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Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
Posted 25 January 2007 - 09:51 PM
Posted 25 January 2007 - 11:10 PM
Posted 26 January 2007 - 04:12 AM
Posted 26 January 2007 - 06:00 AM
Quote:
Original post by erissian
It's discontinuous (it has corners)
Posted 26 January 2007 - 07:48 AM
Quote:
Original post by Zipster
Actually, you just remove the the sum restriction u + t ≤ 1 completely to get the full quadrilateral (keep 0 ≤ u/t ≤ 1). But it describes the entire inner area. You would have more logic to get just the edges.
Posted 26 January 2007 - 08:41 AM
Quote:
Original post by TheAdmiral Quote:
Original post by erissian
It's discontinuous (it has corners)
Not quite. A square is continuous (there are certainly no gaps), but it isn't smooth. If one were to describe it piecewise-implicitly or parametrically, then it's the first derivatives (and hence all that follow) that would be discontinuous. </unnecessary aside>
Admiral
Posted 01 February 2007 - 05:48 PM
Quote:That's a filled square though (or a filled diamond if you prefer)
Original post by Merlz
Here's another one, just for fun:
abs(x) + abs(y) <= 1
Although this one is at 45 degrees rotation to the axis, it's of max radius 1 :-)
Posted 01 February 2007 - 06:46 PM
Quote:Try:
Original post by iMalc
A more generic unfilled version of that would be:
abs(x) + abs(y) = r
Of course you could probably rotate that by 45 degrees, which would probably look something like this:
abs(x*sin(45)+y*cos(45)) + abs(x*cos(45)-y*sin(45)) = r
Posted 13 February 2007 - 04:35 PM
Posted 26 November 2011 - 10:42 AM
Posted 26 November 2011 - 01:18 PM
sawtooth(x){ abs((x-4*floor(0.25*x))-2)-1 } infcos(x){ min(1,max(-1,sawtooth(0.5*x)*2)) } infsin(x){ infcos(x-2) } abscos(x){ x*=sqrt(2)*0.5; sawtooth(x) } abssin(x){ x*=sqrt(2)*0.5; sawtooth(x-1) }
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