- #1
MikeHoulder
- 6
- 0
Perhaps Tensor Calculus holds the answer; but I just can't justify the time for studying that as I know nothing of it.
The end objective is to calculate the mass moment of inertia of the yellow solid parallelepiped about rectangular axes through its centre of mass as in the diagram here.
The determining inputs are:
the length of the parallelepiped projected onto the X axis, length l
the angle of slant, angle a
the slant height of the parallepiped, length h
the depth of the parallelepiped, length d
the volume density which is constant
Specimen inputs are:
length l = 10
slant angle a = 10 degrees
length h = 2
depth d = 6
volume density = k
Using the standard formula for the whole rectangular prism about its centre of mass
Ixc = M 1/12 (d2 + l2 tan2a), Iyc, Izc similarly
The whole rectangular prism is composed of the target parallelepiped and two triangular wedge shapes.
These wedge shapes are symmetrical about the rectangular centre of mass. So only one of them need be considered.
I could not find any standard formula for a triangular wedge shape. But if the mass moments for this could be found about any rectangular axes, using the parallel and perpendicular axis theorems, the mass moments could be found about the Xc, Yc, Zc axes.
Trying to derive these values for the wedge, I always end up with the area moment of a triangle, m.l2/6.
And any attempt to find the mass moment produces inconsistent values.
Please, please, will some kind person sort me out.
Mike
The end objective is to calculate the mass moment of inertia of the yellow solid parallelepiped about rectangular axes through its centre of mass as in the diagram here.
The determining inputs are:
the length of the parallelepiped projected onto the X axis, length l
the angle of slant, angle a
the slant height of the parallepiped, length h
the depth of the parallelepiped, length d
the volume density which is constant
Specimen inputs are:
length l = 10
slant angle a = 10 degrees
length h = 2
depth d = 6
volume density = k
Using the standard formula for the whole rectangular prism about its centre of mass
Ixc = M 1/12 (d2 + l2 tan2a), Iyc, Izc similarly
The whole rectangular prism is composed of the target parallelepiped and two triangular wedge shapes.
These wedge shapes are symmetrical about the rectangular centre of mass. So only one of them need be considered.
I could not find any standard formula for a triangular wedge shape. But if the mass moments for this could be found about any rectangular axes, using the parallel and perpendicular axis theorems, the mass moments could be found about the Xc, Yc, Zc axes.
Trying to derive these values for the wedge, I always end up with the area moment of a triangle, m.l2/6.
And any attempt to find the mass moment produces inconsistent values.
Please, please, will some kind person sort me out.
Mike