scaling targets

[lunicy]

ok, probably a no-brainer but...

If I wanted to scale a 200yd target for 100 yds, its half size/50% right
400yd target at 100yds = 25% right?

100yd target at 400yds = 4x or 400% right?


I thinks its that simple but I'm not sure, someone teach me 2nd grade math. :-D

[AZ-Stew]

All correct. BTW, it's high school Trig.

Regards,

Stew

[lunicy]

at least i don't feel as dumb anymore.:-D

[joeb33050]

I have some targets that work very well with iron sights at 100 yards, and have for at least 25 years. The diameter of the black is 5.5".
The AREA of a black center 5.5" in diameter is 24 square inches.
If we halve the diameter, to 2.75", for 50 yard shooting, the AREA is reduced to 6 square inches.
I don't know if our eyes work on diameter or area.
To halve the area of the black 5.5" target, from 24 to 12 square inches, the diameter would be 3.9".

I remember reading somewhere that our eyes percieve size of an object in a non-linear fashion; the context was astronomy.

An interesting question.
Any astronomers?
joe b.

[joeb33050]

yard targets have a 13" diameter black.
joe b.

[AZ-Stew]

Joe,

In the first part of your post you note that the area for the smaller circle is 1/4 that of the larger circle. This is correct and can be easily shown by drawing a square, then crossing it with horizontal and vertical lines at the midpoints of each side of the square, dividing the original square into four smaller ones. Any one of the four squares that are now visible in the larger square are now half the size linearly (the distance along each side), of the original, but each has only 1/4 the area of the larger original.

The apparent size of an object viewed at a distance is determined by the angle represented by the viewer at the apex of the angle and the legs of the angle going from the viewer to the left and right (or top and bottom) extents of the object. Imagine the letter "A" with the viewer (shooter) at the point at the top of the letter, and the 100yd target at the crossbar of the "A". Now draw a line across the bottom of the "A", at the tip of each leg to represent the 200 yd target. If the distance from the peak of the "A" (shooter location) to the middle cross bar (100 yd target) is the same as the distance from the middle cross bar to the bottom of the "A" (200 yd target), then the bottom crossbar will be twice as long as the middle cross bar. Therefore, a 200 yd target should be twice the diameter of a 100 yd target to appear the same size.

I've never worked for a target manufacturer or taken the time to measure the bullseye sizes on the targets used at various distances or the reduced size targets used at shorter distances to simulate, say, a 600 yard match shot on a 500 yard range, but as far as the optical physics and geometry is concerned, my second paragraph should explain it.

Regards,

Stew

[joeb33050]

Stew;
The diameter of a circle, measured as an angle, varies with distance as the diameter.
A 6" diameter circle at 100 yards subtends ~6 minutes. And, a 3" circle at 50 yards subtends ~6 minutes.
The areas of those circles varies as the square of the diameter, a 3" circle = 1/2 a 6" circle, 1/2 squared = 1/4, a 3" circle area is 1/4 that of a 6" circle.
As you said.
But, I don't know about the eye, if it sees variation as a function of diameter or area, or something else.
Geometry is one thing, vision and eyes is another.
Tomorrow I'll compare them, 5.5" and 2.75".
joe b.














Joe,

In the first part of your post you note that the area for the smaller circle is 1/4 that of the larger circle. This is correct and can be easily shown by drawing a square, then crossing it with horizontal and vertical lines at the midpoints of each side of the square, dividing the original square into four smaller ones. Any one of the four squares that are now visible in the larger square are now half the size linearly (the distance along each side), of the original, but each has only 1/4 the area of the larger original.

The apparent size of an object viewed at a distance is determined by the angle represented by the viewer at the apex of the angle and the legs of the angle going from the viewer to the left and right (or top and bottom) extents of the object. Imagine the letter "A" with the viewer (shooter) at the point at the top of the letter, and the 100yd target at the crossbar of the "A". Now draw a line across the bottom of the "A", at the tip of each leg to represent the 200 yd target. If the distance from the peak of the "A" (shooter location) to the middle cross bar (100 yd target) is the same as the distance from the middle cross bar to the bottom of the "A" (200 yd target), then the bottom crossbar will be twice as long as the middle cross bar. Therefore, a 200 yd target should be twice the diameter of a 100 yd target to appear the same size.

