View Full Version : A very simple task

shineon00

04-08-2003, 09:02 AM

Hello people,

I have a simple question. I am trying to optimize a code and I need a c or c++ library which can make simple geometric calculations. There is no need for this library to employ a user interface by all means, only functions with a good return type is enough.

An example can be:

A function that takes the vertices of a ploygon and a line. It returns the intersection points of these two objects.

This example is actually my starting point.

Thanks in advance,

jebus

04-08-2003, 11:27 AM

a little OT, but here goes:

A

|\

| \

| \

L1___| P_B____L2

| /

| /

| /

|/

C

given: triangle ABC

line L1->L2

find: point P (if it exists)

this implementation assumes all vector

operations (+-*/= operators, length, dot

product, etc.) are already defined. the dot

product operator is defined to use '|'. a

triangle is defined to have an array of 3

vectors for each vertex and a fourth vector

for the normal.

bool intersection(vector *l1, vector *l2, triangle *tri, vector *intersection)

{

vector dir; // direction of line

vector len; // length of line

vector L; // vector from line l1 to

the triangle

vector dist; // distance from l1 to plane

float per; // how much of the line is

in front of the polygon

dir = *l2 - *l1;

len = dir.length();

if (fabsf(len) < 0.001)

{

return false;

}

L = tri->vec[0] - *l1;

dist = L | tri->normal;

per = dist / len;

if (per < 0.0f)

{

// line is completely behind the

polygon

return false;

}

if (per > 1.0f)

{

// line is completely in front of

polygon

return false;

}

*intersection = *l1 + dir * per;

// at this point, the line l1->l2 has been

proven to intersect the plane that our

triangles lies on. the rest of this

function determines if the intersection

point lies inside the triangle.

// vectors from the interseciton point

to each vertex in the triangle

vector v1, v2, v3;

float angle_total = 0;

v1 = *intersection - tri->vec[0];

v2 = *intersection - tri->vec[1];

v3 = *intersection-tri->vec[2];

v1.normalize();

v2.normalize();

v3.normalize();

// if the sum of the angles between these

3 vectors == 360° (2 * PI radians) then

the point is on the triangle

angle_total += acos(v1 | v2);

angle_total += acos(v2 | v3);

angle_total += acos(v3 | v1);

if (fabsf(angle_total - (PI * 2)) < 0.001)

{

return true;

}

else

{

return false;

}

}

hope this works http://www.opengl.org/discussion_boards/ubb/biggrin.gif the if checks on the

fabsf calls may need tweaking, depending on

how tolerant you want your numbers. when

working with floats, it's good practice to

set some value to act as zero and check

against it. very rarely will your floats end

up being exactly zero.

jebus

[This message has been edited by jebus (edited 04-08-2003).]

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