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Jun 11, 2020 at 9:33 history edited CommunityBot
Commonmark migration
Nov 23, 2014 at 4:39 vote accept CommunityBot
Nov 15, 2014 at 18:13 comment added Qmechanic The numerical slope @ 1 is roughly a fifth: $0.2323-2*0.0176=0.1971\approx\frac{1}{5}$.
Nov 15, 2014 at 0:47 comment added André Chalella @Floris there you go: pastebin.com/DuLEhune. I tried to comment it a little bit, but feel free to ask any questions.
Nov 15, 2014 at 0:16 comment added André Chalella @Qmechanic for info: a linear trendline would be of slope .2165 $\left(y=.2165x+.7864\right)$, giving $R^2=.9996$. I wasn't certain if I should convey that or quadratic. Do you think one or the other would be more appropriate?
Nov 14, 2014 at 22:22 comment added Qmechanic +1. The numerical plot seems to agree with the theoretical slope value $\to\frac{1}{5}$ when (minor axis)/(major axis) $\to 1$, cf. my answer.
Nov 14, 2014 at 21:50 history edited André Chalella CC BY-SA 3.0
expanded
Nov 14, 2014 at 21:15 history answered André Chalella CC BY-SA 3.0