TargetTalk

When one downloads NRA type targets for pistol competition do you cut the distance in half being that they are downsized on our printers? If for example, the target says 25yds. do we cut the distance down in half or is there another percentage to cut the distance down? Thank you in advance.

I think you can calculate the size by simple proportion equations. If I remember my math the following is an example of how you would make the calculations.

For example 10 meter range and you want to change the target distance from 10 meters to 8 meters. Assuming that 10 meters is 1 or 100% of the full distance, so x is the new downsize %.

1.00/10 meters=x/8 meters

x=8 m(1.00/10m)=.8

1-.8=.2 or downsize by 20%

Mostly, no. The size of the bullet hole is the same at any distance. The NRA compensates for this by slightly reducing the scoring rings for a 25 yard or 50 foot target, less than the distance % would dictate from the 50 yard scoring ring sizes. This is done to compensate for the average size of the bullet holes, maybe an average of the .22, .38 and .45 calibers used.

Best Regards,
Slo cat

Conventional Pistol Targets:

Slow fire at 50 yards, the 10 ring is 3.36", at 25 yards it is 1.51". 41% at half the distance. At 50 feet it is .9", 26% at 1/3 the distance. So as it gets closer, the target gets proportionally smaller.

At 50 yards, the black is 8", at 25 yards it is 3.82" or 48%. At 50 feet, it is 3.07" or 38% but includes the 7 ring.

If I can ask a silly question, why not just buy the appropriate targets? I've never liked targets printed on bond, laser printer paper or copier papers. The bullet holes are never nice.

If I chose to not use regular targets for practice, I would print or cut out appropriate size black discs with no rings. At any distance, with the rings gone I would not be encumbered with "score"

I scale targets linearly with distance, but I adjust for bullet diameter. for example, the 10 ring of a 10 meter air pistol target is 11.5 mm in diameter. I add the bullet diameter (in this case 4.5 mm) for a total of 16 mm. This is the effective scoring diameter. I scale this linearly: for a 5 meter target the scoring diameter would be 0.5*16 mm = 8 mm. Then I subtract the bullet diameter to get the "equivalent" ring size, in this case 8 - 4.5 = 3.5mm.

in other words:

new ring diameter = (original ring diameter + bullet diameter)*k - bullet diameter

where k = new target distance/original target distance

Hope this helps,

Steve.