Centroid

From Wikipedia, the free encyclopedia

https://en.wikipedia.org/wiki/Centroid 

Centroid of a triangle

In mathematics and physics, the centroid, also known as geometric center or center of figure, of a plane figure or solid figure is the arithmetic mean position of all the points in the surface of the figure. The same definition extends to any object in ${\displaystyle n}$-dimensional Euclidean space.

In geometry, one often assumes uniform mass density, in which case the barycenter or center of mass coincides with the centroid. Informally, it can be understood as the point at which a cutout of the shape (with uniformly distributed mass) could be perfectly balanced on the tip of a pin.

In physics, if variations in gravity are considered, then a center of gravity can be defined as the weighted mean of all points weighted by their specific weight.

In geography, the centroid of a radial projection of a region of the Earth's surface to sea level is the region's geographical center.

History

The term "centroid" is of recent coinage (1814). It is used as a substitute for the older terms "center of gravity" and "center of mass" when the purely geometrical aspects of that point are to be emphasized. The term is peculiar to the English language; the French, for instance, use "centre de gravité" on most occasions, and others use terms of similar meaning.

The center of gravity, as the name indicates, is a notion that arose in mechanics, most likely in connection with building activities. It is uncertain when the idea first appeared, as the concept likely occurred to many people individually with minor differences. Nonetheless, the center of gravity of figures was studied extensively in Antiquity; Bossut credits Archimedes (287–212 BCE) with being the first to find the centroid of plane figures, although he never defines it. A treatment of centroids of solids by Archimedes has been lost.

It is unlikely that Archimedes learned the theorem that the medians of a triangle meet in a point—the center of gravity of the triangle—directly from Euclid, as this proposition is not in the Elements. The first explicit statement of this proposition is due to Heron of Alexandria (perhaps the first century CE) and occurs in his Mechanics. It may be added, in passing, that the proposition did not become common in the textbooks on plane geometry until the nineteenth century.

Properties

The geometric centroid of a convex object always lies in the object. A non-convex object might have a centroid that is outside the figure itself. The centroid of a ring or a bowl, for example, lies in the object's central void.

If the centroid is defined, it is a fixed point of all isometries in its symmetry group. In particular, the geometric centroid of an object lies in the intersection of all its hyperplanes of symmetry. The centroid of many figures (regular polygon, regular polyhedron, cylinder, rectangle, rhombus, circle, sphere, ellipse, ellipsoid, superellipse, superellipsoid, etc.) can be determined by this principle alone.

In particular, the centroid of a parallelogram is the meeting point of its two diagonals. This is not true of other quadrilaterals.

For the same reason, the centroid of an object with translational symmetry is undefined (or lies outside the enclosing space), because a translation has no fixed point.

Examples

The centroid of a triangle is the intersection of the three medians of the triangle (each median connecting a vertex with the midpoint of the opposite side).

For other properties of a triangle's centroid, see below.

Determination

See also: Center of mass § Determination

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 ${\displaystyle {\mathbf{x}}_1,{\mathbf{x}}_2,\dots ,{\mathbf{x}}_k}$ in ${\displaystyle {\mathbb{R}}^n}$ is

$${\displaystyle \mathbf{C}=\frac{{\mathbf{x}}_1+{\mathbf{x}}_2+\cdots +{\mathbf{x}}_k}{k}.}$$

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

$${\displaystyle C_x=\frac{\sum _i{C_i}_x{A}_i}{\sum _i{A}_i},C_y=\frac{\sum _i{C_i}_y{A}_i}{\sum _i{A}_i}.}$$

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 integral

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

where the integrals are taken over the whole space ${\displaystyle {\mathbb{R}}^n,}$ and ${\displaystyle g}$ is the characteristic function of the subset, which is ${\displaystyle 1}$ inside ${\displaystyle X}$ and ${\displaystyle 0}$ outside it. 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.

