Rimfire Pistol Accuracy

[jtg]

What is a realistic expectation of accuracy for a RF Bullseye pistol? I tend to get wrapped up in the pursuit of more accuracy and really would like to know when is enough enough.
I know at least one will say that the gun shoots better than i can but if the gun isn't capable of some baseline accuracy then how will a shooter know if he is improving or if the gun is the limitation.
We all have our arc of motion that defines our group size and the gun adds to this. If we want to truly see improvement then where does the gun accuracy need to be and at what cost?

[Steve Swartz]

The "Right Answer:"

. . . as close to a single hole at 50 yards as you can afford . . .


Oh, sorry, you said "realistic expectation!"


A "Better Answer" (but you might not like it):

I don't think the accuracy from a RR is really an issue. Even the old standby Ruger Mark II is capable of a ten shot group entirely enclosed by the X ring at 50 yards, if you use even halfway decent target ammo.

The problem is accuracy from your hand; and that's what you pay the $$$ for- "ergonomics."

Just my opinion- but I think the issue of accuracy for the .22 is somewhat moot; get a decent target gun that fits your hand (most important), then buy good quality target ammo.

Steve Swartz

[p.s. I wholeheartedly agree that the old phrase "the gun shoots better than I do" is not only technically incorrect (there is no such thing) and even worse misleading. But in this case the point doesn't matter.]

[Fred Mannis]

Just my opinion- but I think the issue of accuracy for the .22 is somewhat moot; get a decent target gun that fits your hand (most important), then buy good quality target ammo.

Steve,
Just to flesh out your decent target gun that fits your hand suggestion a little -
yes, grips that fit your hand and allow proper placement of the index finger on the trigger while achieving proper sight alignment

also weight and balance that feels good in your hand

and a trigger that can be adjusted so as to not disturb the sights when released.

That's where the $$$ come in :-)

Fred

[tenex]

Hello JTG,
Here's my litmus test for accuracy. Since the gun and the shooter represent independent random variables (from a statistical point of view), I can take the sizes of the groups and add them using the Pythagorean theorem, overall group size = sqrt(gungroup^2 + shootergroup2^2). Where this leads is the following:

If your gun shoots groups 1/3 the size of your typical offhand group, the gun adds about 5% to the size of your groups.

If your gun shoots groups 1/5 the size of your typical offhand group, the gun adds about 2% to the size of your groups.

It doesn't take long before the gun performance isn't a significant issue. I expect my .22's to shoot a 2" or better 10 shot group at 50 yards (this is just my performance standard, your's may be different). If they do better that's great, but I wouldn't waste any cash to do better. The big fallacy is that a 2 inch gun and a 6 inch shooter makes an 8 inch group, they don't, they make a 6.3 inch group.

You can set your own performance requirements, but this is how the errors add.

Sorry if I didn't do a good job of explaining,
Steve.

[Spencer]

After a series of machine rest tests a year or so ago, we were amazed at how bad the grouping at 25m was for some pistols, and some pistol/ ammunition combinations. The 'tests' were with a variety of pistols (Pardini, Walthers, IZH, Hi-Standard, etc.) and ammunition

For some 40 years I had accepted the adage that a well maintained, reasonable target .22 pistol and good ammunition would deliver acceptable accuracy - and we learnt otherwise!
Some combinations were grouping the size of the 8-ring at 25m - hard to shoot a good score with that type of 'performance'.

After individually matching pistols to ammunition, and re-crowning one of the pistols, all were capable of holding the 50mm 10-ring (or better) with average cost ammunition.
It was interesting that there was no one brand of ammunition that suited all the pistols.

Spencer

As a PS, I am yet to find a target Air Pistol that will not deliver good groups with target pellets (provided the MV is above about 400 fps)
S

[Steve Swartz]

Perhaps we have slighly different definitions of "target" guns and ammo.

My personal "target gun" definition FWIW does not require a Free Pistol per se, just a firearm specifically designed for and marketed as a target shooter.

Ruger Mark II, S&W 41, High Standard, etc. See a lot of them on the line at big Bullseye matches. (heck, the M1911/.22LR conversions even)

My personal "target ammo" definition FWIW does not require Eley Tenex, just ammunition specifically designed and marketed as target shooting ammo.

CCI, Federal, Remington standard velocity etc.

