Smallbore Prone Musings

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bberg7794
Posts: 79
Joined: Tue Apr 22, 2014 12:02 pm
Location: St Lawrence County, NY

Smallbore Prone Musings

Post by bberg7794 »

Group sizes are often discussed in various shooting forums, but rarely does anyone attempt to quantify the degree of skill required to achieve high scores in the smallbore disciplines. I am sure there have been great articles written about this topic somewhere, but I have not been able to find them.

I have often wondered just how close a hold one must have to hit various target centers at different distances and have also wondered about the relative difficulty of different smallbore targets, so I finally attempted to figure it out. I hope that I didn't overlook something obvious in this attempt.

I welcome anyone to check my math and to share their thoughts. Additionally, this is meant to be a commentary on how truly skilled the high master rated shooters are.
Last edited by bberg7794 on Tue Jul 16, 2019 10:54 pm, edited 3 times in total.
bberg7794
Posts: 79
Joined: Tue Apr 22, 2014 12:02 pm
Location: St Lawrence County, NY

Re: Smallbore Prone Musings

Post by bberg7794 »

To determine the largest possible group size to score an ‘X’ or a ‘10’, I took the diameter of the ring and added one bullet diameter (.223 inch), or rather, two half-diameters on either side of the scoring ring, with two exceptions. On the USA-50 I subtracted the diameter of the dot twice from two half-diameters of a bullet since the dot has to be completely covered to count a 10 and with the A-33 metric target I subtracted, instead of added, two half bullet diameters since the outer edge of the bullet cannot go outside the outer edge of the 10 ring to count as an ‘X’.

All units have been converted to inches. I divided the total distance by 26 because that is the approximate length of my barrel (Anschutz 1907), which gives us a ratio for each of the three distances. Then, I took the maximum group size and divided it by the ratio to determine how much movement at the end of my 26-inch barrel would equal that group size. Lastly, I divided that number by 2, which will give the movement on either side of absolute center that the barrel would move for a given group size. That is the number in bold.

This gives us an idea of the relative difficulty of the different targets, but the mathematical numbers would only be truly meaningful if one’s ammo and barrel combination are 100% accurate, which is never the case. In reality, an experienced prone shooter is likely dealing with an ammo/barrel combination that shoots a ¾ inch group at 100 yards in still air. So if we take the 100 yard A-33 target as an example, a shooter already has to have nearly a perfect hold to consistently shoot X's if their rifle/ammo is capable of 3/4" at that distance, since the maximum group size to score an 'X' is .822 inches. And that is before adding wind into the mix.

Indoor
50 feet is 600 inches. 600/26 = 23.077 ratio
A-17 50 feet
10 ring .150 inch (.150 + .223 = .373 inches)
.373/23.077 = .016 inches/2 = .008 inches

USA-50 50 feet
10 ring (dot) .76 mm (.223 – (.03 x 2) = .163 inches)
.163/23.077 = .007 inches/2 = .0035 inches

Conventional Prone
50 yards is 1800 inches. 1800/26 = 69.23 ratio
50 yard A-23
X ring .39 inch (.39 + .223 = .613 inches)
.613/69.23 = .0089/2 = .0045 inches
10 ring .89 inch (.89 + .223 = 1.113 inches)
1.113/69.23 = .016/2 = .008 inches

A-27 50 meter target reduced for firing at 50 yards
X ring .359 inch (.359 + .223 = .582 inches)
.582/69.23 = .008/2 = .0042 inches
10 ring .719 inch (.719 + .223 = .942 inches)
.942/69.23 = .014/2 = .007 inches

A-25 conventional 100 yard
100 yards is 3600 inches. 3600/26 = 138.462 ratio
X-ring 1 inch (1 + .223 = 1.223 inches)
1.223/138.462 = .0088/2 = .0044 inches
10 ring 2 inches (2 + .223 = 2.223 inches)
2.223/138.462 = .0161/2 = .008 inches

Metric Prone
A-51 metric 50 yard
X ring 4.096 mm .161 inches (.161 + .223 = .384 inches)
.384/69.23 = .0055/2 = .00275 inches
10 ring 9.034 mm .356 inches (.356 + .223 = .579 inches)
.579/69.23 = .0084/2 = .0042 inches

A-33 metric 100 yard
X ring 1.045 inches (1.045 - .223 = .822 inches)
.822/138.462 = .0059/2 = .003 inches
10 ring 1.045 inches (1.045 + .223 = 1.268 inches)
1.268/138.462 = .0092/2 = .0046 inches
jhmartin
Posts: 2620
Joined: Mon Nov 29, 2004 2:49 pm
Location: Valencia County, NM USA

Re: Smallbore Prone Musings

Post by jhmartin »

And the above is as if you are shooting indoors/in a tunnel w/ no wind and (essentially) no mirage.

Add wind & mirage into the mix ... well more skill (read thousands of rounds of preparation) and more mental focus that I can imagine.....
Hat's off to ya all!

Oh, and in an International style competition, add in the Michael Jackson music..... ;-)
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