Dynamic Modeling of Sphere, Cylinder, Cone, and their Assembly

Abstract

Dynamic simulation of revolved solids plays an important role in many fields. Aiming at the lacks of solutions in some key aspects, this study establishes governing equation of motion based on theory of variational inequality; designs a compatibility iteration algorithm for solving contact forces; deduces parametric equations of arbitrary cylinder and cone in three-dimensional space; provides corresponding analytical methods to identify contact points between bodies and to calculate volume integral over bodies; proposes a rotation matrix modification approach to conserve volume and shape of rigid body in the case of large rotation. The accuracy, availability, competence, robustness, and application prospects of the presented methodology are demonstrated by several interesting and challenging problems.

Appendices

Appendix 1: Several Parameter Equations

1.1 Parameter Equation of Line Segment

In Fig. 53, any point C(x, y, z) on line AB can be written as

$$\left\{ \begin{array}{l} x = x_{\text{A}} + (x_{\text{B}} - x_{\text{A}} )t \hfill \\ y = y_{\text{A}} + (y_{\text{B}} - y_{\text{A}} )t \hfill \\ z = z_{\text{A}} + (z_{\text{B}} - z_{\text{A}} )t \hfill \\ \end{array} \right.$$

(48)

where t is the parameter associated with point C. If 0 ≤ t ≤ 1, point C is on line segment AB; otherwise, point C is outside of the line segment.

Fig. 53

Point on line AB

1.2 Parameter Equation of Plane

A plane with the normal vector n is determined by points A, B, and C, as shown in Fig. 54. An arbitrary point D(x, y, z) on this plane can be expressed by

$$\left\{ \begin{array}{l} x = x_{\text{A}} + (x_{\text{B}} - x_{\text{A}} )\lambda + m\mu \hfill \\ y = y_{\text{A}} + (y_{\text{B}} - y_{\text{A}} )\lambda + n\mu \hfill \\ z = z_{\text{A}} + (z_{\text{B}} - z_{\text{A}} )\lambda + p\mu \hfill \\ \end{array} \right.$$

(49)

where λ and μ are the parameters. Vector τ(m, n, p) is the direction vector of line BC.

Fig. 54

Point on plane ABC

1.3 Parameter Equation of Circle

In three-dimensional space, as depicted in Fig. 55, point C(x_C, y_C, z_C) is the center of circle, n(n_x, n_y, n_z) = n(cosα, cosβ, cosγ) is the unit normal vector of the circular plane. α, β, and γ are the angles between n and x-, y-, and z-axis, respectively. By taking point C as the origin of coordinates, we can make a local Cartesian coordinate system o*–u–v–w, in which the positive direction of w-axis is the same as that of vector n. The construction of the coordinate system o*–u–v–w is as follows: We move the unit normal vector n to the origin o of the global Cartesian coordinate system (see Fig. 55), where vectors n_x and n_y are the two components of vector n regarding to x- and y-axis, respectively. While vector n_xy is the projection of vector n onto o–x–y plane. By rotating n_xy by 90° around z-axis, we have another vector ξ(ξ_x, ξ_y, ξ_z), as described in Fig. 55. Obviously, through rotating the global Cartesian coordinate system o–x–y–z by γ angle around vector ξ, and then by moving it to point C, we can obtain the local Cartesian coordinate system o*–u–v–w. This is to say that the following coordinate transformation between the two systems holds:

$$\left\{ \begin{array}{l} x = x_{\text{C}} + A_{x} u + B_{x} v + C_{x} w \hfill \\ y = y_{\text{C}} + A_{y} u + B_{y} v + C_{y} w \hfill \\ z = \;\;z_{\text{C}} + \;A_{z} u + B_{z} v + C_{z} w \hfill \\ \end{array} \right.,$$

(50)

where

$$\begin{aligned} A_{x} = \cos \gamma + (1 - \cos \gamma )\xi_{x}^{2} ,\;B_{x} = (1 - \cos \gamma )\xi_{x}^{{}} \xi_{y}^{{}} - \xi_{z}^{{}} \sin \gamma ,\;C_{x} = (1 - \cos \gamma )\xi_{x}^{{}} \xi_{z}^{{}} + \xi_{y}^{{}} \sin \gamma \hfill \\ A_{y} = (1 - \cos \gamma )\xi_{x}^{{}} \xi_{y}^{{}} + \xi_{z}^{{}} \sin \gamma ,\;B_{y} = \cos \gamma + (1 - \cos \gamma )\xi_{y}^{2} ,\;C_{y} = (1 - \cos \gamma )\xi_{y}^{{}} \xi_{z}^{{}} - \xi_{x}^{{}} \sin \gamma \hfill \\ A_{z} = (1 - \cos \gamma )\xi_{x}^{{}} \xi_{z}^{{}} - \xi_{y}^{{}} \sin \gamma ,\;B_{z} = (1 - \cos \gamma )\xi_{y}^{{}} \xi_{z}^{{}} + \xi_{x}^{{}} \sin \gamma ,\;C_{z} = \cos \gamma + (1 - \cos \gamma )\xi_{z}^{2} , \hfill \\ \end{aligned}$$

(51)

where (ξ_x, ξ_y, ξ_z) is the direction cosine of vector ξ. Note that ξ_z = 0, Eq. (51) can be simplified as

Fig. 55

Circle in three-dimensional space

$$\begin{aligned} A_{x} & = \cos \gamma + (1 - \cos \gamma )\xi_{x}^{2} ,\;B_{x} = (1 - \cos \gamma )\xi_{x}^{{}} \xi_{y}^{{}} ,\;C_{x} = \xi_{y}^{{}} \sin \gamma \\ A_{y} & = (1 - \cos \gamma )\xi_{x}^{{}} \xi_{y}^{{}} ,\;B_{y} = \cos \gamma + (1 - \cos \gamma )\xi_{y}^{2} ,\;C_{y} = - \xi_{x}^{{}} \sin \gamma \\ A_{z} & = - \xi_{y}^{{}} \sin \gamma ,\;B_{z} = \xi_{x}^{{}} \sin \gamma ,\;C_{z} = \cos \gamma . \\ \end{aligned}$$

(52)

On the other hand, in the local Cartesian coordinate system the circle can be expressed as

$$\left\{ \begin{array}{l} u = R\cos \theta \hfill \\ v = R\sin \theta ,\;\;\;\theta \in [0,2\pi ] \hfill \\ w = 0, \hfill \\ \end{array} \right.$$

(53)

where R is the radius of the circle. Additionally, the angle parameter θ is measured according to the rule of right-hand. Substitution of Eqs. (53) into (50) yields the parameter equation of circle as follows:

$$\left\{ \begin{array}{l} x = x_{\text{C}} + A_{x} R\cos \theta + B_{x} R\sin \theta \hfill \\ y = y_{\text{C}} + A_{y} R\cos \theta + B_{y} R\sin \theta \hfill \\ z = \;\;z_{\text{C}} + \;A_{z} R\cos \theta + B_{z} R\sin \theta . \hfill \\ \end{array} \right.$$

(54)

1.4 Parameter Equation of Cylinder Surface

Thanks to the parameter equation of circle, given a bottom circle with the unit normal vector n(cosα, cosβ, cosγ) and the radius R, in three-dimensional space a cylinder with the height H (see Fig. 56) can be expressed by

$$\left\{ \begin{array}{l} x\; = \;x_{\text{C}} + A_{x} R\cos \theta + B_{x} R\sin \theta + C_{x} s \hfill \\ y\; = \;y_{\text{C}} + A_{y} R\cos \theta + B_{y} R\sin \theta + C_{y} s,\;\;\;\;\theta \in [0,2\pi ],\;s \in [0,H] \hfill \\ z\; = \;z_{\text{C}} + \;A_{z} R\cos \theta + B_{z} R\sin \theta + C_{z} s, \hfill \\ \end{array} \right.$$

(55)

where the coefficients A_i, B_i, and C_i (i = x, y, z) are defined by Eq. (52). In addition, θ and s are the two parameters. Point C(x_C, y_C, z_C) and R are the center and radius of the bottom circle, while point C* is the center of the upper circle. In addition, the s-axis is local one-dimensional axis pointing the center C to the center C*. If s = 0, Eq. (53) gives us the bottom circle; while if s = H, the upper circle can be obtained.

