In the same time I wanted to ask about Standard Deviation and how does it differ from Spread read on My Chronograph.

Standard deviation for 5-shot groups =
Extreme spread / 2.326. 
Extreme spread for 5-shot groups =
Standard deviation X 2.326.

The number, 2.326, varies with shots per group.

Thus, Standard deviation and Extreme spread are measures of the same thing, dispersion.


Like measuring the contents of a barrel in quarts or gallons, same bucket, different measures.

joeb33050 wrote on Oct 30^th, 2018 at 3:15pm:


The difference is that there is no fixed relationship between ES and SD, as you pointed out.  It is not really like "quarts or gallons" at all, for which there is a fixed relationship.

They are both measures of dispersion but they are not equivalent.

Understanding and analyzing both measures is likely to provide the most insight.

joeb33050 wrote on Oct 30^th, 2018 at 4:27pm:

[/quote]
How did you come up with your constant?

If we have 5 shots with the following velocities:

1,000
1,000
1,000
1,000
700
SD = 134

If they are:
1,000
1,000
700
700
700
SD = 164.

ES is 300 in both cases, so the ES:SD ratios have to be different.

This really isn't a dispute - the theory and math are pretty well settled.

SD, is a assumption or probability that any shot will be within +/- N, 90% of the time. That is how it was explained to me and that is what I use to understand SD.

SD does not mean that in reality, all shots will fall within the stated SD, maybe not even 90% of the time.

You can not dispute that if velocity matters in the accuracy of your load, that ES will effect the accuracy the most. I've chronographed loads since 1986 and calculated both SD & ES, since then. But, I have found that relying on ES has given me better results than assuming things will fall into the SD.

Here is what ES can do to a subsonic target, at 200 yards.

Mean V = 900 fps, ES 10 fps, .45 BC

905 fps, drop 89.77"

895 fps, drop 91.69"

In this case, no matter what your SD or your accuracy potential is, you can not shoot a group of less than 1.92", unless barrel harmonics compensate. 

That is what I discovered when I got into subsonic loads, back in the '90's and that's when I started relying more on ES than SD.

Control your ES and your SD will take care of itself.

I think that a population of numbers can be arranged, so that the SD can exceed ES. In that case, it's misleading.

Frank

Or, try this, less arithmetic:

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joeb33050 wrote on Oct 30^th, 2018 at 3:15pm:


spot on

Since this discussion has devolved into a “dispute” akin to the dispute as to whether or not the earth is flat, I will try one more time then leave it to others.

Standard deviation for a population is calculated per the formula shown below.

To calculate SD, subtract each observed value X from the mean of all the values X̄ (X bar) and square each result.  Add up these squared results and divide the total by the number of observations (N).  Take the square root of the result.  This is the standard deviation.

Ponder the following example.

2 muzzle velocity datasets with 100 observations each.

Set A has 99 observations of 1,000 fps and one observation of 200 fps.  ES = 1,000 – 200 = 800.  The mean = 992, SD = 80 (calculated using Excel SD formula).

Set B has 50 observations of 1,000 fps and 50 observation of 200 fps.  ES = 1,000 – 200 = 800.  The mean = 600, SD = 402 (calculated using Excel SD formula).

It seems rather obvious, intuitively, that set B has much higher variability than set A, despite the fact that the ES is the same.

For those not inclined to do calculations, ask yourself why a complex SD formula exists if one can simply divide the difference between the values of the high and low observations by a single number. 

Alternatively, think about two rifles producing a 10 shot, 5” group.

In the first, 9 bullets went through one hole, the last was a flyer, 5” away.

The second produced a perfect circle (decagon) with a 5” diameter.

Do you believe that the variability of these two groups is the same, as would be the case if we could define SD by the ‘divide ES by 2.326’ (or some other number) approach?

From a practical standpoint, this example also illustrates why both measures might be important to consider when trying to figure out how to improve shooting performance.  The approach to fixing the single flyer group is unlikely to be the same as that for the perfect circle group.

Bottomline e is the extreme is what really matters.  What good does it do to get 4 shots in one hole and have the 5th 2 inches out?  Or, 9 25s and a 23?