I read this before, but it is interesting to review.

Years ago, I shot a Hunter Class Bench Rest rifle in 308 Win. State of the art bullet was the 150gr SMKs. I learned that Sierra used three point up dies to form the final nose shape, but all three fed into one bin for a final tumble to polish them.

Sinclair and others offered a hex nut looking ogive checker fitted to your dial calipers. It had six holes for the six usual BR bullets. You would fit it into the caliper jaws and lock it down. Then, you took the bullets one at a time and fit them into the appropriate hole. When you slid the caliper jaws shut, you would get a reading on the dial. That reading showed bullet base to ogive. Ogive is where the bullet diameter is land diameter. That point on the bullet would touch the origin of rifling.

Others before me found that each final point up die was different in length. There were three different lengths. Two guys from the LaGrande, OR, club and I split a 5000 round box of Sierra's and gauged them all one winter.

The measurements for the short third were set at zero. The middle third was +.004, and the long third was +.009" longer (total). We split them and found much better accuracy by doing so, I was jumping mine .010", that is ten-thousandths of an inch to the origin or rifling. Accuracy testing would show a rifle's preference for the amount of jump. Some few found that jamming a bullet .010" worked. Most found the .010" to .025" area found the node.

Anyway, neat stuff.

Very good example of the question at hand- however, in reality, what is the effect on the down range performance of determining SD, then grouping all of the 3 SD bullets into one random group?

I had some time waiting for my canned tomatoes to cook, so here is what I came up with-

n=182
AV=108.97
SD=3.59

1SD=105.38 to 112.56 (131 bullets)
2SD=101.79 to 116.16 (169 bullets)
3SD= 98.2 to 119.75 ( all but 2 bullets, the .020s)

So, almost all bullets fell into the 95% interval, 92.8% fell into 2 SD, and 71.9% fell into +/- one SD.

What would be interesting would be how well a representative shot group fired with say five bullets from each interval ( a 15 round group) would perform at say 300 or 600 yards? ( short enough to reduce other wind or geographic based elevation variables?

I know from personal experience that a random batch of bullets, brass and thrown powder can clean the 2MOA 10 ring of a 600 yard MR target given a good rifle and a decent shooter- from sling prone with iron sights....

Now, hitting a 9x9 plate first shot at 1200 yds- that may be another game altogether!

25 years ago, and the split was nearly evenly spread.

I guess my point is "is a SD of .00359" for OAL of match bullets good or bad?"

Just because nearly all of the bullets fall into +/- 3SD does not mean the SD is a good example of controlling variation.

Secondly, just how does this sample of bullets perform on target? Does an OAL variation of .020" really matter? And if so, to what degree at what distance?

In lay-man's terms, SD is a fancy way of saying "Probably", when predicting what the next shot(s) will do.

In lay-mans terms extreme spread is a way of saying "Worst case scenario", when predicting what the next shot(s) will do.

Average is average, to a lay-man or anyone else.

Average will appeal to the idealist.

SD will appeal to the realist.

ES will get the attention of a pessimist.

The important measurement for bullets is the length of the bearing surface, yet OAL of the bullet correlates with the bearing surface length.

This is akin the fact that cartridge base to ogive length is more important than OAL cartridge length.

A large ES = a large SD

I think the CBTO is more critical; due to the fact that it sets the distance before the bullet bearing surface engages the rifling.

It is possible to have a fairly large ES, but still maintain a very respectable SD.

For example, look at the following two sets of data.

1st set

3000,3010,2990,3005,2995,2993,3003,3007,2997,3012

Average=3001.2; SD=7.4; ES=22

2nd set

3000,3010,2990,3005,2995,2993,3003,3007,2997,3012,3000,3010,2990,3005,2995,2993,3003,3007,2997,3032

Average=3002.2;SD=9.7;ES=42

The almost twice as large ES for the 2nd data set suggests there is an outlier, which is this set is the 3032 value. Yet, both sets have single digit SDs.

The SD and ES are numerical means of describing the shape of the bell shaped curve of a data set.

quote:

Originally posted by Ackley Improved User:
It is possible to have a fairly large ES, but still maintain a very respectable SD.

For example, look at the following two sets of data.

1st set

3000,3010,2990,3005,2995,2993,3003,3007,2997,3012

Average=3001.2; SD=7.4; ES=22

2nd set

3000,3010,2990,3005,2995,2993,3003,3007,2997,3012,3000,3010,2990,3005,2995,2993,3003,3007,2997,3032

Average=3002.2;SD=9.7;ES=42

The almost twice as large ES for the 2nd data set suggests there is an outlier, which is this set is the 3032 value. Yet, both sets have single digit SDs.

