JP3308685B2 - Object collision simulation method and apparatus - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a method and apparatus for simulating the collision of an object in computer graphics, in particular, in animation or virtual reality.

[0002]

2. Description of the Related Art When simulating collision between virtual objects in a virtual three-dimensional space constructed inside a computer, conventionally, the shape of the objects is regarded as a mass point or a sphere, and a coefficient of restitution and a coefficient of friction between the objects are considered. The calculation was performed without considering it as a rigid body without friction.

[0003]

A conventional method of performing a collision simulation by regarding the shape of an object as a mass point or a sphere is simple in calculation, but cannot accurately calculate an object having a complicated shape. Is clear. Further, the method that does not consider the coefficient of restitution or the coefficient of friction between objects has a drawback that it cannot cope with various actual collision phenomena.

SUMMARY OF THE INVENTION The present invention has been made in view of the above conventional example, and has as its object to provide an object collision simulation system and an apparatus thereof capable of approximating the motion of an object at the time of collision between objects with high accuracy.

[0005]

In order to achieve the above object, a method for simulating collision of an object according to the present invention is a virtual object simulation method.
Simulate collision between virtual objects in 3D space
A method of tio down, based on the state data of the virtual object at the time of collision with the characteristic data of the virtual object, the calculation step of calculating the state data of the virtual object after the collision, of the virtual object after the collision based on the status data, and a display step of displaying the status of the virtual object after the collision, the calculation
Of coefficient of restitution and friction between virtual objects in process
If both are other than 0 or 1, the coefficient of restitution and friction
The impulse is calculated by interpolation using impulse when the coefficients are 0 and 1.
Calculate post-crash state data .

A collision simulation apparatus according to the present invention for achieving the above object has the following configuration.
That is, collision between virtual objects in a virtual three-dimensional space
An apparatus for simulating, on the basis of the state data of the virtual object at the time of collision with the characteristic data of the virtual object, and calculating means for calculating a state of the virtual object after the collision, the state of the virtual object after the collision based on the data, and display means for displaying the status of the virtual object after the collision,
Coefficient of restitution and coefficient of friction between virtual objects in the calculation means
If at least one of is not 0 or 1, the coefficient of restitution
Using impulse when friction coefficient is 0 and 1
To calculate the state data after the collision .

[0007]

SUMMARY OF] According to the above configuration, based on the state data of the virtual object at the time of collision with the characteristic data of the virtual object, is calculated state data of the virtual object after the collision, based on the calculated state data, collision The state of the subsequent virtual object is displayed.
Here, in the above calculation, if at least one of the coefficient of restitution and the coefficient of friction between the virtual objects is other than 0 and 1, after the collision by interpolation using the impulse when the coefficient of restitution and the coefficient of friction are 0 and 1, Is calculated.

[0008]

[0009]

DESCRIPTION OF THE PREFERRED EMBODIMENTS Automatic calculation of the behavior of a group of virtual objects in a virtual three-dimensional space constructed inside a computer is as follows.
This is a useful technique for animation and virtual reality. The present embodiment relates to a method of simulating the behavior when objects collide with each other, and shows a method of calculating the state of the object immediately after the collision from the state of the object at the time of the collision and the state of the collision.

In this embodiment, when simulating a collision between virtual objects, the state (position, inclination, velocity, angular velocity), collision position, and collision direction (tangential plane direction of the collision position) of the virtual object at the time of the collision are considered. And calculate the state of the virtual object after the collision. This method can perform calculations on arbitrary shapes if the mass and moment of inertia of the object are known. Also, the collision time, collision position, and collision direction are determined by the state and shape of the two objects, and by using these, the same processing can be performed regardless of the object shape.

