Centroid

Plumb line method

The centroid of a uniformly dense planar lamina, such as in figure (a) below, may be determined experimentally by using a plumbline and a pin to find the collocated center of mass of a thin body of uniform density having the same shape. The body is held by the pin, inserted at a point, off the presumed centroid in such a way that it can freely rotate around the pin; the plumb line is then dropped from the pin (figure b). The position of the plumbline is traced on the surface, and the procedure is repeated with the pin inserted at any different point (or a number of points) off the centroid of the object. The unique intersection point of these lines will be the centroid (figure c). Provided that the body is of uniform density, all lines made this way will include the centroid, and all lines will cross at exactly the same place.

(a)	(b)	(c)

This method can be extended (in theory) to concave shapes where the centroid may lie outside the shape, and virtually to solids (again, of uniform density), where the centroid may lie within the body. The (virtual) positions of the plumb lines need to be recorded by means other than by drawing them along the shape.

Balancing method

For convex two-dimensional shapes, the centroid can be found by balancing the shape on a smaller shape, such as the top of a narrow cylinder. The centroid occurs somewhere within the range of contact between the two shapes (and exactly at the point where the shape would balance on a pin). In principle, progressively narrower cylinders can be used to find the centroid to arbitrary precision. In practice air currents make this infeasible. However, by marking the overlap range from multiple balances, one can achieve a considerable level of accuracy.

Of a finite set of points

The centroid of a finite set of ${\displaystyle k}$ points ${\displaystyle \mathbf {x} _{1},\mathbf {x} _{2},\ldots ,\mathbf {x} _{k}}$ in ${\displaystyle \mathbb {R} ^{n}}$ is^[1] $${\displaystyle \mathbf {C} ={\frac {\mathbf {x} _{1}+\mathbf {x} _{2}+\cdots +\mathbf {x} _{k}}{k}}.}$$ This point minimizes the sum of squared Euclidean distances between itself and each point in the set.

By geometric decomposition

The centroid of a plane figure ${\displaystyle X}$ can be computed by dividing it into a finite number of simpler figures ${\displaystyle X_{1},X_{2},\dots ,X_{n},}$ computing the centroid ${\displaystyle C_{i}}$ and area ${\displaystyle A_{i}}$ of each part, and then computing

$${\displaystyle C_{x}={\frac {\sum _{i}{C_{i}}_{x}A_{i}}{\sum _{i}A_{i}}},\quad C_{y}={\frac {\sum _{i}{C_{i}}_{y}A_{i}}{\sum _{i}A_{i}}}.}$$

Holes in the figure ${\displaystyle X,}$ overlaps between the parts, or parts that extend outside the figure can all be handled using negative areas ${\displaystyle A_{i}.}$ Namely, the measures ${\displaystyle A_{i}}$ should be taken with positive and negative signs in such a way that the sum of the signs of ${\displaystyle A_{i}}$ for all parts that enclose a given point ${\displaystyle p}$ is ${\displaystyle 1}$ if ${\displaystyle p}$ belongs to ${\displaystyle X,}$ and ${\displaystyle 0}$ otherwise.

For example, the figure below (a) is easily divided into a square and a triangle, both with positive area; and a circular hole, with negative area (b).

(a) 2D Object

(b) Object described using simpler elements

(c) Centroids of elements of the object

The centroid of each part can be found in any list of centroids of simple shapes (c). Then the centroid of the figure is the weighted average of the three points. The horizontal position of the centroid, from the left edge of the figure is $${\displaystyle x={\frac {5\times 10^{2}+13.33\times {\frac {1}{2}}10^{2}-3\times \pi (2.5)^{2}}{10^{2}+{\frac {1}{2}}10^{2}-\pi (2.5)^{2}}}\approx 8.5{\text{ units}}.}$$ The vertical position of the centroid is found in the same way.

The same formula holds for any three-dimensional objects, except that each ${\displaystyle A_{i}}$ should be the volume of ${\displaystyle X_{i},}$ rather than its area. It also holds for any subset of ${\displaystyle \mathbb {R} ^{d},}$ for any dimension ${\displaystyle d,}$ with the areas replaced by the ${\displaystyle d}$-dimensional measures of the parts.

