I do not have access to a 500 yard shooting range.  But I want to see if my shooting skills are good enough to hit a 20" diameter target at a distance of 500 yards.   So by what percentage do I need to reduce the 20" target in order to shoot it at a distance of 50 yards and still keep everything proportionally the same?  How about if I want to shoot at a distance of 36 yards?  Thanks.

2" @ 50= 20" @ 500

I would think 2 inches would be it.

50yd * 10 = 500yd
2in * 10 = 20in

But hey I suck at math, no way its this easy.  I'm sure its some 5 page formula

50 of 500 = 10%.
10% of 20" = 2".

Quoted: A math experts is need to figure a simple ratio?

Better than an engrish experts.

Your a Navy guy, right?

yep 2 inches

but rember you need to figure your holdover also

1.44 in. for 36 yd.

Don't forget you will need several box fans blowing at more than 20 times normal wind gusts to complete the simulation.

target software i use to print my own will convert your target to any distance you want

target express

Quoted: target software i use to print my own will convert your target to any distance you want target express

Sweet

Quoted: 1.44 in. for 36 yd. Don't forget you will need several box fans blowing at more than 20 times normal wind gusts to complete the simulation.

Thats what i was thinking.

Plus drop of the bullet needs to be factored in...

Yeah, its not an exact science, but I'd think it would at least tell him if he can hold the rifle steady enough, and have a good trigger squeeze.

I'm a , so I could be wrong.

Quoted:
1.44 in. for 36 yd.quote]

OK, a 2" target at 50 yards is the same as a 20" target at 500 yards.
And 36 yards is closer than 50 yards, therefore the same target should seem to be bigger when viewed at 36 yards than at 50 yards, correct?  But 1.44" is smaller than 2.00".  So how can that be?  My brain just turns to jelly whenever it comes to math.  I appreciate all your help with this, guys.

The closer the target, the smaller you have to make it in order for it to be the size it would appear at 500 yards.

Quoted: A math experts is need to figure a simple ratio?

A+b=c


SGatr15

Quoted: Quoted: 1.44 in. for 36 yd.quote] OK, a 2" target at 50 yards is the same as a 20" target at 500 yards. And 36 yards is closer than 50 yards, therefore the same target should seem to be bigger when viewed at 36 yards than at 50 yards, correct?  But 1.44" is smaller than 2.00".  So how can that be?  My brain just turns to jelly whenever it comes to math.  I appreciate all your help with this, guys.

As you move closer, the target gets smaller. ie: 1.44"@36=2''@50=20"@500

Quoted: The closer the target, the smaller you have to make it in order for it to be the size it would appear at 500 yards.

OK, I get it now.  You are correct.  Thanks for your help again.

That math gives you bull size data, but don't expect to be able to hit the same spot at the 500 as the 50, even with PERFECT wind because the path of the bullet isn't straight out and drop, it has some what of an arc to it, depending on what cartridge/bullet/primer/barrel/powder/etc. It will give you a good idea. However if you need to sight in to shoot at 500 yards, I usually bore-sight, 50 foot sight with the right size target, and then get as close to the range I am sighting for and adjust with the target size. If you just want to see how well you would shoot at 500 yards but have no where to do it, scale it down and then add 10-15% of the distance, adds some more of the difficulty to it.

Quoted: Quoted: A math experts is need to figure a simple ratio? Better than an engrish experts.

Ouch!....

That has to hurt! ( I wood no)

20in is to 500yd as X is to 50yd.

20/500 = X/50
(20x50)/500 = X
so X=2in

My guess is that wind, bullet stability, and cartridge variances will increase the variance of the trajectory over distance. Better nail the 50 and have some consistent loads.

Quoted: A math experts is need to figure a simple ratio?

You asked for a math expert and, unfortunately, you found one.


All the answers above are WRONG!    

The correct answer is, the correct target size depends upon the caliber of the bullet you are shooting.  Here are a few examples:

At 50 yards, shooting a .223 Rem, the correct target diameter is 1.900"
At 50 yards, shooting a .308 Win, the correct target diameter is 1.861"

At 36 yards, shooting a .223 Rem, the correct target diameter is 1.336"
At 36 yards, shooting a .308 Win, the correct target diameter is 1.297"

You asked, I delivered.  Any other questions?    


P.S. - a 20" target at 500 yards is 4 MOA in size.  You should be able to hit it.  As your shooting skills improve, you will hit it with increasing regularity up to the point of hitting it approximately 100% of the time, even at 500 yards (even considering the increased effects of wind at the true 500 yard distance).  

Hitting a small target at 50' has nothing to do with the skill level it takes to hit a larger target at 500'.  I should know, I can't hit either one.