The state of the virtual object after the collision is calculated by calculating the impulse generated by the collision.
By calculating taking into account the coefficient of restitution and the coefficient of friction between two objects, it is possible to perform calculations corresponding to various collision phenomena. When the coefficient of restitution and the coefficient of friction between two objects are unknown, the coefficient of restitution (0 to 1) and the coefficient of friction (0 to 1) specific to each object are defined, and expressed by the product of each coefficient between the two objects. When the coefficient of restitution or the coefficient of friction between two objects is not 0 or 1, the impulse is calculated when the coefficients are 0 and 1, and the impulse is obtained by interpolation.

When the friction coefficient between the objects is 0, slip occurs at the collision position, and the force acts in the collision direction. When the friction coefficient is 1, rolling occurs at the collision position, and a force acts so that the speeds in the collision tangent plane at the collision position become equal. When the coefficient of restitution between objects is 0, no force acts in the collision direction, and the object moves at the collision position at the same speed in the collision direction. When the coefficient of restitution is 1, it is regarded as a collision between rigid bodies, and when the coefficient of friction is 0 or 1, kinetic energy is preserved.

FIG. 1 shows a state at the moment when two objects, that is, a rectangular parallelepiped object 1 (1) and a spherical object 2 (2) collide. Where the mass is M
It is assumed that 1, M2 and the inertia tensor are [I1] and [I2]. The state of each object at the time of collision is represented by the following parameters. That is, the coordinates of the center of gravity of the object A: O1, the coordinates of the center of gravity of the object B: O2, the speed of the object A: V1, the speed of the object B: V2, the inclination of the object A: [S1] the inclination of the object B: [S2]
], Angular velocity of the object A: W1 angular velocity of the object B: W2, where O1, O2, V1, and V2 are vector quantities in the inertial system, and [S1] and [S2] are converted from the object coordinate system to the inertial system. A rotation matrix of × 3, W1 and W2 are vector quantities in the object coordinate system, and the inertial system and the object coordinate system are [S
1] and [S2], and can be mutually converted. If the collision position P and the collision direction N are given to the inertial system, the collision positions R1 and R2 with respect to the center of gravity in the respective object coordinate systems can be immediately obtained using [S]. That is,

(Equation 1) (Equation 1)

(Equation 2) (Equation 2) Here, [S] -1 is the inverse matrix of [S]. In this embodiment, the object coordinate system is selected so as to match the direction of the principal axis, and the inertia tensor [I] is as follows: Are diagonalized in the object coordinate system.

[0014]

(Equation 3) FIG. 2 shows the state of the object after the collision. The collision forces F and -F act on the two objects, respectively, and the state of motion changes. The collision force is based on the law of action and reaction.
The directions are equal and opposite. Immediately after the collision, the coordinates of the center of gravity and the inclination of each object do not change.
The velocities V'1, V'2 and the angular velocities W'1, W'2 immediately after the collision are obtained. From the changes in momentum and angular momentum before and after the collision, the following equation is obtained.

[0015]

(Equation 4) (Equation 3)

(Equation 5) (Equation 4)

(Equation 6) (Equation 5)

(Equation 7) (Equation 6) Equations (3) and (4) show the change in the momentum of each object with respect to the inertial system. Equations (5) and (6) show the change in the angular momentum of each object with respect to the object coordinate system. Is shown. Ft is an impulse generated at the collision position at the time of collision, and F1 and F1
2 is obtained by converting Ft into an object coordinate system. Each impulse is connected by the following relationship, and the above equation satisfies the law of conservation of momentum and the law of conservation of angular momentum.

[0016]

(Equation 8) (Equation 7) From the equations (3) to (7), V′1, V′2, W′1, and W′2 can be expressed as follows using the impulse Ft.

[0017]

(Equation 9) (Equation 8)

(Equation 10) (Equation 9)

[Equation 11] (Equation 10)

(Equation 12) (Equation 11) Here, [C1] and [C2] are inertial tensors [I1],
[I2], a rotation matrix [S1], [S2], a 3 × 3 angular velocity conversion matrix determined by the collision position R with respect to the center of gravity,

(Equation 13)

[Equation 14]

(Equation 15) When,

(Equation 16) (Equation 12) is calculated.

