Volume formed by a moving triangle

4 次查看（过去 30 天）

Miraboreasu 2022-9-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1810475-volume-formed-by-a-moving-triangle

编辑： Torsten 2022-10-3

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

6 个评论
显示 4更早的评论隐藏 4更早的评论

Miraboreasu 2022-10-3

编辑：Miraboreasu 2022-10-3

@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 2022-10-3

编辑：Torsten 2022-10-3

在 MATLAB Online 中打开

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'.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Chunru 2022-9-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1810475-volume-formed-by-a-moving-triangle#answer_1059095

编辑：Chunru 2022-9-23

在 MATLAB Online 中打开

% 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.