Yeah Mike_Mills, I'd like to understand how the bullet's caliber is a factor in this.  (In layman's terms if possible.)

I knew I'd get someone to take that bait.  Kidding aside, it is an honest answer, and I believe it to be correct.  The simple scaling gets you close, that's for sure, but it is not mathematically correct.  Here's why.

In order to HIT a target, the edge of the bullet only has to graze the edge of the target.  This is true for NRA high power, including service rifle competition.  It is true for IPSC shooting.  It is true in ever shooting discipline of which I am aware, except metallic silhouettes, wherein you have to knock the target completely off the stand.

In order to just graze the edge, the center of the bullet can be one bullet radius outside the diameter of the target.  The target is 20" diameter, therefore, the bullet has to be inside a circle with a diameter of 20" + the bullet's caliber.  For 223 Rem this is a circle of 20.223" diameter (20" + 0.223/2"); for 308 Win, it's 20.308" diameter (20" + 0.308").

If you want to shoot at reduced distances, especially 1/10th the original distance, you have to correct for this in your target size(s).  If you do not make the correction it is analogous, in this case, to shooting with a 2.23" caliber or a 3.08" caliber round - it is easier to hit your target because your bullet is effectively larger in diameter (relative to the target).

When you make the correction, you get the sizes I posted earlier.  I'll leave out the geometry, as you asked for a layman's description.

Now, there is one catch.  Ready?

You need a scoring ring of the slightly smaller diameter as described above but the aiming black must be of the size originally posted by the other guys.  The aiming black must be 1/10th it original size (2.000" diameter at 50 yards and 1.440" at 36 yards).  

By now, I'm sure you think this is total BS.  Think again.  

Have you ever check the sizes of the scoring rings on the 100 yard reduced course targets for high power?  Did you notice they do not exactly scale proportional to the distance?

If you check this you will see that the scoring rings do not simply scale with distance but have an approximate correction for caliber as well.  They probably used 308 caliber and rounded off the results to the nearest hundredth.


Oh well, you asked.  I delivered.  

Good night to all.  It's past my bed time.  

Enjoy your lives.  They are shorter than you know, until it's too late.  God bless.

Should I break into inches = minuites = clicks???????????
<------------------------------------------------------------------------------------------  Constant practicer of Fred's 25m reduced target shooting.  (still not at rifleman stage yet)

Service rifle is shot at 200,300,&600 yds. start over?

Hitting a small target at 50' has nothing to do with the skill level it takes to hit a larger target at 500'. I should know, I can't hit either one.

I know whereof ye speak.

Quoted: Quoted: A math experts is need to figure a simple ratio? You asked for a math expert and, unfortunately, you found one. All the answers above are WRONG!     The correct answer is, the correct target size depends upon the caliber of the bullet you are shooting.  Here are a few examples: At 50 yards, shooting a .223 Rem, the correct target diameter is 1.900" At 50 yards, shooting a .308 Win, the correct target diameter is 1.861" At 36 yards, shooting a .223 Rem, the correct target diameter is 1.336" At 36 yards, shooting a .308 Win, the correct target diameter is 1.297" You asked, I delivered.  Any other questions?     P.S. - a 20" target at 500 yards is 4 MOA in size.  You should be able to hit it.  As your shooting skills improve, you will hit it with increasing regularity up to the point of hitting it approximately 100% of the time, even at 500 yards (even considering the increased effects of wind at the true 500 yard distance).

Uh, what?  You don't shoot highpower do you?  There aren't two different sized targets depending on what caliber rifle you are shooting.  That's complete overkill.

Mike_Mills, thanks for the explaination.  

Quoted: Quoted: Quoted: A math experts is need to figure a simple ratio? Better than an engrish experts. Yikes! I do math better than I spell. I guess we all have our burdens.

Yes, and sometimes I think mine is breathing.

Quoted: A math experts is need to figure a simple ratio?

public schools, what can I say?  

Quoted: Plus drop of the bullet needs to be factored in...

No, it does not.

Once the range is known, the drop is adjusted with the elevation wheel.  Once that is set, one's hold on the target is no different if the target is 50 yards or 500 yards away.

Quoted: Quoted: 1.44 in. for 36 yd.quote] OK, a 2" target at 50 yards is the same as a 20" target at 500 yards. And 36 yards is closer than 50 yards, therefore the same target should seem to be bigger when viewed at 36 yards than at 50 yards, correct?  But 1.44" is smaller than 2.00".  So how can that be?  My brain just turns to jelly whenever it comes to math.  I appreciate all your help with this, guys.