I would be pretty surprised (and irked!) if I bought a "target" 22 pistol and loaded it with "target" ammunition (ANY target pistol, ANY target ammo) and it grouped 8 ring at 25 yards!

A random Ruger Mark II (a low end "plinking" pistol) shooting cheap CCI standard velocity (blue box, not green box) should shoot 3" groups at 50 yards even if you get a "bad" gun and "bad" ammo. I know mine does.

Spencer, I am very surprised that some of the ammo you and your mates tested performed that poorly.

I'm very curious as to what those poorly performing gun/ammo combinations were!

Steve

[jackh]

Steve.
The only target 22s I ever Ransom tested that did not group under 2" at 50yds was the keyholing 41 that later got relined, and one HS that needed a a recrown.

[Spencer]

Steve,

Unfortunately I did not keep a record of our tests; we each sorted out what worked in our pistol/s and have stuck with it.

I remember that for group size my 208 did not like CCI, but was (is) 'happy' with both Winchester Supreme Pistol and Lapua Magazine. These returned groups as good as some of the high-end ($$$) brands of ammunition.

Interesting to note :
- that the pistol that was least affected for group size by the various ammunitions was my venerable old Hi Standard.
- the change in group centre between some brands of ammunition.

Spencer

[Shooting Kiwi]

Interesting question, jtg: I'd also like to know the answer. However, I'm not sure that it's really worth knowing.

As tenex has shown, reducing a small-order error (the gun's inaccuracy) doesn't do much for overall accuracy, given that the shooter is the dominant source of error.

I suspect that the 'accuracy' of a gun depends on how the experiment is conducted. Rifles, with their well-known barrel vibration being a major determinant of (in)accuracy, will perform differently depending upon how they are restrained at the time of firing, and until the bullet has left the barrel. I would expect that barrel vibration is much less of a problem with a pistol, but movement of the entire gun, hand, arm and individual components of the gun will be highly significant.

The test group supplied with my Hammerli 208 is comparable to a reasonable small-bore rifle. How was it produced? I've no idea how the gun was constrained, nor what ammo was used for the test. Without a barrel weight on it, I shoot all over the place. With a suitable barrel weight, my accuracy is quite respectable. Now, has the inherent accuracy of the gun changed? The lower barrel line, grip and different balance of my FAS suits me even better - I'm more accurate, but is the gun's inherent accuracy better than the 208? Unlikely.

Change the trigger characteristics and your accuracy will change, etc., etc.

So I would still like the nice warm feeling of knowing I have an 'accurate' gun, but I also know that its inherent accuracy is completely swamped by all the other variables. I think you have to consider the package as a whole - gun, ammo, shooter, atmospheric conditions and visibility, and a few others I can't think of.

I add this not pretending to be an authority, but to add weight to the frequently given advice that what works for you is all that's important.

[alb]

Steve,

I'm having trouble understanding your analysis.

Since the gun and the shooter represent independent random variables (from a statistical point of view), I can take the sizes of the groups and add them using the Pythagorean theorem, overall group size = sqrt(gungroup^2 + shootergroup2^2).

The big fallacy is that a 2 inch gun and a 6 inch shooter makes an 8 inch group, they don't, they make a 6.3 inch group.

Please explain. A 6 inch group shooter is a shooter whose mean group size is 6 inches, not a shooter whose group size has a standard deviation of 6 inches. Likewise, with the accuracy of the gun. If you are looking for the standard deviation of the resulting group when you add the two sources of error together, then it makes sense to add the two standard deviations geometrically, since the variances add algebraically.

However, the sum of the means of the two sources of error also adds algebraically, so a 6 inch shooter with a 2 inch gun really will shoot 8 inch groups:

X = N(u,s)
Y= N(v,t)
Z = X + Y ~ N(u+v, s+t)

where X is the normally distributed random variable for the shooter's group size, Y is the normally distributed random variable for the gun's group size, Z is the normally distributed random variable of the sum of the two group sizes, s and t are the variances, u is the mean size of the shooter's group and v is the mean size of the gun's group.

If I'm wrong, please help me to understand why.

Thanks,

Al B.

[jr_roosa]

I think the way the math works for why 2" + 6" = not 8" is that to get 8" right of center you'd have to simultaneously have the shooter holding all the way right in his wobble on a shot that the gun was going to throw all the way to the right as well, and that this isn't likely to happen much.

