

Happy Valentine's Day - valuegram
https://www.google.com/search?client=ubuntu&channel=fs&q=5+%2B+(-sqrt(1-x%5E2-(y-abs(x))%5E2))*cos(30*((1-x%5E2-(y-abs(x))%5E2)))%2C+x+is+from+-1+to+1%2C+y+is+from+-1+to+1.5%2C+z+is+from+1+to+6&ie=utf-8&oe=utf-8

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pav3l
You too!

[http://www.wolframalpha.com/input/?i=Plot%5BSqrt%5BCos%5Bx%5...](http://www.wolframalpha.com/input/?i=Plot%5BSqrt%5BCos%5Bx%5D%5D*Cos%5B200*x%5D+%2B+Sqrt%5BAbs%5Bx%5D%5D+-+0.7*%284+-+x*x%29%5E0.01%2C+%7Bx%2C+-2%2C+2%7D%5D)

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jacquesm
<http://mathprospects.com/node/5>

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timerickson
> 3D charts require a web browser and system that support WebGL.

I guess I'm not getting a Valentine this year.

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arasmussen
What browser/OS/gpu are you using?

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timerickson
Latest Safari on OS X 10.8.2 in a Macbook Pro w/ Retina

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dubya
You can enable WebGL in Safari. Go to Preferences -> Advanced and click "Show
Develop menu in menubar". "Enable WebGL" is the last item in the Develop menu.

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shurcooL
I believe it's not enabled by default because there are still some security
vulnerabilities present, is that right?

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dubya
Yes. Here's a DoS for Firefox that forced me to reboot:
[https://cvs.khronos.org/svn/repos/registry/trunk/public/webg...](https://cvs.khronos.org/svn/repos/registry/trunk/public/webgl/sdk/tests/extra/lots-
of-polys-example.html) (it asks permission first). I did not test it in
Safari, because I don't want to restart again :(

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arjn
heh, very neat. You can also play around with the parameters and equation of
the graph.

Check out my "tan" heart : 5 + (-sqrt(1-x^2-(y-abs(x))^2)) _tan(45_
((1-x^2-(y-abs(x))^2))), x is from -1 to 1, y is from -1 to 1.5, z is from 1
to 6

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tarahmarie
Ferris wheels make me so happy. In related news, I think I shall watch Ferris
Bueller's Day Off with My Beloved Husband today.

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booop
Such things never cease to amaze me. How do you start with a blank paper
create an expression which produces that?

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jordigh
It's actually pretty easy...

The actual function being plotted is the two ellipsoids

1 = x^2 + (x - y)^2 + z^2 for x >= 0

1 = x^2 + (x + y)^2 + z^2 for x <= 0

Each of which which is just an ordinary ellipse but rotated a little along the
z-axis. If you expand the square and by standard linear algebra techniques
(see: diagonalising a quadratic form), you can put both of these in standard
form:

1 = x^2/a^2 + y^2/b^2 + z^2

which confirms it's an ellipsoid.

Whoever made the original heart just took these two ellipses and then
multiplied by the cosine of the height in order to give it some oscillations
and make it look more showy, but the basic idea is just that: two ellipsoids,
at opposite slant angles, joined along the middle.

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crusso
Google scares me sometimes.

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snogglethorpe
Me too. But in a good way ... :]

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uglytom
Looks good! Happy Valentine's Day to you too!

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iframe
Works in Chrome Canary ;)

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joshualastdon
Nice. Nice again, I say.

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treycopeland
well done sir. well done.

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scoot
The OP din't claim to have created this (although arguably it's implied by
it's posting without comment). Rather, that's just one of quite a number of
such functions posted here before. I don't know their origin.

So well done for remembering it, and reposting it on an appropriate day; or
more likely someone else remembered, and this is just a cross-post of a cross-
post of a blog of a re-post. But well done anyway.