Fig. 56

Cylinder in three-dimensional space

1.5 Parameter Equation of Cone Surface

Based on the parameter equation of circle, for arbitrary cone with the unit normal vector n(cosα, cosβ, cosγ) of the bottom circular plane, as sketched in Fig. 57, it can be described as

$$\left\{ \begin{array}{l} x = x_{\text{C}} + \left( {1 - \frac{s}{H}} \right)A_{x} R\cos \theta + \left( {1 - \frac{s}{H}} \right)B_{x} R\sin \theta + C_{x} s \hfill \\ y = y_{\text{C}} + \left( {1 - \frac{s}{H}} \right)A_{y} R\cos \theta + \left( {1 - \frac{s}{H}} \right)B_{y} R\sin \theta + C_{y} s,\;\;\;\theta \in [0,2\pi ],\;s \in [0,H] \hfill \\ z = z_{\text{C}} + \left( {1 - \frac{s}{H}} \right)A_{z} R\cos \theta + \left( {1 - \frac{s}{H}} \right)B_{z} R\sin \theta + C_{z} s \hfill \\ \end{array} \right.$$

(56)

where θ and s are the two parameters. Point C(x_C, y_C, z_C) and R are the center and radius of the bottom circle, H is the height of the cone. And the s-axis is local one-dimensional axis along the height direction from center C to cone tip V. If s = 0, Eq. (56) gives us the bottom circle; while if s = H, we can get the cone tip.

Fig. 57

Cone in three-dimensional space

Appendix 2: Volume Integral of Arbitrary Sphere, Cylinder, and Cone

For an arbitrary sphere, cylinder, and cone in three-dimensional space, during the generation of the governing equation of motion in this study, the involved volume integral terms can be boiled down to $\iiint_{{}} {\left[ {{\varvec{T}}(x,y,z)} \right]^{\text{T}} dxdydz}$ and $\iiint_{{}} {\left[ {{\varvec{T}}(x,y,z)} \right]^{\text{T}} {\varvec{T}}(x,y,z)dxdydz}$. The former is related to the body force sub-matrix, while the latter is associated with the mass sub-matrix. The corresponding integrand is the linear function or the quadratic function of x, y, and z, respectively. Indeed, we only need to calculate the volume integrals of ten monomials: 1, x, y, z, x², y², z², yz, zx, and xy with respect to the global Cartesian coordinate system o–x–y–z. For this end, we first calculate the volume integrals of ten monomials: 1, u, v, w, u², v², w², uv, vw, and wu in regard to the local Cartesian coordinate system o*–u–v–w. Then, by means of the coordinate transformation to achieve the original purpose.

For a given sphere in three-dimensional space, by translating the origin of global Cartesian coordinate systemt to the center of the sphere, we get a local Cartesian coordinate system o*–u–v–w, as shown in Fig. 58. By using the local spherical coordinate system, we have

$$\begin{aligned} S_{{}}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {dudvdw} &= 4\pi R^{3} /3 \hfill \\ S_{u}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {ududvdw} &= 0 \hfill \\ S_{v}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {vdudvdw} &= 0 \hfill \\ S_{w}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {wdudvdw} &= 0 \hfill \\ S_{uu}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {u^{2} dudvdw} &= 4\pi R^{5} /15 \hfill \\ S_{vv}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {v^{2} dudvdw} &= 4\pi R^{5} /15 \hfill \\ S_{ww}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {w^{2} dudvdw} &= 4\pi R^{5} /15 \hfill \\ S_{vw}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {vwdudvdw} &= 0 \hfill \\ S_{wu}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {wududvdw} &= 0 \hfill \\ S_{uv}^{{\;{\text{Sphere}}}} &= \iiint_{\text{Sphere}} {uvdudvdw} &= 0, \hfill \\ \end{aligned}$$

(57)

where R is the radius of the sphere.

Fig. 58

Global and local Cartesian coordinate systems of a sphere

For a given cylinder in three-dimensional space, we can construct a local Cartesian coordinate system o*–u–v–w, as displayed in Fig. 59, where origin o* of local coordinate system is coincide with center C of the bottom circle of the cylinder. And the positive direction of w-axis is the same as the unit inner normal vector n(cosα, cosβ, cosγ) of the bottom circle. Acoording to the substitution rule for definite integral, we have

$$\begin{aligned} &\iiint_{\text{Cylinder}} {f(x,y,z)dxdydz}\\ &\quad = \iiint_{\text{Cylinder}} {f[x(u,v,w),y(u,v,w),z(u,v,w)]\left| {\varvec{J}} \right|dudvdw}, \end{aligned}$$

(58)

where J is the Jacobian matrix, reading

$${\varvec{J}} = \frac{\partial (x,y,z)}{\partial (u,v,w)} = \left[ {\begin{array}{*{20}c} {\frac{\partial x}{\partial u}} & {\frac{\partial x}{\partial v}} & {\frac{\partial x}{\partial w}} \\ {\frac{\partial y}{\partial u}} & {\frac{\partial y}{\partial v}} & {\frac{\partial y}{\partial w}} \\ {\frac{\partial z}{\partial u}} & {\frac{\partial z}{\partial v}} & {\frac{\partial z}{\partial w}} \\ \end{array} } \right].$$

(59)

Fig. 59

Global and local Cartesian coordinate systems of a cylinder

Substitution of the coordinate tranformation given by Eq. (50) into Eq. (59) yields $\left| {\varvec{J}} \right| \equiv 1$.

