Volume formed by a moving triangle

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Miraboreasu on 23 Sep 2022
Edited: Torsten on 3 Oct 2022
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
Torsten on 3 Oct 2022
Edited: Torsten on 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'.

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Accepted Answer

Chunru on 23 Sep 2022
Edited: Chunru on 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)
cbase = 1×3
0 0 6
vol = dot(cbase, v*t)
vol = 18
% 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.
Miraboreasu on 23 Sep 2022
How can I make sure integral of pressure (force) and velocity are the same direction?

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