Remember that if we take our "6-inch shooter" and give him a perfect gun, he isn't going to shoot his 10 shots in a circle with every shot exactly 3" from the center. Also, our "2-inch gun" won't place each shot 1" from the center. Most of the shots from each group will be much less than 3" or 1" from the center...in fact by definition all but two shots in a 6" group (one that fits in a circle with a diameter of 6") are less than 3" from the center of the group.

The way I conceptualize it is to imagine that we give the 6" shooter a "perfect" gun (one that gets 0" groups) and record where all ten bullets go, 1-10. Then take the 2" gun in a vise and record where all ten bullets go as well. Then we combine the two lists to try to simulate what we might have had if the shooter had a 2" gun instead of his "perfect" gun. We could take where each offhand shot went and then move it in the direction and distance that the corresponding shot from the vise was from the center.

Say bullet 1 from the shooter is dead center, but bullet 1 from the vise was a flyer, say a full 1" right. When we combine these two shots we get a composite that is 1" from the center...not bad.

Say bullet 2 from the shooter is 2 inches high, and 2 from the vise is 1 inch low. Although these are both inaccurate, we don't get a composite shot 3" from the center, but rather the errors cancel a little and you end up only 1" high.

Maybe bullet 3 was as bad as possible in both realms. It was 3" right offhand and 1" low from the vise. If we combine those, we get a hole that is a little more than 3" from the center, but maybe at around 4:00 on the target...not much different than what we get with a "perfect gun."

If you go on and on for a few hundred shots, you'll find that somtimes the errors add, sometimes they cancel, and the overall effect is that your average 10-shot group is closer to 6 inches (what the shooter would do with a perfect gun) than to 8 inches you get when you add 2+6.

This only happens when the shooter with the perfect gun would make a much bigger group than the imperfect gun in a vise. With most pistols and most pistol shooters this is the usual situation.

On the other hand, a shooter with a perfect rifle can shoot a group A LOT closer to the size that a real gun could make from a vise. In that case, the inaccuracy of the rifle would cause more of a problem, and would be worth paying more money to correct to get good results.

I guess instead of saying "my pistol shoots better than I do" we should say "My pistol isn't contributing to the size of my groups to reduce my scores enough to warrant its replacement at this time. Perhaps when my group size approaches 3 inches I'll consider the purchase of a more accurate pistol" Sure the second way is more correct, but the first way sounds less like a Vulcan. Either way we mean the same thing.

Oh, and the mathematics don't take into account a gun that shoots great from a vise but has a 30-pound trigger and grips made from broken glass. Nobody but the vise could shoot that gun well. A gun that is "easy to shoot well" is not necessarily the same as an "accurate gun."

-J.

[alb]

JR,

Thanks for your reply.

The equations I gave in my previous post are straight out of a math book:

X = N(u,s)
Y= N(v,t)
Z = X + Y ~ N(u+v, s+t)

The average group size for a given number of shots will be normally distributed, whether we are talking about an imperfect shooter with a perfect gun, or a perfect shooter with an imperfect gun, or an imperfect shooter with an imperfect gun.

The size of a shot group is defined by the minimum covering circle, which has either two or three shots on its circumference (it can have more than 3 shots on its circumference, but the size of the minimum covering circle is always defined by either 2 or 3 shots.

Think of these shots as each having a dispersion pattern around them, representing the smaller source of error, i.e., a smaller circle surrounding a point on the circumference of a larger circle, which represents the larger source of error. The diameter of the smaller circle added to the diameter of the larger circle defines the diameter of the sum of the two sources of error.

Remember that we are adding the averages, i.e., the 'means', of the two sources of error. Sometimes, one or both sources of error will be larger than average, sometimes less than average. Sometimes they will cancel each other out, sometimes just the opposite, and sometimes, somewhere in between.

When we calculate the sum of the standard deviations, we do a geometric sum, as Steve indicated. This is because the variances sum algebraically, as stated in the equations from the math book. In the equations that I presented in my previous post, the standard deviations are: sqrt(s) and sqrt (t), where 's' and 't' are the variances of the random variables 'X' anf 'Y'. The standard deviation of the random vairiable representing the sum of the random variables 'X' and 'Y' is the square root of the sum of the variances of 'X' and 'Y', which is equal to the square root of the sum of the squares of the standard deviations of 'X' and 'Y'. This is what I think Steve was alluding to.

It still appears to me that based on the math, as well as as common sense, a 6 inch shooter with a 2 inch gun will shoot 8 inch groups on average.