On the other hand, with respect to the local Cartesian coordinate system o*–u–v–w, the cylinder surface can be expressed as

$$\left\{ \begin{array}{l} u^{2} + v^{2} = R \hfill \\ w \in [0,H], \hfill \\ \end{array} \right.^{2}$$

(60)

where R and H are the radius of the bottom circle and the height of the cylinder, respectively. If the local cylindrical coordinate system is adopted, which means

$$\iiint_{\text{Cylinder}} {dudvdw} = \iiint_{\text{Cylinder}} {\rho d\rho d\theta dw},$$

(61)

where

$$\left\{ \begin{array}{l} u = \rho \cos \theta \hfill \\ v = \rho \sin \theta ,\;\;\;\rho \in [0,R] \hfill \\ w = w \hfill \\ \end{array} \right.,\;\theta \in [0,2\pi ].$$

(62)

Then, we have

$$\begin{aligned} S_{{}}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {dudvdw} = \pi R^{2} H \\ S_{u}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {ududvdw} = 0 \\ S_{v}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {vdudvdw} = 0 \\ S_{w}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {wdudvdw} = \pi R^{2} H^{\;2} /2 \\ S_{uu}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {u^{2} dudvdw} = \pi R^{4} H/4 \\ S_{vv}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {v^{2} dudvdw} = \pi R^{4} H/4 \\ S_{ww}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {w^{2} dudvdw} = \pi R^{2} H^{\;3} /3 \\ S_{vw}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {vwdudvdw} = 0 \\ S_{wu}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {wududvdw} = 0 \\ S_{uv}^{{\;{\text{Cylinder}}}} & = \iiint_{\text{Cylinder}} {uvdudvdw} = 0. \\ \end{aligned}$$

(63)

For a specified cone in three-dimensional space, we can also make a local Cartesian coordinate system o*–u–v–w, as shown in Fig. 60, in which the origin o* of local coordinate system is the center C of the bottom circle of the cone. In addition, the positive direction of w-axis is identical to the unit inner normal vector n(cosα, cosβ, cosγ), see Fig. 59. Obviously, there exsits the coordinate transformation described in Eq. (50) between the global and lobal Cartesian coordinate systems.

Fig. 60

Global and local Cartesian coordinate systems of a cone

In the local Cartesian coordinate system o*–u–v–w, the cone surface can be given by

$$\left\{ \begin{array}{l} u^{2} + v^{2} = \left(1 - \;\frac{w}{H}\right)^{2} R^{2} \hfill \\ w \in [0,H], \hfill \\ \end{array}{l} \right.$$

(64)

where R and H are the radius of the bottom circle and the height of the cone, respectively. By using the formula for definite integration by substitution and the local cylindrical coordinate system, we have

$$\begin{aligned} S_{{}}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {dudvdw} = \pi R^{2} H/3 \\ S_{u}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {ududvdw} = 0 \\ S_{v}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {vdudvdw} = 0 \\ S_{w}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {wdudvdw} = \pi R^{2} H^{\;2} /12 \\ S_{uu}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {u^{2} dudvdw} = \pi R^{4} H/20 \\ S_{vv}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {v^{2} dudvdw} = \pi R^{4} H/20 \\ S_{ww}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {w^{2} dudvdw} = \pi R^{2} H^{\;3} /30 \\ S_{vw}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {vwdudvdw} = 0 \\ S_{wu}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {wududvdw} = 0 \\ S_{uv}^{{\;{\text{Cone}}}} & = \iiint_{\text{Cone}} {uvdudvdw} = 0. \\ \end{aligned}$$

(65)

Ultimately, by substituting Eq. (50) into Eq. (58) and considering Eqs. (63) and (65), one can obtain the ten basic volume integrals over an arbitrary sphere, cylinder, or cone in regard to the global Cartesian coordinate system o–x–y-z, reading

$$\begin{aligned} S_{{}}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {dxdydz} = S_{{}}^{{\;{\text{Body}}}} \\ S_{x}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {xdxdydz} = x_{\text{C}} S_{{}}^{{\;{\text{Body}}}} + C_{x} S_{w}^{{\;{\text{Body}}}} \\ S_{y}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {ydxdydz} = y_{\text{C}} S_{{}}^{{\;{\text{Body}}}} + C_{y} S_{w}^{{\;{\text{Body}}}} \\ S_{z}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {zdxdydz} = z_{\text{C}} S_{{}}^{{\;{\text{Body}}}} + C_{z} S_{w}^{{\;{\text{Body}}}} \\ S_{xx}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {x^{2} dxdydz} = x_{\text{C}}^{ 2} S_{{}}^{{\;{\text{Body}}}} + 2x_{\text{C}} C_{x} S_{w}^{{\;{\text{Body}}}} + A_{x}^{2} S_{uu}^{{\;{\text{Body}}}} + B_{x}^{2} S_{vv}^{{\;{\text{Body}}}} { + }C_{x}^{2} S_{ww}^{{\;{\text{Body}}}} \\ S_{yy}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {y^{2} dxdydz} = y_{\text{C}}^{ 2} S_{{}}^{{\;{\text{Body}}}} + 2y_{\text{C}} C_{y} S_{w}^{{\;{\text{Body}}}} + A_{y}^{2} S_{uu}^{{\;{\text{Body}}}} + B_{y}^{2} S_{vv}^{{\;{\text{Body}}}} { + }C_{y}^{2} S_{ww}^{{\;{\text{Body}}}} \\ S_{zz}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {z^{2} dxdydz} = z_{\text{C}}^{ 2} S_{{}}^{{\;{\text{Body}}}} + 2z_{\text{C}} C_{z} S_{w}^{{\;{\text{Body}}}} + A_{z}^{2} S_{uu}^{{\;{\text{Body}}}} + B_{z}^{2} S_{vv}^{{\;{\text{Body}}}} { + }C_{z}^{2} S_{ww}^{{\;{\text{Body}}}} \\ S_{yz}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {yzdxdydz} = y_{\text{C}} z_{\text{C}}^{{}} S_{{}}^{{\;{\text{Body}}}} + y_{\text{C}} C_{z} S_{w}^{{\;{\text{Body}}}} + z_{\text{C}} C_{y} S_{w}^{{\;{\text{Body}}}} + A_{y}^{{}} A_{z}^{{}} S_{uu}^{{\;{\text{Body}}}} + B_{y}^{{}} B_{z}^{{}} S_{vv}^{{\;{\text{Body}}}} { + }C_{y}^{{}} C_{z}^{{}} S_{ww}^{{\;{\text{Body}}}} \\ S_{zx}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {zxdxdydz} = z_{\text{C}}^{{}} x_{\text{C}} S_{{}}^{{\;{\text{Body}}}} + z_{\text{C}} C_{x} S_{w}^{{\;{\text{Body}}}} + x_{\text{C}} C_{z} S_{w}^{{\;{\text{Body}}}} + A_{z}^{{}} A_{x}^{{}} S_{uu}^{{\;{\text{Body}}}} + B_{z}^{{}} B_{x}^{{}} S_{vv}^{{\;{\text{Body}}}} { + }C_{z}^{{}} C_{x}^{{}} S_{ww}^{{\;{\text{Body}}}} \\ S_{xy}^{{\;{\text{Body}}}} & { = }\iiint_{\text{Body}} {xydxdydz} = x_{\text{C}}^{{}} y_{\text{C}} S_{{}}^{{\;{\text{Body}}}} + x_{\text{C}} C_{y} S_{w}^{{\;{\text{Body}}}} + y_{\text{C}} C_{x} S_{w}^{{\;{\text{Body}}}} + A_{x}^{{}} A_{z}^{{}} S_{uu}^{{\;{\text{Body}}}} + B_{x}^{{}} B_{z}^{{}} S_{vv}^{{\;{\text{Body}}}} { + }C_{x}^{{}} C_{z}^{{}} S_{ww}^{{\;{\text{Body}}}} , \\ \end{aligned}$$

(66)

where the word “Body” stands for “Sphere”, “Cylinder”, “Cone”, or an assembly (to be explained in detail in “Appendix 3”). x_C, y_C, and z_C are the global coordinates of the origin of the local Cartesian coordinate system o*-u-v-w. The coefficients A_i, B_i, and C_i (i = x, y, z) are determined by Eq. (52) (For “Sphere”, γ ≡ 0). With the help of Eq. (66), we can readily generate the body force sub-matrix and the mass sub-matix associated with an arbitrary sphere, cylinder, or cone.

