divided by 2, squared, times 3.14159 (Pi).
• Calculate the area of the square from which this disk is cut = width of strip squared.
• Subtract disk area from square area. The difference is the scrap area.
• Multiply the scrap area by the thickness of the strip. This is scrap volume.
• Multiply scrap volume by density of material (.325 lb/cu-in). This is weight of scrap in pounds for one jacket.
• Multiply the weight of scrap for one jacket times number of jackets made with 50 lbs of material. This is your total scrap.

For any strip width, the number of jackets you can make in any weight of material is simply the length of that coil divided by the width. Normally, you don't know the length but you do know the weight, width, and thickness. Here is how to calculate the length and number of jackets you can make:

• The density of any material is the weight divided by the volume. You know the density = 0.325 lb/cu-in and the weight = 50 lbs.

• Volume, therefore, is weight divided by density. Divide 50 lbs by 0.325 lbs/cu-in. The answer is 153.84615 cubic inches.

• The volume is the width times thickness times length. We know volume, width and thickness. We can find length.

• Length = volume divided by the width times thickness (which is cross-sectional area). L = 153.84615 / (w * t)

• For a 1.25-in wide strip of .050 thickness, multiply .050 x 1.25 = 0.0625 square inches (cross sectional area).

• Now divide the volume (153.84615) by the cross sectional area (0.0625) and you have length of 2,461.5384 inches. For practical purposes, this can be considered 2,461 jackets from the 50-lb coil.

• Divide the cost of the strip coil, less the recovery cost for the scrap, by the number of jackets. That is your material cost per jacket.

• J = Jackets per coil
• A = Cross sectional area (width times thickness)

J = 153.84615 / A

A = Width * Thickness