As described above, if the impulse Ft is determined, the velocities V'1, V'2 and the angular velocities W'1, W'2 immediately after the collision are determined.
Can be requested. Hereinafter, a method of obtaining the impulse Ft will be described.

The kinetic energies E and E 'before and after the collision are

[Equation 17] (Equation 13)

(Equation 18) (Equation 14) In addition, the relative speed dV at the contact point immediately after the collision is determined in consideration of the moving speed and the rotational speed.

[Equation 19] (Equation 15) In order to express all the velocities in the inertial system, the rotational speed is converted from the object coordinate system to the inertial system using [S]. Equations (14) and (15) are obtained from equations (8) to (1).
It can be expressed using impulse Ft as follows using 1).

[0020]

(Equation 20) (Equation 16)

(Equation 21) (Equation 17) Here, E0 is a constant, Ef (Ft) is a quadratic expression relating to Ft, V0 is a vector quantity, and [Vf] is a 3 × 3 relative velocity conversion matrix. Ft is determined using both these equations.

In determining Ft, calculation is performed in consideration of the repulsive force and frictional force between objects. The frictional force gives a friction coefficient μ (0 to 1) to each object, and the frictional force between the objects is represented by the product (0 to 1) of the friction coefficients between the objects. When the coefficient of friction between the objects is 0, slip occurs at the collision position, and the force acts in the collision direction. When the friction coefficient is 1, rolling occurs at the collision position, and a force acts so that the speeds in the collision tangent plane at the collision position become equal.

Similarly to the frictional force, the repulsion force gives a repulsion coefficient ε (0 to 1) to each object, and the repulsion force between the objects is represented by the product (0 to 1) of the restitution coefficients between the objects. When the coefficient of restitution between the objects is 0, no force acts in the collision direction, and the object moves at the same speed in the collision direction after the collision. When the coefficient of restitution is 1, it is regarded as a collision between rigid bodies, and when the coefficient of friction is 0 or 1, kinetic energy is preserved.

The coefficient of restitution or coefficient of friction between two objects is 0
If it is not 1, the impulse is obtained when the respective coefficients are 0 and 1, and the impulse is obtained by interpolation. Hereinafter, a method of obtaining Ft for each combination of the values of the coefficient of restitution and the coefficient of friction will be described. [Case 1] When the friction coefficient μ = 0 and the coefficient of restitution ε = 1 Since the contact point is smooth and there is no friction from the friction coefficient μ = 0, the direction of the impulse Ft matches the collision direction N. Also, since there is no friction and the coefficient of restitution ε = 1, kinetic energy is preserved before and after the collision. Ft = kN (Equation 18) E ′ = E (Equation 19) Here, E ′ is represented by using k by substituting Equation (18) into the energy equation of Equation (16), and from Equation (19) Find k. Ft at this time is set to F01. [Case 2] When the coefficient of friction μ = 1 and the coefficient of restitution ε = 1 Since the coefficient of friction μ = 1, rolling occurs without slipping at the contact point,
The velocities after the collision of the two objects on the collision tangent plane become equal.
That is, the direction of the relative speed of the two objects coincides with the collision direction.
Further, since the rolling coefficient of restitution ε = 1 without slipping, kinetic energy is preserved before and after the collision. dV = kN (Equation 20) E ′ = E (Equation 21) Here, Ft is expressed using k by substituting Equation (20) into the equation of the relative velocity of Equation (17).