Highpower rifle targets are all designed to LOOK the same through your sights regardless of the distance.

That's why the MR-1 600 yard target with a 36" aiming black LOOKS the same through your sights as the SR 200 yard target does.

The closer the target, the smaller the aiming black needs to be to retain the same APPEARANCE through your sights.

I cannot believe something this basic is causing so much pain.....

Quoted: Service rifle is shot at 200,300,&600 yds. start over?

You forgot that CMP and NRA HP matches can be shot at 200, 300, and 500 yards using the MR target instead of the MR-1 target.

QUOTING thelibertarian:

Uh, what? You don't shoot highpower do you? There aren't two different sized targets depending on what caliber rifle you are shooting. That's complete overkill.

Actually, I have been involved in high power rifle competition since 1994.  I am classified as expert working my way up to master class.  In years past I have re-established my club's full course match program and am currently involved in ressurrecting it again.  It petered out after I left.  Yesterday, I was building targets for use at 200 yards.  Does that qualify me to speak, in your opinion?


You misunderstood me.  There are not two different sized targets for competition.  Rather, I am showing that when the NRA scaled the targets down for use at 100 yards, they reduced the size of the scoring rings consistent with the then prevalent bullet diameter.  

Example#1:  
200 yd SR target has 3.00" X ring.  Simple scaling to 100 yards would make the X  ring 1.50".  BUT, it is actually, 1.35".  This is approximately one caliber smaller than the simple scaling would dictate.

Example #2:
600 yard MR-1 target has 6.00" X ring.  Simple scaling to 100 yards would make the X  ring 1.00".  BUT, it is actually, 0.75".  This is approximately one caliber smaller than the simple scaling would dictate.

Example #3:
The size of the aiming black is not always exactly the same size.  It is close but not exact.  

Even the caliber corrections the NRA used are not exact, close but not exact.  Certainly, they did not do simple scaling.  Actually, I'd like to know the basis for the actual target sizes they use.  It would be an interesting trivia tidbit.  heck, I might even learn something.

Remember, he asked about the mathematics.  When it comes to math, it's all in the details.

Quoted: Quoted: Service rifle is shot at 200,300,&600 yds. start over? You forgot that CMP and NRA HP matches can be shot at 200, 300, and 500 yards using the MR target instead of the MR-1 target.

Correct.  The 600 yard target is actually a modification of the 500 yard target.  They don't even make 600 yard targets anymore.  You have to modify a full face, 500 yard MR target to create a 600 yard MR-1.

Quoted: Quoted: A math experts is need to figure a simple ratio? You asked for a math expert and, unfortunately, you found one. All the answers above are WRONG!     The correct answer is, the correct target size depends upon the caliber of the bullet you are shooting.  Here are a few examples: At 50 yards, shooting a .223 Rem, the correct target diameter is 1.900" At 50 yards, shooting a .308 Win, the correct target diameter is 1.861" At 36 yards, shooting a .223 Rem, the correct target diameter is 1.336" At 36 yards, shooting a .308 Win, the correct target diameter is 1.297" You asked, I delivered.  Any other questions?     P.S. - a 20" target at 500 yards is 4 MOA in size.  You should be able to hit it.  As your shooting skills improve, you will hit it with increasing regularity up to the point of hitting it approximately 100% of the time, even at 500 yards (even considering the increased effects of wind at the true 500 yard distance).

I wondered when that would come up...you guys call yourselves shooters????HA...
I didnt know either...but guessed 2in

Quoted: I do not have access to a 500 yard shooting range.  But I want to see if my shooting skills are good enough to hit a 20" diameter target at a distance of 500 yards.   So by what percentage do I need to reduce the 20" target in order to shoot it at a distance of 50 yards and still keep everything proportionally the same?  How about if I want to shoot at a distance of 36 yards?  Thanks.

Shooting a 2in target at 50 yards is not the same thing as shooting 20in target at 500. at 50 you don't have to worry about bullet drop, wind or anything like that. It won't really tell you much by doing that.

