Re: Question for all Y'all Lead Metallurgist
Originally Posted by
Southron, Sr.,
About a week before the last Spring Nationals I fired up the old casting machine to cast some Minie Balls for the Nationals. I was using batch of lead that was supposed to be "Pure" lead.
What I quickly found out was that even though my melt was at the proper temperature, the mould heated up properly, I was running something like 40% rejects. Not a good thing at all.
In desperation I jumped into the old horseless carriage and made a bee line to the local hardware store where I picked up a 1 pound roll of 50/50 solder that was composed only of lead and tin.
I refilled my pot which has a 40 pound capacity. Then I threw in HALF the roll of solder-which I figure contained 4 ounces of Tin.
I mixed the solder into my melt (but did not flux again) and started casting bullets again.
"PRESTO" My bullets started coming almost out "Perfect" with maybe a 2% reject rate. Of course because of the Tin, they were also brighter.
Here is my math: 40 pounds of lead equal 640 ounces of Lead. So divide 4 ounces of Tin into 640 ounces of Lead = a 1 in 160 Tin/Lead ratio.
QUESTION: If Pure Lead has a Brinell hardness of 5.0 then what is the Brinell hardness of my 1 in 160 alloy mixture???
Inquiring Minds Want to Know!
THANKS for your answers in advance!
P.S. I am assured that my Minie Balls, even with a 1 in 160 alloy is "Soft" enough for Minie Balls.
At a ratio of 1 part Sn (Tin) to 160 parts Pb (Lead), the Brinell hardness of the tin-lead composition won't make much of any change from BHN 5. According to the web site link below
http://www.lasc.us/brennan_saeco_table.htm
This is a representation of the chart that Saeco ships with their Bullet Hardness Tester.
The chart demonstrates the relative hardness that may be expected when alloying various ratios of Tin/Lead by using the following formula.
BH = ( 222.5 x R - 7.82 x R x R ) / ( 1 + 23.12 x R ) + 5
Let "R" equal a Ratio which is the weight of the Tin in the alloy divided by the total weight of the alloy.-
Then:
1 part tin plus 19 parts lead (by weight) equals a 1 in 20 alloy.
Then 1/20 = 0.05 ... So then R = 0.05..... Therefore:
BHN = ( 222.5 x 0.05 - 7.82 x 0.05 x 0.05 ) / ( 1 + 23.12 x 0.05 ) + 5
The Brinell Hardness Number of a 1 in 20 Tin-Lead alloy = 10.15
An alloy made of half a pound bar of 50/50 Sn-Pb solder plus 40 lbs of lead equates to 40.25 lbs of lead plus a quarter pound of tin or approximately equal to 1 part tin to 160 parts of lead (or 1750 gns/281750 gns)
1/160 = 0.00625
Therefore,
BHN 1/160 Alloy= [(222.5 x 0.00625 – 7.82 x 0.00625 x 0.00625) / (1+23.12x0.00625)]+5
= [(1.3906 - 0.0003) / (1+0.1445)] + 5
= [(1.3903/1.1445)] + 5
= [1.2148] + 5
= 6.2148
First Cousin (7 times removed) to Brigadier General Stand Watie (1806-1871), CSA
1st Cherokee Mounted Rifles | Principal Chief of the Cherokee Nation 1862-66
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