First dip gold in water to find the amount of displacement.
as written above, this is pretty hard.
second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.
Heh.
No.no he's right. you could toss a little platinum in there to offset the weight to have a net equivalent density of gold in the same volume.
in the end you'd really have to go to more complex testing eg ultrasound or conductivity
Platinum? Let's keep things cheap while we're ripping people off. A proper ratio mixture of lead and tungsten could have the exact density of gold.
Platinum is cheaper than gold (Crazy, right?) and lead is 1/2 the density. Pt is the cheapest element that has a higher density than gold.
The equation that would need to be fulfilled is:
p_Au*f_Au + p_Pt*f_Pt + p_W*f_W = p_Au, for the bar to pass the archimedes (effective density) test.
where p_x is the density of element x and f_x is the volume fraction of element x.
p_Au = 19.282, p_Pt = 21.46 and p_W = 19.25. This equation is subject to the constraint f_Au+f_Pt+f_W = 1.
The price of W is $45/kg, which is $1.28/oz. Gold price = $1771/oz., Platinum price = $1638/oz.
Now we can establish the price of the bar as a function of the volume fractions.
The price per oz. of the forged bars is given by P_bar(f_Au, f_W, f_Pt) = P_Au*f_Au+P_Pt*f_Pt+P_W*f_W, which we would like to minimize given the above density and volume constraints. This arrives at an edge solution (max. amount of W, no Au in bar), at f_W = (p_Au-p_Pt)/(p_W - p_Pt) = 98.4vol.% W, 1.6vol% Pt.
This bar would cost $27.40/oz. and have exactly the same apparent density as a gold bar.
Of course, it would not look like gold, but basically you could add any amount of Pt/W in this proportion (98.4% W, 1.6% Pt) to the gold bar, and you would not change the effective density.