## 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}

Object | Area | Volume | x | y | z | xV | yV | zV |

Rectangle | 4 | 8 | -3 | 1 | 0 | -24 | 8 | 0 |

Triangle | 6 | 12 | 5.3333 | 2 | 0 | 63.9996 | 24 | 0 |

Total | 10 | 20 | 39.9996 | 32 | 0 |

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

Object | Area | Volume | Density | Mass | x | y | z | xM | yM | zM |

Rectangle | 4 | 8 | 2 | 16 | -3 | 1 | 0 | -48 | 16 | 0 |

Triangle | 6 | 12 | 2 | 24 | 5.3333 | 2 | 0 | 127.9992 | 48 | 0 |

Total | 10 | 20 | 40 | 79.9992 | 64 | 0 |

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:

Object | Area | Volume | Density | Mass | x | y | z | xM | yM | zM |

Rectangle | 4 | 32 | 5 | 160 | -3 | 1 | 4 | -480 | 160 | 640 |

Triangle | 6 | 18 | 15 | 270 | 5.3333 | 2 | -6 | 1,439.991 | 540 | -1,620 |

Total | 430 | 959.991 | 700 | -980 |

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.