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I've been looking at how to estimate the amount of gold in a specimen. Chris Ralph has an article in the ICMJ about how to calculate this amount and Doc has a Youtube video showing how he calculates the amount of gold. Both of the methods go through the same steps of measuring the dry weight and the wet weight of the specimen but then after this the calculations differ and you get different results. Both equations use constant values that do not seem to have any description into where they come from. So I decided to sit down and use my thinking cap to derive the equations for determining this. This of course gave an even different result. IF anyone can tell me where the numbers from Chris Ralph or Docs equations come from I would be happy to know. Otherwise below is my algebraic proof for solving the weight of the gold.

 

Chris Ralph article: https://www.icmj.com/magazine/article/determining-the-amount-of-gold-in-a-quartz-specimen-1189/

Doc Youtube: 

 

 

 

 

Let me start here with this equation

Mass(total) = Density(Quartz) * Volume(Quartz) + Density(Gold) * Volume(Gold)

To make it shorter I am going to abbreviate M = mass, D = Density, V = Volume, t = total, q = quartz, g = gold

Mt = Dq * Vq + Dg * Vg

 

If we can assume there are no voids in the specimen, the total volume of the specimen can be written as follows

Vt = Vq + Vg

 

We solve for the volume of the quartz as follows

Vq = Vt - Vg

 

Lets plug this into the first equation to get the following equation

Mt = Dq * (Vt - Vg) + Dg * Vg

 

We can solve the above equation for the volume of gold as

Vg = (Mt - Dq * Vt) / (Dg - Dq)

 

The mass of the gold can be solved by multiplying by the density

Mg = Vg * Dg

 

If we use 2.65g/cm^3 for quartz and 19.32g/cm^3 for gold, we can plug this into the equation from above to get the mass of gold in the specimen

Mg = 19.32 * ( Mt - 2.65 * Vt) / ( 19.32 - 2.65 )

 

The mass Mass total (Mt) is the dry weight (Wt) of the specimen divided by acceleration due to gravity 9.81m/s^2.

The volume can be measured by weighing a cup of water with the specimen suspended in the water. This is possible because water has a density extremely close to 1g/cm^3. The buoyancy force (Bf) is equal to the fluid density (1g/cm^3) * acceleration due to gravity (g) * the volume displaced (V).  Therefore if we set up a scale with a cup of water on it, tare the scale, and then suspend the specimen in the water, we get a measurement of the buoyancy force. (Remember the scale measures force not mass) We can then divide by 1g/cm^3 and from this we are given the volume in cm^3 / 9.81m/s^2.

Vt = Bf / ( 1 * 9.81) or just Bf / 9.81

 

We can also solve for the volume by finding the specific gravity as Chris and Doc do. Just subtract the Wet weight from the Dry weight. This gives the Buoyancy Force above. (It is because of buoyancy that we get these two different weights)

 

The acceleration due to gravity can be removed from this equation as it is carried along with all of the other calculations as well. The values given from the scale is a force and not a mass. To convert from weight to mass we divide by 9.81m/s^2. To more correctly state the equations above the Weight of the gold (Wg) is as follows

Wg = 9.81 * 19.32 * ( Wt / 9.81 - 2.65 * Bf / 9.81) / (19.32 - 2.65) 

 

If we multiply 9.81 through we end up with the following equation for the weight of the gold in the specimen

Wg = 19.32 * ( Wt - 2.65 * Bf) / ( 19.32 - 2.65 )

 

 

 

 

 

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Dan,  1st off my friend, you need to find a specimen of quartz with gold in it. 🤣  Knowing you that is on the up and up so I hope to see it soon.

Yes there are a few ways and here is one that I like (see attached).  it is of one of my finds years back but the math is a little more simple, well at least it is for me.SpecificGravity.thumb.jpg.df6f957f5ab279793b00e0002243ef6c.jpg

 

Looking forward to hearing about your so called specimen find and it's outcome.

 

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Here is a long thread already on the forum.

 

 

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10 hours ago, Detector Dan said:

Let me start here with this equation

O.K. now you just lost me with all this math.

I didn't know that I would have to go through all of that just to go metal detecting

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 As you can see there is a lot of specimens with different impurities, so "dolly pot" is worth while if the specimen does not win the beauty contest. That is what I did to these ones and most others.

p1011640.thumb.JPG.df9564a251d8c13285d771dffcf214e4.JPG

If you want to calculate your gold you can use my 1980 calc formula from my old book. You will have to find the  Specific Gravity of a sample of the Quartz & Gold and Modify my calc. Note:- The Quartz sg can be done on a sample of the specimen with no gold or similar bit looking the same from the location. The Gold sg can be got from known sg of gold in that area. My calc were done 95.5% purity typical for the area I was working in the golden triangle Vic Aust.

2017730308_SpecimenMyCalc.jpg.8226d8cd8cbf2e24cb2d7588e7d058a8.jpg

  I found that my Formula was accurate for specimens that showed a fair amount of Gold

" Weight of Gold" = 3.0717718 * (Weight in Water) - 1.9126176 * (Weight in Air)

Hope this help you decide what to do your specimen.

Sg of Gold           = Sg

 Sg of Quartz      =  Sq

Weight_in_Water =  Ww

Weight_in_Air       =  Ww

Weight Gold   = Sg * Sq * Ww / (Sg - Sq)  +  Sg * Wa *(1-Sq)/ (Sg - Sq)

 

 

 

 
 

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8 hours ago, geof_junk said:

" Weight of Gold" = 3.0717718 * (Weight in Water) - 1.9126176 * (Weight in Air)

 
 

That equation gives the same results as my approach. Just a different approach. Its nice to see that algebra works 😁

 

You could reduce my equation to

Weight of Gold = 1.1589 * (Weight in Air) - 3.0712 * (Volume of Specimen)

These values came from Gold at 19.32g/cm^3 and Quartz at 2.65g/cm^3. We would need to adjust those to get a better approximation, but then its just an approximation. haha

 

I also realized my explanation of the volume was convoluted. If we set up a scale with a cup of water on it, tare the scale, and then suspend the specimen in the water, we get a measurement of the buoyancy force. This is the same as measuring the Volume of the Specimen directly. Then we dont have to set up a fancy rig to get the weight of the specimen itself in the water. Just tie a string around the speci and hold it in the water by hand and record the scale readings.

You could take that Volume and subtract that from your Weight in Air and get your Weight in Water.

 

Thanks Gerry for the way you've done it. Thats the same process that Doc uses in his video. I wish there was an explanation to where the numbers you multiply by come from. 

 

Thanks for the Thread mn90403. That thread gave the same conclusion that geof_junk has. 

 

 

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2 hours ago, Detector Dan said:

Thanks Gerry for the way you've done it. Thats the same process that Doc uses in his video. I wish there was an explanation to where the numbers you multiply by come from. 

Like the constant 25.97?   I was curious but I couldn't find how or where that number came from.

 

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I tried last night to go to the online calculator that I've used for years.  It just required that you put in the dry weight and the wet weight and it would spit out the amount of gold you have in a specimen.  Unfortunately the calculator has been removed and the domain is for sale.

JP ran a contest a few months back or a year or so and I used it and it was the most accurate to his actual gold recovered from a specimen.  I wonder if the wayback site has working calculator pages saved.

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