By integral formula

The centroid of a subset ${\displaystyle X}$ of ${\displaystyle \mathbb {R} ^{n}}$ can also be computed by the vector formula

$${\displaystyle C={\frac {\int xg(x)\ dx}{\int g(x)\ dx}}={\frac {\int _{X}x\ dx}{\int _{X}dx}}}$$

where the integrals are taken over the whole space ${\displaystyle \mathbb {R} ^{n},}$ and ${\displaystyle g}$ is the characteristic function of the subset ${\displaystyle X}$ of ${\displaystyle \mathbb {R} ^{n}\!:\ g(x)=1}$ if ${\displaystyle x\in X}$ and ${\displaystyle g(x)=0}$ otherwise.^[7] Note that the denominator is simply the measure of the set ${\displaystyle X.}$ This formula cannot be applied if the set ${\displaystyle X}$ has zero measure, or if either integral diverges.

Alternatively, the coordinate-wise formula for the centroid is defined as

$${\displaystyle C_{k}={\frac {\int zS_{k}(z)\ dz}{\int g(x)\ dx}},}$$

where ${\displaystyle C_{k}}$ is the ${\displaystyle k}$th coordinate of ${\displaystyle C,}$ and ${\displaystyle S_{k}(z)}$ is the measure of the intersection of ${\displaystyle X}$ with the hyperplane defined by the equation ${\displaystyle x_{k}=z.}$ Again, the denominator is simply the measure of ${\displaystyle X.}$

For a plane figure, in particular, the barycentric coordinates are

$${\displaystyle C_{\mathrm {x} }={\frac {\int xS_{\mathrm {y} }(x)\ dx}{A}},\quad C_{\mathrm {y} }={\frac {\int yS_{\mathrm {x} }(y)\ dy}{A}},}$$

where ${\displaystyle A}$ is the area of the figure ${\displaystyle X,}$ ${\displaystyle S_{\mathrm {y} }(x)}$ is the length of the intersection of ${\displaystyle X}$ with the vertical line at abscissa ${\displaystyle x,}$ and ${\displaystyle S_{\mathrm {x} }(y)}$ is the length of the intersection of ${\displaystyle X}$ with the horizontal line at ordinate ${\displaystyle y.}$

With an integraph

An integraph (a relative of the planimeter) can be used to find the centroid of an object of irregular shape with smooth (or piecewise smooth) boundary. The mathematical principle involved is a special case of Green's theorem.^[11]

Of an L-shaped object

This is a method of determining the centroid of an L-shaped object.

1. Divide the shape into two rectangles, as shown in fig 2. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the shape must lie on this line ${\displaystyle AB.}$
2. Divide the shape into two other rectangles, as shown in fig 3. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the L-shape must lie on this line ${\displaystyle CD.}$
3. As the centroid of the shape must lie along ${\displaystyle AB}$ and also along ${\displaystyle CD,}$ it must be at the intersection of these two lines, at ${\displaystyle O.}$ The point ${\displaystyle O}$ might lie inside or outside the L-shaped object.

Of a triangle

The centroid of a triangle is the point of intersection of its medians (the lines joining each vertex with the midpoint of the opposite side).^[6] The centroid divides each of the medians in the ratio ${\displaystyle 2:1,}$ which is to say it is located ${\displaystyle {\tfrac {1}{3}}}$ of the distance from each side to the opposite vertex (see figures at right).^[12][13] Its Cartesian coordinates are the means of the coordinates of the three vertices. That is, if the three vertices are ${\displaystyle L=(x_{L},y_{L}),}$ ${\displaystyle M=(x_{M},y_{M}),}$ and ${\displaystyle N=(x_{N},y_{N}),}$ then the centroid (denoted ${\displaystyle C}$ here but most commonly denoted ${\displaystyle G}$ in triangle geometry) is

$${\displaystyle C={\tfrac {1}{3}}(L+M+N)={\bigl (}{\tfrac {1}{3}}(x_{L}+x_{M}+x_{N}),{\tfrac {1}{3}}(y_{L}+y_{M}+y_{N}){\bigr )}.}$$

The centroid is therefore at ${\displaystyle {\tfrac {1}{3}}:{\tfrac {1}{3}}:{\tfrac {1}{3}}}$ in barycentric coordinates.

