topic 回复： Algorithm: counter clockwise 1, 2, 3, 4, 5 in ObjectARX Forum

Algorithm: counter clockwise 1, 2, 3, 4, 5

463017170 — Wed, 07 Jul 2021 14:45:48 GMT

All lines have the same short point A. find the other end point and sort it anticlockwise 1, 2, 3, 4, 5

回复： Algorithm: counter clockwise 1, 2, 3, 4, 5

463017170 — Mon, 12 Jul 2021 01:20:46 GMT

My friend, who can provide me with an algorithm.

回复： Algorithm: counter clockwise 1, 2, 3, 4, 5

nick83 — Mon, 12 Jul 2021 07:02:21 GMT

first of all, you need to convert Decart (Cartesian) coordinates (X,Y,Z) to Spherical coordinates (r,phi, theta). Spherical - is a 3D variant of polar coordinates.
then, you have to sort your array of spherical coordinates by "phi" (phi[i] < phi[i+1] = counterclockwise. phi[i] > phi[i+1] = clockwise. if you have two points with the same phi, compare "r" and "theta" values).
and when sort is done, you can convert polar coordinates of your sorted array back to decart.

/* DECART AcGePoint3d::x - X AcGePoint3d::y - Y AcGePoint3d::z - Z SPHERICAL AcGePoint3d::x - r AcGePoint3d::y - phi AcGePoint3d::z - theta */ AcGePoint3d Decart2Spherical(AcGePoint3d c) { AcGePoint3d p; p.x=sqrt(c.x*c.x+c.y*c.y+c.z*c.z); p.y=acos(c.x/sqrt(c.x*c.x+c.y*c.y))*(c.y>=0?1:-1); p.z=acos(c.z/p.x); return p; } AcGePoint3d Spherical2Decart(AcGePoint3d p) { decart d; d.x=p.x*cos(p.y)*sin(p.z); d.y=p.x*sin(p.y)*sin(p.z); d.z=p.x*cos(p.z); return d; }

almost forgot. may be, before convertation you will need to subtract "A" coords from the rest points, to make "A" a zero point and append "A" after Spheriacal2Decart. In most cases it's unnecessary, but who knows 🙂

回复： Algorithm: counter clockwise 1, 2, 3, 4, 5

463017170 — Tue, 13 Jul 2021 14:47:35 GMT

Thank you for your help. It seems that this algorithm doesn't work. Can you give me more details?

回复： Algorithm: counter clockwise 1, 2, 3, 4, 5

nick83 — Wed, 14 Jul 2021 12:55:00 GMT

Hm... i have found 2 mistakes in my post:

1. i forgot to change "decart" variable to "AcGePoint3d" variable at Spherical2Decart function, but it's not hard to understand it looking through the rest code and change it at your side.

2. phi[i] < phi[i+1] = clockwise. phi[i] > phi[i+1] = counterclockwise, that will be correct. this mistake can be fixed at your side too, by changing a sign ">" to "<" after code running and analyze the result.

that's all. so, if "this algorithm doesn't work" can you provide your code?

I've created a sample project for demonstration. May be i did something wrong, but i cant attach a screencast video. So, the link to it is below

https://knowledge.autodesk.com/ru/community/screencast/8b3bdff1-82c2-478e-b9e7-2481ddc03c7c

回复： Algorithm: counter clockwise 1, 2, 3, 4, 5

463017170 — Wed, 14 Jul 2021 23:31:32 GMT

If the plane is rotated to this point, the result of the algorithm is wrong.

回复： Algorithm: counter clockwise 1, 2, 3, 4, 5

nick83 — Thu, 15 Jul 2021 13:08:27 GMT

@463017170 написал (-а):
If the plane is rotated to this point, the result of the algorithm is wrong.

no one told it would be easy :).

you have to transform coordinates according to current view before decart-polar and after polar-decart converting (reverse converting is necessary, if you need points in WCS).