TGOFSA,
The way shotgun gauges are determined is by how many lead balls the same diameter as the bore weigh one pound. A lead ball that weighs 1/10 of a pound (1.6 oz) is the diameter of a 10 gauge bore, a lead ball weighs 1/12 of a pound (1.33 ounces) is the diameter of a 12 gauge bore, a lead ball that weighs 1/16 of a pound (1 oz.) is the diameter of a 16 gauge bore, and a lead ball that weight 1/20 of a pound (0.8 ounces) is the diameter of a 20 gauge bore.
Lead has a density of 11.34 grams/cc, a cubic inch is 16.387 cc, so a cubic inch of lead weighs 185.83 grams or 6.5549 ounces. Since we want 0.8 ounces for 20 gauge, we want the diameter of a ball that is 0.8/6.5549 cubic inches, or 0.12205 in³. The volume of a sphere is 4/3πr³, so
0.12205/(4/3π) = r³ = 0.029137
0.029137^(1/3) = 0.307716 = r
So the ball and your 20 gauge bore diameter are twice that or 0.61543"
This is, of course, ideal (excluding rounding error). Actual numbers come from British tables published in the British Proof Act of 1868, and are still used in the U.S. The table numbers are treated as the minimum, and a tolerance of up to +0.020" oversize conforms to the standard.
So your 20 gauge will have a bore somewhere in that last range. It may be in the middle, but you should measure it when it arrives, as machining capability improves all the time.
The way shotgun gauges are determined is by how many lead balls the same diameter as the bore weigh one pound. A lead ball that weighs 1/10 of a pound (1.6 oz) is the diameter of a 10 gauge bore, a lead ball weighs 1/12 of a pound (1.33 ounces) is the diameter of a 12 gauge bore, a lead ball that weighs 1/16 of a pound (1 oz.) is the diameter of a 16 gauge bore, and a lead ball that weight 1/20 of a pound (0.8 ounces) is the diameter of a 20 gauge bore.
Lead has a density of 11.34 grams/cc, a cubic inch is 16.387 cc, so a cubic inch of lead weighs 185.83 grams or 6.5549 ounces. Since we want 0.8 ounces for 20 gauge, we want the diameter of a ball that is 0.8/6.5549 cubic inches, or 0.12205 in³. The volume of a sphere is 4/3πr³, so
0.12205/(4/3π) = r³ = 0.029137
0.029137^(1/3) = 0.307716 = r
So the ball and your 20 gauge bore diameter are twice that or 0.61543"
This is, of course, ideal (excluding rounding error). Actual numbers come from British tables published in the British Proof Act of 1868, and are still used in the U.S. The table numbers are treated as the minimum, and a tolerance of up to +0.020" oversize conforms to the standard.
Code:
Gauge Min Max
10 0.775" 0.795"
12 0.725" 0.745"
16 0.665" 0.685"
20 0.615" 0.635"
So your 20 gauge will have a bore somewhere in that last range. It may be in the middle, but you should measure it when it arrives, as machining capability improves all the time.