To Break or Not Breakdown A Transformer (Using Some Math)

I have about 400 pounds of transformers in my basement that I have not broken apart (ranging from tiny to 40 pounds). I finally decided to use my training (I teach math) to figure out how much copper must be in a transformer to make money.

Skip this part if you're not interested in the math
--------------------------------------------------------------------------------------------


Lets start with some abbreviations
Let x = the weight of copper
Let y = the weight of steel
Let z = your payout
Let the coefficients be your payout per pound

Thus, using prices from my local scrap yard my equations would be

.28(x + y) = z (amount paid not breaking down the transformer)
2.9x + .09y = z (amount paid by breaking down the transformer)


So the question becomes, how much copper needs to be in a transformer in order to make money.

More formally 2.9x + .09y > .28(x + y)
Or, rounding a bit, y < 13.79x

Let’s start with a 10 pound transformer and find the break even point. Ignoring any loss of weight in the breakdown process we have



X + Y = 10 lbs
Or y = -x + 10


Setting y = -x + 10 equal to 13.79x

-x + 10 = 13.79x
X = .676 pounds

Meaning, out of a 10 pound transformer, if you get .676pounds of copper (leaving 9.324 pounds of steel) you will break even. Anything less than .676 pounds, you will loose money by breaking it down.

--------------------------------------------------------------------------------------------
Thus, for a 10 pounds transformer (at my yards prices) you need 6.76% copper recovery.

Here are the numbers for various weights you need to break even.

1 Pound Transformer = .67% or .0067 pounds of copper
5 Pound Transformer = 3.38% or .338 pounds of copper
10 Pound transformer = 6.76% or .676 pounds of copper
20 Pound Transformer = 13.5% or 1.35 pounds of copper

cool! (i'm just glad i'm not still in school ; )

Originally Posted by parrothead

Well, I think a lot of people have stated that 10% is the norm, so this backs that up. Now the question is, is the size of the unit enough to warrant the time spent. If we can deduce that 10% is the norm, we need an equasion that gives the dollar per hour that one wishes to maintain based on the initial weight of the transformer.

That is your assignment for tomorrow. I am thinking that the time involved versus money earned will somewhat follow the same scale as the size of the transformer. This could be interesting.

Say a 5 ounce transformer takes 1/2 the time as a 5 pound transformer which is 10 times the reward rather than just double. I think we need some more data on this.

I love this stuff. Math, physics, and economics. COOL! Keep going teach. We need more info.

Yep, there is still more work to do here. We also need to adjust any break down for the loss of weight during breakdown. Pullling out glue, paper, loss of metal dust, etc. and then, as you suggested, account for the implicit costs of time.

After that, we need some expiermentation to calculate the average weight of copper in a 1, 5, 10, 20 pound transformer. Maybe we can all chip in to do this. I have 5 different 10 pound transformers that i can find the average. Anyone have 5 different 1 pound, 5pound & 20 pound tranfromers they would be willing to break down and weigh?

I can also do this for finger cards vs cutting fingers, ram vs cutting fingers (this should solidify what everyone has stated that you actually loose money), electric motors, etc. I'm currently in planning stages for writing my next book, this one on e-waste recycling. I want to tackle the majority of every issue that is discuessed on the forums.

I like loose money but I hate to lose money.

it wasnt as scientific, but we done the same thing with acr last week, turned out that per unit unclean went for $21 and clean went for $22 per unit, these were all from hotels and all the same size, so for my time its not worth it for the extra $1

Originally Posted by Jeremiah

I have about 400 pounds of transformers in my basement that I have not broken apart (ranging from tiny to 40 pounds). I finally decided to use my training (I teach math) to figure out how much copper must be in a transformer to make money.

Skip this part if you're not interested in the math
--------------------------------------------------------------------------------------------


Lets start with some abbreviations
Let x = the weight of copper
Let y = the weight of steel
Let z = your payout
Let the coefficients be your payout per pound

Thus, using prices from my local scrap yard my equations would be

.28(x + y) = z (amount paid not breaking down the transformer)
2.9x + .09y = z (amount paid by breaking down the transformer)


So the question becomes, how much copper needs to be in a transformer in order to make money.

More formally 2.9x + .09y > .28(x + y)
Or, rounding a bit, y < 13.79x

Let’s start with a 10 pound transformer and find the break even point. Ignoring any loss of weight in the breakdown process we have

X + Y = 10 lbs
Or y = -x + 10


Setting y = -x + 10 equal to 13.79x

-x + 10 = 13.79x
X = .676 pounds

Meaning, out of a 10 pound transformer, if you get .676pounds of copper (leaving 9.324 pounds of steel) you will break even. Anything less than .676 pounds, you will loose money by breaking it down.

--------------------------------------------------------------------------------------------
Thus, for a 10 pounds transformer (at my yards prices) you need 6.76% copper recovery.

Here are the numbers for various weights you need to break even.

1 Pound Transformer = .67% or .0067 pounds of copper
5 Pound Transformer = 3.38% or .338 pounds of copper
10 Pound transformer = 6.76% or .676 pounds of copper
20 Pound Transformer = 13.5% or 1.35 pounds of copper

Cool!
You will also have to form a separate equation for transformer that have a aluminum wire or aluminum and copper wire

.40z + .09y > .28(z + y) "alu wire"
2.9x + .40z + .09y > .28(x + y + z) "alu and copper wire"

When will your book be ready I would love to check it out.

Ive been meaning to post a thread showing what has more return: selling HD boards separate from HD or selling the whole HD with the board. This has been covered in some other threads saying the board has about 7-8% of the weight so....

