Math wizards get in here

[JGalt]

Ok, I suck at math, so here's a question:

I have some metal bars that are composed of 30% tin and 70% lead. The bars weigh 1 pound, 11 ounces.

I want to create an alloy consisting of 2% tin, 6% antimony, and 92% lead.

Using one of those bars, how many pounds and ounces of lead and antimony do I need to get a 2% tin, 6% antimony, and 92% lead alloy?

Probably a simple question, but I haven't used any math in 35 years.

 

[JGalt]

Well duhh. I just realized we have AI now.

ChatGPT:

To solve this, we need to:

1. Convert the metal bar weight from pounds and ounces to just pounds for easier calculation.
2. Determine the amount of tin in the original bar and adjust for the desired alloy composition.
3. Calculate the amount of lead and antimony to add to the tin from the bar to achieve the target alloy composition.

Step 1: Convert the bar weight to pounds​

The bar weighs 1 pound, 11 ounces. There are 16 ounces in a pound, so:

Total weight in pounds=1+1116=1.6875 pounds\text{Total weight in pounds} = 1 + \frac{11}{16} = 1.6875 \text{ pounds}Total weight in pounds=1+1611=1.6875 pounds

Step 2: Determine the amount of tin in the bar​

Since the bar is composed of 30% tin and 70% lead, the amount of tin in the bar is:

Tin in the bar=1.6875×0.30=0.50625 pounds of tin\text{Tin in the bar} = 1.6875 \times 0.30 = 0.50625 \text{ pounds of tin}Tin in the bar=1.6875×0.30=0.50625 pounds of tin

Step 3: Determine the target weight for the alloy​

The final alloy needs to contain 2% tin, 6% antimony, and 92% lead.

Let’s assume the total final weight of the alloy is WWW pounds. We know:

• 2%2\%2% of WWW is tin.
• 6%6\%6% of WWW is antimony.
• 92%92\%92% of WWW is lead.

Given that we have 0.50625 pounds of tin, and that must represent 2% of the total alloy, we can calculate the total weight WWW of the final alloy:

0.50625=0.02×W0.50625 = 0.02 \times W0.50625=0.02×W
Solving for WWW:

W=0.506250.02=25.3125 poundsW = \frac{0.50625}{0.02} = 25.3125 \text{ pounds}W=0.020.50625=25.3125 pounds

Step 4: Determine the required amounts of lead and antimony​

Now that we know the total weight of the alloy is 25.3125 pounds, we can calculate the amounts of lead and antimony needed:

• Lead: 92%92\%92% of 25.3125 pounds is:

Lead required=0.92×25.3125=23.2875 pounds\text{Lead required} = 0.92 \times 25.3125 = 23.2875 \text{ pounds}Lead required=0.92×25.3125=23.2875 pounds

• Antimony: 6%6\%6% of 25.3125 pounds is:

Antimony required=0.06×25.3125=1.51875 pounds\text{Antimony required} = 0.06 \times 25.3125 = 1.51875 \text{ pounds}Antimony required=0.06×25.3125=1.51875 pounds

Step 5: Adjust for the lead already in the bar​

The original bar contains 70% lead, so:

Lead in the bar=1.6875×0.70=1.18125 pounds of lead\text{Lead in the bar} = 1.6875 \times 0.70 = 1.18125 \text{ pounds of lead}Lead in the bar=1.6875×0.70=1.18125 pounds of lead
The additional lead required is:

Additional lead needed=23.2875−1.18125=22.10625 pounds\text{Additional lead needed} = 23.2875 - 1.18125 = 22.10625 \text{ pounds}Additional lead needed=23.2875−1.18125=22.10625 pounds

Final Summary:​

To create the alloy with the composition of 2% tin, 6% antimony, and 92% lead, using one of your bars:

• You need 1.51875 pounds of antimony.
• You need 22.10625 pounds of additional lead.

Would you like the weight of additional lead and antimony in ounces as well?

Let's convert the amounts of additional lead and antimony into pounds and ounces.

Step 1: Lead​

You need 22.10625 pounds of additional lead.

