Volume formed by a moving triangle
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Hello,
A pressure ff (not force) is applied to the three points of a triangle. The triangle is moving during the time ΔtΔt (t2-t1), and I know the coordinates of each point, namely
t1: p1(x1,y1,z1),p2(x2,y2,z2),p3(x3,y3,z3)
t2: p1(xx1,yy1,zz1),p2(xx2,yy2,zz2),p3(xx3,yy3,zz3)
I also know the velocity as a vector for each point
t1: p1(v1x,v1y,v1z),p2(v2x,v2y,v2z),p3(v3x,v3y,v3z)
t2: p1(vv1x,vv1y,vv1z),p2(vv2x,vv2y,vv2z),p3(vv3x,vv3y,vv3z)
I know this is not 100% right expression, but I want to know how much energy this pressure, p, bring to the system
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1133775/image.png)
6 commentaires
Torsten
le 3 Oct 2022
Modifié(e) : Torsten
le 3 Oct 2022
Is the normal to the triangle always equal to the direction in which the triangle is swept ?
Otherwise, you will have to integrate. Something like
V(t) = A*integral_{t'=0}^{t'=t} dot(n(t'),v(t')) dt'
where A is the area of the triangle, n(t') is the (unit) normal vector to the triangle and v(t') is the velocity vector at time t'.
Réponse acceptée
Chunru
le 23 Sep 2022
Modifié(e) : Chunru
le 23 Sep 2022
% initial triangle
p1 = [0, 0, 0]; p2 = [3, 0, 0]; p3 = [0 4 0];
% the velocity vector should be specified (instead of final triangle since
% final triangle coordinates cannot be arbitrary if volume is going to be
% computed)
v = [0 0 1];
t = 3;
cbase = .5*cross(p2-p1, p3-p1)
vol = dot(cbase, v*t)
% If you know p1, p2, p3 and v vs t, you can consider to use the above
% calculation for each time interva, where the volume can be approximated
% by using the base area and the velocity vector.
3 commentaires
Chunru
le 23 Sep 2022
Last statement should be vol (while v is a vector defined earlier). See update.
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