How much

[Don McDowell]

tin would a person need to add to 5 lbs of 20-1 to come up with 16-1?
I have some 20 and 30-1 a mixed box that rotometals sent, it was supposed to be all 20-1, so they sent me a batch of little tin balls to mix with the 30-1. They sent 4 of those for each bar of 30 to bring it to 20.
Now I want to try the BACO 434470 bullet from 16-1 and see if it helps reach the 1000 yd mark a little more evenly.
So I'ld like to try some 16-1 before I twist off and buy some.

[Freedom]

My math is a little primitive...but just about 1oz. of tin is needed to change 5lbs 20:1 to 16:1. I know there is a perfect formula for this, but I don't know it.

My estimate is based on:.... 16-1 equals ...1lb lead /1oz. tin...so to make 5lbs of 16-1 you need 5oz. tin
20-1 equals 5% tin....so in your 5lbs of alloy, you already have 4oz tin (5lb=80oz.alloy x 5%tin =4oz.)
,

[Mike]

Hey Ranch, Five pounds of 20:1 already holds 4 ounces of tin. If you add just one more ounce of tin you'll have 5 pounds & one ounce of 16:1 (or doggone close to it). Shoot sharp, Mike

[Don McDowell]

Thanks guys. I believe those little tin balls weigh about an ounce. I can weigh them and double check.

[rfd]

i *think* the formula is 1 (representing the percentage of tin) divided by the target pecentage of the lead.

so that a 1:16 tin/lead alloy would be 1 / 16 = .0625 and a 1:20 allloy would be 1 / 20 = .05 and then .0125 of tin added to the 1:20 would yield a 1:16 alloy. phwew!

here's the numbers i came up with if starting off with 5# of pure lead ...

[Old Jim]

Wes and Mike said it the way I would have also !