Winchester 38 Spl +P 158 gr LSWCHP - Chronograph test

[Hi-Speed]

For those fine gentlemen who appreciate our chronographing efforts and for the knowledge derived from such, I chronographed the factory Winchester 38 Spl +P LSWCHP “FBI Load” over the weekend as follows:

Firearm: Ruger 4 5/8 inch Blackhawk
Shot #/MV
1. 826 (yikes)
2. 900
3. 865
4. 886
5. 907
6. 897
7. 910
8. 872
9. 878
10. 906
11. 869
12. 896
13. 918
14. 929
15. 911

Hi 929
Lo 826
Avg 891
ES 103
SD 25

That first shot was unusual but throwing out this Lo (826) and the Hi (929), I get this:

Hi 918
Lo 865
Avg 893
ES 53
SD 17

The Winchester FBI Load above can be duplicated easily in my revolver with 4.7 grs W231 and 158 gr SWC…a sound and accurate load in its own right.

[OS OK]

Why would you want to throw those extremes out?
They are part of the test and reflect the overall extreme spread.

[dverna]

Why would you want to throw those extremes out?
They are part of the test and reflect the overall extreme spread.

It is the same as calling fliers as not being part of the group....also called fooling yourself.

You think like I do. Unless someone can identify specifically why a data point should be ignored, it needs to be included.

[Kosh75287]

Can't remember how to set it up, but a chi-square analysis could determine the "goodness of fit" of the two outliers. I think there's also a "rule of thumb" used in less rigorous statistical analyses, in which it's considered fair to toss a data point if it's more than some number of standard deviations away from the mean. The first round's velocity is 2.6 s.d.s away from the mean. The 14th round is 1.52 s.d.s away from the mean.
The first round MAY qualify for omission from the data pool. I'm not at all certain about the 14th.

[Hi-Speed]

Why would you want to throw those extremes out?
They are part of the test and reflect the overall extreme spread.

Very good point, thanks

[Hi-Speed]

It is the same as calling fliers as not being part of the group....also called fooling yourself.

You think like I do. Unless someone can identify specifically why a data point should be ignored, it needs to be included.

Excellent idea!

[No_1]

I wonder what caused those 2 shots to be so different speed wise? I guess one could blame the low on being the first shot but there has to be something else.

[dverna]

I wonder what caused those 2 shots to be so different speed wise? I guess one could blame the low on being the first shot but there has to be something else.

It could be as simple as variation in powder drop, or cases with enough variation in length to affect crimp. Case length would be easy to verify for the OP if it happens again. Easy to check when there is an outlier while shooting. Either save the cases and mark them, or bring a caliper to the range.

[Hi-Speed]

Thank you Don

[HWooldridge]

I put the entire first data set into a process control calculator we use here at work. It came up with an SD of 27 but regardless, this would not be considered a capable recipe based solely on velocity. However, that doesn't mean it won't suit the purpose of self defense.

ave 891.000000
min 826
max 929
range 103
std dev. 26.992877


6 Sigma 161.9572593
CP 0.636
process mean to spec limit
upper 38
lower 65

CPK 0.469

[Hi-Speed]

Thank you HWooldridge, checked my ProChron Chronograph results off an on-line SD calculator and it comes up with 25 SD as I noted. Your Texan math must be smarter than my humble Nevadan slide rule. Regardless, I appreciate your response since it is detailed and informative…I dig numbers also!

Standard Deviation, σ: 25.159932873961
Count, N: 15
Sum, Σx: 13370
Mean, μ: 891.33333333333
Variance, σ2: 633.02222222222

Steps



σ2 =
Σ(xi - μ)2
N
=
(826 - 891.33333333333)2 + ... + (911 - 891.33333333333)2
15
=
9495.3333333333
15
= 633.02222222222
σ = √633.02222222222
= 25.159932873961

Margin of Error (Confidence Interval)

The sampling mean most likely follows a normal distribution. In this case, the standard error of the mean (SEM) can be calculated using the following equation:

σx̄ =
σ
√N
= 6.4962667341698
Based on the SEM, the following are the margins of error (or confidence intervals) at different confidence levels. Depending on the field of study, a confidence level of 95% (or statistical significance of 5%) is typically used for data representation.

[HWooldridge]

Hi-Speed,

FYI, our calculator uses the STDEV formula found in Excel but I'm good with your conclusions...

Would be interesting to shoot this load at 100 yards out of a Ransom rest to validate vertical dispersion.

HW

[Hi-Speed]

Hi-Speed,

FYI, our calculator uses the STDEV formula found in Excel but I'm good with your conclusions...

Would be interesting to shoot this load at 100 yards out of a Ransom rest to validate vertical dispersion.

HW

…that’s well above my humble pay grade…and my firearm (and skill). Would be cool to try though!

I think those who know will concur that the only sure and meaningful way to arrive at sound, meaningful and accurate data results in chronographing is in a controlled ballistics lab environment…

[Electrod47]

Thank you Highspeed. Good info and good work. I worked in the melt shops of several huge steel mills. We tracked every know variable in the entire process of cost, energy, lbs pr/ton. Every fraction imaginable. We ALWAYS threw out the Highest and Lowest whatever, in our calculations. I say again, Good Work.

[justindad]

Can't remember how to set it up, but a chi-square analysis could determine the "goodness of fit" of the two outliers. I think there's also a "rule of thumb" used in less rigorous statistical analyses, in which it's considered fair to toss a data point if it's more than some number of standard deviations away from the mean. The first round's velocity is 2.6 s.d.s away from the mean. The 14th round is 1.52 s.d.s away from the mean.
The first round MAY qualify for omission from the data pool. I'm not at all certain about the 14th.

The enumerative statistician will agree with you; the analytical statistician will disagree with you.
*
I’ve been designing experiments and analyzing data leading to products you likely own for 20 years now. I’d keep the extreme points, unless there is a special cause assigned to the variance which has also been eliminated from all future ammo. But then again, we can disagree and both be practically successful.

[Hi-Speed]

Thank you Highspeed. Good info and good work. I worked in the melt shops of several huge steel mills. We tracked every know variable in the entire process of cost, energy, lbs pr/ton. Every fraction imaginable. We ALWAYS threw out the Highest and Lowest whatever, in our calculations. I say again, Good Work.

Thank you

[Kosh75287]

The enumerative statistician will agree with you; the analytical statistician will disagree with you.
*
I’ve been designing experiments and analyzing data leading to products you likely own for 20 years now. I’d keep the extreme points, unless there is a special cause assigned to the variance which has also been eliminated from all future ammo. But then again, we can disagree and both be practically successful.

<shrug> Keeping the outliers is prolly best.
The only POSSIBLE confound I can imagine is that, with the first shot, perhaps the propellant was concentrated closer to the projectile than to the primer flash hole, giving erratic ignition and low(er) velocity. Recoil and muzzle rise sends the charge in the remaining rounds to the back of the cartridges, and higher velocities are obtained.
I don't know how to explain the velocity of round 14, but it's close enough to the rest to probably be attributed to random factors.

[Golfswithwolves]

Thank you HWooldridge, checked my ProChron Chronograph results off an on-line SD calculator and it comes up with 25 SD as I noted. Your Texan math must be smarter than my humble Nevadan slide rule. Regardless, I appreciate your response since it is detailed and informative…I dig numbers also!

Standard Deviation, σ: 25.159932873961
Count, N: 15
Sum, Σx: 13370
Mean, μ: 891.33333333333
Variance, σ2: 633.02222222222

Steps



σ2 =
Σ(xi - μ)2
N
=
(826 - 891.33333333333)2 + ... + (911 - 891.33333333333)2
15
=
9495.3333333333
15
= 633.02222222222
σ = √633.02222222222
= 25.159932873961

Margin of Error (Confidence Interval)

The sampling mean most likely follows a normal distribution. In this case, the standard error of the mean (SEM) can be calculated using the following equation:

σx̄ =
σ
√N
= 6.4962667341698
Based on the SEM, the following are the margins of error (or confidence intervals) at different confidence levels. Depending on the field of study, a confidence level of 95% (or statistical significance of 5%) is typically used for data representation.

Yes, that's exactly what I came up with when I figured it out in my head.