Making power factor

The January/February 2015 issue of Front Sight magazine has an interesting article on the statistics of making “power factor*”.

While I certainly had the background in statistics it never occurred to me to apply them to the chronograph data from my hand loaded ammunition to determine the chances of me failing to “make major” at a match. This is despite very nearly failing to make major in the 1998 Area One USPSA match.

Looking at my log files for the ammunition I made for that match I found the following data:

Mean velocity: 992.6 fps
Standard Deviation: 11.3 fps
Bullet Mass: 180 grains
Power Factor: 178.67

Back then you had to have a power factor of 175 to make major and for some reason I thought I had plenty of margin.

At the 1998 Area One match staff pulled eight cartridges at random from the magazines on my belt and tested them as per USPSA regulations. They pulled the bullet from a cartridge and weighed it. They fired three rounds and found I failed to make major. They, as per procedure, fired another three rounds, used the highest three velocities from the six rounds fired and found I was closer but still failed. They had one round left and, as per procedure, asked me what to do with it, “Fire it or weight the bullet?” I had them fire it and using the highest three velocities from the seven rounds fired I just barely made major power factor.

It wasn’t until I read the title of the article in Front Sight article, “The Power of Statistics How to Meet Power Factor with Confidence” that I felt stupid for my experience at Area One.

The bottom line is that your chance of failing the test procedure depends on how many standard deviations you are away from the velocity threshold for the power factor you want to meet.

Using my example from the Area One match the velocity threshold is 972.22 fps (175,000 / 180). My mean velocity was 992.6 fps or 20.378 fps above the threshold. With a standard deviation of 11.3 fps the ammo was 20.378 / 11.3 or 1.8 standard deviations (commonly called ‘Z’) from the threshold. Using a normal distribution table or the article you will discover my chances of failing were about 13%.

Update:

Using this table from the article you will discover my chances of failing were about 13%:

Z Chance of Failing Power Factor (per USPSA rules)
2.5 5%
2.0 10%
1.9 11%
1.8 13%
1.7 15%
1.5 21%
1.4 26%
1.2 36%
1.1 40%
1.0 44%
0 50%

This table is not a standard distribution table. It is a mapping from Z (number of standard deviations away from the mean) to the chances of failing the PF test under USPSA rules. This was obtained using a t-distribution because of the small sample size used by the USPSA regulations. It is assumed the shooter obtained the mean velocity and standard deviation with a sample size of eight.

End update.

I’m going to range today to measure the velocities of a new load I plan to use for competition. I’m going to make sure I’m about 2.5 standard deviations away from the threshold which would put my odds of failing to make major at about 5%.


* Power Factor is defined as the mass of the bullet in grains multiplied by the velocity in feet per second divided by 1000. Or:

Power Factor = bullet weight (grains) x average velocity (feet per second) / 1000

In many competitions your targets are scored differently depending on the power factor of the ammunition you are shooting. For example if you are shooting Limited Class USPSA you “make major” with a power factor of 165 or greater and “make minor with a power factor of 125. For major power factor ‘B’ and ‘C’ zones hit are scored as 4 points and ‘D’ zone hits are scored as two points. If you “make minor ‘B’ and ‘C’ zones hit are scored as three points and ‘D’ zone hits are scored as one point. If you don’t have ammunition which gives you a power factor of 125 or greater all zones are scored as zero. I.E. you aren’t participating in the competition.

25 thoughts on “Making power factor

  1. Reading the title made me immediately think of matching impedences. Darn discipline-specific jargon.

  2. In line with what Robert said, “making major” caught my eye and I wondered what Power Factor had to do with being promoted from Captain to Major.
    And it’s only Tuesday!

    • And thanks for the essay. When I get a chronograph I’ll take the advantage to brush up on what little I learned of statistics in the University.

  3. The area from 0 to z, where z= -2.5 standard deviations is .4941 Then we add .5 because anything over is good (only one side of the curve is bad). 99.41% chance of any single cartridge making power factor. (.9941)^8 is 95.38% or close to the 5% in the article. You don’t need all 8 or even all 7, they weigh 1, to make PF. You need 3 of 7 to pass. To put it in other terms if 5 of your 7 fail then you fail to make PF. The probability of that is a lot less than 5%.

    • Yeah, there is something funny about the math, but I cannot take the time to figure it out now. The chances of getting one cartridge in that lot of 8 that does not make PF is about 5%, the chance of you failing the test is a combinatorial function that should put your chance of failure at well below 1% if my gut serves me right.

      • Got it, the probability of getting one “dud” is 0.59%, you need 5 to fail, so .59^5 is ~.07% you will fail to make PF. Significantly better than ~5% I would say.

      • Oh I see, so we are not assuming a normal distribution. I was looking at it from the perspective of “I know the average PF of my load, as well as Z, so I will calculate my chance to fail this test” not “I will determine the average PF of this lot using these 8 cartridges.”

        Granted it has been a long time since I have done this kind of math (and I am sure it shows), so there is probably a bit I am still missing, but at least I see where your numbers are coming from now.

        Since you are the loader, and have a knowledge about the lot that exceeds those 8 cartridges, shouldn’t we be using a normal distribution and not a t distribution to calculate your actual chance of failure? When you work up a load, you take 10 cartridges, throw the powder, weigh the powder, then shoot all 10 for velocity, for perfect knowledge of that lot. When you finalize the load, you load 100 cartridges assuming he powder throw is on a normal distribution (calculated from the average and deviation of the 10 from the test lot), which means you should have a much better idea of how many of that 100 would fall under your target PF than a t distribution would give you right?

