I started reading hoke

[boutselis]

I have no real college education but I have watched 30 episodes of the big bang theory. I have started reading Hoke and I think science may not be as fun as it looks on tv. Could Tv have lied to me?

Any way. to the heart of my question. Is reading all of the 362 pages of hoke necessary? I have been poking around the forum and came across a post where it was basically answered "you don't know what you are doing. You will end up coming here and asking 'what do I do next?' and the you will ask 'What do I do next?' and then you will come back again and ask 'And what do i do next?' I suggest you read Hoke."

I want to read the entire book but I fell asleep around the part where it started speaking of dentures. I know this process isn't going to be as easy as teaching myself welding or wood working or figuring out how to hook up an old phase convertor. When I went looking for that book I was actually hoping for a long pamphlet. 360plus pages of tech info is a little daunting. If its a must then its a must.

 

[jimdoc]

I have no real college education but I have watched 30 episodes of the big bang theory. I have started reading Hoke and I think science may not be as fun as it looks on tv. Could Tv have lied to me?

Any way. to the heart of my question. Is reading all of the 362 pages of hoke necessary? I have been poking around the forum and came across a post where it was basically answered "you don't know what you are doing. You will end up coming here and asking 'what do I do next?' and the you will ask 'What do I do next?' and then you will come back again and ask 'And what do i do next?' I suggest you read Hoke."

I want to read the entire book but I fell asleep around the part where it started speaking of dentures. I know this process isn't going to be as easy as teaching myself welding or wood working or figuring out how to hook up an old phase convertor. When I went looking for that book I was actually hoping for a long pamphlet. 360plus pages of tech info is a little daunting. If its a must then its a must.

It is a must, as well as studying the forum. If you can't handle that, then this isn't the hobby for you.
You can accumulate items to process as you study, so when you are ready to process some materials, hopefully you will not lose any values and know how to do it safely. Starting in without enough materials to make a decent size BB of gold isn't the best way to go. I didn't really process anything for quite a while. Lazer Steve's videos gave me the confidence, and that is after reading all of Hoke's book as well as many others. Now there are other members with Youtube videos as well that show you the right way to do things. That is always a help if you have trouble reading or following the whole book.
I could just imagine what a C.M. Hoke refining video would be like.

Jim

 

[boutselis]

I have no real college education but I have watched 30 episodes of the big bang theory. I have started reading Hoke and I think science may not be as fun as it looks on tv. Could Tv have lied to me?

Any way. to the heart of my question. Is reading all of the 362 pages of hoke necessary? I have been poking around the forum and came across a post where it was basically answered "you don't know what you are doing. You will end up coming here and asking 'what do I do next?' and the you will ask 'What do I do next?' and then you will come back again and ask 'And what do i do next?' I suggest you read Hoke."

I want to read the entire book but I fell asleep around the part where it started speaking of dentures. I know this process isn't going to be as easy as teaching myself welding or wood working or figuring out how to hook up an old phase convertor. When I went looking for that book I was actually hoping for a long pamphlet. 360plus pages of tech info is a little daunting. If its a must then its a must.

It is a must, as well as studying the forum. If you can't handle that, then this isn't the hobby for you.
You can accumulate items to process as you study, so when you are ready to process some materials, hopefully you will not lose any values and know how to do it safely. Starting in without enough materials to make a decent size BB of gold isn't the best way to go. I didn't really process anything for quite a while. Lazer Steve's videos gave me the confidence, and that is after reading all of Hoke's book as well as many others. Now there are other members with Youtube videos as well that show you the right way to do things. That is always a help if you have trouble reading or following the whole book.
I could just imagine what a C.M. Hoke refining video would be like.

Jim

I started reading again. Having gotten further along It is getting a little less boring and starting to get to the meat of the matter. Only 340 some odd pages to go. Should I get some or all of the items he is speaking of. like a bunsin burner , frying pan, test tubes and such right away? Can you still buy Wire gauze with asbestos center?

