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Target size to distance formula?

Pastor T

I'm sure there is a geometry or trig formula to figure this out, but I can't seem to locate one, and I may be over complicating it in my head but here goes...

I am shooting a silhouette style target that would size up proportionally to be a person approximately 3 feet tall (I use that number because the silhouette is the top half of a body and is 18 inches). If I am shooting the target at 10 yards, what would be the equivalent distance to a full size person that is 6 feet tall?

I hope the way I'm wording the question makes sense...lol I also have some "hostage" targets on 8.5x11 paper and I'd like to figure out the real world distance equivalent to my 7 yard efforts to not shoot the hostage!

Thanks in advance for any input!
 
Geometrically speaking, any angle will subtend twice as much space if the distance doubles. For example, if you are looking through a straw at a beach ball and it just fills the field of view, doubling the size of the ball (diameter) will cause the same picture at twice the distance.

Since bullets diverge in straight lines, the answer to your question is that the half scale target allows you to practice at half the distance with an equivalent scoring.

With all of that in mind, there is a heck of a lot more going on in pistol shooting than that. As distance increases, depth of field becomes too short to allow some of the visual techniques we use at 20 feet, etc.. So, if you want to know what you would be able to do at 20 yards, shoot 20 yards.
 
It's very simple. Just use the ratio of size to distance.

3' target at 10yd = 3/10
That's the same as a 6ft target at 20yd.

Twice as far, twice as big.

If you shot a 6' target at 100yd, it would be like shooting a 0.6' target at 10yd (1/10th the distance, 1/10th the size).

If you shoot a 1" circle at 100yd, it looks the same as a 10" circle at 1000yd. But the 1000yd 10" circle is much harder to hit...

And so on...
 
Center of mass is still center of mass, whether it's ten yards or 100 yards. Don't worry about the size of the target, concern yourself with the size of the group and how close it is to COM.
 
Makes sense to me. We used the reduced B27 target due to range restrictions. Te only loss was our POI was not affected at the closer ranges
 
Pastor T said:
I'm sure there is a geometry or trig formula to figure this out, but I can't seem to locate one, and I may be over complicating it in my head but here goes...

I am shooting a silhouette style target that would size up proportionally to be a person approximately 3 feet tall (I use that number because the silhouette is the top half of a body and is 18 inches). If I am shooting the target at 10 yards, what would be the equivalent distance to a full size person that is 6 feet tall?

I hope the way I'm wording the question makes sense...lol I also have some "hostage" targets on 8.5x11 paper and I'd like to figure out the real world distance equivalent to my 7 yard efforts to not shoot the hostage!

Thanks in advance for any input!
Click to expand...

Being a long time Bulls Eye shooter, I remembered my coach saying the 50' indoor target was "harder to score on" than the 25 & 50 yd tgts. My personal scores were always a bit better outdoors as an anecdotal.

However, being retired now, I took the three targets plus a 20 yd target ALL FROM AMERICAN TARGET, DENVER and all being NRA certified and did some ratio, proportioning and figured the AREAS a bullet could land in to score. It didn't seem to be proportional when I figured it - but maybe I didn't understand all I thought I knew. Durned if I know where those figures are now. :\
 
Interesting thread. I have a similar issue, but I need to go in the other direction. I'm wanting to make some custom targets.

I am wanting to simulate a 9ft target at different distances while using a 100 yard range. So for example, how tall does a 9ft target need to be at 100 yards while simulating a distance of 500 yards?
 
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