I am attempting to plot a PV curve using two functions that rely of time, one for pressure and one for volume. When I run my code I only get a straight line along one axis. I am doing something fundamentally wrong?

v[t_] := Power[106216343/(106216343/(6361.725*(285 - (40*Sin[90 - (630*t)] + Sqrt[
    240^2 - (40*Sin[630*t])^2])))^1.4), (1.4)^-1];

p[t_] := 106216343/(6361.725*(285 - (40*Sin[90 - (630*t)] + Sqrt[
    240^2 - (40*Sin[630*t])^2])))^1.4;

ParametricPlot[{p[t], v[t]}, {t, 0, 0.00997331}]

enter image description here

  • $\begingroup$ the other answer is far better than mine $\endgroup$ – ItamarG3 Feb 13 '18 at 11:42

Just use AspectRatio to show your axes as intended

ParametricPlot[{p[t], v[t]}, {t, 0, 0.00997331}, AspectRatio -> 1/3]

enter image description here

  • $\begingroup$ Your answer is much better. (I gave it an upvote). I didn't know about AspectRatio. You learn something new everyday :D $\endgroup$ – ItamarG3 Feb 13 '18 at 10:15

Notice your coefficient for v.

Numerically, it's 31808.6.

On the other hand, for p, it's around 52.

So your y axis (v is in the vertical axis because it is the second argument in the ParametricPlot) has a much larger scale, compared to the horizontal axis.

In the region of time you wanted to plot, the volume increases much, much more than the pressure, which is why you got such a result.

To get a qualitative plot, you can ask Mathematica to plot v[t]/1000, or something similar. By doing so, you downscale v and you'll be able to see the relation between p and v.

I don't know where those numbers came from, so I can't be sure if you calculated them correctly...

  • $\begingroup$ Eureka! Changed atm to Pascals and I have a decent looking graph now! $\endgroup$ – dirtcrazy Feb 8 '18 at 10:23
  • $\begingroup$ @dirtcrazy happy to help :D $\endgroup$ – ItamarG3 Feb 8 '18 at 10:25

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