Appendix 3: Volume Integral of Some Representative Assemblies

In this study, an assembly is a combination of spheres, cyliners, or cones. Next, we will illustrate the calculation of the volume integral of several typical assemblies.

For an assembly (abbreviated as As-1) from one sphere and one cylinder, as shown in Fig. 61, in which R_S and R_C (≤ R_S) are the radiuses of the sphere and the cylinder, respectively. We can set up a local Cartesian coordinate system o*–u–v–w with the center of the sphere as the origin of system (see Fig. 61). The volume integral of the assembly can be calculated by

$$\begin{aligned} S_{{}}^{{\;{\text{As{-}1}}}} & = S_{{}}^{{\;{\text{Sphere}}}} + \pi \left( - \;\frac{2}{3}R_{{\;{\text{S}}}}^{\;3} + \frac{2}{3}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + R_{{\;{\text{C}}}}^{\;2} (H + h)\right) \\ S_{u}^{{\;{\text{As{-}1}}}} & = S_{v}^{{\;{\text{As{-}1}}}} = 0 \\ S_{w}^{{\;{\text{As{-}1}}}} & = S_{w}^{{\;{\text{Sphere}}}} + \pi \left(\frac{1}{4}R_{{\;{\text{C}}}}^{\;4} + \frac{1}{2}R_{{\;{\text{C}}}}^{\;2} ((H + h)^{2} - R_{{\;{\text{S}}}}^{\;2} )\right) \\ S_{uu}^{{\;{\text{As{-}1}}}} & = S_{uu}^{{\;{\text{Sphere}}}} + \pi \left( - \;\frac{2}{15}R_{{\;{\text{S}}}}^{\;5} + \frac{1}{5}R_{{\;{\text{C}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + \frac{2}{15}R_{{\;{\text{S}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + \frac{1}{4}R_{{\;{\text{C}}}}^{\;4} (H + h)\right) \\ S_{v\;v}^{{\;{\text{As{-}1}}}} & = S_{uu}^{{\;{\text{As{-}1}}}} \\ S_{ww}^{{\;{\text{As{-}1}}}} & = S_{ww}^{{\;{\text{Sphere}}}} + \pi \left( - \;\frac{1}{15}R_{{\;{\text{S}}}}^{\;5} + \frac{1}{15}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{5/2} + \frac{1}{6}R_{{\;{\text{C}}}}^{\;2} (H^{\;3} + h^{\;3} ) + \frac{1}{2}R_{{\;{\text{C}}}}^{\;2} (H^{\;2} + h^{\;2} )\right) \\ S_{vw}^{{\;{\text{As{-}1}}}} & = S_{wu}^{{\;{\text{As{-}1}}}} = S_{uv}^{{\;{\text{As{-}1}}}} = 0, \\ \end{aligned}$$

(67)

where H is the height of the cylinder, and h is the distance between the center of the sphere and the center of the bottom circle of the cylinder.

Fig. 61

Assembly (As-1) from one sphere and one cylinder

Consider an assembly (abbreviated as As-2) from one sphere and one cone, as exhibited in Fig. 62, where R_S and R_C (≤ R_S) are the radiuses of the sphere and the cone, respectively. Point A is the cone tip. We can make a local Cartesian coordinate system o*–u–v–w taking the center of the sphere as the origin of system (see Fig. 62). The volume integral of the assembly can be obtained from

$$\begin{aligned} S_{{}}^{{\;{\text{As{- }2}}}} & = S_{{}}^{{\;{\text{Sphere}}}} + \pi \left( - \;\frac{2}{3}R_{{\;{\text{S}}}}^{\;3} + \frac{2}{3}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + \frac{1}{3}R_{{\;{\text{C}}}}^{\;2} H + R_{{\;{\text{C}}}}^{\;2} h\right) \\ S_{u}^{{\;{\text{As{- }2}}}} & = S_{v}^{{\;{\text{As{- }2}}}} = 0 \\ S_{w}^{{\;{\text{As{- }2}}}} & = S_{w}^{{\;{\text{Sphere}}}} + \pi \left(\frac{1}{4}R_{{\;{\text{C}}}}^{\;4} + \frac{1}{4}R_{{\;{\text{C}}}}^{\;2} H^{2} - \frac{2}{3}R_{{\;{\text{C}}}}^{\;2} H(H + h) + \frac{1}{2}R_{{\;{\text{C}}}}^{\;2} ((H + h)^{2} - R_{{\;{\text{S}}}}^{\;2} )\right) \\ S_{uu}^{{\;{\text{As{- }2}}}} & = S_{uu}^{{\;{\text{Sphere}}}} + \pi \left( - \;\frac{2}{15}R_{{\;{\text{S}}}}^{\;5} + \frac{1}{20}R_{{\;{\text{C}}}}^{\;4} H + \frac{2}{15}R_{{\;{\text{S}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + \frac{1}{5}R_{{\;{\text{C}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + \frac{1}{4}R_{{\;{\text{C}}}}^{\;4} h\right) \\ S_{v\;v}^{{\;{\text{As{- }2}}}} & = S_{uu}^{{\;{\text{As{- }2}}}} \\ S_{ww}^{{\;{\text{As{- }2}}}} & = S_{ww}^{{\;{\text{Sphere}}}} + \pi \left( - \;\frac{2}{15}R_{{\;{\text{S}}}}^{\;5} + \frac{1}{30}R_{{\;{\text{C}}}}^{\;2} H^{\;3} + \frac{2}{15}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{5/2} + \frac{1}{3}R_{{\;{\text{C}}}}^{\;2} h^{\;2} (H + h) + \frac{1}{6}R_{{\;{\text{C}}}}^{\;2} H^{\;2} h\right) \\ S_{vw}^{{\;{\text{As{- }2}}}} & = S_{wu}^{{\;{\text{As{- }2}}}} = S_{uv}^{{\;{\text{As{- }2}}}} = 0, \\ \end{aligned}$$

(68)

where H is the height of the cylinder, and h is the distance between the center of sphere and the center of the bottom circle of the cone.