I've never worked for a target manufacturer or taken the time to measure the bullseye sizes on the targets used at various distances or the reduced size targets used at shorter distances to simulate, say, a 600 yard match shot on a 500 yard range, but as far as the optical physics and geometry is concerned, my second paragraph should explain it.

Regards,

Stew

[runfiverun]

i just try to use targets that appear the same size as my front sight at the distance that i am shooting.
sometimes this is determined with some tape and a two way radio.

[45nut]

reminder of the MANY printable targets available from our host, www.gunloads.com , look on the left hand column.

[joeb33050]

I brought a 100 yard target to the UPS store, and the guy reduced it in size to 50%, and made copies.
Today we put a 100 yard target up at 100 yards, and a half size target up at 50 yards. Close to each other when viewed from the line, left and right.
The unanimous decision is that the 100 yard target looks bigger than the half size 50 yard target. It is clearly bigger.
Is this understandable?
Half size targets at half range DO NOT look like full size targets at the set range.
Something is going on, maybe area, maybe our eyes.
joe b.

[AZ-Stew]

Joe,

How did it look through a ring-type sight, such as you'd find on a stock Garand? Do they cover the same amount of sight opening? Or in the case of more open sights, do they appear as the same size dot above the front sight? I'd say that would be the true test.

Thanks for your persistence on this issue. I'd do the testing myself, but I still have to work for a livning.

Regards,

Stew

[357maximum]

The easiest way to scale your target is place one large dead carp about 5 feet in front of said target.[smilie=1:

[joeb33050]

Joe,

How did it look through a ring-type sight, such as you'd find on a stock Garand? Do they cover the same amount of sight opening? Or in the case of more open sights, do they appear as the same size dot above the front sight? I'd say that would be the true test.

Thanks for your persistence on this issue. I'd do the testing myself, but I still have to work for a livning.

Regards,

Stew

We just looked, I didn't have an iron sighted gun. Sometime later I'll look through rear aperture and front aperture sights, but no longer know what the question is.
The original question was about scaling targets as angular relationships; EX: 6" at 100 yards = 3" at 50 yards.
I've decided that the answer is NO, cause they don't look the same.
I don't know what the relationship should be, don't know how to make a 50 yard target look like a 100- yard target, and believe that an astronomer or ??scientist knows the answer.
What are the sizes of iron sight targets at various ranges?
joe b.

[KYCaster]

Well, you guys made me curious...so here's my take on it.

The simple and logical answer is that when the distance doubles then the size of the target should double in order to appear the same through the sights.

Now I'm not a high power shooter, nor am I a mathematician, but I can remember my high school geometry teacher talking about "proportional triangles". If the angles stay the same then the sides are always in the same proportion....double the height and the length of the base doubles. So if you want the same sight picture at 200 yds. that you have at 100 then you need to double the size of the target.

And to answer Joe's question about iron sight target sizes I went to the NRA web site to get the dimensions of the hi-power targets, and I found that the "aiming black", as they call it, for a 200 yd. target is 13 in., 300 yd. is 19 in. and 600 is 36 in.

Now divide the size of the aiming black (the base of the triangle) by the distance to the target (the height of the triangle) and I get .065, .063 and .060, respectively....not exactly the same proportion, but pretty darn close. And just to see what size the long range targets should be to keep the same proportion as the 200 yd. target, it looks like the 300 yd. aiming black should be 18.9 in. and the 600 should be 37.8 in. Again, not exactly what the target dimensions are, but so close I doubt you can tell the difference looking over the sights.

Like r5r said, the most precise sight picture will have an aiming point that appears to be the same width as the front sight. Seems to me that's how the NRA targets are designed...find the right front sight width and use the same "hold" at any distance, and that agrees with what I've been told by guys who shoot hi-power.

And Joe, I can't say for sure, but I suspect that your half size 50 yd. target looks smaller than the full size target at 100 yds. simply because it IS smaller. You made it smaller on purpose and regardless of proportions or relative distances, you know it's smaller so it's going to LOOK smaller..."the mind is a terrible thing..."

I hope this makes some sense.

Jerry

[joeb33050]

And Joe, I can't say for sure, but I suspect that your half size 50 yd. target looks smaller than the full size target at 100 yds. simply because it IS smaller. You made it smaller on purpose and regardless of proportions or relative distances, you know it's smaller so it's going to LOOK smaller..."the mind is a terrible thing..."

I hope this makes some sense.

