Knowing that there are 7000 grains in a pound, we can simply divide 7000 by 28 and see that 28 gauge balls weigh 250 grains each. If we measure one of those balls, we see that it is .550 inches in diameter. In modern usage that is the same as '.55 caliber'. The same relationship holds true for any size bore, of course.

Gauge/ Caliber Weight/gr. Balls/lb. 8 .835 875 10 .776 700 12 .730 583 13 .710 538 14 .693 500 16 .663 438 20 .615 350 24 .579 292 28 .550 250

An interesting observation is that there is an inverse (opposite) relationship between weights and gauges. In other words, as the gauge number goes down, say from 20 gauge to 10 gauge, the size and the weight of the ball goes up. A 20 gauge ball weighs 350 grains, and a 10 gauge ball weighs exactly twice that, 700 grains. Or, going the other direction, a 14 gauge ball weighs 500 grains, so a 28 gauge ball, twice the gauge number, weighs half as much, 250 grains.

Caliber = cube root ( 4.664 divided by gauge)

Example for a 20 gauge:

Caliber = cube root (4.664 divided by 20)

Caliber = cube root .2332

Caliber = .6155

In earlier times, the amount of black powder in a shotgun shell was marked in 'Drams'. After the switch to smokeless powder, this marking was continued, because it was so familiar to shooters, but the designation became 'Drams Equivalent', or the amount of smokeless necessary to give the same power as the stated number of drams of black powder. That designation is still used.

Here are a few equal volume loads, useful in a range of gauges.

Oz. Shot Dr. Powder Gr. Powder 3/4 2 55 7/8 2 1/4 62 1 2 1/2 68 1 1/8 2 3/4 75 1 1/4 3 82 1 3/8 3 1/4 89 1 1/2 3 1/2 96 1 5/8 3 3/4 102 1 3/4 4 109 1 7/8 4 1/4 116 2 4 1/2 123 2 1/8 4 3/4 130 2 1/4 5 137

1 grain = 0.037 dram

1 pound = 7000 grains

1 pound = 16 ounces

1 ounce = 437.5 grains

1 ounce = 16 drams

1 dram = 27.344 grains

Density of lead is 11.34 gram/cc

1 gram =15.4324 grains

1 inch = 2.54 centimeters

1 meter = 39.37 inches, 3.28 feet and 1.09 yards

pi = 3.14159

Volume of a sphere = 4/3pi R3 (4/3 x pi x radius of sphere cubed)

Radius of a sphere = cube root of (3Volume/4 pi)

Volume x density = weight

1) Convert grains to grams:

92.5 divided by 15.4324 grains/gram = 5.9939 grams

2) Find Volume:

5.9939 grams divided by 11.34 grams per cubic centimeter = .5286 cubic centimeter

3) Since:

Radius = cube root of (3Volume) divided by (4 pi)

Radius = cube root of (3 x .5286) divided by (4 x 3.14159)

Radius = .5016 centimeters

4) Find diameter:

Diameter = 2 x Radius

Diameter = 2 x .5016 cm

Diameter = 1.0032 cm

5) Convert centimeters to inches:

Diameter = 1.0032 cm divided by 2.54 cm/in

Diameter = .3950 in.

The end result is a ball perfect for my .40 caliber rifle!

1) Convert inches to centimeters:

Diameter = .395 in. x 2.54 centimeters per inch

Diameter = 1.0033 centimeters

2)Extract the radius:

Radius = Diameter divided by 2

Radius = 1.0033 divided by 2

Radius = .5017 cm

3) Cube the Radius:

Radius cubed = (.5017)3

Radius cubed = .1263 cubic centimeters

4) Then:

Radius = cube root of .1263

Radius = Cube root of (3 x Vol.) divided by (4 x pi)

.1263 = (3 x Vol.) divided by (4 x pi)

.1263 = (3 x Vol.) divided by 12.566

3Vol. = .1263 x 12.566

3Vol = 1.5868

Vol = 1.5868 divided by 3

Vol. = .5289 centimeters cubed

5) Find the Weight in grams:

Since Volume x Mass = Weight

.5289cc x 11.34 gm/cc = 5.9982 grams

6) Convert grams to grains:

5.9982 grams x 15.4324 grains/gram = 92.57 grains

W = 1502.6 x (Diameter x Diameter x Diameter)

Example for a .395 roundball: W = 1502.6 x .395 x .395 x .395 = 92.6 gr.

Weight in pounds = Wt. in grains divided by 7000, so the formula is:

Energy = (Wt. grs. divided by 7000) x Velocity squared divided by 64.32.

MRT = true drop at the target range (D1) divided by 2, then minus the drop at 1/2 the target range range (D2), or MRT = (D1 divided by 2) - D2. Most people then subtract 1/2 the height of the sight above the centerline of the bore (1/2S) to make the figures look better. This gives exactly what will be measured by actually shooting. So the formula becomes:

MRT = (D1 divided by 2) - D2 - 1/2S.

For a roundball traveling more than 1300 fps:

B.C. = Ball Wt. in grains divided by (10640 x ball dia. x ball dia.)

Example: For a .535 ball weighing 230 grains, 230 divided by (10640 x .535 x .535) = a BC of .0755. Lyman's Black Powder Handbook gives a BC of .075 for a .535 in. ball, so the agreement is good. This formula courtesy of "Lee in Denver"

150 x diameter squared divided by bullet length = required spin

Example for a .45 caliber bullet .60 inches long:

150 x .45 x .45 divided by .60 = 50.6 inches

So, for the example bullet, a spin rate of 1:50.6 or faster is required

The formula can also provide us with the maximum bullet length which can be stabilized by a given barrel twist. The formula becomes:

150 x diameter squared divided by twist rate

Example for a .50 caliber barrel of 1:48 twist:

150 x .50 x .50 divided by 48 = .78 inches

The barrel will stabilize a bullet .78 inches long, or shorter.

©1997 B. E. Spencer

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