        The t distribution is for the judges to say, based on only 8 cartridges, “I am x% confident that this lot of ammo exceeds this PF.” Is this correct?

        • you should have a much better idea of how many of that 100 would fall under your target PF than a t distribution would give you right?

          Correct.

          The t distribution is for the judges to say, based on only 8 cartridges, “I am x% confident that this lot of ammo exceeds this PF.” Is this correct?

          The way I understand it the t-distribution was used by the article author to see how likely the USPSA power factor test procedure is to fail an eight sample collection of ammo that came from the lot I was using at the match. This is based on the assumption that I drew eight samples from the same lot to obtain mean velocities and the standard deviation.

          • Ok, see that last assumption is where we differ I think. You are saying that if you take 8 samples from your lot, come up with your t distribution based on that sample that puts your mean 2.5 standard deviations above the target PF, you would still have a 5% chance to fail another random selection of 8, correct?

            Sorry, I would read the article to understand better, but I cannot find it on the site (I am not a member).

          • As near as I can tell the article is not online. I have the physical magazine in front of me.

            You are still not quite correct. The author advises that we take 8 samples and assuming normal distribution compute the mean and standard deviation. Other than that you have it correct.

          • We hit the nesting limit, so I will reply here.

            Ok, this makes sense now, though I would prefer to operate from a position of more knowledge. If you are the lot manufacturer you should be able to come up with a valid normal distribution, since you have knowledge and control over many of the variables. This wold let you calculate near exact odds of failure given a random sampling of 8. If you were not the manufacturer, then hopefully you would be able to take a much larger sample size, allowing you to reduce your uncertainty by a significant margin.

  4. Do you get even more points for using a 50 AE or a 500 Smith? What if the muzzle blast alone rips the target to pieces? Are hits from hand grenade fragments and secondary projectiles scored the same as pistol bullets? Inquiring minds want to know. I haven’t been to an organized shooting match in a while, so I could be behind the times.

    • Sure – larger diameter bullet=better chance of breaking the line for a higher scoring zone. No extra points for collateral damage in the form of rips, tears, or flash burns on the paper.

      • So what your saying is you should switch to one of those fancy bolo rounds to maximize your ability to hit high score zones?

  5. Nitpick: The footnote uses “grams” in the first sentence, but uses “grains” (correctly) in the following formula.

    • Fixed. Thanks.

      I’ll address the other comments after I get some time, investigate more, and read the entire original article rather than just scanning and reading the end.

  6. A question of statistics you may want to watch out for: conventional calculations of probabilities and standard deviations apply to a “normal” (Gaussian) distribution of observations. When you’re dealing with the products of manufacturing processes, the distributions may be rather different.
    I remember an article in an electronics magazine that pointed this out: when you buy electronic components with a given tolerance spec, you’re not getting parts with a normal distribution and that tolerance value at 2 or 3 sigma. What you get is a two-peak distribution around plus and minus the given tolerance. You’re not getting parts much closer to the nominal value; those are sold instead as tighter tolerance parts.
    This probably doesn’t apply to the powder load of cartridges, which is a measured value in a dispensing device. But it might apply to the bullet weight. If so, your velocity distribution is likely to be bimodal as a result.

    • Yup. Standard deviation might still apply if there were many variables of approximately equal contribution and independent but don’t count on that in many cases.

      In this situation the dominate factors are probably powder mass and shell case variations. My powder mass probably has an error of about +/- 0.1 grains with what I expect is something close to a normal distribution. This corresponds to a velocity variation of about +/- 18 fps (from my log files for various powder weights using the same bullets).

      I don’t know the effect of the shell casings but I don’t sort my brass and have a variety of manufactures represented. The distribution is probably not normal.

  7. USPSA uses the average of your 3 best. I don’t know how to do that. Requiring 3 passing velocities is as close as I can calculate. Here is the result. I find using 2 SDs and only failing to make USPSA PF once in 8 million matches to be adequate.

    z Failures
    —- —————–
    0.00 1 in 4
    0.50 1 in 31
    1.00 1 in 628
    1.25 1 in 4,359
    1.50 1 in 40,143
    1.75 1 in 491,547
    2.00 1 in 8 million
    2.50 1 in 5.25 billion

    • Here’s the thing that bothers me about this: you’re already nearly 2 standard deviations from the minimum velocity needed to make major in your original data. The fact that you very nearly didn’t make major when they actually tested your rounds suggests at least a couple of possibilities:

      1) Calibration error in chronograph, yours or theirs
      2) The loads got “cooler” between the time you chrono’ed them and the match. (Storage conditions or temperature at the match?)

      Like Ken said above, if you were really ~1.8 SDs above major, the chances of you so nearly missing major are already incredibly slim. Rather than simply raising the power a little, you may want to check and see if something else is going on.

      • 1) Is a valid concern.
        2) Probably is not valid unless there is a negative correlation with temperature. My chronograph data was obtained on March 29th, while in north central Idaho with a temperature (from the log file) of ~45F. The match was in June (?) in Reno. I recall it being warm (short sleeve shirt weather) but not uncomfortably hot when doing the power factor test.

        Other concerns:

        3) USPSA states the distance to the chronograph is at least 10 feet from the gun muzzle. I did my tests at eight feet.
        4) I had a sample size of only four. This is really too small. I found another log entry from June 24th for the same load using five samples (still too small) with the following results:
        Velocity samples: 1015.8, 967.9, 921.6, 976.5, 990.9
        Mean: 974.6
        Standard Deviation: 31.1

  8. In regards to power factors and shootin’ matches, I’m still working to pass the first rule: only hits count.

Comments are closed.