Jump to content

arc from three points (vex)


Recommended Posts

Please, do you have a vex code for creating Arc from 3 points?
From time to time it would be useful to have it.

AndrewVK shared this C++ source here:

I could rewrite it as an exercise and then I will share it ... but if anybody has it done, it would be a great time saver.

Link to comment
Share on other sites

Tesan, thank you very much for your hip file! Please, do you anybody have also a solution for a circular arc?


Here is a screenshot from the file, for a future reference:



And the complete wrangle is:

// Convert the umap range from 0-1 to 0-1-0
float curveToHalf(
        float value;
        ) {

        return clamp(value, 0, 0.5) * clamp(fit(value, 0, 1, 1, 0) , 0, 0.5) * 4;
// Complement function
float complement(
        float value;
        ) {

        return 1 + (value * -1);
// Main interpolation function
float interpolate_curve(
        float value;
        string sign;
        ) {

        float pi_half = 1.57079632679;

        value = cos(value * pi_half);

        if ( sign == "-" )
                return 1 + (value * -1);
                return value;

// Get user parameters from the UI
string  sign            = ch("sign");
int             complement      = int(ch("complement"));
int             curve_half      = int(ch("curve_half"));

// Create our classic umap attribute ranging over the points from 0-1
float   umap = float(@ptnum) / (npoints(@OpInput1) - 1);

// Convert the umap range
if ( curve_half == 1)
        umap = curveToHalf(umap);

// Do the complement for the umap
if ( complement == 1)
        umap = complement(umap);

// Interpolate curve based on the parameters
@P.y = interpolate_curve(umap, sign);


Link to comment
Share on other sites

Hi Jiri,

to get started I just wrote a script that turns a bunch of lines with three points into circles:


It's basically calculating the intersection point and radius of rectangular vectors starting from the lines' midpoints. To be run in a primitive wrangle:

// https://stackoverflow.com/questions/10551555/need-an-algorithm-for-3d-vectors-intersection
function vector intersection(vector r1, r2, e1, e2){
    float u = dot(e1, e2);
    float t1 = dot(r2 - r1, e1);
    float t2 = dot(r2 - r1, e2);
    float d1 = (t1 - u * t2) / (1 - u * u);
    float d2 = (t2 - u * t1) / (u * u - 1);
    vector p1 = r1 + e1 * d1;
    vector p2 = r2 + e2 * d2;
    vector pos_center = (p1 + p2) * 0.5;
    return pos_center;

int pt_0 = primpoint(0, i@primnum, 0);
int pt_1 = primpoint(0, i@primnum, 1);
int pt_2 = primpoint(0, i@primnum, 2);

vector pos_A = point(0, 'P', pt_0);
vector pos_B = point(0, 'P', pt_1);
vector pos_C = point(0, 'P', pt_2);

vector mid_AB = (pos_A + pos_B) * 0.5;
vector mid_BC = (pos_B + pos_C) * 0.5;

vector dir_BA = normalize(pos_B - pos_A);
vector dir_BC = normalize(pos_B - pos_C);
vector dir_rect = normalize(cross(dir_BA, dir_BC));
vector dir_BA_rect = normalize(cross(dir_BA, dir_rect));
vector dir_BC_rect = normalize(cross(dir_BC, dir_rect));

vector pos_center = intersection(mid_AB, mid_BC, dir_BA_rect, dir_BC_rect);
float radius = distance(pos_center, pos_A);
int pt_add = addpoint(0, pos_center);
int circle_add = addprim(0, 'circle', pt_add);
matrix3 xform_circle = dihedral({0,0,1}, dir_rect);
scale(xform_circle, radius);
setprimintrinsic(0, 'transform', circle_add, xform_circle);

removeprim(0, i@primnum, 1);



  • Like 2
  • Thanks 2
Link to comment
Share on other sites

  • 2 years later...

If you take the first point as center, you can take the second point to define a radius (for example). The third point should have the same distance from center as the second point, or the arc will not intersect with this point.

You can derive the angle and the rotation axis from both direction (P1 - P0) and (P2 - P0). Now you can just rotate the first point along the angle.

If you a have guide axis, you can check if the rotation axis is conform (if not -> axis need to be inverted and the angle should be subtracted from 2*PI (360 degrees))


  • Like 1
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...