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September 01, 2014, 10:37:21 AM
TIGSource ForumsDeveloperTechnical (Moderators: Glaiel-Gamer, ThemsAllTook)Project point onto a line segment? (Solved+Interactive)
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Author Topic: Project point onto a line segment? (Solved+Interactive)  (Read 4106 times)
Aquin
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« on: December 18, 2010, 06:52:35 PM »



Interactive App:
http://aquinhasa.com/scratch/projection.swf
You can move T1, T2, and P around with the mouse.  Give it a try!

Source code for the app provided:
http://aquinhasa.com/scratch/Main.as
« Last Edit: December 19, 2010, 03:56:00 PM by Aquin » Logged

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Aquin
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« Reply #1 on: December 18, 2010, 07:03:32 PM »

Never mind, I have it figured out.  My brain decided to unstupid for a second, thankfully.

If anyone wants the answer, just let me know.  Otherwise, ignore this thread I guess.
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BorisTheBrave
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« Reply #2 on: December 19, 2010, 03:21:29 AM »

It would be better to write the answer anyway - how many times have you got to an archived thread via Google with the question you want, and then a statement that the answer exists.

In that spirit, here it is.

Code:
e = p2 - p1;
normalize(e);
return p1 + e * dot(e, p3 - p1);

Where p1, p2, and p3 are 2d input vectors (i.e. pairs of co-ordinates) and
Code:
function normalize(v)
{
  d = sqrt(v.x*v.x + v.y*v.y)
  v.x /= d;
  v.y /= d;
}
function dot(a,b):float
{
  return a.x*b.x+a.y*b.y;
}
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Aquin
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« Reply #3 on: December 19, 2010, 10:03:38 AM »

Yep, that's pretty much it!  I have another solution (one that doesn't involve the computationally expensive sqrt and relies on a certain dot product being zero), but your write-up is spot-on and much simpler to understand  Beer!
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BorisTheBrave
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« Reply #4 on: December 19, 2010, 01:02:27 PM »

Hm, you've got a point there. How about:

Code:
e = p2 - p1
return p1 + e * dot(e, p3 - p1) / len2(e)

where
Quote
function len2(v):float
{
  return v.x*v.x + v.y*v.y;
}
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Aquin
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« Reply #5 on: December 19, 2010, 01:51:28 PM »

Yeah, your second method is more accurate than the first.  A quick note:

P' = T1 + xT1T2

The dot product should be zero (the lines being perpendicular to each other in the image)

T1T2 PP' = 0
T1T2 <P' - P> = 0
T1T2 <T1 + xT1T2 - P> = 0
T1T2 <T1 - P> + x || T1T2 ||^2 = 0
x || T1T2 ||^2 = -T1T2 <T1 - P>
x || T1T2 ||^2 = T1T2 <P - T1> = T1T2 T1P

x = (T1T2 T1P) / || T1T2 ||^2

P' = T1 + ((T1T2 T1P) / || T1T2 ||^2) T1T2

Which is exactly what you said.  So anyone who cares about the pure math, there she is.

Also, this finds a projected point on the line that runs through T1T2; not a line segment.  So you have to clamp P' to the endpoints if it falls off the line segment.  Simple stuff.

I'll post up the flash app in an hour or so.
« Last Edit: December 19, 2010, 03:48:58 PM by Aquin » Logged

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Aquin
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« Reply #6 on: December 19, 2010, 03:54:04 PM »

Source code and test app are now provided.  The source code runs through the why and how of the maths. 

Thanks Boris!  Grin
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