my math is all kaput

[Phillip Allen]

to think I used to tutor college algebra and have studied trig and calculus...

I'm trying to figure out how to get 1 part tin and 40 parts lead mix using pure lead and solder which is 63% tin and 37% lead...

my mind is all a swim and this stuff used to be so easy... it's very discouraging

My furnace holds 20 pounds of mix… max

[Paul Pless]

This has got to be the most feeble attempt ever at a gun thread.

[Phillip Allen]

This has got to be the most feeble attempt ever at a gun thread.

Paul... I serious... I can't even hardly begin... it's the math that troubling me, really... I haven't done this stuff in 30 years

no doubt I could wag it but I want to be able to repeat the mix

[Phillip Allen]

okay... I just started writing knowns on paper... wrote down all I could think of and came up with an answer... will give me 1lb tin in 39 lb lead... close enough so long as it's repeatable

[Glen Longino]

It does not have to be precise measurement does it?
19 pounds lead + 9 ounces solder.

[Phillip Allen]

now the next problem is weighing the stuff without a proper scale

[Phillip Allen]

It does not have to be precise measurement does it?
19 pounds lead + 9 ounces solder.

at 63/37 % tin/lead... a wag is 2/3 tin/1/3 lead
to get 1 pound of tin I must use 1 1/2 pounds solder... okay I think I have it close enough... the brain cells needed were pretty rusty so what should have been easy was somewhat daunting

getting old

[Phillip Allen]

When you say 1 part do you mean volume or weight? To me a part implies volume not weight. I think lead is much heavier than tin so you can not get the correct quantities by weighing.

it's by weight... but I don't mind approximations so long as it's repeatable

I have bullet/powder scales but nothing to weigh heavier stuff... gonna cast a lot of tin bullets and weigh in enough to get my pound and a half and trust to the manufacturs assertion that the lead pot holds 20 pounds

actually I need to cut my 1 1/2 pounds of solder in half for 20 pounds of lead

[JBreeze]

OK - I get 12.69 oz of solder and 19lb. 3.3oz of lead will give you 20 lbs of mix with a 40:1 ratio of lead to tin.

Is the lead in ingot form? What are the weights of the ingot? Can you use a bathroom scale, compared to something known, like a 5 lb bag of flour, or a gallon of water or something ( to get reasonable accuracy)?

Is the solder on a spool? Unroll the 1 lb spool, measure the length and snip off the correct length, i.e. 12.69/16 = 79.3% of the length.

NB: This gives you an approximate 40:1 ratio by weight of lead to tin ....if you need more accuracy, let me know. Also you won't get 40 atoms of lead to 1 atom of tin with this estimate.

[Phillip Allen]

the solder (S) is cast into muffin sized moulds

I will cast a lot of bullets from the solder then weigh in 12 oz of (S) bullets and mark the pot at the full point and add lead until I reach that point... should be close enough I think

[Iceboy]

Pretty pricey using eutectic solder to make bullets out of. Did you get it surplus?

[htom]

You have 63 parts tin and 37 parts lead. How many parts of lead do you need to add to have 63 parts tin and (40*63) parts lead?

.
.
.
.

40*63-37=2483 parts lead

[P.I. Stazzer-Newt]

Interflux sell "Lead Free anti-oxidant pellets"

They are pure tin.

[Nicholas Carey]

To make up 20 lbs of alloy at 1 part tin to 40 parts lead by weight using pure lead and solder that's 63% tin and 37% lead, I get 1.3184 lbs of solder and the remainder (18.6816 lbs) pure lead. Here's how I got there. I'm pretty sure my maths are right, but I'm open to criticism:

1. Set up your equation:

        37         1
    ---------- = ----
    ( 63 + X )    40
    
2. Cross-multiply and simplify:

    37 * 40 = 1 * ( 63 + X )
    
    1480    = 63 + X
    
    1417    = X
    
3. So, for a given weight of solder, you need to add 14.17 times that weight in lead to get to the desired 1 part tin/40 parts lead. 