Another formula for the centroid is

$${\displaystyle C_k=\frac{\int z{S}_k(z)dz}{\int g(x)dx},}$$

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

$${\displaystyle C_{\mathrm{x}}=\frac{\int x{S}_{\mathrm{y}}(x)dx}{A},C_{\mathrm{y}}=\frac{\int y{S}_{\mathrm{x}}(y)dy}{A},}$$

where ${\displaystyle A}$ is the area of the figure ${\displaystyle X,}$ ${\displaystyle S_y(x)}$ is the length of the intersection of ${\displaystyle X}$ with the vertical line at abscissa ${\displaystyle x,}$ and ${\displaystyle S_x(y)}$ is the analogous quantity for the swapped axes.

Of a bounded region

The centroid ${\displaystyle (\overline{x},\overline{y})}$ of a region bounded by the graphs of the continuous functions ${\displaystyle f}$ and ${\displaystyle g}$ such that ${\displaystyle f(x)\ge g(x)}$ on the interval ${\displaystyle [a,b],}$ ${\displaystyle a\le x\le b}$ is given by

$${\displaystyle \begin{array}{rl}\overline{x}& =\frac{1}{A}{\int }_a^{b}x(f(x)-g(x))dx,\\ \overline{y}& =\frac{1}{A}{\int }_a^b\frac{1}{2}(f(x)+g(x))(f(x)-g(x))dx,\end{array}}$$

where ${\displaystyle A}$ is the area of the region (given by ${\int }_a^b(f(x)-g(x))dx$).

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.

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

Main article: Triangle center

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). The centroid divides each of the medians in the ratio ${\displaystyle 2:1,}$ which is to say it is located ${\displaystyle \frac{1}{3}}$ of the distance from each side to the opposite vertex (see figures at right). 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

$${\displaystyle C=\frac{1}{3}(L+M+N)=(\frac{1}{3}(x_L+x_M+x_N),\frac{1}{3}(y_L+y_M+y_N)).}$$

The centroid is therefore at ${\displaystyle \frac{1}{3}:\frac{1}{3}:\frac{1}{3}}$ 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 ${\displaystyle L,M,N}$:

$${\displaystyle \begin{array}{rl}C& =\frac{1}{a}:\frac{1}{b}:\frac{1}{c}=bc:ca:ab=\csc L:\csc M:\csc N\\ & =\cos L+\cos M\cdot \cos N:\cos M+\cos N\cdot \cos L:\cos N+\cos L\cdot \cos M\\ & =\sec L+\sec M\cdot \sec N:\sec M+\sec N\cdot \sec L:\sec N+\sec L\cdot \sec M.\end{array}}$$

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 ${\displaystyle \frac{3}{2}}$ times the length of any side times the perpendicular distance from the side to the centroid.

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:

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

In addition, for the incenter ${\displaystyle I}$ and nine-point center ${\displaystyle N,}$ we have

$${\displaystyle \begin{array}{rl}\overline{CH}& =4\overline{CN},\\ \overline{CO}& =2\overline{CN},\\ \overline{IC}& <\overline{HC},\\ \overline{IH}& <\overline{HC},\\ \overline{IC}& <\overline{IO}.\end{array}}$$

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

$${\displaystyle (\text{Area of }\mathrm{△}ABG)=(\text{Area of }\mathrm{△}ACG)=(\text{Area of }\mathrm{△}BCG)=\frac{1}{3}(\text{Area of }\mathrm{△}ABC).}$$

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 ${\displaystyle \frac{5}{9}}$ that of the original triangle.

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 ${\displaystyle G}$:

$${\displaystyle P{A}^2+P{B}^2+P{C}^2=G{A}^2+G{B}^2+G{C}^2+3P{G}^2.}$$

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:

$${\displaystyle A{B}^2+B{C}^2+C{A}^2=3(G{A}^2+G{B}^2+G{C}^2).}$$

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

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 ${\displaystyle ABC,}$

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

Of a polygon

The centroid of a non-self-intersecting closed polygon defined by ${\displaystyle n}$ vertices ${\displaystyle (x_0,y_0),}$${\displaystyle (x_1,y_1),\dots ,}$${\displaystyle (x_{n-1},y_{n-1}),}$ is the point ${\displaystyle (C_x,C_y),}$ 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, 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.)

