Linstrum
06-04-2007, 08:23 AM
A few days back I was reading about the muzzleloader musket buck and ball loads that were used for close range deer hunting and defensive/guard loads. I already knew that most large multi-ball loads have the balls stacked in layers and arranged in groups of threes or sevens, and that the ball groups fit fairly snuggly in the bores either inside of large paper sabots or with just a single large round ball beneath that pushes the smaller balls out. Shotguns with chokes of course can’t be loaded this way because the balls would hit the choke and split the barrel and/or deform the balls.
I took a look at the photos of how the three and seven balls per layer loads sit in the barrel and I could see that determining what size of buck ball to use for a three balls per layer load was nowhere near as easy as arriving at the ball size needed for the seven balls per layer load. For the seven-ball load arrangement the ball size is determined by simply dividing bore diameter by three since there are three balls lying across the diameter line of the bore. In the case of the three balls per layer load like is used for the traditional 12 gauge shotgun 00-buckshot loading, though, the balls instead meet to form a triangular pattern that makes it impossible to directly measure or even easily calculate the size ball that will fit snuggly! To discover what size ball it takes does not require a math wizard if a little thought is used. It can be found pretty fast with acceptable accuracy by simply drawing a precise picture with calipers and a draftsman’s compass and then taking measurements directly from the blueprint. This will give a value somewhere around 45% the bore diameter. To find it mathematically requires a little simple trigonometry where the actual dimension wanted turns out to be the radius of the buck ball plus the radius of the slightly larger circle formed if the buck ball circle is used to generate a hexagon around itself. A picture is worth a thousand words so take a look at the drawing. The trigonometry uses the geometry of Pythagoras’ Theorem to calculate the lengths of the legs of a 30° - 60° - 90° triangle that is formed as part of the hexagon. Another way is to use 1 plus the cosecant of 60° to find the ratio of the gun’s bore diameter and the buck ball diameter to determine the ultimate values. After plugging in the numbers and “turning the math crank” the mathematics basically distills down to the ratio of the two diameters being the square root of 3 divided by the quantity of the square root of 3 plus 2, end quantity. That is 1.732 / (1.732 + 2) to give the value of 0.4641, which is multiplied by the gun’s bore diameter (measured at the lands for a rifled barrel) to find what size of ball will snuggly fit three to a layer in the gun’s bore. Of course going the other way to find what unknown size bore three balls of known size will fit in, then the ball diameter is DIVIDED by the value just given. A common smooth bore musket that was widely used in the United States until about 1885 was the .69 caliber Springfield and a common load used in it was the buck and ball combination. To find what size buck ball that fits three to the caliber, just take 0.69” and multiply it by 0.4641, which gives a diameter of 0.32”, which is right around the size of 00 buck.
Have fun!
I took a look at the photos of how the three and seven balls per layer loads sit in the barrel and I could see that determining what size of buck ball to use for a three balls per layer load was nowhere near as easy as arriving at the ball size needed for the seven balls per layer load. For the seven-ball load arrangement the ball size is determined by simply dividing bore diameter by three since there are three balls lying across the diameter line of the bore. In the case of the three balls per layer load like is used for the traditional 12 gauge shotgun 00-buckshot loading, though, the balls instead meet to form a triangular pattern that makes it impossible to directly measure or even easily calculate the size ball that will fit snuggly! To discover what size ball it takes does not require a math wizard if a little thought is used. It can be found pretty fast with acceptable accuracy by simply drawing a precise picture with calipers and a draftsman’s compass and then taking measurements directly from the blueprint. This will give a value somewhere around 45% the bore diameter. To find it mathematically requires a little simple trigonometry where the actual dimension wanted turns out to be the radius of the buck ball plus the radius of the slightly larger circle formed if the buck ball circle is used to generate a hexagon around itself. A picture is worth a thousand words so take a look at the drawing. The trigonometry uses the geometry of Pythagoras’ Theorem to calculate the lengths of the legs of a 30° - 60° - 90° triangle that is formed as part of the hexagon. Another way is to use 1 plus the cosecant of 60° to find the ratio of the gun’s bore diameter and the buck ball diameter to determine the ultimate values. After plugging in the numbers and “turning the math crank” the mathematics basically distills down to the ratio of the two diameters being the square root of 3 divided by the quantity of the square root of 3 plus 2, end quantity. That is 1.732 / (1.732 + 2) to give the value of 0.4641, which is multiplied by the gun’s bore diameter (measured at the lands for a rifled barrel) to find what size of ball will snuggly fit three to a layer in the gun’s bore. Of course going the other way to find what unknown size bore three balls of known size will fit in, then the ball diameter is DIVIDED by the value just given. A common smooth bore musket that was widely used in the United States until about 1885 was the .69 caliber Springfield and a common load used in it was the buck and ball combination. To find what size buck ball that fits three to the caliber, just take 0.69” and multiply it by 0.4641, which gives a diameter of 0.32”, which is right around the size of 00 buck.
Have fun!