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Let us assume that it is factual that the military replaces bolts every 5k rounds. Let us assume, for sake of argument, that there are exactly 10,000 rifles in the current service to be used by every troop. Let us further assume, for simplicity of argument, that each bolt costs just a small amount of 25 bucks. Let us further assume that each year, half of the rifles need to replace bolts. That is 5,000 * 25 bucks = $125,000. That's per year, just simply in weapon maintenance. That's just a baby estimate. In the active military, there are probably way more rifles in circulation and use. Does it make sense, from the stand point of use, for the military to continue and pay hundreds of thousands of dollars in just a simple part to a weapon? Or... does it make more sense that it's purely made up by internet folk tales? View Quote Source of the 1,800 number. |
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Those are the exact factors that would cause it to fall into a normal distribution, as we are talking about the entire population of bolts in the Army. View Quote Of course, this is all rather futile, as in order to come up with data for any sort of distribution, a statistically significant sample of bolts (likely thousands, certainly hundreds at least) would have to be tested to failure under each of a large number of sets of controlled circumstances. |
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I believe the OP was ALL AR-type bolts, not just military (at least that's how I read it). In the subpopulation of Army weapons, you may very well see something close to a normal distribution, as the number of weapons actually deployed and actively used in combat is likely very small compared to the total inventory, and would not influence the total distribution too much; and all are (theoretically) subject to the same ammunition, maintenance and training procedures. When you add in the other subpopulations, though- civilian, other service branches, police- you have distinct groups encountering different sets of stressors. If different sub-populations have different stress factors, you won't see a normal distribution for the entire population. Within each sub-population, assuming every bolt experienced the same mix of stressors, you would indeed see a normal distribution. However, because the separate subunits may have unique factors that affect bolt life, the expected distribution of the entire population would be multimodal, provided the sub-populations are of a statistically significant size. It probably wouldn't be clean enough to see separate peaks, but it would likely distort the distribution curve beyond anything that could be called normal. If it did come up with a clean normal distribution it would be very interesting, as it would imply that differing deployment conditions, ammunition, and maintenance had no significant effect on bolt life. Of course, this is all rather futile, as in order to come up with data for any sort of distribution, a statistically significant sample of bolts (likely thousands, certainly hundreds at least) would have to be tested to failure under each of a large number of sets of controlled circumstances. View Quote It was based on the Army data someone provided that stated the average bolt life was 12,000 rounds for the Army. (Besides, Army sort of keeps track of bolt life, and can estimate how long they last. How would you even start to gather that data for the civilian population of bolts.) |
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Very interesting topic, I love technical discussions like this.
But have to ask the probably stupid question, what about the carriers? Under what circumstances would that be replaced? Speaking strictly within .mil |
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View Quote View All Quotes View All Quotes Quoted:
Quoted:
Let us assume that it is factual that the military replaces bolts every 5k rounds. Let us assume, for sake of argument, that there are exactly 10,000 rifles in the current service to be used by every troop. Let us further assume, for simplicity of argument, that each bolt costs just a small amount of 25 bucks. Let us further assume that each year, half of the rifles need to replace bolts. That is 5,000 * 25 bucks = $125,000. That's per year, just simply in weapon maintenance. That's just a baby estimate. In the active military, there are probably way more rifles in circulation and use. Does it make sense, from the stand point of use, for the military to continue and pay hundreds of thousands of dollars in just a simple part to a weapon? Or... does it make more sense that it's purely made up by internet folk tales? Source of the 1,800 number. The number of bolts, breech, P/N 8448510, (stripped bolt) delivered in the last ten years was 255, at a total contract price of $13,333.95 ($59.26 each). The number of Bolt Assembly, P/N 13004787, (complete bolt) delivered in the last ten years was 50,000, at a total contract price of $2,209000 ($44.18 each). If they were going through bolts at a rate of 1,800 a month, the last ten years of bolt deliveries would have been consumed in just over two years. |
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For combat arms units. Replace bolt carrier at 60K, ref SOFWEP-07-G12P-00032-00 Rev 1
SW370-CC-TRS-010 Rev 1 TECHNICAL REPAIR STANDARD FOR CARBINE, 5.56 MM, M4A1 Appendix F, Table F-1 COMPONENT PN 10K 15K 30K 45K 60K 75K 90K 105K Upper Receiver and Barrel Assembly 12997148 I I I I I I R Bolt Carrier Assembly 13004788 I I I R I I Pin, Firing 8448503 I I I I I I I I Pin, Firing, Pin Retainer 8448504 I I I I I I I I Pin, Grooved, Headed 8448502 I I I I I I I I Carrier, Assembly, Key and Bolt 8448505 I I I I I I I I Notes are I is Inspect, R is replace CD |
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Very interesting topic, I love technical discussions like this. But have to ask the probably stupid question, what about the carriers? Under what circumstances would that be replaced? Speaking strictly within .mil View Quote Generally carriers don't wear out as quickly as other parts. |
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Wiebull distributions can be normal, depends on the value of k, low cycle fatigue usually is fairly normal, at least not highly skewed to the right like high cycle fatigue. In the absence of better data, a normal distribution is pretty good assumption. View Quote |
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Actually most low cycle fatigue failure is skewed to the left more often than not. If your stressing the part hard enough that failure occurs relatively quickly than exploitation of flaws will skew this left. Go get better data this is not an assumption you can make and still have a good discussion. View Quote For new manufactured M4 Carbines, the 10% failure is 7000 rounds, the mean part life is 11,300 rounds, and the 90% failure is 17,000 rounds. For new manufactured M16 Rifles, the 10% failure is 11,000 rounds, the mean part life is 15,000 rounds, and the 90% failure is 19,000 rounds. That's a pretty normal distribution. The only time it gets skewed to the left (in this case) is when you get remanufacturing into the mix, as some parts that still have life are reused, but are nearer the end of their life, so they fail soon after the remanufacture. |
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Let me say that I do appreciate the detailed explanation
I just wanted to offer a perspective on your Pop Quiz. Quoted:
POP Quiz: When do you want to replace your bolt? (refer to the image) View Quote Don't care. Before I bought an AR a couple years ago, I had a Mini-14 that I had bought shortly after I was old enough to legally buy it and it has been with me for nearly 40 years. In that time, it, and another Mini-14 I subsequently bought for my father, have gone through less than 5,000 rounds, total. I figure the bolt life of my ARs will be of concern only to my heirs - and I the same can be said of every other AR owner I know. |
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For combat arms units. Replace bolt carrier at 60K, ref SOFWEP-07-G12P-00032-00 Rev 1 SW370-CC-TRS-010 Rev 1 TECHNICAL REPAIR STANDARD FOR CARBINE, 5.56 MM, M4A1 Appendix F, Table F-1 COMPONENT PN 10K 15K 30K 45K 60K 75K 90K 105K Upper Receiver and Barrel Assembly 12997148 I I I I I I R Bolt Carrier Assembly 13004788 I I I R I I Pin, Firing 8448503 I I I I I I I I Pin, Firing, Pin Retainer 8448504 I I I I I I I I Pin, Grooved, Headed 8448502 I I I I I I I I Carrier, Assembly, Key and Bolt 8448505 I I I I I I I I Notes are I is Inspect, R is replace CD View Quote |
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For combat arms units. Replace bolt carrier at 60K, ref SOFWEP-07-G12P-00032-00 Rev 1 SW370-CC-TRS-010 Rev 1 TECHNICAL REPAIR STANDARD FOR CARBINE, 5.56 MM, M4A1 Appendix F, Table F-1 COMPONENT PN 10K 15K 30K 45K 60K 75K 90K 105K Upper Receiver and Barrel Assembly 12997148 I I I I I I R Bolt Carrier Assembly 13004788 I I I R I I Pin, Firing 8448503 I I I I I I I I Pin, Firing, Pin Retainer 8448504 I I I I I I I I Pin, Grooved, Headed 8448502 I I I I I I I I Carrier, Assembly, Key and Bolt 8448505 I I I I I I I I Notes are I is Inspect, R is replace CD View Quote Bolt Carrier Assembly, P/N 13004788 is the entire bolt carrier assy to include the bolt, firing pin, retaining pin, cam pin, and bolt carrier (ie, all the other parts noted in this table). If you replace the entire assembly at 45,000 rounds, why do all the other sub-components of that assembly say "inspect"? By replacing the higher assembly, all the sub-components would get replaced as well. If it 60,000 rounds, you missing an "I", 60K is at the fifth place, the "R" is at the forth place or 45K. |
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Question - Bolt Carrier Assembly, P/N 13004788 is the entire bolt carrier assy to include the bolt, firing pin, retaining pin, cam pin, and bolt carrier (ie, all the other parts noted in this table). If you replace the entire assembly at 45,000 rounds, why do all the other sub-components of that assembly say "inspect"? By replacing the higher assembly, all the sub-components would get replaced as well. If it 60,000 rounds, you missing an "I", 60K is at the fifth place, the "R" is at the forth place or 45K. View Quote |
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That is 5,000 * 25 bucks = $125,000. That's per year, just simply in weapon maintenance. Does it make sense, from the stand point of use, for the military to continue and pay hundreds of thousands of dollars in just a simple part to a weapon? Or... does it make more sense that it's purely made up by internet folk tales? View Quote Single agencies may misplace - as in lose entirely - billions in a single year. |
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OP, what is your justification for assuming the standard deviation is 1300 rounds?
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OP, what is your justification for assuming the standard deviation is 1300 rounds? View Quote Although, the point of the original post was to show that the "average life of 12,000 rounds" and the often quoted "5,000 round life" are not incompatible, depending on how you want to deal with possible bolt failures. |
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It's a formatting thing. It's 60,000 for the bolt carrier assembly. View Quote What is the bolt's inspection/replacement schedule? |
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It a reasonable round number that fits. (See the post with the 10% and 90% failure counts.) Although, the point of the original post was to show that the "average life of 12,000 rounds" and the often quoted "5,000 round life" are not incompatible, depending on how you want to deal with possible bolt failures. View Quote First, a Weibull curve is much more appropriate, since it more easily models events vs. time. a Normal distribution is better suited to events in a population. Yes, one can define "events" as "failures at X rounds", but that in itself is a derived number. A different way to look at it is that, per your fit, while a tiny number of bolts fail at <5000 rounds, some do. And per the curve, that means a VERY tiny number of bolts will fail at <0 rounds. That is an impossibility in this context - we are looking at failures vs. round count where the lowest possible value is 1. And a model, while it doesn't have to be perfect, CANNOT allow for the impossible - if it does, the model is inappropriate. My second issue is that mean and standard deviation and other parameters are supposed to be derived from the data itself. From the looks of it, there are 2 pieces of actual data - average round count at bold failure=12000, and another less well attributed number of 5000 round life. So if we *assume* that 5000 rounds is .01% failure rate, you can back calculate the rest of the curve. But you say that number is the "safe fatigue life." That isn't a defined number - it is different per application, industry, code, etc. So if 5000 is actually 99.9% (the "Six Sigma" number if I recall correctly), that will change the SD and the number at the upper end. I appreciate the thought and effort that went into this. And I'd really love to see the dataset used to get the 12000, etc. But the curve and numbers you've posted just seem too loose. |
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Is that 12k with 855a1? What's it with the lower pressure stuff?
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No, what he's describing is the mean (which is the most common form of average). The median is the middle number out of the set. The median of 1, 2, 3, 4, 1000 is 3. The mean is the sum of the numbers divided by how many there are. The mean of 1, 2, 3, 4, 1000 is 202. Fatigue cracking isn't a rapid event, it's generally something you can spot well in advance of catastrophic failure. http://www.leancrew.com/all-this/images2011/arm-fatigue.jpg It grows slowly (generally polishing itself to some extent) for a long time, and then eventually fails rapidly (leaving a rough fracture). Is that just for bolts or are there other failures in there too? I'm a little surprised at how much shorter bolt life is for M4s, I guess the unlocking adds significantly to the stresses on the bolt. View Quote |
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For anyone wanting to gamble on better materials, SRC has their S7 DLC bolts for $45.
http://store.srcarms.com/product-p/xpb.htm |
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