To make a 20 lb batch, then, set up your equation and simplify to get the weight of solder required:

    X + ( 14.17 * X ) = 20
    
    15.17 * X = 20
    
         20
    X = -----
        15.17
        
    X = 1.3184
    
It follows then that to make your 20 lb batch, you'll need 1.3184 lbs of solder and 18.6816 lbs of lead.

[Phillip Allen]

To make up 20 lbs of alloy at 1 part tin to 40 parts lead by weight using pure lead and solder that's 63% tin and 37% lead, I get 1.3184 lbs of solder and the remainder (18.6816 lbs) pure lead. Here's how I got there. I'm pretty sure my maths are right, but I'm open to criticism:

1. Set up your equation:

        37         1
    ---------- = ----
    ( 63 + X )    40
    
2. Cross-multiply and simplify:

    37 * 40 = 1 * ( 63 + X )
    
    1480    = 63 + X
    
    1417    = X
    
3. So, for a given weight of solder, you need to add 14.17 times that weight in lead to get to the desired 1 part tin/40 parts lead. 

To make a 20 lb batch, then, set up your equation and simplify to get the weight of solder required:

    X + ( 14.17 * X ) = 20
    
    15.17 * X = 20
    
         20
    X = -----
        15.17
        
    X = 1.3184
    
It follows then that to make your 20 lb batch, you'll need 1.3184 lbs of solder and 18.6816 lbs of lead.

thank you... I haven't set up a problem like that since 1975

[Glen Longino]

thank you... I haven't set up a problem like that since 1975

That makes my head hurt!

[Phillip Allen]

To make up 20 lbs of alloy at 1 part tin to 40 parts lead by weight using pure lead and solder that's 63% tin and 37% lead, I get 1.3184 lbs of solder and the remainder (18.6816 lbs) pure lead. Here's how I got there. I'm pretty sure my maths are right, but I'm open to criticism:

1. Set up your equation:

        37         1
    ---------- = ----
    ( 63 + X )    40
    
2. Cross-multiply and simplify:

    37 * 40 = 1 * ( 63 + X )
    
    1480    = 63 + X
    
    1417    = X
    
3. So, for a given weight of solder, you need to add 14.17 times that weight in lead to get to the desired 1 part tin/40 parts lead. 

To make a 20 lb batch, then, set up your equation and simplify to get the weight of solder required:

    X + ( 14.17 * X ) = 20
    
    15.17 * X = 20
    
         20
    X = -----
        15.17
        
    X = 1.3184
    
It follows then that to make your 20 lb batch, you'll need 1.3184 lbs of solder and 18.6816 lbs of lead.

I think I found a problem:

is the first equation reversed?

it looks like it's set up to show lead over tin in the first half then tin over lead in the second half

I don't know how to put the math on the screen

(63+X) 1
___ = ___
37 40
tried but can't make the typing work

[ron ll]

Please excuse a slight thread divergence, but I've often wondered why math is sometime singular and sometimes plural. Phillip says, "...my math is" and Nicholas says, "...my maths are". Seems like I usually see Americans using singular and others using plural. Is there a better reason for the difference?

[JBreeze]

Nick - I think there is an error there somewhere.....1.3 lbs of solder that is 63% tin will give you 13.3 oz Tin. 13.3/320 = ~ 4.2% solution. 1:40 is ~ 2.5%

Long, long time ago, I'd get infrequent calls in the middle of the night asking "I have small amounts of D50W and plenty of D5W. How much D50W do I add to 100ml of D5W to make a D7.5W solution?" After I gave the answer and hung up, I'd scurry to my reference book to dbl. check the math

[Phillip Allen]

well, this is good excercise for me...

I think my brain will be sore tomorrow

[P.I. Stazzer-Newt]

to think I used to tutor college algebra and have studied trig and calculus...

I'm trying to figure out how to get 1 part tin and 40 parts lead mix using pure lead and solder which is 63% tin and 37% lead...

my mind is all a swim and this stuff used to be so easy... it's very discouraging

My furnace holds 20 pounds of mix… max

N.B. Lots of he solder now on sale is lead free - and and most of the rest is not the classic 63/37 brew.

