The Mathematical Shooter
The sport of black powder shooting, whether hunting or competitive, exposes the shooter to the need to understand some simple mathematical equivalents, formulae and relationships. This is true no matter how casual our shooting, and we will enjoy the shooting more if we understand a few of the basics. Charles Winthrop Sawyer, who wrote the book _Our Rifles_ early in this century, expressed this well when he said, "Rifled arms are the delight of mathematicians and scientists, affording never ending allurement in the attempt to make them do what they often fail to do and yet are theoretically capable of doing." One need not be a mathematician or scientist to understand that challenge and that allure.
Gauge One of the earliest ways to designate the size of the bore of a gun was to figure how many bullets for it could be made from a pound of lead. Since at that time all bullets were roundballs of lead, this made a good standard. In time this designation of 'balls to the pound' became synonymous with gauge, so that what our forefathers called '28 to the pound', we now call '28 gauge'. The British commonly used the term 'bore' for 'gauge', so their '28 bore' is the same as '28 gauge' or '28 balls to the pound'.
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 to Gauge Conversion A short method for finding caliber if gauge is known involves the use of a constant.
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
Equal Volume Loads One of the proven rules of thumb in black powder shotgun shooting is that most guns will pattern the shot well if the gun is loaded with equal volumes of powder and shot. Volume, not weight. Simply measure them both with the same dipper/measure.
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
Some useful values:
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
Diameter of a Roundball If the weight of a roundball is known, the diameter can be calculated. Using a 92.5 grain ball as an example:
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!
Weight of a Roundball (Long Method) If the diameter of a roundball is known, the weight can be calculated. Using the .3950 in. ball from the exercise above, we proceed to:
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
Weight of Round Ball (Short Method) A short-cut method of determining the weight (W) of a roundball when the diameter is known involves the use of a constant.
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.
Patch Thickness Most shooters find the proper patch thickness for their patched roundball by simply trying several until they discover one that shoots well. A slightly less random method is very simple. The designated caliber of a rifle is equal to the maximum land to land distance. The groove to groove distance is greater than that, by twice the groove depth. A proper patch will usually be found to equal maximum groove diameter minus the size of the ball, divided by 2. As an example, a .54 caliber rifle with an actual groove to groove measurement of .560 in., shooting a .530 in. ball. .560 in. minus .530 in. equals .030 in. That difference will be filled with the patch, one half on each side of the ball, so the patch should be .030 divided by 2, or .015 in.
Energy The energy still retained by the projectile at target range is very important to the hunter and silhouette shooter. Retained energy is simply the bullet's Weight in pounds x Velocity squared divided by 64.32 (the acceleration of gravity x 2).
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.
Mid-Range Trajectory As the bullet travels toward the target, it describes an arching curve upward, and then downward. The maximum distance the bullet travels above the line of sight in this path is called the Mid-Range Trajectory (MRT). The high point is usually at about 60 % of the distance to the target.
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.
Ballistic Coefficient In order to calculate the trajectory of a bullet in flight, its ability to push aside the air and retain energy must be known. This property is known as the bullet's Ballistic Coefficient (BC). To calculate an accurate BC for any given bullet requires actually shooting it many times at various velocities, and measuring it's change in velocity over range. There is a simple way to approximate the BC for a roundball, though, so we can play around with theoretical trajectories.
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"
Rate of Spin Whether roundball or conical, any bullet fired from a rifled barrel spins as it travels to the target. Spin is responsible for stability in flight, a very important factor in accuracy. Think of a ball fired at 2000 fps from a barrel with 1: 66 twist. To calculate spin, assume the bullet will travel a distance equal to the velocity, eg: a ball at 2000 fps will travel 1 full second and go 2000 feet. Then, convert twist rate to feet, so that 1:66 = 66 divided by 12 in. = 5.5 feet. If the ball travels 2000 feet in one second, and spins around once every 5.5 feet, it will spin 2000 divided by 5.5, or 363.63 times in one second, rps. We then convert to rpm by multiplying rps by 60, so 363.63 x 60 equals 21,818 rpm.
Bullet Stabilization An elongated bullet flying through the air without spinning will be unstable and inaccurate. The longer the bullet is in relation to its diameter, the more spin is required to stabilize it. How much spin is required? This relationship is expressed in the Greenhill formula, a simplified verson of which is:
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.
