# Exp[-690255.2204176334] is too small to represent as a normalized machine number; precision may be lost [closed]

I'm trying to visualize the relationship of maximum temperature $$(T_{m})$$ with the heating rate with Randall-Wilkins theory for an energy depth of 0.85 eV. The Randall-Wilkins expression used is Equation 3 from https://www.lumipedia.org/index.php?title=Theory_of_Thermoluminescence.

I have already tried everything else, but I still can't plot it with Mathematica.

Here's my code:

p = (Exp[-N[((0.85/(y*8.62/10^5))), $MachinePrecision]]) eqn = 0 == 0.85/(y^2*(8.62/10^5)) - (10^9)*(1/x)*(p) ContourPlot[Evaluate@eqn, {x, 0, 20}, {y, 0, 200}  • You are dealing with very very small numbers with your exponentials so it may be difficult to obtain a reliable result. Can you provide some context to your equations? Where do they come from? Can you change the units of measurement so you will deal with more reasonable numbers? Jul 20 at 16:19 • Thank you for responding, I added context in my original post. I've already considered converting the units, but it still won't work. – D A. Jul 20 at 17:13 • You can avoid machine underflow by using exact numbers. So you can Rationalize your input. As for why no plot shows, this is different issue. Could be it can't find contour lines that satisfies the equation Jul 20 at 17:23 • What about plotting the Log of your terms? Jul 20 at 18:23 • You have energy values in your expression; you can surely find an appropriate unit of energy, or combined unit, that allows you to avoid the ridiculously large / small numbers you currently have. For instance, if you now use Joules, think of eV instead, or$cm^{-1}$, or whatever other unit is common and appropriate; since the energy always appears as a ratio with with the Boltzmann constant$k$, you could generate a new variable representing$E/k$, which will probably range over much more reasonable values. Jul 20 at 18:33 ## 1 Answer It's a pretty simple explanation. The region plotted by the OP contained none of the solution curve, which is found further up: p = (Exp[-N[((0.85/(y*8.62/10^5))),$MachinePrecision]])

eqn = 0 == 0.85/(y^2*(8.62/10^5)) - (10^9)*(1/x)*(p)
ContourPlot[Evaluate@eqn,
{x, 0, 20}, {y, 0, 1000}] (* <-- change 200 to 1000 *)