To convert 0.10625 pounds to ounces:

0.10625×16=1.7 ounces0.10625 \times 16 = 1.7 \text{ ounces}0.10625×16=1.7 ounces
So, the total additional lead required is:

22 pounds and 1.7 ounces22 \text{ pounds and } 1.7 \text{ ounces}22 pounds and 1.7 ounces

Step 2: Antimony​

You need 1.51875 pounds of antimony.

To convert 0.51875 pounds to ounces:

0.51875×16=8.3 ounces0.51875 \times 16 = 8.3 \text{ ounces}0.51875×16=8.3 ounces
So, the total antimony required is:

1 pound and 8.3 ounces1 \text{ pound and } 8.3 \text{ ounces}1 pound and 8.3 ounces

Final Summary:​

• You need 22 pounds, 1.7 ounces of additional lead.
• You need 1 pound, 8.3 ounces of antimony.

 

[shockedcanadian]

Maybe I am confused and I am not to proud to look foolish so i ask. Are you able to.separate the two compounds.after you melt the bars down? If you aren't able to separate the two compounds then the resulting compounds will remain 30/70 even after you melt them down, no?

 

[JGalt]

Maybe I am confused and I am not to proud to look foolish so i ask. Are you able to.separate the two compounds.after you melt the bars down? If you aren't able to separate the two compounds then the resulting compounds will remain 30/70 even after you melt them down, no?

The 30/70 tin and lead in the bars are inseparable, since they're an alloy. I wanted to proportionally increase the amount of lead and add antimony to get a 2%, 6%, and 92% alloy of tin, antimony, and lead.

 

[shockedcanadian]

Yeah, ok, as I suspected. Based on your posting you explain my confusion. The missing factor is that you are going to buy additional.weight of each separate compound in order to achieve the right mixture not separate. The last question would be is that enough or too much for your purpose? If you only needed a bar or two worth of this mix it could be a massive overrun to obtain all you need in many pounds in order to get the exact mix.

 

[JGalt]

Yeah, ok, as I suspected. Based on your posting you explain my confusion. The missing factor is that you are going to buy additional.weight of each separate compound in order to achieve the right mixture. That is the critical information I needed. The last question would be is that enough or too much for your purpose? If you only needed a bar or two worth of this alloy it would be a massive overrun to obtain all you need in many pounds in order to get the exact mix.

I have 6 bars of the 30/70 tin/lead alloy and about 250 pounds of pure lead, plus probably 20 pounds of linotype which is 15% antimony. By dividing ChatGPT's answer in half, I might be able to fill a cast iron melting pot. I'm not sure how many pounds it holds, but my Lee bullet casting pot supposedly holds 10 pounds.

 

[shockedcanadian]

The 30/70 tin and lead in the bars are inseparable, since they're an alloy. I wanted to proportionally increase the amount of lead and add antimony to get a 2%, 6%, and 92% alloy of tin, antimony, and lead.

I suppose I didnt consider how.many bars you already had and the purpose. I instinctively knew the factors in the equation (not to the extent GPT chat provided.lol) but didnt know the purpose (big job, small job etc).

 

[shockedcanadian]

I have 6 bars of the 30/70 tin/lead alloy and about 250 pounds of pure lead, plus probably 20 pounds of linotype which is 15% antimony. By dividing ChatGPT's answer in half, I might be able to fill a cast iron melting pot that holds 10 pounds.

Omg thats alot of lead. Yeah Chat GPT is great. Trying to figure.out that result isnt too difficult as you suspect but we dont do it everyday,.just a fractional equation translated into a whole by weight (if you are adding not.separating like a.chemist). However, you'd have to be very accurate obviously. I would double check that with a different chat program to be absolutely sure if you need that exact mix, just my opinion. I had to do the same for a concrete mix years ago for our house and had so much of a couple of compounds left over that it hardened in the garage until we threw it out. Such a waste but what could I do?

