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Estimating gold weight


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Hi all,

This may be a two part post if I can not post all the pictures in one.

After detecting my first gold, I wanted to go about trying to determine how much gold was in the samples I found. Out came the books and a few internet searches confirmed the formula I wanted to use.

Because I routinely test meteorites (and wrongs) I knew the procedure to determine bulk density of a stone...even did a video on the procedure a while back.

So, here is what I did and I was wondering if anyone can catch an error in the procedure??? Note that I really DO NOT know what type of rock the gold is in. I do see a little quartz and some kind of rock that looks like basalt. I have a friend that is a mineralogist and when I see him, I will get a better idea.

Now, I know the density of gold is 2.6g/cc and basalt is about 2.9g/cc. Because more of this rock looks like basalt, I just choose basalt as my rock density in my formula.

This is always an estimate anyways, but it does give me some what of an idea on what to offer up for sale, with an honest attempt at getting the weight right and it is documented.

So, the first thing I did was fire up the oven in the lab. I wanted to ensure any significant moisture was removed from the rock. The over was set to increment it's heat range starting at 100 degrees F to 300 degrees F and stay at 300 for a period of time and then increment back down to 100 degrees. Ambient humidity was measured at 13% throughout the heating and cooling process.

Here it is in the oven

th_Drying.jpg

Here it is after being cooled in oven, allowing it to become same temp as ambient at 13% humidity.

th_Cooling.jpg

Part two to follow

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Part two.

So, after the samples cool to ambient, I weigh them. Note that I only use the big sample for this thread. It's dry weight turns out to be 65 grams.

th_dryweight.jpg

I then fill a glass with distilled water and tare my scale with it on the scale. This zero's the scale with the weight of the glass and water on it.

I tie a thread around the rock and then submerge the rock in the water. The wet weight is 17.7 grams.

th_wetweight.jpg

Now I have the numbers I need to calculate the sample density and the numbers I need to key into the formula I have for estimating the amount of gold in the rock.

th_goldestimate.jpg

So it comes out to 0.52 ounces.

Now, because I do see quartz and because the rock still has some cliche on it, the density factor is likely going to be less than that of pure basalt, which would increase the weight of the gold.

If it was pure quartz, it would calculate out to 0.71 ounces of gold. So, I figure it's between those two numbers.

I am considering selling this, but not having ever done it before, I am really looking into it. Any suggestions on my procedure and selling this sort of gold would be appreciated.

Jim

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65 grams dry and 17.7 grams in water seems wrong to me. Plus I think your specific gravity of the basalt is way to high. I don't know how to compute grams per cubic CC, but it's specific gravity in my book is 3.01, so I am thinking 1 CC of solid basalt should only weigh 3.01 grams and a CC of pure gold should weigh 19.3 grams ( 1 CC of water weight 1 gram)

You have much more gold in that rock than you think

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Jim-

I think the 2.6 gram cc weight for gold you listed is way off. Basalt weighs more. No way...

Steve

Hi!

Typo Steve...should have been 19.3g/cc. See picture of calculation on Excel Spread sheet that I posted.

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I considered what El Dorado said. Obviously, I do want it to measure more! Increasing the density of the host rock will LOWER the amount of gold when you calculate it. For example, if I use 3.01g/cc as suggested, it comes out to about 0.45 ounces. If I use pure quartz at 2.6g/cc it comes to 0.71 ounces.

I used a different scale to quickly measure again. However, for the wet weight, I want to point out that the rock was wet prior to be dipping it....which gives the increase in wet weight...a little.

th_secondtest.jpg

So, I will call my Dry weight of 65 grams and my wet weight weight of 17.7g good.

Steve, Give your formula you suggest a try. I come out with Negative numbers. And, the density of gold, related to this formula, can vary greatly and not change the ounce reading when rounded off to two digits.

The density of the host rock is most important. Not knowing this number is the real bummer as it appears to be very critical to the end result.