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 ${\displaystyle \frac{1}{4}}$ 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 ${\displaystyle \frac{1}{3}}$ 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. 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 ${\displaystyle v_0,\dots ,v_n,}$ 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 ${\displaystyle \frac{3}{8}}$ 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.

Center of mass

From Wikipedia, the free encyclopedia
https://en.wikipedia.org/wiki/Center_of_mass

This toy uses the principles of center of mass to keep balance when sitting on a finger.

In physics, the center of mass of a distribution of mass in space (sometimes referred to as the barycenter or balance point) is the unique point at any given time where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration. Calculations in mechanics are often simplified when formulated with respect to the center of mass. It is a hypothetical point where the entire mass of an object may be assumed to be concentrated to visualise its motion. In other words, the center of mass is the particle equivalent of a given object for application of Newton's laws of motion.

In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid. The center of mass may be located outside the physical body, as is sometimes the case for hollow or open-shaped objects, such as a horseshoe. In the case of a distribution of separate bodies, such as the planets of the Solar System, the center of mass may not correspond to the position of any individual member of the system.

The center of mass is a useful reference point for calculations in mechanics that involve masses distributed in space, such as the linear and angular momentum of planetary bodies and rigid body dynamics. In orbital mechanics, the equations of motion of planets are formulated as point masses located at the centers of mass (see Barycenter (astronomy) for details). The center of mass frame is an inertial frame in which the center of mass of a system is at rest with respect to the origin of the coordinate system.

History

The concept of center of gravity or weight was studied extensively by the ancient Greek mathematician, physicist, and engineer Archimedes of Syracuse. He worked with simplified assumptions about gravity that amount to a uniform field, thus arriving at the mathematical properties of what we now call the center of mass. Archimedes showed that the torque exerted on a lever by weights resting at various points along the lever is the same as what it would be if all of the weights were moved to a single point—their center of mass. In his work On Floating Bodies, Archimedes demonstrated that the orientation of a floating object is the one that makes its center of mass as low as possible. He developed mathematical techniques for finding the centers of mass of objects of uniform density of various well-defined shapes.

Other ancient mathematicians who contributed to the theory of the center of mass include Hero of Alexandria and Pappus of Alexandria. In the Renaissance and Early Modern periods, work by Guido Ubaldi, Francesco Maurolico, Federico Commandino, Evangelista Torricelli, Simon Stevin, Luca Valerio, Jean-Charles de la Faille, Paul Guldin, John Wallis, Christiaan Huygens, Louis Carré, Pierre Varignon, and Alexis Clairaut expanded the concept further.

Newton's second law is reformulated with respect to the center of mass in Euler's first law.

Definition

The center of mass is the unique point at the center of a distribution of mass in space that has the property that the weighted position vectors relative to this point sum to zero. In analogy to statistics, the center of mass is the mean location of a distribution of mass in space.

A system of particles

In the case of a system of particles P_i, i = 1, ..., n , each with mass m_i that are located in space with coordinates r_i, i = 1, ..., n , the coordinates R of the center of mass satisfy the condition

$${\displaystyle \sum _{i=1}^n{m}_i({\mathbf{r}}_i-\mathbf{R})=\mathbf{0}.}$$

Solving this equation for R yields the formula

$${\displaystyle \mathbf{R}=\frac{\sum _{i=1}^n{m}_i{\mathbf{r}}_i}{\sum _{i=1}^n{m}_i}.}$$

A continuous volume

If the mass distribution is continuous with the density ρ(r) within a solid Q, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass R over the volume V is zero, that is

$${\displaystyle {\iiint }_Q\rho (\mathbf{r})\left(\mathbf{r}-\mathbf{R}\right)dV=0.}$$

Solve this equation for the coordinates R to obtain

$${\displaystyle \mathbf{R}=\frac{1}{M}{\iiint }_Q\rho (\mathbf{r})\mathbf{r}dV,}$$

Where M is the total mass in the volume.