 

[jimdoc]

I started reading again. Having gotten further along It is getting a little less boring and starting to get to the meat of the matter. Only 340 some odd pages to go. Should I get some or all of the items he is speaking of. like a bunsin burner , frying pan, test tubes and such right away? Can you still buy Wire gauze with asbestos center?

He is a she, Calm Morrison Hoke. In answer to what you should get, what material do you have to process now? Are you really ready to do any of this right away? What exactly will you be processing?

You can accumulate what you need as you learn, as you build up your stash to refine. Then you will know what you are doing, before you do it. As in avoiding mistakes, and having to ask "what do I do now?" and "where did my gold go?".

Studying the forum will give you better advice on what you need, as times have changed since 1940 and Ms. Hoke didn't process computer scrap. And some of the newer techniques are far better for computer scrap, since nitric is harder to find nowadays.

There is no read this today and you will be ready to refine tomorrow. You need to put some time in.
You found the best place to put that time in, now it is study time. And build up material to process, it runs out fast and then you will have more time to study. So you don't really lose anything by taking your time to learn right while you are accumulating your stash. You actually win by not losing your values avoiding mistakes. And mistakes can be deadly with this hobby.
Do you understand all the safety concerns yet?

Jim

 

[boutselis]

Good thoughts. I have about 50 lbs of mirror like 4" hard drive discs. I have about 40 lbs of hard drive logic boards and 2 lbs of pins that still have plastic on them but are not connected to the boards. I also have a few thousand lbs of hard drives, Must be at least 1000 drives, but I do not know how many still have the logic boards on them but I do know all of them have at least one platter. some have up to 6 platers. Some of the reader arms from the drives have a silver looking wire for the coil. most of the arms have copper, a few have some colored wire (like green or red) and i have been separating the silver looking ones. off course my hope is that it is a silver coated wire but I probably only have 2 lbs of the silver wire so thats no big deal.

I was told by some one who had bought a lot of the boards that they were hi yield (He may have said high grade) and I was getting upwards of $12 per pound from them. There are about 14 boards to the pound. So I am thinking there must be some gold in there some place.

And there should be more on the way.

 

[jimdoc]

Good thoughts. I have about 50 lbs of mirror like 4" hard drive discs. I have about 40 lbs of hard drive logic boards and 2 lbs of pins that still have plastic on them but are not connected to the boards. I also have a few thousand lbs of hard drives, Must be at least 1000 drives, but I do not know how many still have the logic boards on them but I do know all of them have at least one platter. some have up to 6 platers. Some of the reader arms from the drives have a silver looking wire for the coil. most of the arms have copper, a few have some colored wire (like green or red) and i have been separating the silver looking ones. off course my hope is that it is a silver coated wire but I probably only have 2 lbs of the silver wire so thats no big deal.

I was told by some one who had bought a lot of the boards that they were hi yield (He may have said high grade) and I was getting upwards of $12 per pound from them. There are about 14 boards to the pound. So I am thinking there must be some gold in there some place.

And there should be more on the way.

I would forget about the hard drive platters for now, I sell them with my scrap aluminum.
There isn't platinum in all of them, and only a small percentage of the magnetic layer in the ones that do. How are you going to tell which ones contain any platinum? Many on the forum have come to this conclusion; they aren't worth the time and chemicals to process. Of course this is your decision, but I think you spoke of not wanting to waste money in another post.

I sell my hard drive boards and buy silver or palladium with the money. That saves time and chemicals and I still get my precious metals. If you get into trying whole boards you will probably run into problems.

Jim

 

[NobleMetalWorks]

Your hard drives have a lot more value in Al than in Platinum, there is maybe, if you are really lucky, $100 worth of platinum in your 1000 hard drive platters. That might sound like a gross under-estimate, but the sad part is, it's probably an over estimate. Platinum only makes up a small part of the metal used to plate the Al the platter is made of. It's the hard drive housing, made out of Al, that is worth money.