Fig. 62

Assembly (As-2) from one sphere and one cone

As for the assembly (referred to as As-3) from one sphere and one cone-tip, as presented in Fig. 63. Here, R_S and R_C are the radiuses of the sphere and the cone, respectively. While point A is the cone tip. Choosing the center of the bottom circle of the cone as the origin, we can construct a local Cartesian coordinate system o*–u–v–w (see Fig. 63). The volume integral of the overlap region (abbreviated as Ov-1) is given by

$$\begin{aligned} S_{{}}^{{\;{\text{Ov{-}1}}}} & = \pi \left( - \;\frac{2}{3}R_{{\;{\text{S}}}}^{\;3} + \frac{2}{3}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} - \frac{2}{3}R_{{\;{\text{C}}}}^{\;2} H - R_{{\;{\text{C}}}}^{\;2} h\right) \\ S_{u}^{{\;{\text{Ov{-}1}}}} & = S_{v}^{{\;{\text{Ov{-}1}}}} = 0 \\ S_{w}^{{\;{\text{Ov{-}1}}}} & = \pi \left( - \frac{2}{3}R_{{\;{\text{S}}}}^{\;3} (H + h) - \frac{5}{12}R_{{\;{\text{C}}}}^{\;2} H^{2} - \frac{1}{2}R_{{\;{\text{C}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} + h^{2} ) + \frac{2}{3}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} (H + h) + \frac{1}{4}R_{{\;{\text{C}}}}^{\;4} - R_{{\;{\text{C}}}}^{\;2} Hh\right) \\ S_{uu}^{{\;{\text{Ov{-}1}}}} & = \pi \left( - \;\frac{2}{15}R_{{\;{\text{S}}}}^{\;5} - \frac{1}{5}R_{{\;{\text{C}}}}^{\;4} H + \frac{2}{15}R_{{\;{\text{S}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} + \frac{1}{5}R_{{\;{\text{C}}}}^{\;2} (R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} - \frac{1}{4}R_{{\;{\text{C}}}}^{\;4} h\right) \\ S_{v\;v}^{{\;{\text{Ov{-}1}}}} & = S_{uu}^{{\;{\text{Ov{-}1}}}} \\ S_{ww}^{{\;{\text{Ov{-}1}}}} & = \pi \left( - \;\frac{2}{15}R_{{\;{\text{S}}}}^{\;5} - \;\frac{2}{3}R_{{\;{\text{S}}}}^{\;3} (H + h)^{\;2} + \frac{2}{15}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{5/2} + \frac{2}{3}(R_{{\;{\text{S}}}}^{\;2} - R_{{\;{\text{C}}}}^{\;2} )^{3/2} (H + h)^{\;2} - \frac{1}{3}R_{{\;{\text{C}}}}^{\;2} h^{\;3}\right. + \\ & \quad \left.\frac{1}{2}R_{{\;{\text{C}}}}^{\;4} (H + h) - \frac{3}{10}R_{{\;{\text{C}}}}^{\;2} H^{\;3} - R_{{\;{\text{C}}}}^{\;2} Hh^{\;2}\right) \\ S_{vw}^{{\;{\text{Ov{-}1}}}} &= S_{wu}^{{\;{\text{Ov{-}1}}}} = S_{uv}^{{\;{\text{Ov{-}1}}}} = 0, \\ \end{aligned}$$

(69)

where H is the height of the cylinder, and h is the distance between the center of sphere and the cone tip. Further, the volume integral of the assembly can be expressed as

$$\begin{aligned} S_{{}}^{{\;{\text{As{-}3}}}} & = S_{{}}^{{\;{\text{Cone}}}} + S_{{}}^{{\;{\text{Sphere}}}} - S_{{}}^{{\;{\text{Ov{-}1}}}} \\ S_{u}^{{\;{\text{As{-}3}}}} & = S_{v}^{{\;{\text{As{-}3}}}} = 0 \\ S_{w}^{{\;{\text{As{-}3}}}} & = S_{w}^{{\;{\text{Cone}}}} + S_{w}^{{\;{\text{Sphere}}}} - S_{w}^{{\;{\text{Ov{-}1}}}} \\ S_{uu}^{{\;{\text{As{-}3}}}} & = S_{uu}^{{\;{\text{Cone}}}} + S_{uu}^{{\;{\text{Sphere}}}} - S_{uu}^{{\;{\text{Ov{-}1}}}} \\ S_{v\;v}^{{\;{\text{As{-}3}}}} & = S_{uu}^{{\;{\text{As{-}3}}}} \\ S_{ww}^{{\;{\text{As{-}3}}}} & = S_{ww}^{{\;{\text{Cone}}}} + S_{ww}^{{\;{\text{Sphere}}}} - S_{ww}^{{\;{\text{Ov{-}1}}}} \\ S_{vw}^{{\;{\text{As{-}3}}}} & = S_{wu}^{{\;{\text{As{-}3}}}} = S_{uv}^{{\;{\text{As{-}3}}}} = 0, \\ \end{aligned}$$

(70)

Fig. 63

Assembly (As-3) from one sphere and one cone-tip

Now, let us see the assembly (called As-4) in Fig. 64, which is the combination of one sphere and two cylinders. The radiuses of the three bodies are equal to each other. For the two cylinder, there is a one of the bottom circles that just coincides with a certain center section of the sphere (see Fig. 64). We can make two tangent planes of the sphere, which are respectively perpendicular to the two cylinders (see the right-hand side of Fig. 64). Then, we can construct two shorter cylinders, i.e. the shorter cyliner1 (referred to as Sc-1) and the shorter cyliner2 (referred to as Sc-2). Thus, the volume integral of the assembly can be approximated by

$$\begin{aligned} S_{{}}^{{\;{\text{As{-}4}}}} & = S_{{}}^{{\;{\text{Sc{-}1}}}} + S_{{}}^{{\;{\text{Sc{-} 2}}}} + S_{{}}^{{\;{\text{Sphere}}}} \\ S_{u}^{{\;{\text{As{-}4}}}} & = S_{v}^{{\;{\text{As{-}4}}}} = 0 \\ S_{w}^{{\;{\text{As{-}4}}}} & = S_{w}^{{\;{\text{Sc}} - 1}} + S_{w}^{{\;{\text{Sc}} - 2}} + S_{w}^{{\;{\text{Sphere}}}} \\ S_{uu}^{{\;{\text{As{-}4}}}} & = S_{uu}^{{\;{\text{Sc}} - 1}} + S_{uu}^{{\;{\text{Sc}} - 2}} + S_{uu}^{{\;{\text{Sphere}}}} \\ S_{v\;v}^{{\;{\text{As{-}4}}}} & = S_{uu}^{{\;{\text{As{-}4}}}} \\ S_{ww}^{{\;{\text{As{-}4}}}} & = S_{ww}^{{\;{\text{Sc}} - 1}} + S_{ww}^{{\;{\text{Sc}} - 2}} + S_{ww}^{{\;{\text{Sphere}}}} \\ S_{vw}^{{\;{\text{As{-}4}}}} & = S_{wu}^{{\;{\text{As{-}4}}}} = S_{uv}^{{\;{\text{As{-}4}}}} = 0, \\ \end{aligned}$$

(71)