If a continuous mass distribution has uniform density, which means that ρ is constant, then the center of mass is the same as the centroid of the volume.

Barycentric coordinates

Further information: Barycentric coordinate system

The coordinates R of the center of mass of a two-particle system, P₁ and P₂, with masses m₁ and m₂ is given by

$${\displaystyle \mathbf{R}=\frac{m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2}{m_1+m_2}.}$$

Let the percentage of the total mass divided between these two particles vary from 100% P₁ and 0% P₂ through 50% P₁ and 50% P₂ to 0% P₁ and 100% P₂, then the center of mass R moves along the line from P₁ to P₂. The percentages of mass at each point can be viewed as projective coordinates of the point R on this line, and are termed barycentric coordinates. Another way of interpreting the process here is the mechanical balancing of moments about an arbitrary point. The numerator gives the total moment that is then balanced by an equivalent total force at the center of mass. This can be generalized to three points and four points to define projective coordinates in the plane, and in space, respectively.

Systems with periodic boundary conditions

For particles in a system with periodic boundary conditions two particles can be neighbours even though they are on opposite sides of the system. This occurs often in molecular dynamics simulations, for example, in which clusters form at random locations and sometimes neighbouring atoms cross the periodic boundary. When a cluster straddles the periodic boundary, a naive calculation of the center of mass will be incorrect. A generalized method for calculating the center of mass for periodic systems is to treat each coordinate, x and y and/or z, as if it were on a circle instead of a line. The calculation takes every particle's x coordinate and maps it to an angle,

$${\displaystyle {\theta }_i=\frac{x_i}{x_{max}}2\pi }$$

where x_max is the system size in the x direction and ${\displaystyle x_i\in [0,x_{max})}$. From this angle, two new points ${\displaystyle ({\xi }_i,{\zeta }_i)}$ can be generated, which can be weighted by the mass of the particle ${\displaystyle x_i}$ for the center of mass or given a value of 1 for the geometric center:

$${\displaystyle \begin{array}{rl}{\xi }_i& =\cos ({\theta }_i)\\ {\zeta }_i& =\sin ({\theta }_i)\end{array}}$$

In the ${\displaystyle (\xi ,\zeta )}$ plane, these coordinates lie on a circle of radius 1. From the collection of ${\displaystyle {\xi }_i}$ and ${\displaystyle {\zeta }_i}$ values from all the particles, the averages ${\displaystyle \overline{\xi }}$ and ${\displaystyle \overline{\zeta }}$ are calculated.

$${\displaystyle \begin{array}{rl}\overline{\xi }& =\frac{1}{M}\sum _{i=1}^n{m}_i{\xi }_i,\\ \overline{\zeta }& =\frac{1}{M}\sum _{i=1}^n{m}_i{\zeta }_i,\end{array}}$$

where M is the sum of the masses of all of the particles.

These values are mapped back into a new angle, ${\displaystyle \overline{\theta }}$, from which the x coordinate of the center of mass can be obtained:

$${\displaystyle \begin{array}{rl}\overline{\theta }& =\mathrm{atan2}\left(-\overline{\zeta },-\overline{\xi }\right)+\pi \\ x_{\text{com}}& =x_{max}\frac{\overline{\theta }}{2\pi }\end{array}}$$

The process can be repeated for all dimensions of the system to determine the complete center of mass. The utility of the algorithm is that it allows the mathematics to determine where the "best" center of mass is, instead of guessing or using cluster analysis to "unfold" a cluster straddling the periodic boundaries. If both average values are zero, ${\displaystyle \left(\overline{\xi },\overline{\zeta }\right)=(0,0)}$, then ${\displaystyle \overline{\theta }}$ is undefined. This is a correct result, because it only occurs when all particles are exactly evenly spaced. In that condition, their x coordinates are mathematically identical in a periodic system.

Center of gravity

"Center of gravity" redirects here. For other uses, see Center of gravity (disambiguation).

