got some 50/50 solder, help with the math

[corey012778]

I was cleaning up around the house the other day and found some 50/50 solder, looks like half a roll, I don't know how long it has been there but it has been sometime, I say the age would be over 10 years, I have not cleaned the closet I found it in my self so I really don't know how long it has been in there.

I am wanting to added to about 10 lbs of ww ingots, just see if I like how my boolits does. it looks like there is about 1/2 of an 1/2 lbs roll.

thanks, and if yall like I got pics of the roll if that will help anything.

[NVcurmudgeon]

I usually add 2% tin to my WW ingots. For convenience I add 140 gr. every time I add a WW ingot to the casting pot. That would be 280 gr. of 50/50. If you intend to smelt it all together, add 2,800 gr. to the entire 10 lbs. of ingots. Yes, you are also adding 1,400 gr. of lead, but that shouldn't make a great deal of dfference, as WW are mostly lead.

[NVcurmudgeon]

2,800 gr.=64 oz.=4 lb. Sorry if I'm preaching to the choir, but you may have a !@#$%^&* cold like I do, so that will make it easier.

[corey012778]

so pritty much throw the whole roll in [smilie=1:

[JesterGrin_1]

NVcurmudgeon is saying to add 4 POUNDS of the 50/50 with 10 pounds of WW I think? So 14 pounds total of melt?

[corey012778]

oh well, I got is a little bit. no biggie, I like how my bullets are coming out anyway.

[whisler]

NV curmudgen: I think your cold messed you up. 2800 gr. /7000 gr per pound = 0.4 pounds.

[randyrat]

Yeh that dang flu going around about kicked my ****, and i had a flu shot.
WAIT, why add tin? Only add it if you need to or need some tougher hunting bullets right? If your filling your molds out ok then just use WWs. Save the tin for when you really need it. Or did i miss something.

[corey012778]

the flu has hit my family hard to, I think I am starting to get it now. I am going to hold off on using it until I can get hold of somemore. them I may just hold on until the ww ingots I have run low.

[Tom Myers]

I am wanting to added to about 10 lbs of ww ingots, just see if I like how my boolits does. it looks like there is about 1/2 of an 1/2 lbs roll.

thanks, and if yall like I got pics of the roll if that will help anything.

Corey,

If you do decide to add your 1/4 lb of 50:50 solder to enough lead to make the standard 2% Tin / 98% Wheel-weight alloy, just add the 1/4 lb of 50:50 solder to 5.75 pounds of wheel-weights for a total alloy weight of 6.25 lbs.

You have 0.125 pounds of tin and 0.125 pounds of lead in the 1/2 of the 1/2 pound roll of 50:50 solder.

2%(0.02) * ( [X lb of ww]+ [0.25 lb of solder] ) = 0.125 lb of tin

0.02 * X + (0.02 * 0.5) = 0.125
0.02 * X + 0.01 = 0.125
0.02 * X = 0.125 - 0.01
0.02 * X = 0.115
X = 0.115 / 0.02
X = 5.75 lb of w/w
5.75 lb of wheel-weights + 0.125 lbs of lead + 0.125 lbs of tin = 6.25 lbs of alloy

0.125 lbs of tin divided by 6.25 lbs of alloy = 0.02 = 2%

Hope this helps
Tom Myers
Precision Ballistics and Records (http://ww.tmtpages.com)

[MT Gianni]

If I think my alloy needs tin. 15"-18" of 50-50 solder in a Lee 20 lb pot is usually enough for me.

[calaloo]

I mix my own lead tin alloy, usually 20:1 or 25:1. On the couple of occasions that I've used solder (50:50) for its tin content I've gotten something I didn't want. Solder has other metals other than lead and tin. This probably would not matter for pistol bullets etc. but be aware.

[NVcurmudgeon]

NV curmudgen: I think your cold messed you up. 2800 gr. /7000 gr per pound = 0.4 pounds.

Whisler, thanks for correcting me at a time I'm mathematically chalenged, and for preventing the waste of precious tin.

[grumpy one]

If I think my alloy needs tin. 15"-18" of 50-50 solder in a Lee 20 lb pot is usually enough for me.

Personally I like 1% or so of tin so I can use a very low casting temperature, thus speeding up the casting process (too much mould-quenching between casts gives uneven mould temperatures, especially with iron moulds).

