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 Help finding the Y axis angle of a plane?  (Read 2882 times) 0 Members and 1 Guest are viewing this topic.
CyanPrime
 « Posted 2012-05-27 23:42:05 »

Alright, so I need to know the y axis angle of a plane made by 4 points in 3d space. basically I want code to where if it's _ (flat) I can walk on it and if its | (a wall) I run into it. So could anyone explain to me a good way to figure this out?
theagentd

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 « Reply #1 - Posted 2012-05-27 23:51:09 »

Calculate the normal using 3 of the vertices (a triangle) and check the value of y?

Myomyomyo.
CyanPrime
 « Reply #2 - Posted 2012-05-28 05:45:47 »

Okay, this is what I got so far. Assume quadToCheck is 4 Vector3f's with random values.

 1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20 `public float getNormal(Vector3f[] quadToCheck){      Vector3f ab = new Vector3f(0,0,0);      Vector3f.sub(quadToCheck[1], quadToCheck[0], ab);            Vector3f ac = new Vector3f(0,0,0);      Vector3f.sub(quadToCheck[2], quadToCheck[0], ac);            Vector3f crossed = new Vector3f(0,0,0);      Vector3f.cross(ac, ab, crossed);      Vector3f backup = new Vector3f(crossed);      try{         crossed.normalise();      }            catch(Exception e){         //e.printStackTrace();         crossed = backup;      }      return crossed.y;   }`

Am I doing this right? Cause I'm not seeing good results, and it tends to hit that catch like every time, or every other time :<
theagentd

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 « Reply #3 - Posted 2012-05-28 09:29:24 »

This is some GLSL code I have to calculate the normal of a triangle in a geometry shader:
 1  2  3  4  5  6 `vec3 calculateNormal(){    vec3 u = gl_in[1].gl_Position.xyz - gl_in[0].gl_Position.xyz;    vec3 v = gl_in[2].gl_Position.xyz - gl_in[0].gl_Position.xyz;        return normalize(cross(u, v));}`

Shouldn't it be Vector3f.cross(ab, ac, crossed)? Mine works as it should.

EDIT: I love how clean GLSL code looks for this kind of math. =>

Myomyomyo.
Roquen
 « Reply #4 - Posted 2012-05-28 09:58:38 »

There's no need to normalize unless you really want the normal.  The original question only needs the direction.
theagentd

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 « Reply #5 - Posted 2012-05-28 13:08:29 »

There's no need to normalize unless you really want the normal.  The original question only needs the direction.
But to calculate the angle to y+ you need to do a dot product with a vector pointing up --> you need normalize it and just look at y. Maybe that's not the best way of doing it...

Myomyomyo.
Roquen
 « Reply #6 - Posted 2012-05-28 13:17:26 »

Really this just isn't the way to handle the problem.  If polys are always in XZ (floor/ceiling) and X^Z=0 (walls), then just have three lists floors, ceilings and walls and be done with it...which I assume is the case since only the Y component is being returned.  But if you don't need distance from test point, then no normalization is needed...the sign is the answer.
Danny02
 « Reply #7 - Posted 2012-05-28 13:35:24 »

ofcourse you can always check if some plane/triangle/quad is walkable with the normal(angle between the normal and the 90° flat plane). Another possibility is to generate a navigation mesh(something in the direction of Roquens idea).

Just generate a list of areas/triangles ... which are wakable. By hand or an algorithm, there are also nice opensource tools for that see recast
theagentd

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 « Reply #8 - Posted 2012-05-28 14:01:39 »

I was assuming that OP wanted something similar to most shooters/RPGs, where you can walk on everything that isn't too steep. My idea was to look at the y value of the normal and check if it was above a certain threshold...

Myomyomyo.
Danny02
 « Reply #9 - Posted 2012-05-28 14:10:44 »

I was assuming that OP wanted something similar to most shooters/RPGs, where you can walk on everything that isn't too steep. My idea was to look at the y value of the normal and check if it was above a certain threshold...

sure that works, the y value of a normalized plane normal should be the cosinus angle between the plane and the ground, right?
Roquen
 « Reply #10 - Posted 2012-05-28 16:38:32 »

Doesn't work.  For arbitrary surfaces you need the full direction and not a single component.  Consider rotating the surface about 'y'.  You can find the slope but not the maximum direction of change.
Icecore

Senior Devvie

Medals: 6

 « Reply #11 - Posted 2012-05-28 17:25:19 »

Check this out
Where Y and X 2 axis distance
Let sta you have 1.3 and 2.4
X = 2 – 1
Y = 4 – 3
Play with dif axis to find what you need

 1  2  3  4  5 `static public float Angle360(float y, float x){      float ret = (float) (Math.atan2(y, x) * 180 / Math.PI);      if(ret < 0)ret += 360;      return ret;}`

http://polygeek.com/1819_flex_exploring-math-atan2

You live in so cool and insane world, what you f**king doing here?
Danny02
 « Reply #12 - Posted 2012-05-28 18:02:38 »

Doesn't work.  For arbitrary surfaces you need the full direction and not a single component.  Consider rotating the surface about 'y'.  You can find the slope but not the maximum direction of change.

sry, I don't get what you are saying. A first thought how to handle player movement would be to check if the surface the player is standing on has a slope below a given threshold. When the slope is high the player starts sliding according to the slope.
Roquen
 « Reply #13 - Posted 2012-05-28 18:20:51 »

In what direction does the player start to slide?
Danny02
 « Reply #14 - Posted 2012-05-28 18:36:41 »

to get direction where to slide you need of course the normal, but to check if you slide is another calculation(the one above)

to get the direction:
slideDir = normal x (normal x up)

slope:
cos(a) = up * normalize(normal)
cos(a) = (normalize(normal)).y // because up = (0, 1, 0)

Roquen
 « Reply #15 - Posted 2012-05-28 19:28:04 »

That was a rhetorical question...you require all three components to know the direction.  cross gives direction of plane, project back into plane is direction of max slope..normalize if needed.
theagentd

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 « Reply #16 - Posted 2012-05-29 12:37:07 »

That was a rhetorical question...you require all three components to know the direction.  cross gives direction of plane, project back into plane is direction of max slope..normalize if needed.
I don't understand the math behind that, so I'll just tell you how I would do it:

 1  2  3  4  5  6  7  8  9  10  11  12  13  14 `vec3 t0, t1, t2; //The triangle (three sides of a quad)Vec3 u = t1 - t0;Vec3 v = t2 - t0;Vec3 n = cross(u, v); //Unnormalized normal of the planefloat cosSlopeAngle = n.y / sqrt(n.x*n.x + n.y*n.y + n.z*n.z);if(cosSlopeAngle > cos(maximumAngle){    vec2 slidingDirection = vec2(n.x, n.z);    slidingDirection.normalize();    //Use sliding direction to modify the objects velocity...}`

Excuse my psuedo-GLSL...

Myomyomyo.
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