The SD and ES are numerical means of describing the shape of the bell shaped curve of a data set.

In your first data set your SD is 7.4 which would suggest an extreme spread of 44.4, which is very close to the ES you saw when you shot ten more shots (the first data set is the first half of the second data set). The 99.7% confidence limit for the second set is an ES of 58.2.

AAZW,

What's your point?

The ES for the first set is 22 (i.e., the difference of 3012 minus 2990), for the second 42 (i.e., 3032 minus 2990).

The ES (extreme spread, data extent, or range) is the difference between the lowest and highest observed velocities.

44.4 = 6 x SD (7.4)
58.2 = 6 x SD (9.7)

You're saying ES is entirely a function of SD or 99.7% confidence limit as defined by SD.

You're wrong.

In LR shooting circles, ES is the difference between the highest and lowest velocity (or range or extent of velocities).

ES is an observation, not a calculation, except for simple subtraction. One cannot determine the SD from the ES, because each data point must be included in the SD determination.

A useful analogy in shooting is that extreme spread is how 100 yd bench rest shooters determine group size and, thus the winners and losers. Shots falling between don't count.

In long-range score shooting putting as many shots as possible within the 10 ring for the high accumulated score matters most (i.e., smallest SD from the center), one outlier may not cause you to lose the match. The shots that fall between matter.

Here is a nice link defining how these terms are used and defined with chronographs and shooting...

http://www.pilkguns.com/Chrony.shtml

What Ackley Improved User said, plus one!

The only thing that matters is the size of the target group at the intended yardage...

Agreed, however, IOT achieve such results, it is a given that your ES pretty much must indicated a range or spread of velocity that creates the condition wherein all of you shots indeed can fall into that required group size on target.

Additionally, there must be accouting for the capabilitiy of the gun and shooter, so all three variables must compute into a goup size of less than the required bullet group.

Example- ammo is theoretically able to clean the target, .5MOA, the rifle can shoot .5 MOA, the shooter can hold .5 MOA, the wind or environmental conditions can be assessed at .5 MOA. What is the expected group size?

The solution is the root sum of squares of the errors.

thank you both...

Whoops, sorry , the answer didn't make it. While not specifically the full production management answer, as I do not have ES, mean or SD of each variable, but a layman's answer would be the best this shooter given above could expect is an average group size of 1 MOA.

square root of the sum of each variable squared

.5 sq= .25, x 4= 1, sq root of 1 is (+/-) 1.....

Again, this is "hip shoot" production management math.

quote:

Originally posted by Ackley Improved User:
AAZW,

What's your point?

The ES for the first set is 22 (i.e., the difference of 3012 minus 2990), for the second 42 (i.e., 3032 minus 2990).

The ES (extreme spread, data extent, or range) is the difference between the lowest and highest observed velocities.

44.4 = 6 x SD (7.4)
58.2 = 6 x SD (9.7)

You're saying ES is entirely a function of SD or 99.7% confidence limit as defined by SD.

You're wrong.

In LR shooting circles, ES is the difference between the highest and lowest velocity (or range or extent of velocities).

ES is an observation, not a calculation, except for simple subtraction. One cannot determine the SD from the ES, because each data point must be included in the SD determination.

A useful analogy in shooting is that extreme spread is how 100 yd bench rest shooters determine group size and, thus the winners and losers. Shots falling between don't count.

In long-range score shooting putting as many shots as possible within the 10 ring for the high accumulated score matters most (i.e., smallest SD from the center), one outlier may not cause you to lose the match. The shots that fall between matter.

Here is a nice link defining how these terms are used and defined with chronographs and shooting...

http://www.pilkguns.com/Chrony.shtml

AIU:

I am fully aware of what ES is; technically, in the world of statistics, ES is the same as the range. I used the word "suggest" not "calculate."

I simply pointed out that it is interesting that the 99.7% confidence level of your first ten shots was close to the observed range of the second data set.

SD is useful for estimating the distribution of the population. The fact the ES in your first data set was 22 does not mean that load would never deviate more than 11 fps from the mean; the 99.7% confidence level generated by the SD practically guarantees that the true ES for the expected population was greater than 22, and you proved that when you shot ten more times.

Statisticians would say any sample less than 30 is too small to generate meaningful estimates of the population, which at the end of the day, is what we are trying to do with ES and SD.