Jerry

Jerry;
I understand the arithmetic. Why don't you try the half size 50 yard test, and tell us what you see. Every shooter I asked, most of whom didn't know what the experiment was about, said that the 100 yard target looked bigger.
joe b.

[C A Plater]

Jerry;
I understand the arithmetic. Why don't you try the half size 50 yard test, and tell us what you see. Every shooter I asked, most of whom didn't know what the experiment was about, said that the 100 yard target looked bigger.
joe b.

The reason the 100 yard circle looks bigger than the reduced 50% at 50 yards is because the area circles do not scale the same way. A 12 inch circle has an area about 113 square inches. To look the same size at 50 yards it would be half the area making it 56.55 square inches or 8.49 inches in diameter. So in round numbers, scale it 70% to look like the same area at half the distance.

[buck1]

Funny this seems like it would be simple at first. But I am lost. LOL!

[Doc Highwall]

First off 20/20 vision is based off one minute of angle. What this means is a person with 20/20 vision can see a 1" black dot on a white piece of paper at 100 yards and a person with 20/15 vision can see a 3/4" dot, and a person with 20/10 can see a 1/2" dot at 100 yards. Now how to find out what a minute of angle is and what it means. A circle has 360 degrees and a degree has 60 minutes, 360 degrees x 60 minutes = 21,600 minutes in a circle. The circumference of a circle is it's outside measurement, and is found by the math formula, ( 2 * PI * Radius ) or ( Two times pi times the radius ). Now lets say we want to find out how big a minute of angle is at 100 yards, ( 2 x 3.141592654 x 3600" = 22619.46711" divided by 21,600 minutes = 1.047197551" @ 100 yards). Now this means that at 1000 yards a true minute of angle is 10.47197551", in layman's terms 10" is close enough at 1000 yards or 1" @ 100 yards, 2" @ 200 yards etc. The size of the high power targets are 13" @ 200 yards or 6.5 minutes of angle, 19" @ 300 yards or 6.33 minutes of angle, 36" @ 600 yards or 6.0 minutes of angle. The difference is .5 minutes of angle between 200 to 600 yards, in other words unless you have better then 20/20 vision like 20/10 you cannot see the difference on these targets. In joeb33050 case if the distance was not exactly 50 yards and 100 yards with a 6" and 3" targets they will not look the same. ANGLES ARE PROPORTIONAL, DOUBLE THE DISTANCE AND DOUBLE THE HEIGHT. Now using the math formula for minute of angle lets apply it to our front sight using 1.0472" as a minute of angle at 100 yards. Example your front sight is 27" from your EYE, and is .070" wide and you want to know how much this covers at 100 yards. (3600" divided by 27" = 133.333 times .070" = 9.33" at 100 yards divided by 1.0472 = 8.912 minutes of angle that your sight covers at all distances, remember ratio and proportion your front sight now covers 18.666" at 200 yards. You can also use the one inch equals one minute at 100 yards with your scope for target size. Say a person has 20/20 vision and they have a 10 power scope on their gun, this means they can see 1/10 minute of angle or .100" at 100 yards, 20 power means .050" again ratio and proportional. I hope this helps and was not too boring.

[JCherry]

Doc Highwall,

Lets say I want to practice my offhand shooting for the Quigley Bucket Shoot with my Sharps at 100 yds. (I have a 22LR insert barrel for my Sharps.)

I'm told the bucket is shot at 350 yards and is 32"X 22" which equates basically to 9 x 6 MOA.

So if I understand everything correctly a 9" X 6" target at 100 yards would be visually the same for the purposes of practice as the full sized target at 350 yds?

Have Fun,

JCherry

[Doc Highwall]

JCherry, yes that is correct. Now to get the best benefit you need to scale the target at a distance where the wind drift at 350 yards for your bullet matches the 22lr in inches or minutes of angle and this will help you to read the wind. The difference between a person with a masters classification and a person with a long range masters classification is knowing how to read the wind. Here is a chart that should help you out, now get your butt out and practice, the only way to learn to read the wind is to shoot in the wind.

[JCherry]

Doc Highwall

Practice to better my offhand shooting, including the reading of wind is exactly why I got the .22LR insert for the Sharps.

A good ballistic calculator I have used is;

http://www.handloads.com/calc/index.html

The .22LR ammo I'm using now is the Federal 36 gr Hollow Point you can buy in boxes of 550 at Walmart. While it is not match ammo it seems to do more than well enough for my offhand practice.

