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Re: Surface-Grid Intercept

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ArbiterOfDeath is not online. ArbiterOfDeath
Joined: 20 Jun 2011
Total Posts: 1458
20 Dec 2013 02:44 PM
I have a part, a list of all the points on the part, and a list grouping those points into surfaces. For every surface, I want to find every y for whole number x and whole number z. For those who are visual, that is like looking down directly onto the surface, and viewing where the grid lines for x and z cross, and finding the height of the surface at that point.

I've tried to define a scalar plane using the normal of the surface, but while that can give me the height, it doesn't tell me if that point still exists on the surface, it tells me if it is on the same plane. That means that I can't tell if I'm looking too far in my x's or z's.

Now, I could just rotate and position points using the parts CFrame, but that takes way to much computer power when I'm planning on using this on entire places in real time.

And yes, I'm working with rotated parts, so this isn't easy.

In question format, how do I find the height of a surface at a given x and z and determine if that point is on the surface still, without preforming expensive rotations?
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ArbiterOfDeath is not online. ArbiterOfDeath
Joined: 20 Jun 2011
Total Posts: 1458
20 Dec 2013 03:09 PM
Bump. Only five people even viewed this. Do you guys understand what I'm asking, or just don't have any suggestions? I can go to scripter's, and try my luck there xD
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ArbiterOfDeath is not online. ArbiterOfDeath
Joined: 20 Jun 2011
Total Posts: 1458
20 Dec 2013 06:53 PM
For those interested, I asked it on scripters, and on the math stack exchange website. The answer is here, just remove the semicolons (";"):

ht;tp:/;/math;.stacke;xchange.c;om/questions/614407/height-at-2d-coordinate-on-a-3d-rectangular-surface
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Jetta765214 is not online. Jetta765214
Joined: 22 Oct 2008
Total Posts: 1855
20 Dec 2013 06:59 PM
I can come up with a formula if the question is still unanswered.
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ArbiterOfDeath is not online. ArbiterOfDeath
Joined: 20 Jun 2011
Total Posts: 1458
20 Dec 2013 07:08 PM
It's answered now... and look at the detailed explanation the guy gave me xD
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