target reductions for 25' dry fire practice

I have a 2 1/2" black circle to simulate slow fire at 25 feet.   I started to research what the formula would be to convert the 50yd black to 25'.   

I couldn't find it.   Can someone point me to the answer?

For dry fire I just divide by the distance. Live fire, you need to compensate the scoring rings for bullet diameter. I haven't yet figured out how they did the standard pistol targets. HP rifle is add .308, divide, subtract .308.

I found a 10M air rifle bull, colored in to the outer ring, is about right for a 50 yd at 10M. I originally did it to simulate a 50 ft Free Pistol. It's about 1-3/4".

If you want them, here are the true to size TQ-6 and TQ-7 targets, which are are official NRA 25 feet targets.

TQ-6 - 25' Slow Fire Target:

https://drive.google.com/open?id=1oiYtkYcIVuGTMHi6v5Fbn_Q510vY9DoL

TQ-7 - 25' Timed Fire Target

https://drive.google.com/open?id=1UwEN529NZ9NGMMOhqKLHwrfuoTGGgvGE

Thanks Oleg!  Exactly what I needed.

Ian, this is the formula I use and it's pretty accurate. I use it for the black bull area for dry fire training in my basement. Just measure the black area on the target whether it's a 50' or 50 yard then convert the distance to inches also. Divide the black area in inches by that targets distance in inches and take that sum and multiply by the new distance in inches and that will be your reduced black bull size. Ian your answer is 1.25". If you wanted to convert to a 10 foot distance the answer would be .167"

  ( full size target" / distance") X ( new distance") = reduced target in inches 
 Stan

You can also download the SCATT software.  It will reduce any target in it's system to any distance you want.

I measured the diameter of the nine ring on the TQ-7 target and it is about 1.535".