On your logic boards, dependent upon the board, there might be Au, Ag and some other PGMs not to mention the copper and tin. The pin connecters should be gold plated. There is information if you use the search function, on all this in great detail.

Scott

 

[boutselis]

What did all those people buy the platters for then? Any ideas what they are good for? I figured the percentage of platinum and palladium was probably pretty small but I have sold the platters at up to $3 per pound. I suppose some one may have needed 100's of doughnut shaped mirrors but when some one was willing to spend 20 to $30 plus shipping on them in any condition I figured they had to be getting money out of them. One guy purchased about 100 lbs over time in 10lb lots.

Finding out there isn't anything of value in them is a bummer.

 

[jimdoc]

Finding out there isn't anything of value in them is a bummer.

Imagine how the buyer feels.

Jim

 

[NobleMetalWorks]

What did all those people buy the platters for then? Any ideas what they are good for? I figured the percentage of platinum and palladium was probably pretty small but I have sold the platters at up to $3 per pound. I suppose some one may have needed 100's of doughnut shaped mirrors but when some one was willing to spend 20 to $30 plus shipping on them in any condition I figured they had to be getting money out of them. One guy purchased about 100 lbs over time in 10lb lots.

Finding out there isn't anything of value in them is a bummer.

eBay and reality tend to differ, and not by any small amount. I can sell Pentium Pro's on eBay for around $40 each, I have seen them go for as much as $55 each a few months ago when gold was at it's peak earlier this year.

Gold has an amazing ability to lead people into believing, against information that states otherwise very clearly, that there is more gold then there actually is. Every time I look at a Pentium Pro I think to myself, damn that looks like a lot of gold. The reality is that it's not that much gold at all. The same thing happens with things like your hard drive platters, they just look like shiny doughnut mirrors with holes in them until someone tells you they have Pt and Pd in them. Then all the sudden, they look like they are all platinum. Have you ever held platinum in your hand? Even a small piece? Think to yourself this, if only a small piece of platinum weighs so much, and it's only the smallest of quantities, how should a hard drive platter feel then?

And it's much worse on the newer, less expensive hard drives.

HOWEVER, they are worth the Al. You can choose to sell them on eBay and some poor smuck that doesn't know any better, will come along and eventually purchase them. Or you can sell them to a scrap yard for the Al.

Scott

 

[boutselis]

Finding out there isn't anything of value in them is a bummer.

Imagine how the buyer feels.

Jim

Thats a good point. has me laughing too.

 

[boutselis]

What did all those people buy the platters for then? Any ideas what they are good for? I figured the percentage of platinum and palladium was probably pretty small but I have sold the platters at up to $3 per pound. I suppose some one may have needed 100's of doughnut shaped mirrors but when some one was willing to spend 20 to $30 plus shipping on them in any condition I figured they had to be getting money out of them. One guy purchased about 100 lbs over time in 10lb lots.

Finding out there isn't anything of value in them is a bummer.

eBay and reality tend to differ, and not by any small amount. I can sell Pentium Pro's on eBay for around $40 each, I have seen them go for as much as $55 each a few months ago when gold was at it's peak earlier this year.


HOWEVER, they are worth the Al. You can choose to sell them on eBay and some poor smuck that doesn't know any better, will come along and eventually purchase them. Or you can sell them to a scrap yard for the Al.

Scott

The scrap yard is out. I can easily clear $1.50 a pound on ebay. And like I said i usually got more till I started hoarding them. I was just loving all that platinum. I would take out a bunch of platters, shuffle them around the table like cards and just think of all those little ingots of platinum I was going to learn how to make. When I finally made a 1/2 oz I was going to carry it around and show people what I had done. Foolish idea. I know. But it just seemed so cool.

I will say one thing. When I compare the weight of the platers by volume to the cast aluminum HD cases the platters seem to be closer to the weight of stainless steel then aluminum. I may just be adding them together wrong but the platters deffinitly seem heavier than the cast aluminum to me.

Now that some of my dreams are shattered I am off to read more of hoke and hope it isn't more hokum than chemistry.

