问题
I'm researching how to find the inertia for a 2D shape. The contour of this shape is meshed with several points, the x and y coordinate of each point is already known.
I know the expression of Ixx, Iyy and Ixy but the body has no mass. How do I proceed?
回答1:
For whatever shape you have, you need to split it into triangles and handle each triangle separately. Then in the end combined the results using the following rules
Overall
% Combined total area of all triangles
total_area = SUM( area(i), i=1:n )
total_mass = SUM( mass(i), i=1:n )
% Combined centroid (center of mass) coordinates
combined_centroid_x = SUM( mass(i)*centroid_x(i), i=1:n)/total_mass
combined_centroid_y = SUM( mass(i)*centroid_y(i), i=1:n)/total_mass
% Each distance to triangle (squared)
centroid_distance_sq(i) = centroid_x(i)*centroid_x(i)+centroid_y(i)*centroid_y(i)
% Combined mass moment of inertia
combined_mmoi = SUM(mmoi(i)+mass(i)*centroid_distance_sq(i), i=1:n)
Now for each triangle.
Consider the three corner vertices with vector coordinates, points A, B and C
a=[ax,ay]
b=[bx,by]
c=[cx,cy]
and the following dot and cross product (scalar) combinations
a·a = ax*ax+ay*ay
b·b = bx*bx+by*by
c·c = cx*cx+cy*cy
a·b = ax*bx+ay*by
b·c = bx*cx+by*cy
c·a = cx*ax+cy*ay
a×b = ax*by-ay*bx
b×c = bx*cy-by*cx
c×a = cx*ay-cy*ax
The properties of the triangle are (with t(i) the thickness and rho the mass density)
area(i) = 1/2*ABS( a×b + b×c + c×a )
mass(i) = rho*t(i)*area(i)
centroid_x(i) = 1/3*(ax + bx + cx)
centroid_y(i) = 1/3*(ay + by + cy)
mmoi(i) = 1/6*mass(i)*( a·a + b·b + c·c + a·b + b·c + c·a )
By component the above are
area(i) = 1/2*ABS( ax*(by-cy)+ay*(cx-bx)+bx*cy-by*cx)
mmoi(i) = mass(i)/6*(ax^2+ax*(bx+cx)+bx^2+bx*cx+cx^2+ay^2+ay*(by+cy)+by^2+by*cy+cy^2)
Appendix
A little theory here. The area of each triangle is found using
Area = 1/2 * || (b-a) × (c-b) ||
where × is a vector cross product, and || .. || is vector norm (length function).
The triangle is parametrized by two variables t and s such that the double integral A = INT(INT(1,dx),dy) gives the total area
% position r(s,t) = [x,y]
[x,y] = [ax,ay] + t*[bx-ax, by-zy] + t*s*[cx-bx,cy-by]
% gradient directions along s and t
(dr/dt) = [bx-ax,by-ay] + s*[cx-bx,cy-by]
(dr/ds) = t*[cx-bx,cy-by]
% Integration area element
dA = || (dr/ds)×(dr/dt) || = (2*A*t)*ds*dt
%
% where A = 1/2*||(b-a)×(c-b)||
% Check that the integral returns the area
Area = INT( INT( 2*A*t,s=0..1), t=0..1) = 2*A*(1/2) = A
% Mass moment of inertia components
/ / / | y^2+z^2 -x*y -x*z |
I = 2*m*| | | t*| -x*y x^2+z^2 -y*z | dz ds dt
/ / / | -x*z -y*z x^2+y^2 |
% where [x,y] are defined from the parametrization
来源：https://stackoverflow.com/questions/41592034/computing-tensor-of-inertia-in-2d