In trilinear coordinates the centroid can be expressed in any of these equivalent ways in terms of the side lengths ${\displaystyle a,b,c}$ and vertex angles ${\displaystyle L,M,N}$:^[14]

$${\displaystyle {\begin{aligned}C&={\frac {1}{a}}:{\frac {1}{b}}:{\frac {1}{c}}=bc:ca:ab=\csc L:\csc M:\csc N\\[6pt]&=\cos L+\cos M\cdot \cos N:\cos M+\cos N\cdot \cos L:\cos N+\cos L\cdot \cos M.\end{aligned}}}$$

The centroid is also the physical center of mass if the triangle is made from a uniform sheet of material; or if all the mass is concentrated at the three vertices, and evenly divided among them. On the other hand, if the mass is distributed along the triangle's perimeter, with uniform linear density, then the center of mass lies at the Spieker center (the incenter of the medial triangle), which does not (in general) coincide with the geometric centroid of the full triangle.

The area of the triangle is ${\displaystyle {\tfrac {3}{2}}}$ times the length of any side times the perpendicular distance from the side to the centroid.^[15]

A triangle's centroid lies on its Euler line between its orthocenter ${\displaystyle H}$ and its circumcenter ${\displaystyle O,}$ exactly twice as close to the latter as to the former:^[16][17]

$${\displaystyle {\overline {CH}}=2{\overline {CO}}.}$$

In addition, for the incenter ${\displaystyle I}$ and nine-point center ${\displaystyle N,}$ we have $${\displaystyle {\begin{aligned}{\overline {CH}}&=4{\overline {CN}},\\[5pt]{\overline {CO}}&=2{\overline {CN}},\\[5pt]{\overline {IC}}&<{\overline {HC}},\\[5pt]{\overline {IH}}&<{\overline {HC}},\\[5pt]{\overline {IC}}&<{\overline {IO}}.\end{aligned}}}$$

If ${\displaystyle G}$ is the centroid of the triangle ${\displaystyle ABC,}$ then

$${\displaystyle ({\text{Area of }}\triangle ABG)=({\text{Area of }}\triangle ACG)=({\text{Area of }}\triangle BCG)={\tfrac {1}{3}}({\text{Area of }}\triangle ABC).}$$

The isogonal conjugate of a triangle's centroid is its symmedian point.

Any of the three medians through the centroid divides the triangle's area in half. This is not true for other lines through the centroid; the greatest departure from the equal-area division occurs when a line through the centroid is parallel to a side of the triangle, creating a smaller triangle and a trapezoid; in this case the trapezoid's area is ${\displaystyle {\tfrac {5}{9}}}$ that of the original triangle.^[18]

Let ${\displaystyle P}$ be any point in the plane of a triangle with vertices ${\displaystyle A,B,C}$ and centroid ${\displaystyle G.}$ Then the sum of the squared distances of ${\displaystyle P}$ from the three vertices exceeds the sum of the squared distances of the centroid ${\displaystyle G}$ from the vertices by three times the squared distance between ${\displaystyle P}$ and ${\displaystyle G}$:^[19]

$${\displaystyle PA^{2}+PB^{2}+PC^{2}=GA^{2}+GB^{2}+GC^{2}+3PG^{2}.}$$

The sum of the squares of the triangle's sides equals three times the sum of the squared distances of the centroid from the vertices:^[19]

$${\displaystyle AB^{2}+BC^{2}+CA^{2}=3(GA^{2}+GB^{2}+GC^{2}).}$$

A triangle's centroid is the point that maximizes the product of the directed distances of a point from the triangle's sidelines.^[20]