Fig. 64

Assembly (As-4) from one sphere and two cylinders

For the assembly from two completely holing-through cylinders, whose axes intersect with each other at point A with an angle φ, as demonstrated in Fig. 65. Here, R₁ and R₂ are their radiuses, respetively. And R₁ ≤ R₂. By choosing the center of bottom circle of the cylinder with the radius R₁ as the origin, we have a local Cartesian coordinate system o*–u–v–w (see Fig. 65). H is the distance between points A and o*. The volume integral of the overlap region (referred to as Ov-2) is given by

$$\begin{aligned} S_{{}}^{{\;{\text{Ov - 2}}}} & = \;\frac{16}{3\cos \varphi }R_{\;1}^{\;3} \\ S_{u}^{{\;{\text{Ov - 2}}}} & = S_{v}^{{\;{\text{Ov - 2}}}} = 0 \\ S_{w}^{{\;{\text{Ov - 2}}}} & = \frac{16}{3\cos \varphi }R_{\;1}^{\;3} H \\ S_{uu}^{{\;{\text{Ov - 2}}}} & = \frac{64}{45\cos \varphi }R_{\;1}^{\;5} \\ S_{v\;v}^{{\;{\text{Ov - 2}}}} & = \frac{64}{15\cos \varphi }R_{\;1}^{\;5} \\ S_{ww}^{{\;{\text{Ov - 2}}}} & = \frac{16}{{45\cos^{3} \varphi }}R_{\;1}^{\;3} \left(15H^{2} \cos^{2} \varphi + 4R_{\;1}^{\;2} (1 + \sin^{2} \varphi )\right) \\ S_{wu}^{{\;{\text{Ov - 2}}}} & = - \frac{64\sin \varphi }{{45\cos^{2} \varphi }}R_{\;1}^{\;5} \\ S_{vw}^{{\;{\text{Ov - 2}}}} & = S_{uv}^{{\;{\text{Ov - 2}}}} = 0. \\ \end{aligned}$$

(72)

Fig. 65

Assembly (As-5) from two cylinders (completely holing-through)

Then, according to the addition rule we can easily calculate the volume integral of the assembly.

If the cylinder 1 is just pasted on the suface of the cylinder 2, as presented in Fig. 66. We can make a shorter cylinder 1 with height H = |AB| − h = |AB| − R₁ctgφ − R₂sinφ (see the right-hand side of Fig. 66). Here, point A and angle φ are the intersection point and the intersection angle between their axes, respectively. Obviously, the volume integral of the assembly (referred to as As-6) is approximated to the sum of that of the shorter cylinder 1 and the cylinder 2.

Fig. 66

Assembly (As-6) from two cylinders (R₁ ≤ R₂)

Figure 67 shows an assembly (named as As-7) from one sphere and one cylinder. Here, R₁ and R₂ (R₁ ≥ R₂/2) are the radiuses of the cylinder and the sphere, respectively. For this assembly, the center O of the sphere is always on the surface of the cylinder in current study. According to the geometrical relationship, we can determine the central angle φ (see the right-hand side of Fig. 67). After this, the volume integral of the assembly can be computed approximately by

$$\begin{aligned} S_{{}}^{{\;{\text{As{-}7}}}} & = S_{{}}^{{\;{\text{Cylinder}}}} + 0.5\varphi S_{{}}^{{\;{\text{Sphere}}}} /\pi \\ S_{u}^{{\;{\text{As{-}7}}}} & = S_{v}^{{\;{\text{As{-}7}}}} = 0 \\ S_{w}^{{\;{\text{As{-}7}}}} & = S_{w}^{{\;{\text{Cylinder}}}} + 0.5\varphi S_{w}^{{\;{\text{Sphere}}}} /\pi \\ S_{uu}^{{\;{\text{As{-}7}}}} & = S_{uu}^{{\;{\text{Cylinder}}}} + 0.5\varphi S_{uu}^{{\;{\text{Sphere}}}} /\pi \\ S_{v\;v}^{{\;{\text{As{-}7}}}} & = S_{uu}^{{\;{\text{As{-}7}}}} \\ S_{ww}^{{\;{\text{As{-}7}}}} & = S_{ww}^{{\;{\text{Cylinder}}}} + 0.5\varphi S_{ww}^{{\;{\text{Sphere}}}} /\pi \\ S_{vw}^{{\;{\text{As{-}7}}}} & = S_{wu}^{{\;{\text{As{-}7}}}} = S_{uv}^{{\;{\text{As{-}7}}}} = 0, \\ \end{aligned}$$

(73)

Fig. 67

Assembly (As-7) from one sphere and one cylinder (R₁ ≥ R₂/2)

Consider the assembly (referred to as As-8) from one sphere and one cone, as indicated in Fig. 68. For this asssebmly, we always set the center O of the sphere to be on the surface of the cone. Passing through point O a smaller circular section can be obtained. Similar to the assembly As-7, after getting the central angle φ (see the right-hand side of Fig. 68), the volume integral of the assembly can be determined approximately by

$$\begin{aligned} S_{{}}^{{\;{\text{As{-}8}}}} & = S_{{}}^{{\;{\text{Cone}}}} + 0.5\varphi S_{{}}^{{\;{\text{Sphere}}}} /\pi \\ S_{u}^{{\;{\text{As{-}8}}}} & = S_{v}^{{\;{\text{As{-}8}}}} = 0 \\ S_{w}^{{\;{\text{As{-}8}}}} & = S_{w}^{{\;{\text{Cone}}}} + 0.5\varphi S_{w}^{{\;{\text{Sphere}}}} /\pi \\ S_{uu}^{{\;{\text{As{-}8}}}} & = S_{uu}^{{\;{\text{Cone}}}} + 0.5\varphi S_{uu}^{{\;{\text{Sphere}}}} /\pi \\ S_{v\;v}^{{\;{\text{As{-}8}}}} & = S_{uu}^{{\;{\text{As{-}8}}}} \\ S_{ww}^{{\;{\text{As{-}8}}}} & = S_{ww}^{{\;{\text{Cone}}}} + 0.5\varphi S_{ww}^{{\;{\text{Sphere}}}} /\pi \\ S_{vw}^{{\;{\text{As{-}8}}}} & = S_{wu}^{{\;{\text{As{-}8}}}} = S_{uv}^{{\;{\text{As{-}8}}}} = 0, \\ \end{aligned}$$

(74)

Fig. 68

Assembly (As-8) from one sphere and one cone

Now, let su see the two simple assemblies that is respectively from one cylinder + one cone, or one clinder + one cone, as shown in Fig. 69. For these assemblies, the overlap cones can determined with easy. Based on this, we can compute the volume integrals of the assemblies by applying the addition rule of definite integral.

Fig. 69

Two simple assemblies

After getting S, S_u, S_v, S_w, S_uu, S_vv, S_ww, S_uv, S_vw, and S_wu, one can easily compute S, S_x, S_y, S_z, S_xx, S_yy, S_zz, S_xy, S_yz, and S_zx acoording to Eq. (66).

Appendix 4: Some Sub-Functions

When two bodies get closer and closer, the contact points depend on the closest points between them. To find the relevant closest points, the algebraic or geometric algorithms will be given in this section.

4.1 Closest Point Between Point and Sphere Surface

The problem of the closest points between two sphere surfaces can be reduced to this topic.

As shown in Fig. 70, point A(x_A, y_A, z_A) is outside of sphere O(x_O, y_O, z_O). Point B(x_B, y_B, z_B) is the undetermined closest point. The algebraic algorithm to find point B as follows:

$$\left\{ \begin{array}{l} x_{\text{B}} = x_{\text{O}} + (x_{\text{A}} - x_{\text{O}} )t_{\text{B}} \hfill \\ y_{\text{B}} = y_{\text{O}} + (y_{\text{A}} - y_{\text{O}} )t_{\text{B}} \hfill \\ z_{\text{B}} = z_{\text{O}} + (z_{\text{A}} - z_{\text{O}} )t_{\text{B}} \hfill \\ t_{\text{B}} = R/\left| {\text{OA}} \right|. \hfill \\ \end{array} \right.$$

(75)

where t_B is the parameter associated with point B. R is the radius of the sphere.