The practical significance of all of this is whether or not it is important to use an accurate gun with accurate ammunition, even though shooter error may be siginificantly larger than the error contribution of the gun and ammunition. Steve suggests that the accuracy of the gun isn't so important, I would suggest that it is.

Regards,

Al B.

[Steve Swartz]

Al:

Part of the misunderstanding lies in definitions and perspective. The true unit of analysis is not the ES of the group, but the distribution of individual shots from the centroid.

Each individual shot contains error (in direction and distance) from the desired mean point of aim.

If we assume the desired mean point of aim and desired mean point of impact are the same (gun properly sighted in), then the errors we need to add are the errors from the shooter and the errors from the gun for each individual shot.

Take one source of error as the "primary" error and the other as the "secondary" error. The primary error will always be "away" from the DMPI. Sometimes larger, sometimes smaller, dispersed around the DMPI. Assume uniform distribution for direction of error, and gaussian for the magnitude (centered on distance=0). Assume the same for the secondary error; except that the secondary error is with respect to the "endpoint" of the first error.

Roughly half the time, the secondary error will be away from the DMPI (positive) and half the time the secondary error will be toward the DMPI (negative); with magnitude away/toward DMPI. The secondary error therefore "moderates" the primary error.

So a +- 3 standard deviation ES between shots in the group (resulting from the primary error) will not simply add to the secondary error.

I *think* that's what Steve/Tenex was implying.

[alb]

Take one source of error as the "primary" error and the other as the "secondary" error. The primary error will always be "away" from the DMPI. Sometimes larger, sometimes smaller, dispersed around the DMPI. Assume uniform distribution for direction of error, and gaussian for the magnitude (centered on distance=0). Assume the same for the secondary error; except that the secondary error is with respect to the "endpoint" of the first error.

Roughly half the time, the secondary error will be away from the DMPI (positive) and half the time the secondary error will be toward the DMPI (negative); with magnitude away/toward DMPI. The secondary error therefore "moderates" the primary error.

Steve,

Let me try to put it another way. The shooter is always trying to hit the center of the target, but, being human, he doesn't always point the gun at exactly the same spot on the target, with exactly the same velocity and acceleration across the target face, so his shot won't go in exactly the same place each time. We call this 'shooter error', what you referred to as the primary error.

The gun itself, being a mechanical system, isn't perfect either. So, regardless of where the shooter points the gun when the shot breaks, there will be some dispersion around this point on the target. This is what you refer to as secondary error. So far, so good.

Both the shooter error and the gun error follow a 2-dimensional Gaussian distribution. However, if you are going to express this in terms of polar coordinates, i.e., magnitude and direction, then the direction will indeed be uniformly distributed (unless there is wind or some mechanical defect). However, the magnitude of the error will be 'Rayleigh' distributed.

What Tenex was referring to, however, was average group size, which follows a Gaussian distribution. The gun itself has an average group size, or dispersion, which also follows a Gaussian or 'normal' distribution. Likewise for the shooter. Both random variables, gun group size and shooter group size, are independent of each other. The average group sizes, i.e., the means of these two independent normally distributed random variables, sum algebraically. For an on-line discussion of this, see:

http://en.wikipedia.org/wiki/Sum_of_nor ... _variables

The size of a shot group is defined by its covering circle, i.e., the smallest circle that will cover all of the shots. This circle is determined by either 2 or 3 of the shots in the group. Extreme spread is not as accurate or as efficient a measure, since a good portion of the time, the extreme spread only measures a chord of the covering circle of the group (See "Statistical Measures of Accuracy for Riflement and Missle Engineers", Frank E. Grubbs, Ph.D, 1964, pgs. 14-16). Nearly every discussion that I have read on the relationship between number of shots in a group and group size starts off with a Monte Carlo simulation that measures extreme spread rather than diameter of the covering circle, and then the author can't do the math to explain his erroneous results.

Be that as it may, when you add the dispersion of the gun to the dispersion of the shooter, the average magnitude of the dispersion is the same in every direction, as you pointed out yourself. Now, if the gun has an average group size of 2 inches with a variance of 0.5 inches, and the shooter has an average group size of 6 inches with a variance of 2 inches, when we put the two together we get an average group size of 8 inches with a variance of 2.5 inches. But the standard deviation for the gun's group size is 0.707 inches and the standard deviation for the shooter's group size is 1.414 inches. When we add these two together, however, the standard deviation for the resulting group size is 1.58 inches, not 0.707 + 1.141 inches, as Tenex pointed out in his post.

