broggesy9

2022-05-02

How do you find the midpoint of several 3D points?

Dexter Conner

Beginner2022-05-03Added 15 answers

Step 1

If you are using a computer graphics system that uses homogeneous coordinates (4 values per 3D point) then you can just add all the components together.

For example

$\left(\begin{array}{c}{x}_{1}\\ {y}_{1}\\ {z}_{1}\\ 1\end{array}\right)+\left(\begin{array}{c}{x}_{2}\\ {y}_{2}\\ {z}_{2}\\ 1\end{array}\right)+\left(\begin{array}{c}{x}_{3}\\ {y}_{3}\\ {z}_{3}\\ 1\end{array}\right)=\left(\begin{array}{c}{x}_{1}+{x}_{2}+{x}_{3}\\ {y}_{1}+{y}_{2}+{y}_{3}\\ {z}_{1}+{z}_{2}+{z}_{3}\\ 3\end{array}\right)$

The cartesian coordinates of the result is the "mid-point" (also known as the barycenter).

$\left(\begin{array}{c}\frac{{x}_{1}+{y}_{1}+{z}_{1}}{3}\\ \frac{{x}_{2}+{y}_{2}+{z}_{2}}{3}\\ \frac{{x}_{3}+{y}_{3}+{z}_{3}}{3}\end{array}\right)$

If you are using a computer graphics system that uses homogeneous coordinates (4 values per 3D point) then you can just add all the components together.

For example

$\left(\begin{array}{c}{x}_{1}\\ {y}_{1}\\ {z}_{1}\\ 1\end{array}\right)+\left(\begin{array}{c}{x}_{2}\\ {y}_{2}\\ {z}_{2}\\ 1\end{array}\right)+\left(\begin{array}{c}{x}_{3}\\ {y}_{3}\\ {z}_{3}\\ 1\end{array}\right)=\left(\begin{array}{c}{x}_{1}+{x}_{2}+{x}_{3}\\ {y}_{1}+{y}_{2}+{y}_{3}\\ {z}_{1}+{z}_{2}+{z}_{3}\\ 3\end{array}\right)$

The cartesian coordinates of the result is the "mid-point" (also known as the barycenter).

$\left(\begin{array}{c}\frac{{x}_{1}+{y}_{1}+{z}_{1}}{3}\\ \frac{{x}_{2}+{y}_{2}+{z}_{2}}{3}\\ \frac{{x}_{3}+{y}_{3}+{z}_{3}}{3}\end{array}\right)$

Ronnie Porter

Beginner2022-05-04Added 12 answers

Step 1

This will be the center of gravity, or the average of the points. If the points are $x=({x}_{1},{x}_{2},{x}_{3})$ , $\text{}y=({y}_{1},{y}_{2},{y}_{3})$ , and $z=({z}_{1},{z}_{2},{z}_{3})$ , you want to average each coordinate to obtain the point:

$a=\phantom{\rule{0ex}{0ex}}(\frac{{x}_{1}+{y}_{1}+{z}_{1}}{3},\frac{{x}_{2}+{y}_{2}+{z}_{2}}{3},\frac{{x}_{3}+{y}_{3}+{z}_{3}}{3})$

This same solution works for any number of points, not just three. For example, for two points, this gives you the midpoint.

Note that, in vector arithmetic, we have $x+y+z=3a$ . Thus, $(x-a)+(y-a)+(z-a)=0$ . We can think of each of those terms as a force from the point a towards one of the three points. Since it adds to 0, the forces cancel each other out, and so a is in equilibrium. This means it works with your grappling hook intuition.

This will be the center of gravity, or the average of the points. If the points are $x=({x}_{1},{x}_{2},{x}_{3})$ , $\text{}y=({y}_{1},{y}_{2},{y}_{3})$ , and $z=({z}_{1},{z}_{2},{z}_{3})$ , you want to average each coordinate to obtain the point:

$a=\phantom{\rule{0ex}{0ex}}(\frac{{x}_{1}+{y}_{1}+{z}_{1}}{3},\frac{{x}_{2}+{y}_{2}+{z}_{2}}{3},\frac{{x}_{3}+{y}_{3}+{z}_{3}}{3})$

This same solution works for any number of points, not just three. For example, for two points, this gives you the midpoint.

Note that, in vector arithmetic, we have $x+y+z=3a$ . Thus, $(x-a)+(y-a)+(z-a)=0$ . We can think of each of those terms as a force from the point a towards one of the three points. Since it adds to 0, the forces cancel each other out, and so a is in equilibrium. This means it works with your grappling hook intuition.

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