Finding the volume-weighted average centroid of a polyhedron

Question

In order to find the centroid of a polyhedral element (for finite volume method) bounded by a set of vertices, we follow the following procedure (from my textbook):

1 - Calculate the geometric centre $(x_g, y_g, z_g)$ of the polyhedron by summing all the vertices(points) and dividing by their total number, where:

$$ x_g = \frac{1}{n}\sum_n x_i$$

2 - Divide the polyhedron into a set of sub-pyramids, with the geometric centre as the apex with each surface polygon as the base.

3 - Calculate the centroid of each sub-pyramid, measured from the centroid of the base, is situated at 1/4 of the line joining the centroid of the base to the apex of the pyramid.

4 - And finally, the volume-weighted average can be calculated.

The part that I don't understand is the part of calculating the centroid of the sub-pyramids, my textbook uses the following equation:

$$ (x_{CE})_{pyramid} = 0.75(x_{CE})_f + 0.25(x_G)_{pyramid} $$

where $x_{CE}$ is the centroid of the pyramid we want to calculate, $(x_{CE})_f$ if the centroid of the polygonal base and $x_g$ is the geometric centre of the pyramid.

So, how does point number 3 relates to the previous formula?

Answer 1

$$ (x_{CE})_{pyramid}=(x_{CE})_f+{1\over4}(x_{G}-(x_{CE})_f)={3\over4}(x_{CE})_f+{1\over4}x_{G}. $$

Answer 2

Let's call the position vector of the apex of the pyramid $\vec{G}$. Let's call the position vector of the base's center of mass $\vec{K}$. In addition, let's call the height of the pyramid $h$ and the area of the base $A$. The volume of the pyramid is then $V=\frac{1}{3}Ah$. Consider the the position vector $\vec{P}=\vec{G} + \frac{\lambda}{h} (\vec{K}-\vec{G})$, for $0\leq \lambda \leq h$. The area of the horizontal cross section of the pyramid at $\vec{P}$ has area $A_{\vec{P}}=\left(\frac{\lambda}{h}\right)^2A$. To calculate the center of mass of the pyramid, we can calculate the following integral:

\begin{align} \frac{1}{V}\int_0^hA_{\vec{P}} \vec{P}d\lambda &= \frac{3}{Ah}\int_0^h\left(\frac{\lambda}{h}\right)^2 A\left(\vec{G} + \frac{\lambda}{h} (\vec{K}-\vec{G})\right)d\lambda\\ &=\frac{3}{Ah}\int_0^h\left[\left(\frac{\lambda}{h}\right)^2 A\vec{G} + \left(\frac{\lambda}{h}\right)^3 A(\vec{K}-\vec{G})\right]d\lambda\\ &=\frac{3}{Ah}\left[\frac{h^3}{3h^2} A\vec{G} + \frac{h^4}{4h^3} A(\vec{K}-\vec{G})\right]\\ &=\vec{G}+\frac{3}{4} (\vec{K}-\vec{G})\\ &=\frac{1}{4}\vec{G}+\frac{3}{4}\vec{K} \end{align}