So a +- 3 standard deviation ES between shots in the group (resulting from the primary error) will not simply add to the secondary error.

We may very well be in violent agreement on this. But it still appears to me that Tenex was confusing mean and standard deviation.

Another way to think of it is this: Draw a circle to represent the shooter's average group size. Mark two or three points on the circumference of this circle. Now draw a smaller circle around each of these points to represent the average group size or dispersion of the gun. Now draw another circle that encloses everything. The diameter of this circle will be the sum of the diameters of the shooter's dispersion circle and the gun's dispersion circle.

Remember, we are talking about adding the effect of gun dispersion to the effect of shooter dispersion to get total dispersion, i.e., we are adding average group sizes.

Regards,

Al B.

[Steve Swartz]

Al:

I ran a quick simulation in Excel . . . set it up to be convenient so modeled 100 shot strings. Calculated Extreme Spreads for the primary error (set it up for 3" group sizes), secondary error (set it up for 1" group sizes), and the combined error; allowing negative errors to cancel each other out:

Average Extreme Spread, Primary: 2.58
Average Extreme Spread, Secondary: 0.91
Average Combined Extreme Spreads: 2.73

O.K. this is flawed but was fast- one thing I ignored was the angular effect of the secondary error (ie the fact that the offset of the secondary error was radial- this would lessen the impact of secondary error significantly). The flaw actually makes the analysis more "conservative" (likely to magnify the combined error).

Anyhow

We can see the errors do not cumulate- the fact that a primary error could be "tot eh right" and the gun error could be "to the left" (compensating errors) actually makes the combined group size smaller than a straight add of errors.

However

If I get the time today I think it would be interesting to properly model the relationship in two dimensions . . .

[Steve Swartz]

Al:

Barring empirical data- [boy I would *love* to be able to capture/convert positioning stream data directly out of my Rika- anyone know how to do that?]- the following modeling assumptions could apply?


If I remember correctly we agreed a while back that it was valid to model the traversing of the muzzle over the desired mean point of impact (gun properly sioghted in etc. etc.) as a random phenomenon, such that:

- The distribution of time spent over any one portion of the aiming area was "mound shaped" and "symmetrical" in two dimensions
- The distribution of distance away from the dmpi could be modeled as normally distributed in two dimensions (ie standard normal times some "error magnitude constant")
- Therefore to model the distance away from the dmpi in one dimension would simply involve converting to the "absolute value" of the two dimensional values
- It was fair to ignore the "random walk" vs. "independent" nature of the muzzle traversing around in two dimensions, as the distribution captures the "focusing effect" of the athlete trying to keep the muzzle painted on top of the dmpi.

This allows us to sample randomly from the distribution as an approximation of where the muzzle would be pointed at any moment in time as the shot is released. Over a suitably large number of trials, of course.

Is this acceptable?

How could the modeling be improved- other than going to a continuous probabilistic system-state simulation (hate those)- as a discrete event probabilistic simulation?

[alb]

- The distribution of time spent over any one portion of the aiming area was "mound shaped" and "symmetrical" in two dimensions
- The distribution of distance away from the dmpi could be modeled as normally distributed in two dimensions (ie standard normal times some "error magnitude constant")

Hi Steve,

I'm not sure what you mean by your second statement. The magnitude of the 'distance away from the DPMI' is Rayleigh distributed, i.e., Weibull distributed with a shape parameter of 2 and a scale parameter of 1.414. This is well known to antenna engineers, and I've seen the proof in elementary probability and statistics texts. Your first statement is correct though. Shot dispersion is normally distributed in 2 dimensions.

- Therefore to model the distance away from the dmpi in one dimension would simply involve converting to the "absolute value" of the two dimensional values
- It was fair to ignore the "random walk" vs. "independent" nature of the muzzle traversing around in two dimensions, as the distribution captures the "focusing effect" of the athlete trying to keep the muzzle painted on top of the dmpi.

This allows us to sample randomly from the distribution as an approximation of where the muzzle would be pointed at any moment in time as the shot is released. Over a suitably large number of trials, of course.

Is this acceptable?

How could the modeling be improved- other than going to a continuous probabilistic system-state simulation (hate those)- as a discrete event probabilistic simulation?

