How many pounds of .44 caliber lead media to fill a 5" jar?

[jonesy]

Hi there,

 

I've just found a rather cheap source of lead milling media, and i'm wondering how much of it I'd need to buy to fill my jar halfway. My mill jar's capacity is 2.55 litres or 0.673 US gallons. Does anyone who has used .44 caliber or similar sized lead shot know roughly how many pounds of it i'd need?

 

More info: the jar's ID is 5" and it's length is 8" (yes, I know about the rule that Jar ID = Jar length, but I couldn't find a suitably shaped jar, sorry)

 

I'm only asking because I'm not really sure how to approach the question mathematically, and your experience is a probably a better judge of the answer anyway

 

If anyone could help, I'd be grateful.

Thanks

Edited April 16, 2012 by jonesy

[Peret]

I can walk you through the math.

 

Volume of your jar = pi *R² * L, = 3.1416 * (2.5)² * 8, = 157 cubic inches. You would like it half full, or 78 cubic inches.

 

The volume of a 0.44 inch sphere is 4/3 * pi * R³, = 1.33 * (0.22)³ * 3.1416, = 0.045 cubic inches. In metric, which we need in a moment, it's 0.73cc.

 

The maximum packing density of spheres is 0.75, so to half fill the jar you need (78 / 0.045) * 0.75 balls, which works out to 1300 balls.

 

The density of lead is 11.34 grams/cc, so the total weight you need is 1300 * 0.73 * 11.34 = 10762 grams, 10.762 kilograms, 23.7 pounds.

 

Check my math in case I made a stupid mistake, but the numbers are in the right range.

[jonesy]

The numbers are perfect. Ah, I already understood how to do most of that. It was the maximum packing density part that I was unsure of. I probably could have worked it out if I'd done a better google search to be honest, sorry about that. Nonetheless, that was a very detailed and clear response, and it's got the answer I needed!

 

Thanks for the help