What is the center of gravity for humans?
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I am not 100% sure if I understand your question but this might clarify the concept of center of mass (COM) of an object. It is defined by the following equation $$\int_{\mathbb R^3} \rho(\vec r) \cdot (\vec r- \vec r_{com}) \, \mathrm d V =\vec 0 $$ where $\rho(\vec r)$ is the density of the object at a position $\vec r$ from the origin. If you modify this integral you get: $$ \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r \, \mathrm d V = \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r_{com} \, \mathrm d V = \vec r_{com} \int_{\mathbb R^3} \rho(\vec r) \cdot \, \mathrm d V = \vec r_{com }\cdot M $$ $$ \implies \vec r_{com} = \frac 1 M \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r \, \mathrm d V $$ where $M=\int_{\mathbb R^3} \rho(\vec r) \, \mathrm d V$ is the total mass of the object This integral tells you to do one thing: Take density-weighed average of the position vectors. You see however there is one (technical) problem with this definition. The equation tells you if you give me $\rho(\vec r)$ at every point in space. I'll give you $\vec r_{com}$. For a complex object like a human being $\rho(\vec r)$ is a very very complicated function. Your muscles have different density from your bones and your fat tissue etc. There is one more problem and that is you don't always have the same fat tissue or whatever at the same position for every person. Thus calculating that integral is for all practical purposes impossible for a human being. However the nature doesn't care about the complexity of the object. There is still a unique COM at each fraction of time for each person.** Let's take a very rough model of a human being. I.e. a box with height $h$ and a square base with side length $a$. Let the density constant $\rho(\vec r) = \rho$ and zero outside of the box. Let's choose the origin to be on the bottom corner of the box. $$ M \cdot \vec r_{com} = \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r \, \mathrm d V = \iiint\limits_{\text{Box}} \rho \cdot \vec r \, \mathrm d V = \int_{z=0}^h \int_{y=0}^{a} \int_{x=0}^a \rho \cdot \left( \begin{smallmatrix} x \\ y \\ z\end{smallmatrix} \right) \, \mathrm dx \mathrm dy \mathrm dz$$ $$ \implies M \cdot \vec r_{com,\ standing} = \frac \rho 2 \left( \begin{smallmatrix} a^3h \\ a^3h \\ a^2h^2\end{smallmatrix} \right) \implies \vec r_{com} = \frac 1 2 \left( \begin{smallmatrix} a \\ a \\ h\end{smallmatrix} \right)$$ Suppose you have raised your hands above. We can very crudely model this a box wiht a base of a rectangle with sides $a/2$ and $a$ and height $2h$. Note that the volume and thus the mass of the box is the same. If you calculate the integral you'll get $$\vec r_{com,\ hands\ raised} = \left( \begin{smallmatrix} a/2 \\ a/4 \\ h\end{smallmatrix} \right) \neq \vec r_{com,\ standing}$$ Note that in both cases you have a unique center of mass. However the COM of a person raising hands is clearly different from a person just standing. What you can do however is to measure the COM experimentally and get a rough idea of where the COM of a human being might be in certain positions. *In order for this to make physical sense $\rho(\vec r) =0 $ for all $\lVert \vec r \rVert > R$ i.e. your object has to end somewhere. ** Your COM is not the same when your stomach is full and empty. That is why you have to take a small fraction of time to make sure that there is nothing funny going on.
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