[Nicholas Carey]

Nick - I think there is an error there somewhere.....1.3 lbs of solder that is 63% tin will give you 13.3 oz Tin. 13.3/320 = ~ 4.2% solution. 1:40 is ~ 2.5%

Long, long time ago, I'd get infrequent calls in the middle of the night asking "I have small amounts of D50W and plenty of D5W. How much D50W do I add to 100ml of D5W to make a D7.5W solution?" After I gave the answer and hung up, I'd scurry to my reference book to dbl. check the math

I think you're right. I hadn't had my cofee yet. I gotta run out the door this minute, but I think I confused my lead and tin.

1. Set up your equation:

    ( 37 + X )     40
    ---------- = ----
         63       1

To make up 20 lbs of alloy at 1 part tin to 40 parts lead by weight using pure lead and solder that's 63% tin and 37% lead, I get 1.3184 lbs of solder and the remainder (18.6816 lbs) pure lead. Here's how I got there. I'm pretty sure my maths are right, but I'm open to criticism:

1. Set up your equation:

        37         1
    ---------- = ----
    ( 63 + X )    40
    
2. Cross-multiply and simplify:

    37 * 40 = 1 * ( 63 + X )
    
    1480    = 63 + X
    
    1417    = X
    
3. So, for a given weight of solder, you need to add 14.17 times that weight in lead to get to the desired 1 part tin/40 parts lead. 

To make a 20 lb batch, then, set up your equation and simplify to get the weight of solder required:

    X + ( 14.17 * X ) = 20
    
    15.17 * X = 20
    
         20
    X = -----
        15.17
        
    X = 1.3184
    
It follows then that to make your 20 lb batch, you'll need 1.3184 lbs of solder and 18.6816 lbs of lead.

[Phillip Allen]

I look forward to the next episode

[htom]

For every 100g of solder there is 63g of tin and 37g of lead.

So to every 100g of solder you need to add 2483g of lead, making a total of 2583g, about 5.7 pounds total.

[Phillip Allen]

For every 100g of solder there is 63g of tin and 37g of lead.

So to every 100g of solder you need to add 2483g of lead, making a total of 2583g, about 5.7 pounds total.

that is the way I finally worked it out

[John Smith]

Paul... I serious... I can't even hardly begin... it's the math that troubling me, really... I haven't done this stuff in 30 years

no doubt I could wag it but I want to be able to repeat the mix

find a kid who knows the new math. Knowing what he's doing is more important than getting the right answer, so you can't go wrong, no matter how wrong you go.

[JBreeze]

that is the way I finally worked it out

I can use grams, now !?

I was trying to make it easy, with household scales, etc. to get you to 20 lbs.

[Phillip Allen]

I can use grams, now !?

I was trying to make it easy, with household scales, etc. to get you to 20 lbs.

grams or carats or whatever... conversion is easy... no 'system' is more accurate than another

[JBreeze]

grams or carats or whatever... conversion is easy... no 'system' is more accurate than another

Did you end up making 20 lbs total?
How much solder did you end up using?

[The Bigfella]

Hey guys......

..... go metric, eh?

[Phillip Allen]

Did you end up making 20 lbs total?
How much solder did you end up using?

96 drachms solder to 684 scurples of lead

[Phillip Allen]

Hey guys......

..... go metric, eh?

I like barley corns and cubits... you Frenchies are stuck on metric

[Nicholas Carey]

I look forward to the next episode

Yes...I got a mite confused with the lead and the tin. Here's the updated algebraic solution:

1. Set up your equation:

         63        1
    ---------- = ----
    ( 37 + X )    40
    
2. Cross-multiply and simplify:

    63 * 40 = 1 * ( 37 + X )
    
    2520    = 37 + X
    
    2483    = X
    
    
3. So, for a given weight of solder, you need to add 24.83 times that weight in lead to get to the desired 1:40 tin/lead ratio.

To make a batch with a total weight of 20 lbs, then, set up your equation and simplify to get the weight of solder required:

    X + ( 24.83 * X ) = 20
    
    25.83 * X = 20
    
         20
    X = -----
        25.83
        
    X = 0.7743
    
It follows then that to make your 20 lb batch, you'll need 0.7783 lbs of solder and 19.2217 lbs of lead.

