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How Much 45 Acp Brass Fits In A 5 Gallon Bucket ?


BigSlick

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according to my calculations ...

length 45 acp cartridge ~ 0.9 in

diameter of 45 acp cartridge ~ 0.47 in

the volume of a 45 acp cartidge should be:

(pi*r^2)*l or

(pi*(d/2)^2)*l

substituting the above values

(3.14*(0.470/2)^2)*0.9

yielding

0.156 in^3 or

9.03*10^-5 ft^3

using the conversion

ft^3*7.5 = gallons

a five gallon bucket should then be

5/7.5 = 0.667 ft^3

assuming a 20% loss of space since the

brass does not stack perfectly, the amount

of brass should be approximately:

0.667 ft^3 / 9.03*10^-4 * 0.8

5905.2 pieces of brass

Edited by short_round
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The random placment of the empty casings in the bucket would seem to be a big, big variable. Maybe Chaos Theroy has a role to play...

My flat guess, given what the known 40 & 9 volumes were, would have been 5K. Anybody have the time to neatly stack the brass in a bucket to see how many would really fit?

edit by Middle Man to add new method of calculation

Just had a thought while taking the dog out. The way to calculate the number of 45 cases in a 5 gal bucket would be to take the weight of one 45 case and plug in the formula for Weight by Volume, subtract 1lb for the bucket, and divide total weight of the 5gal volume by the single 45 case weight to obtain the number of pieces in the 5gal.

Anyhow off to today's match...I'll have check the math books later.

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  • 1 year later...
The random placment of the empty casings in the bucket would seem to be a big, big variable. Maybe Chaos Theroy has a role to play...

My flat guess, given what the known 40 & 9 volumes were, would have been 5K. Anybody have the time to neatly stack the brass in a bucket to see how many would really fit?

edit by Middle Man to add new method of calculation

Just had a thought while taking the dog out. The way to calculate the number of 45 cases in a 5 gal bucket would be to take the weight of one 45 case and plug in the formula for Weight by Volume, subtract 1lb for the bucket, and divide total weight of the 5gal volume by the single 45 case weight to obtain the number of pieces in the 5gal.

Anyhow off to today's match...I'll have check the math books later.

Yes - but if the brass were mixed head stamp, then wouldn't you need to take into account the variance in weight between manufacturers? I would measure the weigh in grains of a sample of each represented brand and then figure the average of the samples. Thoughts?

By the way, anyone have a rough # of .45 casings that fit in a US Postal Service flat rate box?

Edited by Carlos
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Off topic I know but why did the OP delete ALL of his posts? All 35.

Someone must have pee'd in his Cheerios.

Noticed that too. Dunno the scoop. Mods- if this is a dead horse topic or off limits, just say the word. Suprised someone would leave in a huff - this is a very polite corner of the web.

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If it's who I think it is that plays "King of Reloading" on GT's reloading forum going by the same name then it's better off he is gone anyway. :cheers:

Edited by unclez
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Off topic I know but why did the OP delete ALL of his posts? All 35.

Someone must have pee'd in his Cheerios.

Noticed that too. Dunno the scoop. Mods- if this is a dead horse topic or off limits, just say the word. Suprised someone would leave in a huff - this is a very polite corner of the web.

This is one of the reasons there is now a time-limit on editing and deleting posts.. Beyond that time you have to convince one of the admins to do it.

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