Yes, my butt will be out there practicing.

Thanks,

Have Fun,

JCherry

[Doc Highwall]

JCherry, I have shot that same ammo out to 300 yards in my model 57M Cooper and it does very well out to 200 yards, at 300 yards I tend to notice some vertical stringing from the velocity variations. It is a lucky man that has a nice 22 that will shoot like a match rifle with CHEAP AMMO instead of a cheap 22 that has a appetite for Eley match only.

[joeb33050]

Doc seems to be another arithmetic guy, lots of arithmetic but no answer. Here's the question:
How big must a 50 yard target be to look the same as a 5.5" black target at 100 yards?

Or, if that arithmetic is burning a hole in your pocket: What is the relationship between distance and target size, such that apparent size at D1 = apparent size at D2?

The question has to do with eyes as well as arithmetic.
joe b.








First off 20/20 vision is based off one minute of angle. What this means is a person with 20/20 vision can see a 1" black dot on a white piece of paper at 100 yards and a person with 20/15 vision can see a 3/4" dot, and a person with 20/10 can see a 1/2" dot at 100 yards. Now how to find out what a minute of angle is and what it means. A circle has 360 degrees and a degree has 60 minutes, 360 degrees x 60 minutes = 21,600 minutes in a circle. The circumference of a circle is it's outside measurement, and is found by the math formula, ( 2 * PI * Radius ) or ( Two times pi times the radius ). Now lets say we want to find out how big a minute of angle is at 100 yards, ( 2 x 3.141592654 x 3600" = 22619.46711" divided by 21,600 minutes = 1.047197551" @ 100 yards). Now this means that at 1000 yards a true minute of angle is 10.47197551", in layman's terms 10" is close enough at 1000 yards or 1" @ 100 yards, 2" @ 200 yards etc. The size of the high power targets are 13" @ 200 yards or 6.5 minutes of angle, 19" @ 300 yards or 6.33 minutes of angle, 36" @ 600 yards or 6.0 minutes of angle. The difference is .5 minutes of angle between 200 to 600 yards, in other words unless you have better then 20/20 vision like 20/10 you cannot see the difference on these targets. In joeb33050 case if the distance was not exactly 50 yards and 100 yards with a 6" and 3" targets they will not look the same. ANGLES ARE PROPORTIONAL, DOUBLE THE DISTANCE AND DOUBLE THE HEIGHT. Now using the math formula for minute of angle lets apply it to our front sight using 1.0472" as a minute of angle at 100 yards. Example your front sight is 27" from your EYE, and is .070" wide and you want to know how much this covers at 100 yards. (3600" divided by 27" = 133.333 times .070" = 9.33" at 100 yards divided by 1.0472 = 8.912 minutes of angle that your sight covers at all distances, remember ratio and proportion your front sight now covers 18.666" at 200 yards. You can also use the one inch equals one minute at 100 yards with your scope for target size. Say a person has 20/20 vision and they have a 10 power scope on their gun, this means they can see 1/10 minute of angle or .100" at 100 yards, 20 power means .050" again ratio and proportional. I hope this helps and was not too boring.

[Doc Highwall]

joeb33050, if your 100 yard target is 5.5"divide by 2 = 2.75". Joe the relationship is that the 50 yard target is 1/2 the distance of 100 yards so you divide by 2 to get the answer. In small bore shooting at 100 yards the American target is 8" diameter, the prone metric A33 is 8.25" you cannot see the difference except that the A33 has a blacker black ( I think it is the paper that it is printed on) the 50 yard metric target A51 is 4.0" in diameter and the indoor 50' A36 is 1.396" in diameter. The relationship between these 50 yard and 100 yard targets is they are 7.639 to 7.878 minutes of angle and the 50 foot A36 is 8 minutes of angle. The scoring rings on these targets is also proportional, the A33 at 100 yards has .600 wide rings, the A51 at 50 yards has .300 wide rings and the A36 at 50' has .100 wide rings. If you notice the 50' target is 1/6 the distance of 100 yards and the size of the target is kept in relation along with the scoring rings.

[runfiverun]

wow.. more math the simple solution seems to be mine.
maybe i can get #1 or the b.i.l. to do the tape thing again and pull out the ruler.
it appears to also be dependant on your eyesight somewhat, so what i see will be different for someone else?

