Volume formed by a moving triangle

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

[Walter Roberson]

The velocities are not the same, but are they constant? If so you can express the positions with a simple parametric formula, calculate the parametric area of the triangle, and integrate that area over time.

If the velocities are not constant then in theory you can still do that, but the area might no longer be a simple parametric formula, and you might run into problems with positions overlapping previous positions, leading you to need to make definitions about what it means for a volume to be swept out.

[Miraboreasu]

More detail has been added

[Chunru]

You can use the answer provided below for each time interval, assuming that points are not moving back to the volume formed earlier.

[Bruno Luong]

Not clear why you ask for volume or energy. Energy is the work,

integral of the force x distance = integral of force x velocity dt = integral pressure / area * v dt

Volume is simmpy

integral area * v dt

They doesn't seem equivalent at all to me.

[Miraboreasu]

@Walter Roberson Let's make a simplication regard to the velocity, can you please show me how to "express the positions with a simple parametric formula, calculate the parametric area of the triangle, and integrate that area over time."

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)

[Torsten]

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'.

Answer 1

[Chunru]

% 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]

Thanks, but I didn't get it, velocity should be vector right? this v is which point?

[Chunru]

Last statement should be vol (while v is a vector defined earlier). See update.

[Miraboreasu]

How can I make sure integral of pressure (force) and velocity are the same direction?