[0024]

(Equation 22) (Equation 22) By substituting this into the energy equation of Equation (16), E ′
Is represented using k, and k is obtained from Expression (21). Ft at this time is set to F11. [Case 3] 3 When the friction coefficient μ = 0 and the coefficient of restitution ε = 0 Since the friction point μ = 0, the contact point is smooth and there is no friction, so that the direction of the impulse Ft coincides with the collision direction N. Also, from the coefficient of restitution ε = 0, the speeds in the collision direction match after the collision,
The direction of the relative velocity is perpendicular to the collision direction. Ft = kN (Equation 23) dV · N = 0 (Equation 24) Here, when Equations (23) and (24) are substituted for the relative velocity of Equation (17), k is obtained. k = −V0 · N / ([Vf] N) · N (Equation 25) Let Ft be F00 at this time. [Case 4] When the friction coefficient μ = 1 and the coefficient of restitution ε = 0, the rolling occurs without slipping at the contact point from the friction coefficient μ = 1,
The velocities after the collision of the two objects on the collision tangent plane become equal.
Further, from the coefficient of restitution ε = 0, the velocities in the collision direction coincide after the collision. That is, the velocities of the two objects after the collision at the collision position match. dV = 0 (Equation 26) Here, Ft is obtained by substituting Equation (26) for the relative speed of Equation (17) into the equation.

[0025]

(Equation 23) (Equation 27) Ft at this time is set to F10.

[Case 5] When friction coefficient μ (0 to 1) and coefficient of restitution ε (0 to 1) When friction coefficient μ and coefficient of restitution ε are in the range of 0 to 1,
The impulse Ft is the impulse F00, F01, F10, F
The calculation is performed by using 11 and interpolating as follows. Ft = F00 + (F10-F00) .mu. + (F01-F00) .epsilon. + (F11-F01-F10 + F00) .epsilon..mu.
Can be calculated, and the velocities V′1, V′2 and the angular velocities W′1, W′2 immediately after the collision can be calculated from the equations (8) to (11).

Next, a concrete example of collision of various objects will be described.

[Specific Example 1] Collision between spheres FIG. 3 shows the state at the moment when spheres 3 and 4 collide with each other while rotating on the same straight line while rotating in the same direction. Here, reference numeral 10 denotes a three-dimensional basic orthogonal coordinate axis that defines the coordinates of these spheres, and is composed of an x axis 5, a y axis 7, and a z axis 6. The traveling directions of the spheres 1 (3) and 2 (4) are parallel to the x-axis 5, and their rotation axes are parallel to the z-axis 6. Both objects have a radius L = 1, a mass M = 1, and a rotation speed W = 1.
The velocity of the sphere 1 (3) in the positive direction on the x-axis is V1 = 1,
The sphere 2 (4) is traveling in the negative direction on the x-axis at a speed V2 = -1. At this time, the collision position R viewed from the center of the ball 1 (3)
1 is (1, 0, 0), the collision position R2 viewed from the center of the sphere 2 (4) is (-1, 0, 0), and the collision direction is the x-axis 5 direction.

FIGS. 4 to 8 show impulse F generated when balls collide under various conditions of the coefficient of friction and the coefficient of restitution.
1, F2, the velocities V'1, V'2 after the collision, the angular velocities W'1, W'2 after the collision, and the state of the two objects when the time Δt = 1 has elapsed. For impulse, speed, and angular velocity, the direction is indicated by an arrow, the size is indicated by the length of the arrow and a numerical value, and the sign of the sum of the values in the x, y, and z directions is added to the numerical value. I have.

In the following description, the values of the six components on the z axis are 0
At the time, only the (x, y) component is shown. Furthermore, when the value of the y-axis 7 component is also 0, only the x component is shown.

FIG. 4 shows that the friction coefficient μ = 0 and the restitution coefficient ε = 1.
The state after the collision at the time of is shown. In this state, since there is no friction, the impulse faces in the collision direction (the direction of the x-axis 5), and the sphere 1
The impulse in the object coordinate system of (3) and sphere 2 (4) are respectively
F1 = -2 and F2 = 2. Since the impulse faces the center of the sphere, the angular velocity does not change before and after the collision, W'1 = 1, W'2
= 1. Further, the vehicle travels on the x-axis 5 at a speed of V'1 = -1, V'2 = 1. That is, the angular velocities of the two objects do not change, but advance at the same size as before the collision and in the opposite direction.