Let ${\displaystyle ABC}$ be a triangle, let ${\displaystyle G}$ be its centroid, and let ${\displaystyle D,E,F}$ be the midpoints of segments ${\displaystyle BC,CA,AB,}$ respectively. For any point ${\displaystyle P}$ in the plane of ${\displaystyle ABC,}$^[21]

$${\displaystyle PA+PB+PC\leq 2(PD+PE+PF)+3PG.}$$

Of a polygon

The centroid of a non-self-intersecting closed polygon defined by ${\displaystyle n}$ vertices ${\displaystyle (x_{0},y_{0}),\;}$${\displaystyle (x_{1},y_{1}),\;\ldots ,\;}$${\displaystyle (x_{n-1},y_{n-1}),}$ is the point ${\displaystyle (C_{x},C_{y}),}$^[22] where

$${\displaystyle C_{\mathrm {x} }={\frac {1}{6A}}\sum _{i=0}^{n-1}(x_{i}+x_{i+1})(x_{i}\ y_{i+1}-x_{i+1}\ y_{i}),}$$

and

$${\displaystyle C_{\mathrm {y} }={\frac {1}{6A}}\sum _{i=0}^{n-1}(y_{i}+y_{i+1})(x_{i}\ y_{i+1}-x_{i+1}\ y_{i}),}$$

and where ${\displaystyle A}$ is the polygon's signed area,^[22] as described by the shoelace formula:

$${\displaystyle A={\frac {1}{2}}\sum _{i=0}^{n-1}(x_{i}\ y_{i+1}-x_{i+1}\ y_{i}).}$$

In these formulae, the vertices are assumed to be numbered in order of their occurrence along the polygon's perimeter; furthermore, the vertex ${\displaystyle (x_{n},y_{n})}$ is assumed to be the same as ${\displaystyle (x_{0},y_{0}),}$ meaning ${\displaystyle i+1}$ on the last case must loop around to ${\displaystyle i=0.}$ (If the points are numbered in clockwise order, the area ${\displaystyle A,}$ computed as above, will be negative; however, the centroid coordinates will be correct even in this case.)

The centroid of a non-triangular polygon is not the same as its vertex centroid, considering only its vertex set (as the centroid of a finite set of points; see also: Polygon#Centroid).

Of a cone or pyramid

The centroid of a cone or pyramid is located on the line segment that connects the apex to the centroid of the base. For a solid cone or pyramid, the centroid is ${\displaystyle {\tfrac {1}{4}}}$ the distance from the base to the apex. For a cone or pyramid that is just a shell (hollow) with no base, the centroid is ${\displaystyle {\tfrac {1}{3}}}$ the distance from the base plane to the apex.

Of a tetrahedron and n-dimensional simplex

A tetrahedron is an object in three-dimensional space having four triangles as its faces. A line segment joining a vertex of a tetrahedron with the centroid of the opposite face is called a median, and a line segment joining the midpoints of two opposite edges is called a bimedian. Hence there are four medians and three bimedians. These seven line segments all meet at the centroid of the tetrahedron.^[23] The medians are divided by the centroid in the ratio ${\displaystyle 3:1.}$ The centroid of a tetrahedron is the midpoint between its Monge point and circumcenter (center of the circumscribed sphere). These three points define the Euler line of the tetrahedron that is analogous to the Euler line of a triangle.

These results generalize to any ${\displaystyle n}$-dimensional simplex in the following way. If the set of vertices of a simplex is ${\displaystyle {v_{0},\ldots ,v_{n}},}$ then considering the vertices as vectors, the centroid is

$${\displaystyle C={\frac {1}{n+1}}\sum _{i=0}^{n}v_{i}.}$$

The geometric centroid coincides with the center of mass if the mass is uniformly distributed over the whole simplex, or concentrated at the vertices as ${\displaystyle n+1}$ equal masses.

Of a hemisphere

The centroid of a solid hemisphere (i.e. half of a solid ball) divides the line segment connecting the sphere's center to the hemisphere's pole in the ratio ${\displaystyle 3:5}$ (i.e. it lies ${\displaystyle {\tfrac {3}{8}}}$ of the way from the center to the pole). The centroid of a hollow hemisphere (i.e. half of a hollow sphere) divides the line segment connecting the sphere's center to the hemisphere's pole in half.