The sport of black powder shooting, whether hunting or competitive, exposes the shooter to the need to understand some simple mathematical equivalents, formulae and relationships. This is true no matter how casual our shooting, and we will enjoy the shooting more if we understand a few of the basics. Charles Winthrop Sawyer, who wrote the book _Our Rifles_ early in this century, expressed this well when he said, "Rifled arms are the delight of mathematicians and scientists, affording never ending allurement in the attempt to make them do what they often fail to do and yet are theoretically capable of doing." One need not be a mathematician or scientist to understand that challenge and that allure.
Gauge One of the earliest ways to designate the size of the bore of a gun was to figure how many bullets for it could be made from a pound of lead. Since at that time all bullets were roundballs of lead, this made a good standard. In time this designation of 'balls to the pound' became synonymous with gauge, so that what our forefathers called '28 to the pound', we now call '28 gauge'. The British commonly used the term 'bore' for 'gauge', so their '28 bore' is the same as '28 gauge' or '28 balls to the pound'.
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 to Gauge Conversion A short method for finding caliber if gauge is known involves the use of a constant.
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
Equal Volume Loads One of the proven rules of thumb in black powder shotgun shooting is that most guns will pattern the shot well if the gun is loaded with equal volumes of powder and shot. Volume, not weight. Simply measure them both with the same dipper/measure.
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
Some useful values:
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
Diameter of a Roundball If the weight of a roundball is known, the diameter can be calculated. Using a 92.5 grain ball as an example:
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!
Weight of a Roundball (Long Method) If the diameter of a roundball is known, the weight can be calculated. Using the .3950 in. ball from the exercise above, we proceed to:
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
Weight of Round Ball (Short Method) A short-cut method of determining the weight (W) of a roundball when the diameter is known involves the use of a constant.
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.
Patch Thickness Most shooters find the proper patch thickness for their patched roundball by simply trying several until they discover one that shoots well. A slightly less random method is very simple. The designated caliber of a rifle is equal to the maximum land to land distance. The groove to groove distance is greater than that, by twice the groove depth. A proper patch will usually be found to equal maximum groove diameter minus the size of the ball, divided by 2. As an example, a .54 caliber rifle with an actual groove to groove measurement of .560 in., shooting a .530 in. ball. .560 in. minus .530 in. equals .030 in. That difference will be filled with the patch, one half on each side of the ball, so the patch should be .030 divided by 2, or .015 in.
Energy The energy still retained by the projectile at target range is very important to the hunter and silhouette shooter. Retained energy is simply the bullet's Weight in pounds x Velocity squared divided by 64.32 (the acceleration of gravity x 2).
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.
Mid-Range Trajectory As the bullet travels toward the target, it describes an arching curve upward, and then downward. The maximum distance the bullet travels above the line of sight in this path is called the Mid-Range Trajectory (MRT). The high point is usually at about 60 % of the distance to the target.
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.
Ballistic Coefficient In order to calculate the trajectory of a bullet in flight, its ability to push aside the air and retain energy must be known. This property is known as the bullet's Ballistic Coefficient (BC). To calculate an accurate BC for any given bullet requires actually shooting it many times at various velocities, and measuring it's change in velocity over range. There is a simple way to approximate the BC for a roundball, though, so we can play around with theoretical trajectories.
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"
Rate of Spin Whether roundball or conical, any bullet fired from a rifled barrel spins as it travels to the target. Spin is responsible for stability in flight, a very important factor in accuracy. Think of a ball fired at 2000 fps from a barrel with 1: 66 twist. To calculate spin, assume the bullet will travel a distance equal to the velocity, eg: a ball at 2000 fps will travel 1 full second and go 2000 feet. Then, convert twist rate to feet, so that 1:66 = 66 divided by 12 in. = 5.5 feet. If the ball travels 2000 feet in one second, and spins around once every 5.5 feet, it will spin 2000 divided by 5.5, or 363.63 times in one second, rps. We then convert to rpm by multiplying rps by 60, so 363.63 x 60 equals 21,818 rpm.
Bullet Stabilization An elongated bullet flying through the air without spinning will be unstable and inaccurate. The longer the bullet is in relation to its diameter, the more spin is required to stabilize it. How much spin is required? This relationship is expressed in the Greenhill formula, a simplified verson of which is:
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.