If anyone has a nugget in what they would consider quartz, I would love to see the dry weight and the wet weight (without knowing your calculated weight) to see what I can calculate here. Having one in near pure quartz would be ideal.

Jim

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Reno Chris wrote an article in the icmj a little over a year ago that will give you all the information you need to calculate the gold content. I tested the formula in a later issue and found it to be pretty close. If you would like, I can look up which issues the articles were in.

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Reno Chris wrote an article in the icmj a little over a year ago that will give you all the information you need to calculate the gold content. I tested the formula in a later issue and found it to be pretty close. If you would like, I can look up which issues the articles were in.

I think the formula is pretty much text book stuff, but yes please, would like to see his formula.

Jim

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Here is why I am totally confused!

In Bill's example which uses the basic standard D=m/v formula, His example...

(1) Air Weight in grams = (A) - Water Weight in grams = (W)

(2) (A) - (W) = Volume of specimen in cubic centimeters. Based on Archimedes principle.

(3) Density = (A)/(V) = (D) in grams/cubic centimeter

(4) The Formula!

Proportion of gold = (D - 2.65) x 19.3

(19.3 - 2.65) x D

A worked example

Air weight is 8.7g

Water weight is 7.0g

(2) So Volume is 8.7 - 7.0 = 1.7 cubic centimeters

(3) So Density is 8.7 / 1.7 = 5.12 grams per cubic centimeter

(4) Proportion of gold = (5.12 - 2.65) x 19.3 = 47.67 = 0.5592 gold or 55.92%

(19.3 - 2.65) x 5.12 85.25

Above is the basic formula from Bill's page.

Using formula above, I plug in my numbers below...

Air weight is 65g

Water weight is 17.7g

So volume is 65 - 17.7 = 47.3

So Density is 65 / 47.3 = 1.374g/cc

Proportion of gold = (1.374 - 2.9) x 19.3

= (-1.526) x 19.3 = -29.4518

(19.3-2.9) x 1.374 = 22.5336

-1.307 gold or negative 130% Something serious wrong with my rock!!! :hmmmmm:

I went back to the standard D=M/V which I should not have needed to change in the first place in the program. I plugged Bill's example into the standard goldtester calculator. It squirts out 4.87 grams of gold or 55.97% gold for Bill's example weights.

So the calculator works....but my rock is driving me nuts!

Jim

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Jim one of the problems with your rock is it has many hosts, including what I believe is a very heavy ironstone or magnetite type composition plus it looked like it had a tan or yellowish colored composition also.

The best thing is to probably use the "DOLLY POT" to get an exact weight on the gold content for your sale.

ALSO TRY THIS

wet weight multiply by 3.07

dry weight " " 1.16

Then subtract smaller number from larger and thats the answer.

Using your wet and dry figures it comes to 21.1

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I think the problem is that your wet weight is incorrect because you cannot measure it like that.

The way you are doing it you are just measuring the weight of the water displaced by whatever volume your sample is, not the actual wet weight of the sample itself. But you did succeed in calculating the volume of your sample since water density is 1 gram/cc, so your sample volume is 17.7 cc's.

To get a correct wet weight you need to suspend the rock from a scale into a glass of water (the water itself need not be on a scale).

A beam scale is useful for this, I just rig up a pencil across my digital scale and tie a string around that and then zero the scale and attach the rock.

There is a good program someone wrote and posted on one of the aussie gold forums that I've found to be more accurate than anything else out there. You could probably find it with a quick google.

Interesting thoughts. However, the weight would be the same. I verified this to be sure.

Jim

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Here is an example of what I said. Scales generally take the place of beams and do the same thing. In the picture below, I tared the scale and weighed a nut in the water, with the glass on top of the scale. 6.4 grams

I then taped a pencil to the scale and hung the nut in air. I moved the glass of water up to the nut and weighed. 6.7 grams. Note that the nut was still wet from the first reading, thus the .3g increase in weight.

th_same.jpg

If I had some mercury, I'd be crunching this rock up. Anyways, that glass will have whiskey in it by noon if I keep this up! :brows:

Interesting thought Jason, just did not pan out!

