Seeking your kind help on a VEX question...
I have a vector, and I'd like to find the ortho axis which is nearest to it, and whether it faces in the positive or negative direction on that axis - ie: one of six results.
Can anyone suggest a VEX solution, or point me in the right direction? (pun happily accepted : )
With thanks.
Full Version: Which ortho axis is closest to this vector?
You could simply compare the distance to each ortho axis:
printf('------------ \n'); vector p = point(1, 'P', 0); addpoint(0, 0); addpoint(0, p); addprim(0, 'polyline', 0, 1); vector ort_axis = 0; float min_dist = 0; for (int i = 0; i < 6; ++i) { vector tmp = 0; tmp[i / 2] = (i % 2 * 2) - 1; float tmp_dist = distance(p, tmp); if (min_dist < tmp_dist) { min_dist = tmp_dist; ort_axis = floor(tmp * -1); } } printf('%d \n', ort_axis );
Hi,
You can use the dot product to do this:
You can use the dot product to do this:
vector v = normalize ( chv("vec") ); vector dirs [ ] = array ( {1, 0, 0}, {0, 1, 0 }, {0, 0, 1} ); float maxdv = -1; vector nearestdir = 0; for ( int i = 0; i < len ( dirs ); ++i ) { float dv = dot ( v, dirs[ i ] ); float adv = abs ( dv ); if ( adv > maxdv ) { maxdv = adv; nearestdir = dv > 0 ? dirs [ i ] : -dirs [ i ]; } } int pt = addpoint ( 0, nearestdir ); setpointattrib ( 0, "N", pt, nearestdir );
you can also try checking the longest value and that will be your axis
(I know that == with floats is questionable, but seems to work in VEX at least in this case)
(I know that == with floats is questionable, but seems to work in VEX at least in this case)
vector v = chv("vector"); float longest = max( abs(v) ); vector axis = {0,0,0}; for ( int i=0; i<3; i++){ if ( abs(v[i]) == longest ) { axis[i] = sign( v[i] ); break; } }
Gents - many thanks for your generosity in offering these thoughtful and top quality answers. I really appreciate it. They're both an excellent solution to the problem at hand - and a learning resource!