Hard Drive Boards $10.25/lb
Hard Drives with Board: $1.10/lb
Hard Drives without Board $.42/lb

Let's say that we have 100lbs of HD.

without taking board off: 100 x 1.10 = $110
taking board off: (8 x 10.25) + (92 x .42)= 82 + 38.64= $120.64

So... $10.64 more to separate the board on 100lbs


Jeremiah, keep up the good work!

Your %'s are wrong. They should all be the same.

Also if you want to do any DOE (design of experiments) based on price, you should express weight as a dimensionless ratio (removes a variable). Prices can also probably be expressed dimensionlessly.

This is likely going to be a small sample size perhaps using Bayesian statistics (a prior) is appropriate to increase the sample size.

Actually spend time breaking down what ever you wish to analyze. The proper tools and skill coupled with experience really determine the break even point. You should also include an "entertainment multi tasking coefficient" to include time spent watching T.V. while doing mindless "cleaning" of recyclable materials. Many years of data run through a modal analysis would be the only way to provide a meaningful set of equations by which to evaluate the "tipping point" of a recovery operation.

Cory, are you talking about the PTAC's out of the rooms? I would have to question your math on them if so.

I get a great deal of satisfaction breaking things that someone else made, and trying to figure out how it all worked to achieve the final product. Reverse engineering with a hatchet or hammer.

lol, i keep it fairly simple, as I SUCK at math. if the transformer is the size that comes out of a microwave or bigger, I break it down ... if it isnt, i scrap it whole.

Originally Posted by Amnelson

Your %'s are wrong. They should all be the same.

Not sure why you would think the percentages would be the same for a break even point? The copper/steel ratio may be the same, as this is really a function of transformer type and various transformers may yield various copper/steel ratios. None of this has to do with your break even point. The math is correct.

You would have to elaborate the rest of your post. My statistics is very solid as my graduate studies were in probability and econometrics but I'm not sure why you would want to incorporate a dimensionaliss variable like pi into a model about weight

Originally Posted by Area67

Actually spend time breaking down what ever you wish to analyze. The proper tools and skill coupled with experience really determine the break even point. You should also include an "entertainment multi tasking coefficient" to include time spent watching T.V. while doing mindless "cleaning" of recyclable materials. Many years of data run through a modal analysis would be the only way to provide a meaningful set of equations by which to evaluate the "tipping point" of a recovery operation.

Your right to assume that breakdown time is a function of personal skill but this has nothing to do with your breakeven point as the math above does not incorporate break-down time. The point of the math shown is that it is a fact that if you break-down a 10 pound transformer with less than .676pounds of copper within it you will loose money as compared with selling the unit whole (using the prices stated)

Originally Posted by Jeremiah

Not sure why you would think the percentages would be the same for a break even point? The copper/steel ratio may be the same, as this is really a function of transformer type and various transformers may yield various copper/steel ratios. None of this has to do with your break even point. The math is correct.

You would have to elaborate the rest of your post. My statistics is very solid as my graduate studies were in probability and econometrics but I'm not sure why you would want to incorporate a dimensionaliss variable like pi into a model about weight

Not that it is that important but, I would have to agree with Amnelson. I believe that math on the bottom of your comment is incorrect where it says:

"1 Pound Transformer = .67% or .0067 pounds of copper
5 Pound Transformer = 3.38% or .338 pounds of copper
10 Pound transformer = 6.76% or .676 pounds of copper
20 Pound Transformer = 13.5% or 1.35 pounds of copper"

.0067 is incorrect for the pounds of copper in the 1 pound transformer. I think you meant to put .067 lbs. All of the percentages are incorrect except for the 10lb transformer. So, everything looked right that you wrote except the last portion at the end and correct me if I'm wrong. Oh and your last comment you spelled "your" when I think you meant "you're".

Good thing this is a web forum and things dont have to be exact. I make similar mistakes all the time.

My undergrad was Mech. Engineering so x,y,z confuses me so I changed the variable names so I could do my math easier.

Grad school was AI/MI/Robotics/C++ but mostly was Image Processing (using MI principles ... Bayesian Statistics, Neural Nets etc etc etc). However
it was mostly practical (for research) implement algorithm -> make it fast -> test. My largest success was simplifying a pair of polynomial time algorithms
into linear time algorithms.

Pc = copper price
Ps = steel price
Pt = Transformer price

Wc = copper weight
Ws = steel weight
Wt = transformer weight

Tvb = value transformer broken
Tvu = value transformer unbroken

Tvb = Wc * Pc + Ws * Ps
Tvu = Wt * Pt

Pt = Tvu / Wt

Wt = Wc + Ws
Ws = Wt - Wc

Rc = copper steel ratio
Rc = Wc/Wt

Tvb = Wc * Pc + Ws * Ps
substitution for Ws
Tvb = Wc * Pc + (Wt - Wc) * Ps
Divide equation by Wt
Tvb / Wt = Wc/Wt * Pc + (1 - Wc/Wt) * Ps
substition
Pt = Rc * Pc + (1 - Rc) * Ps
Solve for Rc
Rc = (Pt - Ps) / (Pc - Ps)

Using your prices
Rc = (0.28 - 0.09) / (2.9 - 0.09)
Rc = 0.0676
Rc = 6.76%

Break even point for all weights is 6.76% copper.

When I am saying dimensionless, I mean price and/or weight ratios. The reason I reccommended it was two fold. One, it makes the analysis
of a transformer breakdown independent of its weight or of the current market (useful for future markets). Two, when graphing the results,
its easier to graph with fewer variables, hence my desire to use dimensionless numbers.