Jim

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BTW, I forgot to mention...

The dry weight for this nut was 49.7g With the wet weight at 6.4g I calculate 7.7656g/cc for the bulk density. Steel is typically 7.8g/cc so I would say my method of measurement is spot on. Close enough to call it steel! And it is a steel nut.

Same with measuring quartz, hematite, and other various rocks I've measured in the past.

This is so strange!

Jim

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Jim one of the problems with your rock is it has many hosts, including what I believe is a very heavy ironstone or magnetite type composition plus it looked like it had a tan or yellowish colored composition also.

The best thing is to probably use the "DOLLY POT" to get an exact weight on the gold content for your sale.

ALSO TRY THIS

wet weight multiply by 3.07

dry weight " " 1.16

Then subtract smaller number from larger and thats the answer.

Using your wet and dry figures it comes to 21.1

Hi Frank!

This rock is a mess.

The formula you presented is the same one on the Aussy Gold Nugget pages (pretty much) and is for gold in quartz. However, you can't swap numbers and it actually comes out to -8.23. :cry2:

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Hi Jason!

I used the steel nut to illustrate the measuring technique is correct.. I used a steel nut because we know the density of steel.

It proves the technique is correct with a known standard. Proof!

You have to get the wet weight no matter what formula you use.... Period. "Wet weight" in water, is known as volume. Volume is fundamental to density as density is fundamental to ANY formula used to determine percentages of mixed material. Look at EVERY formula and you will see this clearly.

You seem to be shooting yourself in the foot! However to help illustrate it even better, please take a look at this rough cut video.

(I sure hope I post the link correctly)

The measurement technique is not wrong. If you still do not believe it, I do not know what else I can do other than suggest to do this test yourself with any sample that has a known density...like steel or aluminum or anything you have handy.

There is something strange with the rock I can not explain. No one has been able to explain it. Frank had it pretty close in that the rock is a mixture of some unknown material. The measurements with the rock are no different than with the steel nut.

I do realize the density of the suspect rock is the most important, most critical number there is in any of these equations. Not knowing that actual number, because it's not pure quartz and not pure basalt or whatever is key to not solving this.

It's certainly not the measurement technique.

Anyways, I give up as I am not going to solve this. This rock is getting way too much in the way of my meteorite activities.

Watch the video!

Cheers!

Jim

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Jason,

Try this calculator http://www.nqminersden.com/downloads/goldtester.xls

You can change the numbers to anything you like.

Jim

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I do see my boo-boo. I put the wrong number in the box in the calculator. However, this is the formula used. It is fairly standard and if you break it down and look at it on the calculator link, the math is where the (weight in air) - (weight in water) takes place. It's not done in the measurement process as you are trying to get me to do. I fully understand I am measuring density with my procedure as it calls for.

It is not necessary to hang a sample from these scales. And I understand what they call Density is not density.

G (gold) = (d x N x (1 - q/n)) / (d - q)

G = Weight of gold (grams)

d = Density of gold (approx. 19 gm/cubic cm)

N = Weight of nugget in air (grams)

q = Density of quartz (approx. 2.6 gm/cubic cm)

n = Density of nugget ( gm/cubic cm)

Density = (weight of nugget in air) / (weight in air) - (weight in water)

Ref: http://www.nqminersden.com/specific.htm

Until I know more about the rock and it's density the gold is in, nothing is going to work right.

Jim

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Jason,

My numbers and estimates are in the original post. The program used was modified and the picture was include of it earlier in the original posts. It was suggested that those numbers are really low and I had more gold in there than I think. It is within 0.01ounces of what you will come up with using that calculator. So after it was thought to be low, I changed the program back to original and that is where I did not put the right number it. I left the volume number in.

I have written another script I would like you to take a look at, If you would email me, I'd send it to you. It only requires the density measurement of the rock to be known, no hanging of rocks, etc., and defaults with quartz as the host rock. It also includes the formula FrankC likes to use.

my addr is jimwooddell at gmail.com

Jim

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