Hi Kelvin,
Thank you for the reply. Theoretically, my algorithm is correct, right? If
the vector is not between 2 edge vector, the dot produce should be -1,
otherwise, it should be 1. I don't know why the 2 other cases don't get
value close to 1 or -1.
Since we found the vector (V1) of the intersected point to Vertex 0 is between its 2
edge vectors, because its dot produce is 1. So I change my code to calculate the
intersection between the vectors V1 and V2(Vertex1 to Vertex2) and then I compute the
proportional factor to see if it is between 0 and 1. my test shows it's 1, so it is on
the edge of Vertex1 and Vertex2,
and my conclusion is still that: the intersected point is inside the triangle.
Could you check this for me? Thanks much.
The code to test if the point is inside a triangle:
private boolean isOutofTriangle() {
if (isOnPlane()) {
Vector3d v1 = new Vector3d();
Vector3d v2 = new Vector3d();
Vector3d v3 = new Vector3d();
v1.sub(primitiveCoordinatesVW[1], primitiveCoordinatesVW[0]);
v2.sub(pointCoordinatesVW, primitiveCoordinatesVW[0]);
v3.sub(primitiveCoordinatesVW[2], primitiveCoordinatesVW[0]);
v1.normalize(); v2.normalize(); v3.normalize();
v1.cross(v1, v2);
v3.cross(v2, v3);
v1.normalize(); v3.normalize();
double eps;
eps = v1.dot(v3);
System.out.println("eps = " + eps);
if (eps < 0 && (eps + TOL) < 0)
return true;
v1.sub(primitiveCoordinatesVW[2], primitiveCoordinatesVW[1]);
v1.normalize();
double a1 = (v1.y*primitiveCoordinatesVW[1].x -
v1.x*primitiveCoordinatesVW[1].y) -
(v1.y*primitiveCoordinatesVW[0].x - v1.x*primitiveCoordinatesVW[0].y);
double b1 = (v2.x*v1.y - v2.y*v1.x);
double k = a1/b1;
double x, y, z;
x = v2.x*k + primitiveCoordinatesVW[0].x;
y = v2.y*k + primitiveCoordinatesVW[0].y;
z = v2.z*k + primitiveCoordinatesVW[0].z;
double t1 = (pointCoordinatesVW.x - primitiveCoordinatesVW[0].x)/
(x - primitiveCoordinatesVW[0].x);
double t2= (pointCoordinatesVW.y - primitiveCoordinatesVW[0].y)/
(y - primitiveCoordinatesVW[0].y);
double t3 = (pointCoordinatesVW.z - primitiveCoordinatesVW[0].z)/
(z - primitiveCoordinatesVW[0].z);
if (t1 < 0 && (t1 + TOL) < 0)
return true;
if (t1 > 1 && (t1 - TOL) > 1)
return true;
return false;
}
return true;
}
>More info. on this...
>
>In the test for same vectors, given
>v1, v2, v3 are coplanar vectors it uses 3 check
>with each one similar to the following,
>
> v1.cross(v1, v2);
> v2.cross(v2, v3);
>
> double eps = v1.dot(v2);
> if (eps < 0 && (eps + TOL) < 0)
> return true;
>
>to return true when point out of triangle.
>
>(i) There is a missing v1.normalize() and v2.normalize() after computing
>the cross product.
>
>(ii) The test
>
> if (eps < 0 && (eps + TOL) < 0)
>
>is same as
>
> if (eps < 0 && (eps < - 1.0e-5)
>
>which is equal to
> if (eps < -1.0e-5)
>
>
>Theoretically, eps return is either 1 or -1 (after v1, v2 normalize in
>step 1) given the 3 input vectors are coplaner.
>So the checking is not correct.
>
>In this particular examples, I got
>
>0.9999999999999998
>0.2636325256261517
>0.2622153128979992
>
>In the other two cases it doesn't return a value close
>to 1 or -1 as expected. A close look show that in the last
>two case on of the vector is ~ 0 so normalization fail.
>
>As a result the algorithm is not robust numerically.
>
>- Kelvin
>----------------
>Java 3D Tam
>Sun Microsystems Inc.
>
>
>
>
>
>Kelvin Chung wrote:
>> Hi White,
>>
>> The algorithm isOutofTriangle() you use is not quite correct.
>>
>> A ------------ B
>> \ /
>> \ /
>> \ /
>> \ / x D
>> \ /
>> \ /
>> C
>>
>> It will report D inside triangle if A,B,C,D are coplanar.
>>
>> Since it computes the vector ADxAB which is perpendicular to the
>> plane ABC, then computes the vector ACXAB which is also
>> perpendicular to the plane. The dot produce of them is zero
>> and the algoirthm return true which is not correct.
>>
>> The utility in the case throw RuntimException
>> Interp point outside triangle
>> as expected.
>>
>> - Kelvin
>> -------------
>> Java 3D Team
>> Sun Microsystems Inc.
>>
>>
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