Deleting the shorter edge of a rectangle
2021年5月6日 11:25
Hello dear Houdini artists,
I'm relatively new to Houdini and still struggling with some of the more basic concepts.
My problem is the following: I have a rectangle (created from a bound node) and I want to delete either the shorter or the longer edge of that rectangle, so that I can then create a vector that shows me the orientation of that rectangle.
How would I go about doing that? I have measured the lengths of the rectangle with a convert line node, but how can I compare the 2 lengths and remove one or the other?
Is there a better way to go about this?
I would really appreciate some help with this.
Best regards,
Julian
Attachments:
rectangle_problem.JPG (24.8 KB)
tamte Member
8837 posts Joined: 7月 2007 Offline
2021年5月7日 9:55
julianbragagna You can then use Sort SOP to sort prims by your length attribute
And then Blast SOP to keep or delete primitives: 0 1
Tomas Slancik
tamte Member
8837 posts Joined: 7月 2007 Offline
2021年5月7日 9:56
julianbragagna You can then use Sort SOP to sort prims by your length attribute
And then Blast SOP to keep or delete primitives: 0 1
As those will be the shortest 2
Tomas Slancik
2021年5月7日 10:19
tamte That's a great solution! How would I go about applying this on scale, for an arbitrary number of rectangles?
tamte Member
8837 posts Joined: 7月 2007 Offline
2021年5月7日 10:46
easiest would be to run it through For Each Block that iterates per rectangle
Tomas Slancik
2021年5月7日 10:49
julianbragagna using the rest length prim attribute generating form the convertline can sort multiple rectangles with the same blast or delete node
if u precisely need per rectangle sort then go with tamte suggestion
Edited by Houdini Obsession - 2021年5月7日 10:59:23
Attachments:
use rest length from convert line.png (575.1 KB)
Houdini Fx Artist (Build)
tamte Member
8837 posts Joined: 7月 2007 Offline
2021年5月7日 16:06
Naga Pavan Only if no rectangle has larger edge smaller than other rectangle's smaller one
Tomas Slancik
2021年5月8日 7:28
julianbragagna To obtain the short direction you could unroll your rectangles and subtract their minimal position from the primitive's center:
v@dir = normalize (minpos (1 , v@P ) - v@P );
To obtain the long direction you would calculate the cross product of the short direction and the primitive's normal:
vector nml = prim_normal (0 , i@primnum , vector (0.0 ));
v@perp = normalize (cross (v@dir , nml ));
Attachments:
rect_dir.hipnc (96.9 KB)
2021年5月8日 7:53
Alternatively you could iterate through the primitive's points, and compare the subtractions of their point positions:
int pts [] = primpoints (0 , i@primnum );
vector pos_0 = point (0 , 'P' , pts [0 ]);
vector pos_1 = point (0 , 'P' , pts [1 ]);
vector pos_2 = point (0 , 'P' , pts [2 ]);
vector dir_10 = pos_1 - pos_0 ;
vector dir_21 = pos_2 - pos_1 ;
v@dir_short ;
if (length (dir_10 ) > length (dir_21 )){
v@dir_short = normalize (dir_21 );
}
else{
v@dir_short = normalize (dir_10 );
}
Attachments:
rect_dir_pts.hipnc (100.6 KB)
2021年5月9日 1:47
tamte yeah, if rectangles vary in size it fails, I realized sorting with the attribute and deleting 0 1 per connectivity is safer
Konstantin Magnus thank you sharing your sorting methods
Edited by Houdini Obsession - 2021年5月9日 11:11:03
Houdini Fx Artist (Build)
