## Adding Center of Gravity’s

Let’s Start off with a 2D problem, we need to know the current areas of both shapes and the current CoG’s for both shapes:

We can then do some basic easy math to figure out the combined CoG:

\bold{Step1} \\
Total Area = 4 + 6 = 10 \\
\bold{Step2} \\
Rectangle \space R\space \overline{x}A = 4 * -3 = -12 \\
Triangle \space T\space \overline{x}A = 6 * 5.3333 = 32 \\
R\space \overline{x}A + T\space \overline{x}A = 20 \\
\bold{Step3} \\
Rectangle \space R\space \overline{y}A = 4 * 1 = 4 \\
Triangle \space T\space \overline{y}A = 6 * 2 = 12 \\
R\space \overline{y}A + T\space \overline{y}A = 16 \\
\bold{Step4} \\
\overline{X} = (20/10) = 2 \\
\overline{Y} = (16/10) = 1.6 \\


To summarize, sum (X . Area) and then divided by the sum of the areas, repeat for Y and Z. This equation can be modified for each magnitude, replace Area with Volume, or Mass.

### Volume CoG Addition

So let’s assume that the same problem has symmetrical thickness of +1 -1, we need to calculate one more dimension:

\begin{aligned}
\overline{X} =\dfrac{\sum\overline{x}iAi}{\sum Ai} \\
&
\overline{Y} =\dfrac{\sum\overline{y}iAi}{\sum Ai} \\
&&
\bold{\overline{Z} =\dfrac{\sum\overline{z}iAi}{\sum Ai}} \\
\end{aligned}

So we simply change Area(A) to Volume(V)

\begin{aligned}
\overline{X} =\dfrac{\sum\overline{x}iVi}{\sum Vi} \\
&
\overline{Y} =\dfrac{\sum\overline{y}iVi}{\sum Vi} \\
&&
\overline{Z} =\dfrac{\sum\overline{z}iVi}{\sum Vi} \\
\end{aligned}

In this case since the third dimension is parallel we know that its value will be 0, but lets carry on and see:

\bold{Step1} \\
Total Volume = 8 + 12 = 20 \\
\bold{Step2} \\
Rectangle \space R\space \overline{x}V = 8 * -3 = -24 \\
Triangle \space T\space \overline{x}V = 12 * 5.3333 = 63.9996 \\
R\space \overline{x}V + T\space \overline{x}V = 39.9996 \\
\bold{Step3} \\
Rectangle \space R\space \overline{y}V = 8 * 1 = 8 \\
Triangle \space T\space \overline{y}V = 12 * 2 = 24 \\
R\space \overline{y}V + T\space \overline{y}V = 32 \\
\bold{Step4} \\
Rectangle \space R\space \overline{z}V = 8 * 0 = 0 \\
Triangle \space T\space \overline{z}V = 12 * 0 = 0 \\
R\space \overline{z}V + T\space \overline{z}V = 0 \\
\bold{Step5} \\
\overline{X} = (39.9996/20) = 2 \\
\overline{Y} = (32/20) = 1.6 \\
\overline{Z} = (0/20) = 0 \\


So we were right the third dimension is 0, and the CoG has not changed from the 2D problem, because we added material symmetrically about the mid plane.

### Mass CoG Addition

So let’s repeat this time for Mass

So we simply change Volume(V) to Mass(M)

\begin{aligned}
\overline{X} =\dfrac{\sum\overline{x}iMi}{\sum Mi} \\
&
\overline{Y} =\dfrac{\sum\overline{y}iMi}{\sum Mi} \\
&&
\overline{Z} =\dfrac{\sum\overline{z}iMi}{\sum Mi} \\
\end{aligned}

We perform the exact same math again:

\bold{Step1} \\
Total Mass = 16 + 24 = 40 \\
\bold{Step2} \\
Rectangle \space R\space \overline{x}V = 16 * -3 = -48 \\
Triangle \space T\space \overline{x}V = 24 * 5.3333 = 127.9992 \\
R\space \overline{x}V + T\space \overline{x}V = 79.9996 \\
\bold{Step3} \\
Rectangle \space R\space \overline{y}V = 16 * 1 = 16\\
Triangle \space T\space \overline{y}V = 24 * 2 = 48 \\
R\space \overline{y}V + T\space \overline{y}V = 64 \\
\bold{Step4} \\
Rectangle \space R\space \overline{z}V = 16 * 0 = 0 \\
Triangle \space T\space \overline{z}V = 24 * 0 = 0 \\
R\space \overline{z}V + T\space \overline{z}V = 0 \\
\bold{Step5} \\
\overline{X} = (79.9996/40) = 2 \\
\overline{Y} = (64/40) = 1.6 \\
\overline{Z} = (0/40) = 0 \\

Pretty simple right, we know how to handle uniform shapes, with uniform density, lets try something non-uniform.

Let’s assume the Rectangle has a thickness 8 where is midplane is at +4 with a density of 5 and the triangle has a thickness of 3 where it’s midplane is at -6 with a density of 15:

Since we have all the information in the table lets just do the last step:

\overline{X} = (959.991/430) = 2.2325 \\
\overline{Y} = (700/430) = 1.6279 \\
\overline{Z} = (-980/430) = -2.2790 \\

You can download an excel spreadsheet here with the proof.