If I create a coin and I want its multiple thickness size, to be as the same as the area of this coin, what's the way to calculate the thickness-diameter ratio that needed- considering the specific amount (in grams) and density of the metal (g/cm³) I'm going to use? I'm looking for the thickness and the diameter that their multiplication will fit the area of this coin.

To make it simpler, I'll take one example: I'm taking 1 gram of silver (it has a density of 10.49 g/cm³) and I want one surface area of the coin I'm planning to do, to be the same as 2* diameter * thickness. (for instance: if the diameter is 10mm, and the thickness is 1 mm, then I want the are of the coin to be 20mm^2, which is the multiplication of the thickness size in the diameter *2).

Shortly, I want to find a way to find the thickness and diameter if I have density data and a specific weight of metal of which I want to create the coin:
Thickness (in mm)* diameter (in mm) * 2 = the same as the surface area of one side our of the two sides of the coin (in mm^2).

  • $\begingroup$ Your question is not clear. Are you saying you want $2$ times dia and thickness to be same as the surface area of the coin? Which surface of the coin? Only two faces or the side part too? When you divide weight by density, you get volume which is $(\pi / 4) d^2 t$. $d$ is dia, $t$ is thickness. $\endgroup$ – Math Lover Sep 13 '20 at 12:43
  • $\begingroup$ I tried to say that I want thickness (in mm)* diameter (in mm) * 2 = the same as the area of the coin (in mm^2). I'd like to know what I can improve in my question to make it clearer for anybody. I have only two things in my hand: 1. specific amount of metal (1 gram of silver). 2. data about the density of a silver which is: 10.49 g/cm³. I want to make a coin of which its area is equal to the (diameter*thickness)*2. Is it better now?) $\endgroup$ – Math-Enthusiast Sep 13 '20 at 12:49
  • $\begingroup$ you need to mention which area? total surface area (side area too)? $\endgroup$ – Math Lover Sep 13 '20 at 13:09
  • $\begingroup$ Thank you, edited. I meant to surface area of one side (out of two sides) of the coins. $\endgroup$ – Math-Enthusiast Sep 13 '20 at 13:19

$v = 10^3 \frac{w}{\rho}$ where $v$ is volume in $mm^3$, $w$ is weight in gram and $\rho$ is density in $gm/cm^3$

As you now know volume, equate it to the volume of the coin in terms of its diameter ($d$) and thickness ($t$) -

$\frac {\pi} {4} d^2 t = v$

Or, $d^2 t = \frac {4v}{\pi}$ ...(i)

Area of one of the round faces of the coin $ = \frac{\pi}{4} \times d^2 = 2 \, d \, t, \,$ as per condition given in the question.

So, $d = \frac{8}{\pi} \, t$ ...(ii)

Substituting $d$ from (ii) in (i), you can find $t$ and then $d$.


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