I am a newbie at this but I managed to get my hands on some computers and old cell phones so now i am presently doing some research to determine how much gold is in them. I'm hoping to get some advice for anyone to check out this scientific method used
In order to calculate the amount of gold I was advised if i use the formula from http://www.finishing.com/269/90.shtmlm I should come up with a good estimate...
A typical PCI circuit board used today in a personal computer has gold edge connectors
that are plated fifty millionths of an inch (0.000050")thick with gold.
IN ORDER TO PUT MY NEWLY FOUND CALCULATION TO THE TEST I USE THE A CELL PHONE WHICH
CAN BE SEEN AT http://farm9.staticflickr.com/8242/8660550413_72e3f1ebca_z.jpg
As you can see from the photo, this phone's circuit board comes in two sections ...The first section contains
six blocks of gold estimated to be 2.4sq in. And the keyboard gold connectors is estimated at 2.4sq in as well.
So, multiply the following conversions to get:
SECTION A = 2.4 sq in * .00005 in X 16.39 cu cm per cu in X 19.3 g per cu cm X .03222 troy oz per gram
X $1,659.00 per tr oz = $1.71 for this particular section of gold on the phone.
SECTION B = The area of gold in the keyboard section is also estimated at a ballpark
figure of 2.4sq in (Give or Take)
Therefore SECTION A + SECTION B equals $3.42 worth of platted gold (not including the pins and other gold components.
The conversions used from the calculations are:
The density of gold is 19.3 grams per cubic centimeter
There are 16.39 cubic centimeters in a cubic inch.
$1,659.00 per troy ounce is a crude estimate of the current price of gold.
.03222 troy ounces per gram is the inverse (upside down) 1 tr oz per 31.035 grams.
Can anyone out there tell me if this is a good ballpark figure. I am sure there are many variables that can throw this off like the thickness of the gold plate.
In order to calculate the amount of gold I was advised if i use the formula from http://www.finishing.com/269/90.shtmlm I should come up with a good estimate...
A typical PCI circuit board used today in a personal computer has gold edge connectors
that are plated fifty millionths of an inch (0.000050")thick with gold.
IN ORDER TO PUT MY NEWLY FOUND CALCULATION TO THE TEST I USE THE A CELL PHONE WHICH
CAN BE SEEN AT http://farm9.staticflickr.com/8242/8660550413_72e3f1ebca_z.jpg
As you can see from the photo, this phone's circuit board comes in two sections ...The first section contains
six blocks of gold estimated to be 2.4sq in. And the keyboard gold connectors is estimated at 2.4sq in as well.
So, multiply the following conversions to get:
SECTION A = 2.4 sq in * .00005 in X 16.39 cu cm per cu in X 19.3 g per cu cm X .03222 troy oz per gram
X $1,659.00 per tr oz = $1.71 for this particular section of gold on the phone.
SECTION B = The area of gold in the keyboard section is also estimated at a ballpark
figure of 2.4sq in (Give or Take)
Therefore SECTION A + SECTION B equals $3.42 worth of platted gold (not including the pins and other gold components.
The conversions used from the calculations are:
The density of gold is 19.3 grams per cubic centimeter
There are 16.39 cubic centimeters in a cubic inch.
$1,659.00 per troy ounce is a crude estimate of the current price of gold.
.03222 troy ounces per gram is the inverse (upside down) 1 tr oz per 31.035 grams.
Can anyone out there tell me if this is a good ballpark figure. I am sure there are many variables that can throw this off like the thickness of the gold plate.