Fig. 70

Closest point from point to sphere surface

4.2 Closest Point Between Point and Cylinder Surface

The problem of the closest points between sphere surface and cylinder surface can be boiled down to this topic.

In Fig. 71, points C and C* are the centers of the bottom and upper circles of a cylinder. Point B B(x_B, y_B, z_B) is the undetermined closest point between point A(x_A, y_A, z_A) and the cylinder surface.

Fig. 71

Closest point from point to cylinder surface

The algebraic algorithm

• i. Construct a function of parameters θ and s

  $$f(\theta ,s) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (76)

  where the coordinate components x_B, y_B, and z_B all can be expressed by two parameters θ and s (see Eq. 55).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{\partial f(\theta ,s)}{\partial \theta } = 0 \hfill \\ \frac{\partial f(\theta ,s)}{\partial s} = 0 \hfill \\ \end{aligned} \right.$$

  (77)

  Note that Eq. (77) is a system of nonlinear equations.

• iii. Determine point B by applying parameters θ and s to the parameter equation of cylinder surface (Eq. 55).

The geometric algorithm

• i. Find the foot of perpendicular D of point A on line CC*.

• ii. Calculate parameter t_B associated with point B on line AD.

  $$t_{\text{B}} = R/\left| {\text{DA}} \right|$$

  (78)

  where R is the radius of the bottom circle of the cylinder.

• iii. Determine point B by applying parameter t_B to the parameter equation of point (Eq. 48).

4.3 Closest Point Between Point and Cone Surface

The problem of the closest points between sphere surface and cone surface can be transformed to this item.

Points C, V, radius R, and height H determine a cone surface, as presented in Fig. 72. Points A, C, and V are three known points. Point B(x_B, y_B, z_B) is the undetermined closest point on the cone surface between point A and the cone surface.

Fig. 72

Closest point from point to cone surface

The algebraic algorithm

• i. Construct a function of parameters θ and s

  $$f(\theta ,s) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (79)

  where the coordinate components x_B, y_B, and z_B all can be expressed by two parameters θ and s (see Eq. 56).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{\partial f(\theta ,s)}{\partial \theta } = 0 \hfill \\ \frac{\partial f(\theta ,s)}{\partial s} = 0 \hfill \\ \end{aligned} \right.$$

  (80)

  where Eq. (80) is also a system of nonlinear equations.

• iii. Determine point B by applying parameters θ and s to the parameter equation of cone surface (Eq. 56).

The geometric algorithm

It is easily be known that points A, B, C, D, E, and V are coplanar and ∆ABV is a right triangle.

• i. Calculate angle ∠AVC through two intersection lines VA and VC.

• ii. Estimate two distances |VE| and |AB|

  $$\begin{aligned} \left| {\text{VE}} \right| = \frac{{\left| {\text{VA}} \right|\sin (\pi /2 - \angle {\text{AVC}} + \arctan (R/H))}}{\sin (\pi /2 - \arctan (R/H))} \hfill \\ \left| {\text{AB}} \right| = \left| {\text{VA}} \right|\sin (\angle {\text{AVC}} - \arctan (R/H)) \hfill \\ \end{aligned}$$

  (81)

• iii. Find point E by using parameter t_E = |VE|/|VC| and Eq. (48).

• iv. Determine point B applying parameter t_B = |AB|/|AE| to Eq. (48).

4.4 Closest Point Between Point and Circle

This topic includes the closest point between sphere surface and circle.

In Fig. 73 line CD is perpendicular to the plane of circle C(x_C, y_C, z_C), point A is not on line CD, which is parallel to the unit normal vector n. Point B(x_B, y_B, z_B) is unknown closest point between the circle and point A(x_A, y_A, z_A). The algebraic algorithm is as follows:

Fig. 73

Closest point from point to circle

• i. Construct a function of parameter θ

  $$f(\theta ) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (82)

  where the coordinate components x_B, y_B, and z_B all can be expressed by parameter θ (see Eq. 54).

• ii. Solve the minimum condition

  $$\frac{\partial f(\theta )}{\partial \theta } = 0$$

  (83)

  Here Eq. (83) is a quartic equation of tan(θ/2), which can be solved analytically.

• iii. Determine point B using the parameter equation of circle (Eq. 54) with parameter θ.

4.5 Closest Points Between Circle and Line

Line AE is outside of circle C(x_C, y_C, z_C), as shown in Fig. 74. Points B(x_B, y_B, z_B) and D(x_D, y_D, z_D), which belong to line AE and circle C, respectively, are the two closest points that are undetermined. This means that the distance between them is the shortest. To find points B and D, The algebraic algorithm:

Fig. 74

Closest points from circle to line

• i. Construct a function of parameters θ and t

  $$f(\theta ,t) = (x_{\text{B}} - x_{\text{D}} )^{2} + (y_{\text{B}} - y_{\text{D}} )^{2} + (z_{\text{B}} - z_{\text{D}} )^{2}$$

  (84)

  where the coordinate components x_B, y_B, and z_B all can be expressed by parameter t (see Eq. 48). While the coordinate components x_D, y_D, and z_D can be described by using parameter θ (see Eq. 54).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{\partial f(\theta ,t)}{\partial \theta } = 0 \hfill \\ \frac{\partial f(\theta ,t)}{\partial t} = 0 \hfill \\ \end{aligned} \right.$$

  (85)

  Equation (85) is a system of nonlinear equations.

• iii. Determine point B using parameter t and Eq. (48), determine point D by applying parameter t to Eq. (54).

It is a time-consuming process to solve the system of nonlinear equations. In order to improve compute efficiency, we propose the following geometric algorithm (see Fig. 75):

Fig. 75

Projection iterative method for finding the closest points between circle and line

• i. Find the foot of perpendicular B₁ of point C on line AE.

• ii. Calculate the closest point D₁ between point B₁ and circle C (refer to Fig. 73).

• iii. Find the foot of perpendicular B₂ of point D₁ on line AE.

• iv. Repeat step i–iii until |B_n−1B_n| ≤ 10⁻⁶.

In this study, the geometric algorithm is referred to as “projection iterative method”. Our numerical experiments show that a satisfactory solution can be obtained within ten iterative steps and that the geometric algorithm is faster than the above algebraic algorithm in most cases.

4.6 Closest Points Between Circle and Plane

In Fig. 76, points A(x_A, y_A, z_A) and B(x_B, y_B, z_B) are the undetermined closest points between circle C(x_C, y_C, z_C) and place DEFG.

Fig. 76

Closest points from circle to plane

The algebraic algorithm

• i. Construct a function of parameter θ, λ, and μ

  $$f(\theta ,\lambda ,\mu ) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (86)

  where the coordinate components x_A, y_A, and z_A can be given by parameter θ (see Eq. 54). The coordinate components x_B, y_B, and z_B are related to parameters λ and μ (see Eq. 49).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{\partial f(\theta ,\lambda ,\mu )}{\partial \theta } = 0 \hfill \\ \frac{\partial f(\theta ,\lambda ,\mu )}{\partial \lambda } = 0 \hfill \\ \frac{\partial f(\theta ,\lambda ,\mu )}{\partial \mu } = 0 \hfill \\ \end{aligned} \right.$$

  (87)

  Here, Eq. (87) is a system of nonlinear equations.

