. . . my previous post was useless, here’s something to consider.
You go to a nice restaurant and order a $40 glass of Domaine de Montille Corton Clos du Roi Grand. The sommelier pours you a bit, you swirl, smell, sip and nod your approval. The sommelier pours your wine and leaves you looking at the glass wondering if you’ve been cheated.
The shape of the glass is determined by the function sin(x). Here’s the plot (x=0.0 to 1.5 radians) from the R statistical system. It’s been rotated 90 degrees counter clockwise to make visualization easier and is marked at x=0.75 (half the height) and x=1.111871 (half-full by volume).
Using this as a template draw a cross-section of a nice wine glass and fill it to half the height (fig. 1) and half the volume (fig. 2) like so:
Being a beer kind of guy, if I’m going to spend $40 on a glass of wine, I want a full glass of wine. But, as you can see what appears to be a fairly full glass of wine is still only half full by volume. To prove this we can use the program from the previous post to calculate the volumes bounded by a = 0.0 and b = 1.111871 and a = 1.111871 and b=1.5. Here are the results:
[Walters-iMac:~/desktop] wemrt% python halffull.py lower_bound = 0.000000 upper_bound = 1.500000 height = 1.500000 volume = 2.245359 half-volume = 1.122680 <- This should be the volume of the top half of the glass. Tolerance : 0.000001 Iterations: 20 lower_bound = 0.000000 upper_bound = 1.111871 height = 1.111871 volume = 1.122679
Then using the upper_bound for the lower_bound and 1.5 for the upper bound:
[Walters-iMac:~/desktop] wemrt% python halffull.py lower_bound = 1.111871 upper_bound = 1.500000 height = 0.388129 volume = 1.122676 <- Off by 4 X 10^-6 rounding error. half-volume = 0.561338 Tolerance : 0.000001 Iterations: 19 lower_bound = 1.111871 upper_bound = 1.315996 height = 0.204125 volume = 0.561337
So, yes you were being shorted by quite a lot of wine. Call the sommelier back, show him this post and tell him that for $40 he can bloody well give you half a glass of wine.