Main article: Centers of gravity in non-uniform fields

Diagram of an educational toy that balances on a point: the center of mass (C) settles below its support (P)

A body's center of gravity is the point around which the resultant torque due to gravity forces vanishes. Where a gravity field can be considered to be uniform, the mass-center and the center-of-gravity will be the same. However, for satellites in orbit around a planet, in the absence of other torques being applied to a satellite, the slight variation (gradient) in gravitational field between closer-to and further-from the planet (stronger and weaker gravity respectively) can lead to a torque that will tend to align the satellite such that its long axis is vertical. In such a case, it is important to make the distinction between the center-of-gravity and the mass-center. Any horizontal offset between the two will result in an applied torque.

The mass-center is a fixed property for a given rigid body (e.g. with no slosh or articulation), whereas the center-of-gravity may, in addition, depend upon its orientation in a non-uniform gravitational field. In the latter case, the center-of-gravity will always be located somewhat closer to the main attractive body as compared to the mass-center, and thus will change its position in the body of interest as its orientation is changed.

In the study of the dynamics of aircraft, vehicles and vessels, forces and moments need to be resolved relative to the mass center. That is true independent of whether gravity itself is a consideration. Referring to the mass-center as the center-of-gravity is something of a colloquialism, but it is in common usage and when gravity gradient effects are negligible, center-of-gravity and mass-center are the same and are used interchangeably.

In physics the benefits of using the center of mass to model a mass distribution can be seen by considering the resultant of the gravity forces on a continuous body. Consider a body Q of volume V with density ρ(r) at each point r in the volume. In a parallel gravity field the force f at each point r is given by,

$${\displaystyle \mathbf{f}(\mathbf{r})=-dmg\hat{\mathbf{k}}=-\rho (\mathbf{r})dVg\hat{\mathbf{k}},}$$

where dm is the mass at the point r, g is the acceleration of gravity, and $\hat{\mathbf{k}}$ is a unit vector defining the vertical direction.

Choose a reference point R in the volume and compute the resultant force and torque at this point,

$${\displaystyle \mathbf{F}={\iiint }_Q\mathbf{f}(\mathbf{r})dV={\iiint }_Q\rho (\mathbf{r})dV\left(-g\hat{\mathbf{k}}\right)=-Mg\hat{\mathbf{k}},}$$

and

$${\displaystyle \mathbf{T}={\iiint }_Q(\mathbf{r}-\mathbf{R})\times \mathbf{f}(\mathbf{r})dV={\iiint }_Q(\mathbf{r}-\mathbf{R})\times \left(-g\rho (\mathbf{r})dV\hat{\mathbf{k}}\right)=\left({\iiint }_Q\rho (\mathbf{r})\left(\mathbf{r}-\mathbf{R}\right)dV\right)\times \left(-g\hat{\mathbf{k}}\right).}$$

If the reference point R is chosen so that it is the center of mass, then

$${\displaystyle {\iiint }_Q\rho (\mathbf{r})\left(\mathbf{r}-\mathbf{R}\right)dV=0,}$$

which means the resultant torque T = 0. Because the resultant torque is zero the body will move as though it is a particle with its mass concentrated at the center of mass.

By selecting the center of gravity as the reference point for a rigid body, the gravity forces will not cause the body to rotate, which means the weight of the body can be considered to be concentrated at the center of mass.

Linear and angular momentum

The linear and angular momentum of a collection of particles can be simplified by measuring the position and velocity of the particles relative to the center of mass. Let the system of particles P_i, i = 1, ..., n of masses m_i be located at the coordinates r_i with velocities v_i. Select a reference point R and compute the relative position and velocity vectors,

$${\displaystyle {\mathbf{r}}_i=({\mathbf{r}}_i-\mathbf{R})+\mathbf{R},{\mathbf{v}}_i=\frac{d}{dt}({\mathbf{r}}_i-\mathbf{R})+\mathbf{v}.}$$