 

[goldsilverpro]

The infamous quote from Johnson Matthey says that Pt is used on all newer hard drive disks and that it makes up about 35% by weight of the magnetic layer. The other 65% is cobalt. According to Wikipedia, the magnetic layer is 10-20nm thick.

One nm is equal to 1 billionth of a meter or .00000004". Therefore, 10nm = .0000004" or .4 microinches. Taking into consideration that the density of a 35/65, Pt/Co alloy is about 11.14 and only 35% of this is Pt, the Pt value/in² is

.0000004 x 2.05 x 1477 = $.0012/in².

If the disk is 5.25" in diameter, it has a surface area of 21.6 in², not considering the hole in the middle. Therefore, the Pt value on a disk this size is .0012 x 21.6 = $.026. 1000 disks would have $26 worth of Pt on them. If the thickness is on the high side, 20nm, the Pt value of 1000 disks would be $52 or, about a nickel each.

As always, my disclaimer. If you find an error in my math or in my assumptions, please let me know

 

[dtectr]

if folks could BEGIN here BEFORE they go to you(BS)tube and FeeBay, what a pile of bytes we could save!
But then, we wouldn't retain the quality of regulars with whom we indulge RELEVANT, PERTINENT discussions.
'Nuff said.

 

[goldsilverpro]

I'm sure that the eyes of many members tend to glaze over when they read these math concepts that I post. I'll try to explain how I got these numbers as simply as I can. For those people with no math abilities, there will still be utter confusion. Maybe some math guy can devise a calculator for this although it would be complicated.

First, how I got the density of 11.14 for a 35/65, Pt/Co alloy. The following formula can be used to approximate the density of any alloy. 21.4g/cc is the density of Pt and 8.86g/cc is the density of cobalt - these can be found in any chemical handbook. The .35 and .65 are the decimal equivalents of the metal percentages (35% and 65%) in this particular case. If there are more than 2 metals in the alloy, you can simply add more terms in the denominator. For each metal, you first divide the decimal equivalent of each metal by the density of that metal. You then add all these results together and divide 1 by this total.

1/[(.35/21.4) + (.65/8.86)] = 11.14g/cc

Second, the formula: .0000004 x 2.05 x 1477 = $.0012/in²

The general layout of this 3 term formula can be used to determine the $ value/in² of any plated layer of any metal or alloy. An explanation of the 3 terms in the formula:

-----The 3rd term, 1477, is the spot price, in $/tr.oz., of the metal in question - in this case, Pt.

-----The 1st term, .0000004, is simply the thickness of the layer expressed in inches. This same exact number is equal to the number of cubic inches of the material per square inch of surface area. The reason they're the same is that the formula for volume is Length x Width x Height. For a square inch, both the length and width are equal to 1. The height is the thickness. Therefore, the volume of 1 square inch of .0000004" material = .0000004" x 1" x 1" = .0000004 in³

-----The 2nd term, 2.05, is the weight of Pt, in tr.oz., in a cubic inch of 35/65, Pt/Co alloy. This factor will vary depending on the metal(s) involved. I keep it simple by relating whatever I'm looking for to how many oz of pure gold is in a cubic inch, which is 10.17. I have that number memorized. I also know that the density of gold is 19.3. By knowing these numbers, I can easily calculate the factor for any metal by dividing the density of that metal by 19.3 and multiplying by 10.17.

For the 35/65 alloy, the density is 11.14, as determined above, and only 35% of that is Pt. Therefore, the 2nd term factor = (11.14/19.3) x 10.17 x .35 = 2.05 = number of tr.oz. of Pt in 1 cubic inch of a 35/65, Pt/Co alloy.

