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IronDigger

Gold / Quartz Specific Gravity... Question

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Davsgold ,  thanks for the good news !! I did subtract and got the 171, where I may have messed up is the rest of the formula I will give it a shot until I get the same numbers. Looking at the piece and the amount of surface gold on it I knew it had more than 31 grams. The amount inside the rock was the mystery..appreciate your help.

 

 

 

 

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All the formulas in the world won't help if the specific gravity of the gold/quartz specimen is lower than quartz. If you divide 242 by 171, you get 1.415, which is much lower than quartz. If it was 2.65, it would be 100% quartz, so being 1.415 means either the water weight, as stated, is incorrect, or there is a large portion of lightweight mineral, or air, inside. If it's a gold/quartz specimen it has to have a specific gravity higher than 2.65.

Jim

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So I made a mistake on the wet weight ..I looked at the link on how to properly weigh the specimen piece in water and I now have a weight of 173 for the submerged weight,  242 for the dry weight. In the beginning all I did was place a piece of Tupperware and filled it with water deep enough for the speci to be suspended without hitting the bottom or sides. I than tied a fish line to the speci and zeroed out the scale. I than submerged it by holding the fish line into the zeroed out scale...I got the original figure of 70. This figure did not make any sense comparing it to the dry weight, it was a huge disparity. So,  

D.W. 242 x 1.9 = 459

W.W. 173 x 3.1= 536

according to the video we subtracted these figures 536 - 459 = gives me the approx. weight of gold at 77 grams. or 2.48 ounces troy.

The speci is solid quartz...anyways hope this helps others as it helped me, thanks you all for the math lesson. 

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That's better, though I'm coming up with 88.34 grams of gold....LOL

Jim

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1 minute ago, Jim in Idaho said:

That's better, though I'm coming up with 88.34 grams of gold....LOL

Jim

Hey Jim, I'll take the 88 grams....lol :) I will try a few other formulas and see what it comes to, but thankfully I have figured out how to properly do a submerged weight of a speci. I tried to cut corners....didn't work. 

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Yup...it's a good thing to know how to do, especially for a prospector. It was a good refresher for me on calculating the portion that's gold, too.

Jim

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10 hours ago, IronDigger said:

So I made a mistake on the wet weight ..I looked at the link on how to properly weigh the specimen piece in water and I now have a weight of 173 for the submerged weight,  242 for the dry weight. In the beginning all I did was place a piece of Tupperware and filled it with water deep enough for the speci to be suspended without hitting the bottom or sides. I than tied a fish line to the speci and zeroed out the scale. I than submerged it by holding the fish line into the zeroed out scale...I got the original figure of 70. This figure did not make any sense comparing it to the dry weight, it was a huge disparity. So,  

D.W. 242 x 1.9 = 459

W.W. 173 x 3.1= 536

according to the video we subtracted these figures 536 - 459 = gives me the approx. weight of gold at 77 grams. or 2.48 ounces troy.

The speci is solid quartz...anyways hope this helps others as it helped me, thanks you all for the math lesson. 

Iron Digger,

You were actually on the right track the first time; your mistake was simply plugging the data into the wrong equation. Allow me to explain:

When lowering a specimen into a zeroed container of water, the resultant weight isn't the weight of the submerged specimen, but rather is the weight of the water that is displaced by the volume of the specimen; in your case 70 grams. One of the cool properties of water is that at room temperature, a volume of 1 cubic centimeter of water weighs exactly 1 gram. From this we can deduce that your specimen has a volume of 70 cubic centimeters. The formula you need to calculate the weight of gold in the specimen with this data is:

gold weight = (1.213 X dry weight of speci) - (3.1 X volume of speci)

Plugging in your data we have:

gold weight = (1.213 X 242) - (3.1 X 70)

gold weight = 293.546 - 217

gold weight = 76.546 grams, rounded to 77 grams

This method is less complicated to perform than the one in the video. Just make sure to gently tap the specimen against the bottom of the water container to remove any entrapped air bubbles before suspending it.

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Lunk, 

Interesting ! I read that somewhere but didn't understand it , thought I had  to get a special container with centimeters lines and than had to count the small lines to figure out the volume displaced. Actually this formula seems more accurate...thanks a bunch....I have sure learned a lot ..appreciate all the input from you all. 

 

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On 1/7/2019 at 10:08 PM, mn90403 said:

I always use this calculator.

http://desert-gold-diggers.org/gold/specgrav.htm

Recently I came up with a negative weight on a specimen like you are doing with this one.  I had to re-weigh the specimen more accurately making sure the wet weight was not touching any sides of a container with the water.

As has already been stated, your specimen should weigh more in water than what you are reporting to us.  Something is wrong if this is a host of quartz.

We still haven't seen a photo of the specimen so no further help can be given.

Mitchel

Common Variables
(Enter these values first and then proceed to either Method 1 or Method 2 below)
Variable Value
Specific gravity of the gold or the first mineral (Sg)
Specific gravity of the second mineral (So)
Dry Weight of the sample (Wsa)
Significant decimal digits in answer, default 1

The number of significant decimal digits should be set to the least number of digits after the decimal point in the input values. For instance, the value of 18.6 has only 1 number after the decimal point. Thus the number of significant decimal digits should be set to one. Calculations are set to use one more digit than this entry.

The Common Variables above must be completed before proceeding. For either method below, fill in the first value and then press the Calculate button. The bottom 2 values are then computed.

Method 1
Variable Value
Weight of the displaced water (Ww)
Calculate the weights of both minerals
Weight of the gold or the first mineral (Wg)
Weight of the second mineral (W2)

 

Method 2
Variable Value
Weight of the sample in water (Wsw)
Calculate the weights of both minerals
Weight of the gold or the first mineral (Wg)
Weight of the second mineral (W2)

 

Method 2 

242 g for dry weight

173 g for wet weight

69.9 g Gold

I like this calculator because you only need two values and you don't do any math.  That has all been done.

My copy and past of the formula from the link does not allow for calculating but you can plug in the numbers.  The way you 'play' with the formula is to determine a closer density for the host quartz.  All of these assumptions are covered in the link.

Mitchel

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