FIG. 5 shows a state after the collision when the friction coefficient μ = 1 and the restitution coefficient ε = 1. In this state,
Because of the friction, the impulse also includes a component in the y-axis direction,
The impulse in the object coordinate system of (3) and sphere 2 (4) are respectively
F1 = (− 2.13, −0.29), F2 = (2.13,
0.29). Since the impulse is not directed to the center of the sphere, the magnitude of the angular velocity changes before and after the collision, and the angular velocities of the balls 3 and 4 are W′1 = 0.29 and W′2 = 0.29, respectively. V′1 = (− 1.13, −0.2)
9), V'2 = (1.13, 0.29). That is, the directions of the angular velocities do not change for both objects, but the magnitudes decrease by the same amount, and the two objects move at the same speed but in opposite directions.

FIG. 6 shows that the coefficient of friction μ = 0 and the coefficient of restitution ε = 0.
The state after the collision at the time of is shown. In this state, since there is no friction, the impulse is in the collision direction (the direction of the x-axis 7).
And the impulse of the sphere 1 (3) and the sphere 2 (4) in the object coordinate system are F1 = −1 and F2 = 1, respectively. Since the impulse faces the center of the sphere, the angular velocity does not change before and after the collision, and W ′ 1 = 1 and W ′ 2 = 1, respectively. Further, since the speed does not repel in the collision direction, both objects become 0. That is, after the collision, the two objects only rotate at the collision position and do not change their positions.

FIG. 7 shows that the coefficient of friction μ = 1 and the coefficient of restitution ε = 0.
The state after the collision at the time of is shown. In this state, the impulse also includes the y-axis 7 component due to friction, and the ball 1
The impulse in the object coordinate system of (3) and sphere 2 (4) are respectively
F1 = (-1, -0.29) and F2 = (1, 0.29). Since the impulse does not point toward the center of the sphere, the magnitude of the angular velocity changes before and after the collision, and W ′ 1 = 0.29,
W ′ 2 = 0.29, and the velocity does not rebound in the collision direction, so there are only seven components on the y-axis. V ′ 1 = −0.29, V ′ 2 =
It becomes 0.29. That is, the two objects that have traveled on the x-axis 5 have their rotation speed reduced by the same amount and travel in parallel with the y-axis 7 and in opposite directions.

FIG. 8 shows a coefficient of friction μ = 0.5 and a coefficient of restitution ε.
= 0.5 after the collision. In this state, the impulse is F1 = (-1.53, -0.1, respectively).
4), F2 = (1.53, 0.14). The angular velocities vary in magnitude, W'1 = 0.64 and W'2 =
0.64, and the speed is V '1 = (− 0.53, −
0.14), V '2 = (0.53, 0.14). That is, the angular velocity of the two objects decreases by the same amount, and the velocities are equal in magnitude and travel in opposite directions. [Specific Example 2] Collision between a rectangular parallelepiped and a sphere FIG. 9 shows a sphere 9 rotating while rotating on a stationary rectangular solid 8.
Shows the state at the moment of collision. Here, the rectangular parallelepiped 8 is arranged such that the x-axis 5, the y-axis 7, and the z-axis 6 and the respective sides of the rectangular parallelepiped 8 corresponding to their respective axes are parallel to each other. The direction of the rotation axis of the sphere 9 is the direction of the z-axis 6, and the traveling direction is parallel to the x-axis 5. Rectangular parallelepiped 8
Are Lx = 2, Ly = 3, Lz = 1 in the x, y, and z-axis directions, respectively, and the mass is M1 = 1.
It is. On the other hand, the sphere 9 has a radius L2 = 1, a mass M2 = 1, an angular velocity W2 = 1, and a velocity V2 = -1. At this time, the collision position R1 viewed from the center of the rectangular parallelepiped is (1,1,0),
The collision position R2 viewed from the center of the sphere is (-1, 0, 0), and the collision direction is the x-axis direction.

