Resultant vector from three different location and normal vector of a plane

v1 = [0.000326  -0.00018  -0.00017 ];
v2 = [0.000365  -0.00016  -0.00017 ];
v3 = [-0.00015  -0.00018  -0.00039 ];
% compute resultant
R = v1 + v2 + v3
R = 1×3
1.0e-03 *

    0.5410   -0.5200   -0.7300
P1 = [2.53493  2.47587  2.47282 ];
P2 = [2.54089  2.48301  2.47585 ];
P3 = [2.4791  2.47583  2.46097 ];
% compute normal
N = cross(P2-P1,P3-P1)/norm(cross(P2-P1,P3-P1))
N = 1×3
   -0.2016   -0.2352    0.9508

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

William Rose 2022-9-30

在 MATLAB Online 中打开

@Miraboreasu,

That was a nice fast answer fron @Torsten.

In case it is onf interest to you (and I think it is of interest, based on your other posts):

The resultant vector is not aligned with the normal.

v1 = [0.000326  -0.00018  -0.00017 ];
v2 = [0.000365  -0.00016  -0.00017 ];
v3 = [-0.00015  -0.00018  -0.00039 ];
R = v1 + v2 + v3;                       % resultant
P1 = [2.53493 2.47587 2.47282 ];
P2 = [2.54089 2.48301 2.47585 ];
P3 = [2.47910 2.47583 2.46097 ];
N = cross(P2-P1,P3-P1)/norm(cross(P2-P1,P3-P1)); % unit normal

The angle between them (in degrees) is

a=acos(dot(R,N)/(norm( R)*norm(N)))*180/pi
a = 130.5708

An angle a=180-130.6 is equally valid, since the direction of N depends on the side order chosen when computing the cross product, and that choice seems arbitrary.

If you are computing Work = force * displacement = pressure * area * displacement, as you said in a separate thread, you should use the dot product (inner product):

Pr=1;                           %pressure
A=0.5*norm(cross(P2-P1,P3-P1)); %area
W=Pr*A*dot(N,R)                 %work (up to a sign)
W = -1.4265e-07

where Pr=pressure, A=area, N=unit normal, R=displacement. Pr and A are scalars, N and R are vectors. You will determine the sign for W based on your understanding of the physics.

请先登录，再进行评论。