The total linear momentum and angular momentum of the system are

$${\displaystyle \mathbf{p}=\frac{d}{dt}\left(\sum _{i=1}^n{m}_i({\mathbf{r}}_i-\mathbf{R})\right)+\left(\sum _{i=1}^n{m}_i\right)\mathbf{v},}$$

and

$${\displaystyle \mathbf{L}=\sum _{i=1}^n{m}_i({\mathbf{r}}_i-\mathbf{R})\times \frac{d}{dt}({\mathbf{r}}_i-\mathbf{R})+\left(\sum _{i=1}^n{m}_i\right)\left[\mathbf{R}\times \frac{d}{dt}({\mathbf{r}}_i-\mathbf{R})+({\mathbf{r}}_i-\mathbf{R})\times \mathbf{v}\right]+\left(\sum _{i=1}^n{m}_i\right)\mathbf{R}\times \mathbf{v}}$$

If R is chosen as the center of mass these equations simplify to

$${\displaystyle \mathbf{p}=m\mathbf{v},\mathbf{L}=\sum _{i=1}^n{m}_i({\mathbf{r}}_i-\mathbf{R})\times \frac{d}{dt}({\mathbf{r}}_i-\mathbf{R})+\sum _{i=1}^n{m}_i\mathbf{R}\times \mathbf{v}}$$

where m is the total mass of all the particles, p is the linear momentum, and L is the angular momentum.

The law of conservation of momentum predicts that for any system not subjected to external forces the momentum of the system will remain constant, which means the center of mass will move with constant velocity. This applies for all systems with classical internal forces, including magnetic fields, electric fields, chemical reactions, and so on. More formally, this is true for any internal forces that cancel in accordance with Newton's Third Law.

Determination

See also: Centroid § Determination

Plumb line method

The experimental determination of a body's centre of mass makes use of gravity forces on the body and is based on the fact that the centre of mass is the same as the centre of gravity in the parallel gravity field near the earth's surface.

The center of mass of a body with an axis of symmetry and constant density must lie on this axis. Thus, the center of mass of a circular cylinder of constant density has its center of mass on the axis of the cylinder. In the same way, the center of mass of a spherically symmetric body of constant density is at the center of the sphere. In general, for any symmetry of a body, its center of mass will be a fixed point of that symmetry.

In two dimensions

An experimental method for locating the center of mass is to suspend the object from two locations and to drop plumb lines from the suspension points. The intersection of the two lines is the center of mass.

The shape of an object might already be mathematically determined, but it may be too complex to use a known formula. In this case, one can subdivide the complex shape into simpler, more elementary shapes, whose centers of mass are easy to find. If the total mass and center of mass can be determined for each area, then the center of mass of the whole is the weighted average of the centers. This method can even work for objects with holes, which can be accounted for as negative masses.

A direct development of the planimeter known as an integraph, or integerometer, can be used to establish the position of the centroid or center of mass of an irregular two-dimensional shape. This method can be applied to a shape with an irregular, smooth or complex boundary where other methods are too difficult. It was regularly used by ship builders to compare with the required displacement and center of buoyancy of a ship, and ensure it would not capsize.

In three dimensions

An experimental method to locate the three-dimensional coordinates of the center of mass begins by supporting the object at three points and measuring the forces, F₁, F₂, and F₃ that resist the weight of the object, ${\displaystyle \mathbf{W}=-W\hat{\mathbf{k}}}$ (${\displaystyle \hat{\mathbf{k}}}$ is the unit vector in the vertical direction). Let r₁, r₂, and r₃ be the position coordinates of the support points, then the coordinates R of the center of mass satisfy the condition that the resultant torque is zero,

$${\displaystyle \mathbf{T}=({\mathbf{r}}_1-\mathbf{R})\times {\mathbf{F}}_1+({\mathbf{r}}_2-\mathbf{R})\times {\mathbf{F}}_2+({\mathbf{r}}_3-\mathbf{R})\times {\mathbf{F}}_3=0,}$$

or

$${\displaystyle \mathbf{R}\times \left(-W\hat{\mathbf{k}}\right)={\mathbf{r}}_1\times {\mathbf{F}}_1+{\mathbf{r}}_2\times {\mathbf{F}}_2+{\mathbf{r}}_3\times {\mathbf{F}}_3.}$$