 

[JGalt]

Omg thats alot of lead. Yeah Chat GPT is great. Trying to figure.out that result isnt too difficult as you suspect but we dont do it everyday,.just a fractional equation translated into a whole (which was part of my confusion if you were separating like a.chemist not adding). However, you'd have to be very accurate obviously. I would double check that with a different chat.program to be absolutely sure if you need that exact mix, just my opinion.

I have a hardness tester that determines the Brinell hardness of metals. I can always add more lead to soften the alloy up, or more antimony to harden it. Plus when I drop the cast bullets out of the mold into a bucket of cold water, that also hardens them.

Then there's the fact that when I powder-coat them, they're baked in a small toaster over for 20 minutes at 400 degrees. That softens the metal somewhat, and they have to be re-quenched in water when they come out of the oven.

Bullet casting isn't an exact science, there are too many variables.

 

[shockedcanadian]

I have a hardness tester that determines the Brinell hardness of metals. I can always add more lead to soften the alloy up, or more antimony to harden it. Plus when I drop the cast bullets out of the mold into a bucket of cold water, that also hardens them.

Then there's the fact that when I powder-coat them, they're baked in a small toaster over for 20 minutes at 400 degrees. That softens the metal somewhat, and they have to be re-quenched in water when they come out of the oven.

Bullet casting isn't an exact science, there are too many variables.

Oh I see. That's an interesting science I suppose. A smaller single use purpose but in higher volume of casting based on the amount of metal you have. Again, I presume.

 

[Votto]

Ok, I suck at math, so here's a question:

I have some metal bars that are composed of 30% tin and 70% lead. The bars weigh 1 pound, 11 ounces.

I want to create an alloy consisting of 2% tin, 6% antimony, and 92% lead.

Using one of those bars, how many pounds and ounces of lead and antimony do I need to get a 2% tin, 6% antimony, and 92% lead alloy?

Probably a simple question, but I haven't used any math in 35 years.

That depends, am I on a train going 60 mph? If so, for how many miles?

 

[luiza]

Well duhh. I just realized we have AI now.

ChatGPT:

To solve this, we need to:

1. Convert the metal bar weight from pounds and ounces to just pounds for easier calculation.
2. Determine the amount of tin in the original bar and adjust for the desired alloy composition.
3. Calculate the amount of lead and antimony to add to the tin from the bar to achieve the target alloy composition.

Step 1: Convert the bar weight to pounds​

The bar weighs 1 pound, 11 ounces. There are 16 ounces in a pound, so:

Total weight in pounds=1+1116=1.6875 pounds\text{Total weight in pounds} = 1 + \frac{11}{16} = 1.6875 \text{ pounds}Total weight in pounds=1+1611=1.6875 pounds

Step 2: Determine the amount of tin in the bar​

Since the bar is composed of 30% tin and 70% lead, the amount of tin in the bar is:

Tin in the bar=1.6875×0.30=0.50625 pounds of tin\text{Tin in the bar} = 1.6875 \times 0.30 = 0.50625 \text{ pounds of tin}Tin in the bar=1.6875×0.30=0.50625 pounds of tin

Step 3: Determine the target weight for the alloy​

The final alloy needs to contain 2% tin, 6% antimony, and 92% lead.

Let’s assume the total final weight of the alloy is WWW pounds. We know:

• 2%2\%2% of WWW is tin.
• 6%6\%6% of WWW is antimony.
• 92%92\%92% of WWW is lead.

Given that we have 0.50625 pounds of tin, and that must represent 2% of the total alloy, we can calculate the total weight WWW of the final alloy:

0.50625=0.02×W0.50625 = 0.02 \times W0.50625=0.02×W
Solving for WWW:

W=0.506250.02=25.3125 poundsW = \frac{0.50625}{0.02} = 25.3125 \text{ pounds}W=0.020.50625=25.3125 pounds

Step 4: Determine the required amounts of lead and antimony​

Now that we know the total weight of the alloy is 25.3125 pounds, we can calculate the amounts of lead and antimony needed:

• Lead: 92%92\%92% of 25.3125 pounds is:

Lead required=0.92×25.3125=23.2875 pounds\text{Lead required} = 0.92 \times 25.3125 = 23.2875 \text{ pounds}Lead required=0.92×25.3125=23.2875 pounds

• Antimony: 6%6\%6% of 25.3125 pounds is:

Antimony required=0.06×25.3125=1.51875 pounds\text{Antimony required} = 0.06 \times 25.3125 = 1.51875 \text{ pounds}Antimony required=0.06×25.3125=1.51875 pounds

Step 5: Adjust for the lead already in the bar​

The original bar contains 70% lead, so:

Lead in the bar=1.6875×0.70=1.18125 pounds of lead\text{Lead in the bar} = 1.6875 \times 0.70 = 1.18125 \text{ pounds of lead}Lead in the bar=1.6875×0.70=1.18125 pounds of lead
The additional lead required is:

Additional lead needed=23.2875−1.18125=22.10625 pounds\text{Additional lead needed} = 23.2875 - 1.18125 = 22.10625 \text{ pounds}Additional lead needed=23.2875−1.18125=22.10625 pounds

Final Summary:​

To create the alloy with the composition of 2% tin, 6% antimony, and 92% lead, using one of your bars:

• You need 1.51875 pounds of antimony.
• You need 22.10625 pounds of additional lead.

Would you like the weight of additional lead and antimony in ounces as well?

Let's convert the amounts of additional lead and antimony into pounds and ounces.

Step 1: Lead​

You need 22.10625 pounds of additional lead.

To convert 0.10625 pounds to ounces:

0.10625×16=1.7 ounces0.10625 \times 16 = 1.7 \text{ ounces}0.10625×16=1.7 ounces
So, the total additional lead required is:

22 pounds and 1.7 ounces22 \text{ pounds and } 1.7 \text{ ounces}22 pounds and 1.7 ounces

Step 2: Antimony​

You need 1.51875 pounds of antimony.

To convert 0.51875 pounds to ounces:

0.51875×16=8.3 ounces0.51875 \times 16 = 8.3 \text{ ounces}0.51875×16=8.3 ounces
So, the total antimony required is:

1 pound and 8.3 ounces1 \text{ pound and } 8.3 \text{ ounces}1 pound and 8.3 ounces

Final Summary:​

• You need 22 pounds, 1.7 ounces of additional lead.
• You need 1 pound, 8.3 ounces of antimony.

Similar to the thread here where some visionary nutter tried to convince us that multipling 21x 13 need not be mentally computed in two seconds when you can combine a Japanese art class with geometry and design and get the same answer in a day .

 

[wamose]

Ok, I suck at math, so here's a question:

I have some metal bars that are composed of 30% tin and 70% lead. The bars weigh 1 pound, 11 ounces.

I want to create an alloy consisting of 2% tin, 6% antimony, and 92% lead.

Using one of those bars, how many pounds and ounces of lead and antimony do I need to get a 2% tin, 6% antimony, and 92% lead alloy?

Probably a simple question, but I haven't used any math in 35 years.

Thanks for the reality check. I used to think I was good. But I did come up with something. Add 22 pounds 3.2 ounces of lead and 1 pound 5.2 ounces of antimony.

 

[wamose]

Wow, I racked my brain for over an hour only to find out you posted the answer an hour ago. But it was good to know I did finally figure out the methodology. Thanks for the brain teaser.

 

[JGalt]

Wow, I racked my brain for over an hour only to find out you posted the answer an hour ago. But it was good to know I did finally figure out the methodology. Thanks for the brain teaser.

There's another variable I forgot to throw in with the mix: For antimony, I'm using Linotype, which is also 4% tin, 12% antimony, and 84% lead. Or Monotype, which is 9% tin, 19% antimony, and 72% lead.

I have some of both but I'm not sure which is which. The only way to tell is to cast a small sample and do a hardness test.

 

[wamose]

There's another variable I forgot to throw in with the mix: For antimony, I'm using Linotype, which is also 4% tin, 12% antimony, and 84% lead. Or Monotype, which is 9% tin, 19% antimony, and 72% lead.