WRT to units, they don't really apply. In computing the right factor, you're dealing with ratios and percentages, not weights. Weight only comes into play in deciding on your target batch size.

FWIW, to get the fancy dancy fixed-pitch font I used, use the [code]...[/code] markup tags.

[Phillip Allen]

Yes...I got a mite confused with the lead and the tin. Here's the updated algebraic solution:

1. Set up your equation:

         63        1
    ---------- = ----
    ( 37 + X )    40
    
2. Cross-multiply and simplify:

    63 * 40 = 1 * ( 37 + X )
    
    2520    = 37 + X
    
    2483    = X
    
    
3. So, for a given weight of solder, you need to add 24.83 times that weight in lead to get to the desired 1:40 tin/lead ratio.

To make a batch with a total weight of 20 lbs, then, set up your equation and simplify to get the weight of solder required:

    X + ( 24.83 * X ) = 20
    
    25.83 * X = 20
    
         20
    X = -----
        25.83
        
    X = 0.7743
    
It follows then that to make your 20 lb batch, you'll need 0.7783 lbs of solder and 19.2217 lbs of lead.

WRT to units, they don't really apply. In computing the right factor, you're dealing with ratios and percentages, not weights. Weight only comes into play in deciding on your target batch size.

FWIW, to get the fancy dancy fixed-pitch font I used, use the [code]...[/code] markup tags.

I ended up using 12 oz solder and because I can't weigh that much lead (I only have a powder scale) I assumed the pot holdes what it's advertised to hold... I measured the finished level at 1/2 inch from the top with a straight edge laid across to measure to

this is repeatable even though It won't be exactly 1/40. The point is to make a very soft bullet but pure lead is hard to cast and the addition of tin is like adding soap to water so it pours better

[Todd D]

1) The correct answer is 0.7743 lbs solder to 19.2257 lbs lead. 2) 0.7743 lbs of solder contains 0.63 x 0.7743 = 0.4878 lbs tin and 0.7743-0.4878 = 0.2865 lbs lead. 3) So the total tin is 0.4878 lbs and the lead is 19.2257 + 0.2865 = 19.5122 lbs. 4) 0.4878/19.5122 = 0.0249997. 5) 1/40 = 0.025. 6) QED

[Phillip Allen]

The correct answer is 0.7743 lbs solder to 19.2257 lbs lead0.7743 lbs of solder contains 0.63 x 0.7743 = 0.4878 lbs tin and 0.7743-0.4878 = 0.2865 lbs leadSo the total tin is 0.4878 lbs and the lead is 19.2257 + 0.2865 = 19.5122 lbs0.4878/19.5122 = 0.02499971/40 = 0.025QED

looks aproximately right to me

[BrianW]

Well Phillip. Your math may be kaput, but your problem solving skills are still excellent.

[PhaseLockedLoop]

Please excuse a slight thread divergence, but I've often wondered why math is sometime singular and sometimes plural. Phillip says, "...my math is" and Nicholas says, "...my maths are". Seems like I usually see Americans using singular and others using plural. Is there a better reason for the difference?

"Maths" is correct in British English, and "math" correct in American English. I have no idea why.

[ron ll]

"Maths" is correct in British English, and "math" correct in American English. I have no idea why.

Okay Nicholas, you got some 'splainin' to do. Are you a Brit transplanted to Seattle?

[Nicholas Carey]

Okay Nicholas, you got some 'splainin' to do. Are you a Brit transplanted to Seattle?

No, but I grew up with a bunch of Anglos and Scots and I used to work (here in Seattle) for an English software company out of Cambridge...though the office I worked with was in Alderley Edge, just south of Manchester.

[Phillip Allen]

don't feel bad Nick... I love to play with the language(s)

[Nicholas Carey]

"Maths" is correct in British English, and "math" correct in American English. I have no idea why.

I think "maths" is plural because there is more than one "math": you've got arithmetic, algebra, geometry, calculus, trignometry, etc. That's why its "mathematics".