• iii. Determine point A by using parameter θ (Eq. 54) and point B though parameters λ and μ (see Eq. 49).

The geometric algorithm I (“Projection iterative method”, see Fig. 77)

Fig. 77

Geometric algorithm I for finding the closest points between circle and plane

• i. Find the foot of perpendicular B₁ of point C on plane DEFG.

• ii. Calculate the closest point A₁ between point B₁ and circle C (refer to Fig. 73).

• iii. Find the foot of perpendicular B₂ of point A₁ on plane DEFG.

• iv. Repeat step i–iii until |B_n−1B_n| ≤ 10⁻⁶.

It has been shown that a satisfactory solution can be obtained within twenty iterative steps and that the geometric algorithm I is always faster than the corresponding algebraic algorithm.

The geometric algorithm II (see Fig. 78)

Fig. 78

Geometric algorithm II for finding the closest points between circle and plane

• i. Construt a point C₁ through point C along normal vector n_P of plane DEFG.

• ii. Construt a point C₂ through point C along normal vector n_C of circle C.

• iii. Calculate two intersection points A* and A between plane CC₁C₂ and circle C. Here, point A is just the one of the closest points.

• iv. Find the foot of perpendicular B of point A on plane DEFG.

Points A and B are the closest points between the circle and the plane.

4.7 Closest Points Between Two Circles

Points A(x_A, y_A, z_A) and B(x_B, y_B, z_B) are the undetermined closest points between circles C₁ and C₂, as depicted in Fig. 79.

Fig. 79

Closest points from circle to circle

The algebraic algorithm

• i. Construct a function of parameters θ₁ and θ₂

  $$f(\theta_{1} ,\theta_{2} ) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (88)

  where the coordinate components x_A, y_A, and z_A can be given by parameter θ₁, and the coordinate components x_B, y_B, and z_B are related to parameter θ₂ (see Eqs. 3 or 4).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{{\partial f(\theta_{1} ,\theta_{2} )}}{{\partial \theta_{1} }} = 0 \hfill \\ \frac{{\partial f(\theta_{1} ,\theta_{2} )}}{{\partial \theta_{2} }} = 0 \hfill \\ \end{aligned} \right.$$

  (89)

  Here, Eq. (90) is a system of nonlinear equations.

• iii. Determine points A and B by using parameters θ₁ and θ₂, respectively (see Eqs. 3 or 4).

The geometric algorithm (“Projection iterative method”, see Fig. 80)

Fig. 80

Projection iterative method for finding the closest points between two circles

• i. Find the foot of perpendicular D of point C₂ on the plane of circle C₁.

• ii. Calculate the closest point B₁ between point D and circle C₁ (refer to Fig. 73).

• iii. Calculate the closest point A₁ between point B₁ and circle C₂ (refer to Fig. 73).

• iv. Calculate the closest point B₂ between point A₁ and circle C₁(refer to Fig. 73).

• v. Repeat step iii–iv until |B_n−1B_n| ≤ 10⁻⁶.

It has been shown that a satisfactory solution can be obtained within about ten iterative steps and that the projection iterative method is always faster than the above algebraic algorithm.

4.8 Closest Points Between Circle and Cylinder Surface

In Fig. 81, points A(x_A, y_A, z_A) and B(x_B, y_B, z_B) are the two undetermined closest points between circle C₁ and a cylinder surface.

Fig. 81

Closest points from circle to cylinder surface (algebraic algorithm)

The algebraic algorithm

• i. Construct a function of parameters θ₁, θ₂, and s

  $$f(\theta_{1} ,\theta_{2} ,s) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (90)

  where the coordinate components x_A, y_A, and z_A can be given by parameter θ₁ (see Eq. 54). The coordinate components x_B, y_B, and z_B are associated with parameters θ₂ and s (see Eq. 55).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{{\partial f(\theta_{1} ,\theta_{2} ,s)}}{{\partial \theta_{1} }} = 0 \hfill \\ \frac{{\partial f(\theta_{1} ,\theta_{2} ,s)}}{{\partial \theta_{2} }} = 0 \hfill \\ \frac{{\partial f(\theta_{1} ,\theta_{2} ,s)}}{\partial s} = 0 \hfill \\ \end{aligned} \right.$$

  (91)

  Equation (91) is a system of nonlinear equations.

• iii. Determine points A and B by using parameter θ₁ (see Eq. 54), and θ₂ and s, respectively (see Eq. 55).

The geometric algorithm (see Fig. 82)

Fig. 82

Closest points from circle to cylinder surface (geometry algorithm)

• i. Find the closest point A between circle C₁ and axis line C₂C* of the cylinder surface (refer to Figs. 74 or 75).

• ii. Calculate the closest point B between point A and the cylinder surface (refer to Fig. 71).

• iii. This geometric algorithm is very efficient.

4.9 Closest Points Between Circle and Cone Surface

Points A(x_A, y_A, z_A) and B(x_B, y_B, z_B) are the two undetermined closest points between circle C₁ and a cone surface, as shown in Fig. 83.

Fig. 83

Closest points from circle to cone surface

The algebraic algorithm

• i. Construct a function of parameters θ₁, θ₂, and s

  $$f(\theta_{1} ,\theta_{2} ,s) = (x_{\text{B}} - x_{\text{A}} )^{2} + (y_{\text{B}} - y_{\text{A}} )^{2} + (z_{\text{B}} - z_{\text{A}} )^{2}$$

  (92)

  where the coordinate components x_A, y_A, and z_A can be given by parameter θ₁ (see Eq. 54). The coordinate components x_B, y_B, and z_B are associated with parameters θ₂ and s (see Eq. 56).

• ii. Solve the minimum condition

  $$\left\{ \begin{aligned} \frac{{\partial f(\theta_{1} ,\theta_{2} ,s)}}{{\partial \theta_{1} }} = 0 \hfill \\ \frac{{\partial f(\theta_{1} ,\theta_{2} ,s)}}{{\partial \theta_{2} }} = 0 \hfill \\ \frac{{\partial f(\theta_{1} ,\theta_{2} ,s)}}{\partial s} = 0 \hfill \\ \end{aligned} \right.$$

  (93)

  Equation (93) is a system of nonlinear equations.

• iii. Determine points A and B by using parameter θ₁ (see Eq. 54), and parameters θ₂ and s (see Eq. 55), respectively.

The geometric algorithm (“Projection iterative method”, see Fig. 84)

Fig. 84

Projection iterative method for the closest points between circle and cone surface

• i. Find the closest point A₁ between circle C₁ and axis C₂V of the cone surface.

• ii. Calculate the closest point B₁ between point A₁ and the cone surface.

• iii. Find the closest point A₂ between point B₁ and circle C₁.

• iv. Repeat step ii–iii until |B_n−1B_n| ≤ 10⁻⁶.

Numerical experiments show that a satisfactory solution can be obtained within about ten iterative steps. The projection iterative method is faster than the above algebraic algorithm.