I have some of both but I'm not sure which is which. The only way to tell is to cast a small sample and do a hardness test.

Yeah, I had a little contact with metals and Brinell testing during my brief time at Ingersoll Rand, pump division. But I don't remember shit about it. So anyway, good luck with your project. I think I'll stick to something I can still handle like mowing the grass and recaulking the toilet. Adios amigo. MAGA

 

[zaangalewa]

Ok, I suck at math, so here's a question:

I have some metal bars that are composed of 30% tin and 70% lead. The bars weigh 1 pound, 11 ounces.

I want to create an alloy consisting of 2% tin, 6% antimony, and 92% lead.

Using one of those bars, how many pounds and ounces of lead and antimony do I need to get a 2% tin, 6% antimony, and 92% lead alloy?

Probably a simple question, but I haven't used any math in 35 years.

1 pound 11 ounces are 0.76543712437 kg.

tin = 30% = 0.229631137311 kg.
antimon (3 times more than tin)=0.688893411933 kg.
lead (41times more than tin)=9.414876629751 kg.

In your bar are 0,76543712437 - 0,229631137311 kg = 0,535805987059 kg lead.
So you need additional 9,414876629751 - 0,535805987059 = 8,879070642692 kg lead.

1 of your bars
+ 0.688893411933 kg antimon
and additional 8,879070642692 kg lead
should give a mix of 2% tin, 6% antimon and 92% lead.

 

[Deplorable Yankee]

Ok, I suck at math, so here's a question:

I have some metal bars that are composed of 30% tin and 70% lead. The bars weigh 1 pound, 11 ounces.

I want to create an alloy consisting of 2% tin, 6% antimony, and 92% lead.

Using one of those bars, how many pounds and ounces of lead and antimony do I need to get a 2% tin, 6% antimony, and 92% lead alloy?

Probably a simple question, but I haven't used any math in 35 years.

1 +1= 5
And is racist

 

[Captain Caveman]

Well duhh. I just realized we have AI now.

ChatGPT:

To solve this, we need to:

1. Convert the metal bar weight from pounds and ounces to just pounds for easier calculation.
2. Determine the amount of tin in the original bar and adjust for the desired alloy composition.
3. Calculate the amount of lead and antimony to add to the tin from the bar to achieve the target alloy composition.

Step 1: Convert the bar weight to pounds​

The bar weighs 1 pound, 11 ounces. There are 16 ounces in a pound, so:

Total weight in pounds=1+1116=1.6875 pounds\text{Total weight in pounds} = 1 + \frac{11}{16} = 1.6875 \text{ pounds}Total weight in pounds=1+1611=1.6875 pounds

Step 2: Determine the amount of tin in the bar​

Since the bar is composed of 30% tin and 70% lead, the amount of tin in the bar is:

Tin in the bar=1.6875×0.30=0.50625 pounds of tin\text{Tin in the bar} = 1.6875 \times 0.30 = 0.50625 \text{ pounds of tin}Tin in the bar=1.6875×0.30=0.50625 pounds of tin

Step 3: Determine the target weight for the alloy​

The final alloy needs to contain 2% tin, 6% antimony, and 92% lead.

Let’s assume the total final weight of the alloy is WWW pounds. We know:

• 2%2\%2% of WWW is tin.
• 6%6\%6% of WWW is antimony.
• 92%92\%92% of WWW is lead.

Given that we have 0.50625 pounds of tin, and that must represent 2% of the total alloy, we can calculate the total weight WWW of the final alloy:

0.50625=0.02×W0.50625 = 0.02 \times W0.50625=0.02×W
Solving for WWW:

W=0.506250.02=25.3125 poundsW = \frac{0.50625}{0.02} = 25.3125 \text{ pounds}W=0.020.50625=25.3125 pounds

Step 4: Determine the required amounts of lead and antimony​

Now that we know the total weight of the alloy is 25.3125 pounds, we can calculate the amounts of lead and antimony needed:

• Lead: 92%92\%92% of 25.3125 pounds is:

