Exactly how elliptical is Mercury's orbit, visually, without exaggeration? Reposting question from https://www.reddit.com/r/askscience/comments/5e1zhy/what_does_mercurys_orbit_really_look_like/ which reads:


  First the full question: What does the orbit of Mercury really look
  like visually, without artistic liberties taken to amplify the
  eccentricity of the orbit, or simplifying it to appear as a perfect
  circle? What eccentricities are needed for orbits that readily appear
  elliptical?


I originally did this as a self-answer, but my answer was of such poor quality I've now deleted it. I hope to do a better answer backed by some calculations, but, as always, anyone else is free to answer.
Some things we know about Mercury's orbit:


Semi-major axis: 0.387 AU, about 57.9 million km
Eccentricity: 0.205


We can calculate the semi-minor axis, $b$ from the equation
$$e=\sqrt{1-\frac{b^2}{a^2}}\to b=0.379\text{ AU}$$
We can also calculate the distance to the focus from the center of the ellipse, $f$, as
$$f=ae=0.078\text{ AU}$$
I used Mathematica's ParametricPlot to plot Mercury's orbit using these parameters, along with Earth's. Earth's semi-major axis is 1 AU, and with an eccentricity of 0.016, its semi-minor axis is about 1 to three significant figures; I've treated it as a circle. Mathematica gives



ParametricPlot[{{0.387*Sin[x] - 0.0793, 0.379*Cos[x]}, {Sin[x],
  Cos[x]}}, {x, 0, 2*Pi}, AxesLabel -&gt; {AU, AU}, PlotLegends -&gt; 
  Placed[LineLegend[{Red, Blue}, {"Mercury", "Earth"}], {1, 0.25}], 
  PlotStyle -&gt; {Red, Blue}]


I'd consider the deviation from a perfect circle to be noticeable insofar as the focus is shifted from the point $(0, 0)$, but it's not incredible.