FIGS. 10 to 14 show the respective impulse F1 and F2 generated when the rectangular parallelepiped 8 and the ball 9 collide under various conditions of the coefficient of friction and the coefficient of restitution, and the respective speeds V'1 after the collision. , V'2, the respective angular velocities W'1, W'2 after the collision,
And the state of the two objects when the time Δt = 1 has elapsed. For impulse, velocity, and angular velocity, the direction is indicated by an arrow, the size is indicated by the length of the arrow and a numerical value, and the numerical value is obtained by adding the values in the x-axis, y-axis, and z-axis directions. Is added.

In the following description, when the value of the z-axis component is 0, only the (x, y) component is shown. Further, when the value of the y-axis component is also 0, only the x component is shown.

FIG. 10 shows that the coefficient of friction μ = 0 and the coefficient of restitution ε =
The state after the collision at the time of 1 is shown. In this state, since there is no friction, the impulse is in the collision direction (the direction of the x-axis 5).
And F1 = -0.68 and F2 = 0.68. Since the impulse is directed toward the center with respect to the sphere 9, the angular velocity does not change before and after the collision and W′2 = 1, but a rotational force is applied to the rectangular parallelepiped 8 and the angular velocity Around W'1 = 0.6
It becomes 3. The velocities of the two objects proceed in the negative direction in parallel with the x-axis, and V′1 = −0.68 and V′2 = −0.32, respectively.

FIG. 11 shows the coefficient of friction μ = 1 and the coefficient of restitution ε = 1.
The state after the collision at the time of is shown. In this state, the impulse also includes a component in the y-axis 7 direction due to friction, and F
1 = (-0.92, -0.34), F2 = (0.92,
0, 34). Since the direction of the impulse is off the center of both objects, the magnitude of the angular velocity changes before and after the collision and W'1 =
0.53, W'2 = 0.15, and the speed is V'1 = (-
0.92, -0.34), V'2 = (-0.08, 0.3
4).

FIG. 12 shows a coefficient of friction μ = 0 and a coefficient of restitution ε = 0.
The state after the collision at the time of is shown. In this state, since there is no friction, the impulse is in the collision direction (the direction of the x-axis 5).
And F1 = -0.34, F2 = 0.34. Since the impulse is directed toward the center with respect to the sphere 9, the angular velocity does not change before and after the collision and W'2 = 1, but a rotational force is applied to the rectangular parallelepiped 8, and the angular velocity is adjusted around the z axis. W'1 = 0.32
Becomes The velocities of the two objects proceed in the negative direction in parallel with the x-axis 5, and V′1 = −0.34, V′2 = −0.66. Here, both objects of the sphere 9 and the rectangular parallelepiped 8 are displayed in an overlapping manner, which indicates that the ball 9 keeps pushing the rectangular parallelepiped 8 for a certain period of time in order not to rebound in the collision direction. In order to calculate accurately, the collision state between the rectangular parallelepiped 8 and the sphere 9 after a short time must be obtained, and the calculation of the collision must be repeated. For example, if the short time is set to 1 in this embodiment, the collision direction is N = (0.95, 0.31), the collision position is R1 = (0.73, 0, 69) when viewed from the center of the rectangular parallelepiped, and The calculation may be performed with R2 = (-0.95, -0.32) as viewed from the center of the sphere.

FIG. 13 shows that the coefficient of friction μ = 1 and the coefficient of restitution ε =
This shows the state after the collision at the time of 0. In this state, the impulse also includes 7 components on the y-axis due to friction, and F1 =
(−0.42, −0.26), F 2 = (0.42, 0,
26). Since the direction of the impulse is off the center of both objects, the magnitude of the angular velocity changes before and after the collision, and W′1 = 0.
15, W′2 = 0.36, and the speed is V′1 = (− 0.4
2, -0.26), V'2 = (-0.58, 0.26).