I have software that computes the minimum covering circle for a group of shots. I wrote it a while back because I was interested in the effect of dropping outliers on the variance of the resulting group size. When I set my software to not drop any shots, the results exactly match the analytical results obtained by H. E. Daniels in his 1952 paper, "The Covering Circle of a Sample From a Circular Normal Distribution." In order to compute the minimum covering circle, I used the Elzinga-Hearn minimum covering circle algorithm. I use a database of 150,000 randomly generated shots generated using Excel. I could generate an additional 150,000 random shots to represent secondary errors, or divide my current database in half, to give 75,000 primary errors and 75,000 secondary errors that would sum to 75,000 random shots (I would apply a scale factor to the secondary errors to represent gun errors, e.g., divide the secondary errors by 3).

I don't like the use of extreme spread to compute group size, for reasons that I previously described -- this approach also requires that the secondary errors in the two extreme spread primary shots to be 180 degrees opposite one another in order for the means to sum algebraically. In the case of the simulation that you just described, geometric addition of the average extreme spreads does indeed appear to work.

I'll try a simulation using minimum covering circle tomorrow, and let you know when I get some results. I'll be suprised and intrigued if the means don't sum algebraically.

In the mean time, it's 9 pm, and this is causing my brain to hurt, so I'm going to go back to watching the Phillies game.

Regards,

Al B.

[alb]

Steve,

I ran a simulation using the covering circle method. First I generated 65,536 random shots with a mean of 0 and a std. deviation of 6. This produced 6,553 10-shot groups with a mean size of 11.6068 and a standard deviation of 5.2459.

Next, I generated 65,536 random shots with a mean of 0 and a std. deviation of 2. This produced 6,553 10-shot groups with a mean size of 3.87069 and a standard deviation of 0.590017.

Last, I generated 65,536 random shots by adding the values for the first two strings of shots together. This produced 6,553 10-shot groups with a mean size of 12.2482 and a standard deviation of 5.79657.

The bottom line is that the means do appear to sum geometrically. Also, the standard deviations appear to sum algebraically.

I ran the results through Statistica. The group sizes appear to follow a gamma distribution, not a normal distribution as I had assumed. So the equations that I referenced in my previous posts don't apply.

In any event, Steve/tenex appears to be correct. At least I learned something.

Regards,

Al B.

[tenex]

Hi Guys,
This has turned in to a pretty entertaining thread, mostly for the guys at work abusing me for my math skills (thanks John & Sean).

The short story is that if every shot is the product of 2 random uncorrelated events (the shooter error + the gun error) then the variances of the two probability distributions add, and the total standard deviation of the group is equal to the square root of the sum of the individual standard deviations squared.

Measuring group size is just an easy way to get an estimate of standard deviation without doing a lot of measuring. It's really not a measurement of the mean of the group, which is just the location of the center of the cloud of shots.

Rather than kill everyone with equations (I'm saving that for the malcontents at work), here's a simple visualization you can use to see the effect of gun performance on your score. Imagine you aim a shot right on the edge of the 7 ring. now add a small circle around that point to represent the spread of the shots due to the gun. if the circle is small, about half the area is on the 7 side, and half is on the 6 side and you average to a 6.5. now make the circle a little bigger, and because of the curvature of the 7 ring you get a few more 6's than 7's. If you put a 1" diameter circle on the 7 ring of a B2 target you'll see that you don't get that many more 6's even though the gun's group size is 1/3 of the target size (7 ring, in this case). If you look at it long enough, you'll see that the reduction in score isn't linear with the group size of the gun, it rapidly goes to 0 with smaller and smaller gun group sizes. At some point, the gun performance just isn't a significant factor. Sure, the gun can turn a 7 into a 6, but it can also turn a 6 into a 7, just not quite as often.

Hope this sheds some light on a subject with much uncertainty,

Steve.

[Steve Swartz]

Well, not to "stir the pot" furhter (of course that's exactly what I intend!) but somewhere lost in the mists of time is the issue of "how much accuracy do I need" from the gun itself.

Related to one of my pet peeves: the ridiculousness of the statement "The Gun Shoots Better Than I Do!"

So what *is* an acceptable limit of gun error?

I think it is related tot he skills of hte shooter . . . but not exaclty in the way one might think!

How quickly will a shooter learn proper shooting fundamentals when the outcome of each shot is a random event?

The problem goes beyond the direct effect on your system group size . . .