How to obtain Barycentric Coordinates of a point in a triangle?
19 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
WellEatenPanda
am 2 Aug. 2013
Beantwortet: Atia Najafi
am 19 Sep. 2018
Hello,
I'm trying to find barycentric coordinates of a point p in the triangle with coordinates shown in matrix P, where coordinates of the point are in the last row of the matrix. I have used function cartesianToBarycentric to do this, but I get B = 0 0 1 as the output, which is the barycentric coordinate of the third vertex. There is my code:
P = [-1.17300000000000 -0.952000000000000
-0.199000000000000 1.04800000000000
1.05800000000000 -0.952000000000000
-0.100000000000000 -0.443000000000000]
T = [1 4 3
1 4 2
3 4 2]
TR = triangulation(T,P)
ti = 1
PC = TR.Points(TR.ConnectivityList(1,3),:)
B = cartesianToBarycentric(TR, ti, PC)
Output: B = 0 0 1
I have tried changing the second parameter of the function PC = TR.ConnectivityList(1,3) from 1 to 3, but I still get the initial barycentric coordinates of the three corners. Zero and four doesn't work for the second parameter even though there are 4 vertices.
Can anyone help me with this as I'm completely new with MATLAB. I would be grateful.
0 Kommentare
Akzeptierte Antwort
Roger Stafford
am 3 Aug. 2013
Bearbeitet: Roger Stafford
am 3 Aug. 2013
I've read over the 'cartesianToBarycentric' documentation. Here's a couple of observations. a) Your matrix T does not contain the triangle [1 2 3], so 'cartesianToBarycentric' cannot reference it in finding barycentric coordinates. You need to revise T to contain that triangle before calling on 'triangulation' and set ti to that triangle. b) For your point in PC, in TR.ConnectivityList(1,3) you selected the third vertex of the first triangle, which is ID 3 in P, so naturally you will get barycentric coordinates of 0 0 1. You need to either give that fourth point of P directly without referencing TR.Points or else reference it as the ID = 4 vertex in any one of those first three triangles of T - that is, you could write TR.ConnectivityList(1,2) since that is ID 4 in your present T, for example. Here's what I would try:
T = [1 4 3
1 4 2
3 4 2
1 2 3]
TR = triangulation(T,P);
ti = 4; % The triangle in T you are referencing
PC = TR.Points(TR.ConnectivityList(1,2),:) % The point in question
or
PC = P(4,:); % Give the point directly without using TR.Points
I hope all this makes sense. This is something I have not worked with.
Weitere Antworten (2)
Roger Stafford
am 3 Aug. 2013
I am not familiar with the use of the 'cartesianToBarycentric' function, but I do know how to find the barycentric coordinates of a point in a triangle. In your case where the vertices are the first three rows of P and the point in question the fourth row, it would be the solution to the following three linear equations in three unknowns:
-1.173*b1 - 0.199*b2 + 1.058*b3 = -0.1
-0.952*b1 + 1.048*b2 - 0.952*b3 = -0.443
b1 + b2 + b3 = 1
In matlab this can be solved using the matrix division operator:
b = [P(4,:),1]/[P(1:3,:),ones(3,1)];
and the solution is
b = 0.37565822501121 , 0.25450000000000 , 0.36984177498879
2 Kommentare
Atia Najafi
am 19 Sep. 2018
Hi Roger Stafford can you please elaborate on your solution? I need to implement this function in simulink, so have to write 'triangulation' and ''cartesianToBarycentric' from scratch. I look forward to hearing back from you. Atia
0 Kommentare
Siehe auch
Kategorien
Mehr zu Surface and Mesh Plots finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!