However I'm not sure castability is the only issue here - there are also some metallurgical mysteries to be solved. Having equal percentages of Sb and Sn results in a microstructure that consists of crystals of SbSn (which will only combine chemically if in equal proportions) in a matrix of lead. Any alloy with equal Sn and Sb is a "pseudo-binary", with its own eutectic at 10% Sn and 10% Sb, which has a really pretty lamellar microstructure; it looks much nattier than pure lino. I want to know more about the pseudo-binary eutectic because if it has high impact resistance it would be the ultimate bullet alloy. Unfortunately I don't have any data on its physical properties except that it is about twice as scratch-resistant as lino, and is 21.5 BHN compared with 22 for lino, air-cooled in both cases. I'd really like to see an impact test on it (Izod or Charpy) compared with lino, but haven't seen anything like that yet. However if you look at the microstructures, you can easily see why any high-antimony hypereutectic alloy is likely to have lousy impact resistance; both antimony and SbSn crystals have sharp corners.

[American]

…found some 50/50 solder…
…I am wanting to added to about 10 lbs of ww ingots…
…it looks like there is about 1/2 of an 1/2 lbs roll.


When considering the (precious) tin already in the WW, I'm getting an answer of using the solder (if its exactly 4 oz) with 8 pounds of smelted WW metal to make an alloy with exactly 2% tin.

Of course I understand the amount of tin in WW can vary, and the solder may not be exactly 1/4 pound, but thought I'd give it a shot based on provided info. Could this logic and figuring be correct? Been a long time since high-school algebra:

The facts:
- Smelted WW metal is .5% tin already (as per LASC reference) and tin is precious
- You want an alloy with 2% tin
- You wish to use all of your solder, that is, 1/2 of a 1/2# roll of 50/50 solder, or .25#, which will provide .125# tin & .125# lead, with an undetermined amount of WW to make the new alloy

Let:
W = pounds of smelted WW metal
Tw = pounds of tin already in WW = 1/2% of W = .005 W
Ts = pounds of tin in solder = .125
2% = .02
S = pounds of solder = .25

And:
The objective is (I think) to make the 'total tin' (tin in the WW, plus the tin in the solder) equal 2% of the the total alloy (which will consist of the WW and the solder).

So:
Total tin = 2% of total alloy

Algebratized (is that a word?):
Tw + Ts = .02 (W + S)

With values plugged in:
(.005 W) + .125 = .02 (W + .25)

All terms with variable moved to one side of equation:
.125 = [.02 (W + .25)] - (.005 W)

Multiplied:
.125 = (.02 W) + .005 - (.005 W)

Combined like terms:
.125 = (.015 W) + .005

Isolated term with variable:
.125 - .005 = .015 W

Combined like terms:
.12 = .015 W

Solved for variable:
8 = W

[grumpy one]

American, I agree with your answer of 8 pounds. However I use the WW composition stated on Page 11 of the RCBS Cast Bullet Manual (1986): 0.25% tin, 3% antimony, following a series of reductions in alloy content over the years. If you use that composition it would be 6.85 pounds of WW to 0.25 pounds of solder.

These calculations are required frequently, and it's convenient to set up a spreadsheet that lists all the common raw material compositions, so you can just enter a weight for each raw material and get the overall composition of the final alloy.

[American]

American, I agree with your answer of 8 pounds. However I use the WW composition stated on Page 11 of the RCBS Cast Bullet Manual (1986): 0.25% tin, 3% antimony, following a series of reductions in alloy content over the years. If you use that composition it would be 6.85 pounds of WW to 0.25 pounds of solder.

These calculations are required frequently, and it's convenient to set up a spreadsheet that lists all the common raw material compositions, so you can just enter a weight for each raw material and get the overall composition of the final alloy.

Hey Grump, thanks for the info - I don't have that manual. I'll probably use that .25% figure when calculating for my own use. So new to this, I haven't even used tin in with my WW yet.

Next time I'm going to give it a try, but I'm kinda wondering, given the expense of tin. On one hand, my boolits mostly came out OK from WW without added tin (some rounded bases, but I'm thinking venting and heat...) . On the other, in my ignorance, I maybe don't have such a high standard for them yet...

How important is that 'magic' 2% tin specification? 1% better than none? Or is there some kind of a 'breakpoint' that makes 2% just the thing?

