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Thread: Winchester 38 Spl +P 158 gr LSWCHP - Chronograph test

  1. #1
    Boolit Buddy Hi-Speed's Avatar
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    Winchester 38 Spl +P 158 gr LSWCHP - Chronograph test

    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.

  2. #2
    Boolit Grand Master OS OK's Avatar
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    Why would you want to throw those extremes out?
    They are part of the test and reflect the overall extreme spread.
    a m e r i c a n p r a v d a

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  3. #3
    Boolit Grand Master
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    Quote Originally Posted by OS OK View Post
    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.
    Don Verna


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    Boolit Master
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    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.
    For in much wisdom is much grief: and he that increaseth knowledge increaseth sorrow. Ecclesiastes 1:18
    He that troubleth his own house shall inherit the wind: and the fool become servant to the wise of heart. Proverbs 11:29
    ...Verily I say unto you, Inasmuch as ye have done it unto one of the least of my brethren, ye have done it unto me. Matthew 25:40


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    Boolit Buddy Hi-Speed's Avatar
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    Quote Originally Posted by OS OK View Post
    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

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    Boolit Buddy Hi-Speed's Avatar
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    Quote Originally Posted by dverna View Post
    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!

  7. #7
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    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.
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  8. #8
    Boolit Grand Master
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    Quote Originally Posted by No_1 View Post
    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.
    Don Verna


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    Boolit Buddy Hi-Speed's Avatar
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    Thank you Don

  10. #10
    Boolit Master
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    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

  11. #11
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    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.

  12. #12
    Boolit Master
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    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

  13. #13
    Boolit Buddy Hi-Speed's Avatar
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    Quote Originally Posted by HWooldridge View Post
    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…

  14. #14
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    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.
    “You should tell someone what you know. There should be a history, so that men can learn from it.

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  15. #15
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    Quote Originally Posted by Kosh75287 View Post
    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.

  16. #16
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    Quote Originally Posted by Electrod47 View Post
    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

  17. #17
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    Quote Originally Posted by justindad View Post
    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.
    For in much wisdom is much grief: and he that increaseth knowledge increaseth sorrow. Ecclesiastes 1:18
    He that troubleth his own house shall inherit the wind: and the fool become servant to the wise of heart. Proverbs 11:29
    ...Verily I say unto you, Inasmuch as ye have done it unto one of the least of my brethren, ye have done it unto me. Matthew 25:40


    Carpe SCOTCH!

  18. #18
    Boolit Buddy
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    Quote Originally Posted by Hi-Speed View Post
    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.

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