This equation yields the coordinates of the center of mass R* in the horizontal plane as,

$${\displaystyle {\mathbf{R}}^{\ast }=-\frac{1}{W}\hat{\mathbf{k}}\times ({\mathbf{r}}_1\times {\mathbf{F}}_1+{\mathbf{r}}_2\times {\mathbf{F}}_2+{\mathbf{r}}_3\times {\mathbf{F}}_3).}$$

The center of mass lies on the vertical line L, given by

$${\displaystyle \mathbf{L}(t)={\mathbf{R}}^{\ast }+t\hat{\mathbf{k}}.}$$

The three-dimensional coordinates of the center of mass are determined by performing this experiment twice with the object positioned so that these forces are measured for two different horizontal planes through the object. The center of mass will be the intersection of the two lines L₁ and L₂ obtained from the two experiments.

Applications

Engineering designs

Automotive applications

Engineers try to design a sports car so that its center of mass is lowered to make the car handle better, which is to say, maintain traction while executing relatively sharp turns.

The characteristic low profile of the U.S. military Humvee was designed in part to allow it to tilt farther than taller vehicles without rolling over, by ensuring its low center of mass stays over the space bounded by the four wheels even at angles far from the horizontal.

Aeronautics

Main article: Center of gravity of an aircraft

The center of mass is an important point on an aircraft, which significantly affects the stability of the aircraft. To ensure the aircraft is stable enough to be safe to fly, the center of mass must fall within specified limits. If the center of mass is ahead of the forward limit, the aircraft will be less maneuverable, possibly to the point of being unable to rotate for takeoff or flare for landing. If the center of mass is behind the aft limit, the aircraft will be more maneuverable, but also less stable, and possibly unstable enough so as to be impossible to fly. The moment arm of the elevator will also be reduced, which makes it more difficult to recover from a stalled condition.

For helicopters in hover, the center of mass is always directly below the rotorhead. In forward flight, the center of mass will move forward to balance the negative pitch torque produced by applying cyclic control to propel the helicopter forward; consequently a cruising helicopter flies "nose-down" in level flight.

Astronomy

Main article: Barycenter

Two bodies orbiting their barycenter (red cross)

The center of mass plays an important role in astronomy and astrophysics, where it is commonly referred to as the barycenter. The barycenter is the point between two objects where they balance each other; it is the center of mass where two or more celestial bodies orbit each other. When a moon orbits a planet, or a planet orbits a star, both bodies are actually orbiting a point that lies away from the center of the primary (larger) body. For example, the Moon does not orbit the exact center of the Earth, but a point on a line between the center of the Earth and the Moon, approximately 1,710 km (1,062 miles) below the surface of the Earth, where their respective masses balance. This is the point about which the Earth and Moon orbit as they travel around the Sun. If the masses are more similar, e.g., Pluto and Charon, the barycenter will fall outside both bodies.

Rigging and safety

Knowing the location of the center of gravity when rigging is crucial, possibly resulting in severe injury or death if assumed incorrectly. A center of gravity that is at or above the lift point will most likely result in a tip-over incident. In general, the further the center of gravity below the pick point, the safer the lift. There are other things to consider, such as shifting loads, strength of the load and mass, distance between pick points, and number of pick points. Specifically, when selecting lift points, it's very important to place the center of gravity at the center and well below the lift points.

Body motion

Main article: Kinesiology

In kinesiology and biomechanics, the center of mass is an important parameter that assists people in understanding their human locomotion. Typically, a human's center of mass is detected with one of two methods: the reaction board method is a static analysis that involves the person lying down on that instrument, and use of their static equilibrium equation to find their center of mass; the segmentation method relies on a mathematical solution based on the physical principle that the summation of the torques of individual body sections, relative to a specified axis, must equal the torque of the whole system that constitutes the body, measured relative to the same axis.

Optimization

The Center-of-gravity method is a method for convex optimization, which uses the center-of-gravity of the feasible region.