For something simpler like pure silver, whose density is 10.49, the factor = (10.49/19.32) x 10.17 = 5.53

For pure gold, the factor is, of course, 10.17

In all cases, the formula is $/in² = (thickness in inches) X (Factor) X (spot price)

I didn't simplify this very well, but it's about the best I can do. All I can hope for is that some of you that are good in math can follow along and pick up on the logic I've provided. I feel that it's a very important formula and one that I use constantly. I know it's complicated and if there are any specific questions, don't hesitate to ask them

 

[goldsilverpro]

In my previous 2 posts, I assumed the hard drive disks were 5.25" in diameter. This was evidently true in the old days but not now. After a search, I found that the more modern disks are about 3.75" (called 3.5" disks) and 2.5" (some laptops). Therefore, the Pt value for 1000 disks of 3.75" dia. would be about $25 (2.5 cents each) and about $10 for 1000 2.5" disks (a penny each). In both cases, I assumed a 1" hole and deducted that from the surface area of the disks. These figures are for those with the maximum magnetic layer thickness of 20nm.

Since I have never processed these (and certainly never will), all the data I used to calculate the values was obtained from the internet.

 

[Palladium]

Wow !!!! :shock:

 

[its-all-a-lie]

I'm sure that the eyes of many members tend to glaze over when they read these math concepts that I post. I'll try to explain how I got these numbers as simply as I can. For those people with no math abilities, there will still be utter confusion. Maybe some math guy can devise a calculator for this although it would be complicated.

First, how I got the density of 11.14 for a 35/65, Pt/Co alloy. The following formula can be used to approximate the density of any alloy. 21.4g/cc is the density of Pt and 8.86g/cc is the density of cobalt - these can be found in any chemical handbook. The .35 and .65 are the decimal equivalents of the metal percentages (35% and 65%) in this particular case. If there are more than 2 metals in the alloy, you can simply add more terms in the denominator. For each metal, you first divide the decimal equivalent of each metal by the density of that metal. You then add all these results together and divide 1 by this total.

1/[(.35/21.4) + (.65/8.86)] = 11.14g/cc

Second, the formula: .0000004 x 2.05 x 1477 = $.0012/in²

The general layout of this 3 term formula can be used to determine the $ value/in² of any plated layer of any metal or alloy. An explanation of the 3 terms in the formula:

-----The 3rd term, 1477, is the spot price, in $/tr.oz., of the metal in question - in this case, Pt.

-----The 1st term, .0000004, is simply the thickness of the layer expressed in inches. This same exact number is equal to the number of cubic inches of the material per square inch of surface area. The reason they're the same is that the formula for volume is Length x Width x Height. For a square inch, both the length and width are equal to 1. The height is the thickness. Therefore, the volume of 1 square inch of .0000004" material = .0000004" x 1" x 1" = .0000004 in³

-----The 2nd term, 2.05, is the weight of Pt, in tr.oz., in a cubic inch of 35/65, Pt/Co alloy. This factor will vary depending on the metal(s) involved. I keep it simple by relating whatever I'm looking for to how many oz of pure gold is in a cubic inch, which is 10.17. I have that number memorized. I also know that the density of gold is 19.3. By knowing these numbers, I can easily calculate the factor for any metal by dividing the density of that metal by 19.3 and multiplying by 10.17.

For the 35/65 alloy, the density is 11.14, as determined above, and only 35% of that is Pt. Therefore, the 2nd term factor = (11.14/19.3) x 10.17 x .35 = 2.05 = number of tr.oz. of Pt in 1 cubic inch of a 35/65, Pt/Co alloy.

For something simpler like pure silver, whose density is 10.49, the factor = (10.49/19.32) x 10.17 = 5.53

For pure gold, the factor is, of course, 10.17

In all cases, the formula is $/in² = (thickness in inches) X (Factor) X (spot price)

I didn't simplify this very well, but it's about the best I can do. All I can hope for is that some of you that are good in math can follow along and pick up on the logic I've provided. I feel that it's a very important formula and one that I use constantly. I know it's complicated and if there are any specific questions, don't hesitate to ask them

I need an Advil.

 

[goldsilverpro]

I need an Advil.

When I re-read it, I needed one too.

 

[zzz]

A lot of bullxxxx in this thread.