[grumpy one]

Hey Grump, thanks for the info - I don't have that manual. I'll probably use that .25% figure when calculating for my own use. So new to this, I haven't even used tin in with my WW yet.

Next time I'm going to give it a try, but I'm kinda wondering, given the expense of tin. On one hand, my boolits mostly came out OK from WW without added tin (some rounded bases, but I'm thinking venting and heat...) . On the other, in my ignorance, I maybe don't have such a high standard for them yet...

How important is that 'magic' 2% tin specification? 1% better than none? Or is there some kind of a 'breakpoint' that makes 2% just the thing?

Many people on this list use straight WW, and just cast at a high enough temperature to achieve the required level of fluidity to get good castings. Mould temperature is at least as important as alloy temperature. I use 1% tin, or just over, so that I can use a low alloy temperature and still have fluidity. The advantage of the low temperature is faster cycle time.

Separate from the issue of getting good castings, is the subject of appropriate bullet metallurgy. Some alloys cause bullets to shatter if they hit bone, and some may act rather like a file on the inside of the barrel. I don't think you have a problem at the low alloying levels we are talking about, though there is very little information available so far.

[crabo]

Wow! Too much math for me. I have 3# ingot molds that are 8" long. When I add an ingot of WW, I add an 8" piece of 95/5 solder. It works for me, but I don't have a clue of what I have done as far as the alloy, but I like the way the bullets fill out the molds.

At some point I will start studying metalurgy, I'm still learning a lot of other things that seem more important at this time.

Crabo

[Hobo Of The Lake]

If I think my alloy needs tin. 15"-18" of 50-50 solder in a Lee 20 lb pot is usually enough for me.

Amy here, Really Amy's dad! I looked around for som 50-50 bar solder like my lyman cast bullet book said to add to wheel weights. 9 pound ww + 1 pound solder. It costs about $13.00 a pound for bar solder. It sound like you are using a role of the stuff. Is it cheeper that way? My book says that tin just makes the mix flow better and fill the mold better. Some of the guys in the chat room say that I can just use pure ww and gas checks. What you think Thanks

[Maven]

amy, Yes, you can use wheelweights without adding tin, but fill-out may be a problem. If that happens, buy that bar of 50/50 solder, but don't use it all at once. Try dividing it into eighths and add 1/8 of the bar by length to 10lb. of wheelweights. By contrast, Lyman's #2 alloy is 5% tin, which is not necessary for good results.

[Hobo Of The Lake]

amy, Yes, you can use wheelweights without adding tin, but fill-out may be a problem. If that happens, buy that bar of 50/50 solder, but don't use it all at once. Try dividing it into eighths and add 1/8 of the bar by length to 10lb. of wheelweights. By contrast, Lyman's #2 alloy is 5% tin, which is not necessary for good results.

Thanks Maven, can I get 50-50 solder at the hardware store where they sell wire solder. Someone said he used 15 to 18 inches to 10 pounds ww. Makes me think you can buy it that way. Thanks again

[John Boy]

Corey - download this alloy calculator and it will be your friend figuring out alloy mixes ...
http://www.castbulletassoc.org/downloads/alloycalculator.zip

PS: Compliments of The Cast Bullet Association

[MT Gianni]

Yes I have some old rolls of 50-50. I paid about $3.50 a roll so that should let you know how old they are. I only use it if I do not get good fill out in certain molds. Gianni

[American]

Amy here, Really Amy's dad! I looked around for som 50-50 bar solder like my lyman cast bullet book said to add to wheel weights. 9 pound ww + 1 pound solder. It costs about $13.00 a pound for bar solder. It sound like you are using a role of the stuff. Is it cheeper that way? My book says that tin just makes the mix flow better and fill the mold better. Some of the guys in the chat room say that I can just use pure ww and gas checks. What you think Thanks

The Lyman book must have been written before the common use of lead-free solder, which is usually 95% tin and 5% antimony, and costs about $13 a pound at plumber's supply store, bar or wire form.

If you buy solder for the tin, and you buy a pound of 50/50 solder you are getting 8 oz of tin for $13 (I'm guessing that if you are casting, the 8 oz of lead is not worth much consideration). If you buy a pound of lead-free solder you are getting 15.2 oz of tin for your $13. Don't know about you, but $13 a pound seems like a lot for lead... [smilie=1: