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Catshooter
05-03-2012, 10:29 PM
Gentlemen,

I want to add some 25 BHN alloy to some pure (BHN 5 lead, let's say one pound/nine pounds.

Is there a formula for calculating what the resulting alloy will have for BHN? There has to be but for whatever reasons I just can't wrap my wits around it and figure it out.

I just know I'm gonna get embaressed.


Cat

gbrown
05-03-2012, 10:56 PM
I think this is it. It is in the stickys at the top of this forum.

http://www.lasc.us/CastBulletNotes.htm

Also, this is an old archive piece that might have the answer you are looking for:

http://castboolits.gunloads.com/archive/index.php/t-67401.html

John Boy
05-03-2012, 11:21 PM
I want to add some 25 BHN alloy to some pure (BHN 5 lead, let's say one pound/nine pounds.
BHN OF MIX 9.64 of 1 lb mono to 9 lbs of pure lead

Catshooter
05-03-2012, 11:23 PM
Thank you gbrown.

For the first link, do I really have to read Glen's whole book again to get my answer? :)

The second link is the answer I was both looking for and afraid of. It depends. Figures. I guess testing is the only way to go.

Thanks.


Cat

gbrown
05-04-2012, 12:10 AM
Glad to help, if I did. I have had the same question. I keep forgetting the answer and have to go look it up.

bumpo628
05-04-2012, 12:17 AM
From the Rotometals site:

Basic Rules for Harding Lead
For every 1% additional tin, Brinell hardness increases 0.3.
For every 1% additional antimony, Brinell hardness increases 0.9.

For a simple equation,
Brinell = 8.60 + ( 0.29 * Tin ) + ( 0.92 * Antimony )

My alloy calculator uses this formula to estimate whatever alloys you mix together.

sqlbullet
05-04-2012, 09:35 AM
As you have discovered, it doesn't work his way.

If you were to mix two alloys of known composition, say monotype and pure lead, then a BHN could be determined for the resulting alloy.

But, since the alloy of your BHN 25 alloy is not specified and I assume unknown, the the results will be unknown. Isotope lead, older WW alloy and monotype are all capable of reaching a BHN of 25 depending on heat treat.