Lead required=0.92×25.3125=23.2875 pounds\text{Lead required} = 0.92 \times 25.3125 = 23.2875 \text{ pounds}Lead required=0.92×25.3125=23.2875 pounds

• Antimony: 6%6\%6% of 25.3125 pounds is:

Antimony required=0.06×25.3125=1.51875 pounds\text{Antimony required} = 0.06 \times 25.3125 = 1.51875 \text{ pounds}Antimony required=0.06×25.3125=1.51875 pounds

Step 5: Adjust for the lead already in the bar​

The original bar contains 70% lead, so:

Lead in the bar=1.6875×0.70=1.18125 pounds of lead\text{Lead in the bar} = 1.6875 \times 0.70 = 1.18125 \text{ pounds of lead}Lead in the bar=1.6875×0.70=1.18125 pounds of lead
The additional lead required is:

Additional lead needed=23.2875−1.18125=22.10625 pounds\text{Additional lead needed} = 23.2875 - 1.18125 = 22.10625 \text{ pounds}Additional lead needed=23.2875−1.18125=22.10625 pounds

Final Summary:​

To create the alloy with the composition of 2% tin, 6% antimony, and 92% lead, using one of your bars:

• You need 1.51875 pounds of antimony.
• You need 22.10625 pounds of additional lead.

Would you like the weight of additional lead and antimony in ounces as well?

Let's convert the amounts of additional lead and antimony into pounds and ounces.

Step 1: Lead​

You need 22.10625 pounds of additional lead.

To convert 0.10625 pounds to ounces:

0.10625×16=1.7 ounces0.10625 \times 16 = 1.7 \text{ ounces}0.10625×16=1.7 ounces
So, the total additional lead required is:

22 pounds and 1.7 ounces22 \text{ pounds and } 1.7 \text{ ounces}22 pounds and 1.7 ounces

Step 2: Antimony​

You need 1.51875 pounds of antimony.

To convert 0.51875 pounds to ounces:

0.51875×16=8.3 ounces0.51875 \times 16 = 8.3 \text{ ounces}0.51875×16=8.3 ounces
So, the total antimony required is:

1 pound and 8.3 ounces1 \text{ pound and } 8.3 \text{ ounces}1 pound and 8.3 ounces

Final Summary:​

• You need 22 pounds, 1.7 ounces of additional lead.
• You need 1 pound, 8.3 ounces of antimony.

If you converted the values to kilograms or grams (base 10) from the start, then you could work it easily and in a fraction of the time.

It's done in lbs and ozs and then choose percentages in base 10?

Whatever I'm doing, if it's gonna work out easier in imperial or metric, then I convert my figures to that system first. Most of the time, base 10 is easier and faster.

 

[Sherlock Holmes]

Ok, I suck at math, so here's a question:

I have some metal bars that are composed of 30% tin and 70% lead. The bars weigh 1 pound, 11 ounces.

I want to create an alloy consisting of 2% tin, 6% antimony, and 92% lead.

Using one of those bars, how many pounds and ounces of lead and antimony do I need to get a 2% tin, 6% antimony, and 92% lead alloy?

Probably a simple question, but I haven't used any math in 35 years.

Describe those bars as: 30% tin (Sn), 70% lead (Pb) and 0% antimony (Sb), call the bar's weight W (which is 27 ounces)

We must change the ratio of Sn/Pb from 0.3/0.7 to 0.02/0.92, i.e. we can add lead alone, to get this new ratio.

To increase the Pb concentration we must add lead, the bar already contains 0.7 * 27 = 18.9 ounces Pb and 8.1 ounces Sn.

The ratio of Pb/Sn in a bar is 2.3.

We want the Pb/Sn ratio to now be 0.92/0.02 which is 46.0, that's 20 times what it is already in a bar, therefore we need to add 19 times the lead that's already in a bar, that is: 19 * 18.9 = 359.1 ounces, if we did that the ratio of Pb/Sn will be

20*18.9 / 8.1 = 46 - which is what we want, so there's the first part, you must add 359 (22 lbs) ounces of Pb to a single bar, to get the new Pb/Sn ratio.