Quoted: I knew I'd get someone to take that bait.  Kidding aside, it is an honest answer, and I believe it to be correct.  The simple scaling gets you close, that's for sure, but it is not mathematically correct.  Here's why. In order to HIT a target, the edge of the bullet only has to graze the edge of the target.  This is true for NRA high power, including service rifle competition.  It is true for IPSC shooting.  It is true in ever shooting discipline of which I am aware, except metallic silhouettes, wherein you have to knock the target completely off the stand. In order to just graze the edge, the center of the bullet can be one bullet radius outside the diameter of the target.  The target is 20" diameter, therefore, the bullet has to be inside a circle with a diameter of 20" + the bullet's caliber.  For 223 Rem this is a circle of 20.223" diameter (20" + 0.223/2"); for 308 Win, it's 20.308" diameter (20" + 0.308"). If you want to shoot at reduced distances, especially 1/10th the original distance, you have to correct for this in your target size(s).  If you do not make the correction it is analogous, in this case, to shooting with a 2.23" caliber or a 3.08" caliber round - it is easier to hit your target because your bullet is effectively larger in diameter (relative to the target). When you make the correction, you get the sizes I posted earlier.  I'll leave out the geometry, as you asked for a layman's description. Now, there is one catch.  Ready? You need a scoring ring of the slightly smaller diameter as described above but the aiming black must be of the size originally posted by the other guys.  The aiming black must be 1/10th it original size (2.000" diameter at 50 yards and 1.440" at 36 yards).   By now, I'm sure you think this is total BS.  Think again.   Have you ever check the sizes of the scoring rings on the 100 yard reduced course targets for high power?  Did you notice they do not exactly scale proportional to the distance? If you check this you will see that the scoring rings do not simply scale with distance but have an approximate correction for caliber as well.  They probably used 308 caliber and rounded off the results to the nearest hundredth. Oh well, you asked.  I delivered.   Good night to all.  It's past my bed time.   Enjoy your lives.  They are shorter than you know, until it's too late.  God bless.

He could just score his target based upon the center of his bullets impact. That will make his scoring more difficult, but then he will be able to directly convert his 50 yard results to 500 yard results, windage issue ignored, at least.

Actually, I think this is more of a "first cut" at estimating. We are assuming that all errors are linear with distance, which isn't necessarly the case. Wind effects are an obvious example, but I believe that sight alignment (as opposed to sight picture) and inconsistent hold (position, position, position) can lead to nonleanear results.

Quoted: Quoted: 1.44 in. for 36 yd.quote] OK, a 2" target at 50 yards is the same as a 20" target at 500 yards. And 36 yards is closer than 50 yards, therefore the same target should seem to be bigger when viewed at 36 yards than at 50 yards, correct?  But 1.44" is smaller than 2.00".  So how can that be?  My brain just turns to jelly whenever it comes to math.  I appreciate all your help with this, guys.

36yards is 25meters.

Quoted: Quoted: Quoted: 1.44 in. for 36 yd.quote] OK, a 2" target at 50 yards is the same as a 20" target at 500 yards. And 36 yards is closer than 50 yards, therefore the same target should seem to be bigger when viewed at 36 yards than at 50 yards, correct?  But 1.44" is smaller than 2.00".  So how can that be?  My brain just turns to jelly whenever it comes to math.  I appreciate all your help with this, guys. 36yards is 25meters.

No.  36 yards is closer to 30M.  25M is 27.33 yards.  A meter is 8% longer than a yard.

Quoted: Quoted: Plus drop of the bullet needs to be factored in... No, it does not. Once the range is known, the drop is adjusted with the elevation wheel.  Once that is set, one's hold on the target is no different if the target is 50 yards or 500 yards away.

Adjust for the drop of a 308 at 500 yards, than shoot a target at 50 holding dead on as if you were shooting at 500 and you will not touch the target.  When I shoot at 100 with my 308 and than make the roughly 2in adjustment for 200 yards(ok it's more like an inch and a half) and fire holding dead on as if I was shooting at 200 guess what, my bullet hits 2 inches higher than point of aim. While the hold may be no different the POA/POI will be different.

Quoted: Quoted: 36yards is 25meters. No.  36 yards is closer to 30M.  25M is 27.33 yards.  A meter is 8% longer than a yard.

I see more math is required:  

36 yards = 32.92 meters

Quoted: Quoted: Quoted: 36yards is 25meters. No.  36 yards is closer to 30M.  25M is 27.33 yards.  A meter is 8% longer than a yard. I see more math is required:   36 yards = 32.92 meters

Mike note the 'Closer to' it was an approximation.  The 25M was an exact figure hence the use of 'is'.

None of this matters, because you can hit shit at 500yds based on how you shoot at 50 yds.  You have almost nothing to factor at 50 yds, no windage, elevation or climate/light.  Every month I get to meet another retard who shot a "good score" or "qualified Expert" in the military on a reduced course and can hit his ass with both hands on a bullseye course or farther than 100 yds.  Go shoot 500 yds already.  

Quoted: target software i use to print my own will convert your target to any distance you want target express

on the m16a2 25m zeroing target i see that it refer back to ar15.com for sighting instructions.does the site get any revenue from the program or does ar15.com own the program?