[joeb33050]

Doc;
If so, why does a 1/2 size target at 50 yards look substantially smaller than a full size target at 100 yards? You should look.
joe b.



joeb33050, if your 100 yard target is 5.5"divide by 2 = 2.75". Joe the relationship is that the 50 yard target is 1/2 the distance of 100 yards so you divide by 2 to get the answer. In small bore shooting at 100 yards the American target is 8" diameter, the prone metric A33 is 8.25" you cannot see the difference except that the A33 has a blacker black ( I think it is the paper that it is printed on) the 50 yard metric target A51 is 4.0" in diameter and the indoor 50' A36 is 1.396" in diameter. The relationship between these 50 yard and 100 yard targets is they are 7.639 to 7.878 minutes of angle and the 50 foot A36 is 8 minutes of angle. The scoring rings on these targets is also proportional, the A33 at 100 yards has .600 wide rings, the A51 at 50 yards has .300 wide rings and the A36 at 50' has .100 wide rings. If you notice the 50' target is 1/6 the distance of 100 yards and the size of the target is kept in relation along with the scoring rings.

[Doc Highwall]

It is impossible unless the targets are the wrong size or at the wrong distance or both. I have been working with angles all my life as a cabinet maker, tool maker, machinist, tool designer and photographer. Angles are ratio and proportion. Some times the juxtaposition between objects throws off the look of how things are perceived, if the back ground is the same for both targets they should appear the same. The back ground has a influence on how object appear, take a look at silhouette targets some are painted black and some are painted white. If you put a 6" target on a 12" square at 100 yards and a 3" target on a 12" square at 50 yards it will make the 50 yard target appear to be smaller because of the back ground of the 12" square only. If the back ground is reduced also you will have a better match.

[joeb33050]

It is impossible unless the targets are the wrong size or at the wrong distance or both. I have been working with angles all my life as a cabinet maker, tool maker, machinist, tool designer and photographer. Angles are ratio and proportion. Some times the juxtaposition between objects throws off the look of how things are perceived, if the back ground is the same for both targets they should appear the same. The back ground has a influence on how object appear, take a look at silhouette targets some are painted black and some are painted white. If you put a 6" target on a 12" square at 100 yards and a 3" target on a 12" square at 50 yards it will make the 50 yard target appear to be smaller because of the back ground of the 12" square only. If the back ground is reduced also you will have a better match.

Did you look? Want me to mail you some targets?
joe b.

[Doc Highwall]

joeb33050, yes you can mail me some targets. But I must confess that I have shot 1000's of shots at these targets in matches and practice with my Anschutz with aperture sights both front and rear and I use the same size aperture in the front sight at both distances for the same light conditions. I have even used a light meter so as to get repeatable results and like all good shooters do I keep a shooting diary for each rifle that I use in competition. This goes for both scope and iron sights, everything is written down, place of match, shooting point to take into account of light and wind conditions etc. etc. I have a 10 meter range in my basement where I can control things like light and with my Anschutz 2002CA that are capable of shooting 10 shots into a .010" .020" group I have ran tests to determine what size aperture I need for my eyes. These were performed with a machine rest and the tests proved conclusively that I was using too small of a front aperture. My rear aperture has a diopter for focusing plus 5 colored filters and two polarizers, and the front and rear apertures are infinitely adjustable from min to max and all of the data was put into my diary with charts. Conclusion, my scores and X count have gone up, I have even shot 100 with 10X's with Irons on the 100 yard small bore target in the prone position. Later today I will take some pictures and post them here.

[RayinNH]

I'm like buck1, my head is starting to hurt :veryconfu...Ray

[AZ-Stew]

Joe,

Doc and I agree on this, though we've said it differently.

Instead of looking at the targets with just your eyes, do as I suggested and look at them through some rifle sights that use an aperture, such as a Garand or a rifle with aperture target sights. If the two targets are at 50 and 100 yards, respectively, and the 50 yard target is half the diameter of the 100 yard target, they should appear the same size in the aperture.

If you're not comparing them through an aperture, you may be mistaking your eyesight's ability to focus equally at the two distances for a difference in apparent size of the bullseyes.

Regards,

Stew

[Shotgun Luckey]

I believe the reason the "scaled" target @ 100 yards looks larger than the target @ 50 yards it the fact that both targets are on the same sized paper. If the scaled target was on scaled paper.

For a 1" target on a 3x3 inch square paper placed at 50 yards to look the same as a 2" target at 100 yards, the paper needs to be on a 6x6 paper