FIG. 14 shows the state after the collision when the coefficient of friction μ = 0.5 and the coefficient of restitution ε = 0.5. In this state, the impulse is F1 = (− 0.59, −0.15),
F2 = (0.59, 0.15). The angular velocity changes in magnitude, W′1 = 0.41, W′2 = 0.63, and the velocity is V′1 = (− 0.59, −0.15), V′2 = (− 0. 4
1,0.15).

Finally, the coefficient of friction μ and the coefficient of restitution ε are 0 to 1
The kinetic energy change before and after the collision is examined to check the validity of the interpolation performed in the case of the value between. The kinetic energies E and E 'before and after the collision are given by Equation (13),
It can be obtained using equation (14), and the mass of two objects,
It is determined by the moment of inertia, speed, and angular speed.

FIG. 15 shows the kinetic energy after the collision when the spheres shown in FIG. 3 collide with each other and the friction coefficient μ and the restitution coefficient ε are variously changed. FIG. 16 shows the kinetic energy after the collision when the rectangular parallelepiped and the ball shown in FIG. 9 collide.

Referring to FIG. 15, the kinetic energy before the balls collide with each other is 2.8, and when the coefficient of restitution ε is 1 and the friction coefficient μ is 0 or 1, the kinetic energy is conserved.
It can be seen that the kinetic energy is most lost when ε = 0 and μ = 1. Further, even if the coefficient of restitution ε is 1, the friction coefficient μ is 0.
In the case of a value between １1 and 衝突 1, the kinetic energy decreases before and after the collision. This is in good agreement with natural phenomena because kinetic energy is not preserved when the ball actually rolls while sliding.

Referring to FIG. 16, the kinetic energy before the rectangular parallelepiped collides with the sphere is 1.4, and when the coefficient of restitution ε is 1 and the friction coefficient μ is 0 or 1, the kinetic energy is conserved. It can be seen that the kinetic energy is most lost when ε = 0 and μ = 1. In addition, when the coefficient of friction μ is a value between 0 and 1 even when the coefficient of restitution ε is 1, the kinetic energy decreases before and after the collision. This is in good agreement with natural phenomena because kinetic energy is not preserved when the ball actually rolls while sliding.

Next, an outline of the data processing apparatus for performing the simulation of this embodiment will be described below with reference to FIG.

The CPU 200 controls the entire data processing apparatus, interprets and executes a simulation program stored in the memory 202 in advance. The memory 200 stores the simulation program of the present embodiment and various necessary data in advance. The keyboard 203 and the pointing device 204 input data on an object to be simulated, an execution command command, and the like. The display monitor 201 displays simulation result data, a motion image of an object, and the like.

Next, a procedure for calculating the motion state immediately after the two objects collide will be described with reference to the flowcharts of FIGS.

In step S100, various data expressing the two objects themselves and their motion state are input to the keyboard 203.
And input using the pointing device 204 and the like.
As data of the object, masses M1, M2, inertia tensor [I
1], [I2], barycentric coordinates of object 1: O1, barycentric coordinates of object 2: O2, speed of object 1: V1, speed of object 2: V2,
The inclination of the object 1 is [S1], the inclination of the object 2 is [S2], the angular velocity of the object 1 is W1, the angular velocity of the object B is W2, and data of friction coefficient μ and restitution coefficient ε are input.

In step S101, collision positions R1 and R2 with respect to the center of gravity of two objects are calculated based on equations (1) and (2).

In step S102, the data values of the coefficient of friction μ and the coefficient of restitution ε are checked. And the coefficient of friction μ =
If 0 and the restitution coefficient ε = 1, the routine proceeds to step 103. If the coefficient of friction μ = 1 and the coefficient of restitution ε = 1, step 106
Proceed to. If the coefficient of friction μ = 0 and the coefficient of restitution ε = 0, the routine proceeds to step 109. If the coefficient of friction μ = 1 and the coefficient of restitution ε = 0, the routine proceeds to step 112. If the coefficient of friction μ is a value between 0 and 1 and the coefficient of restitution ε is a value between 0 and 1, the routine proceeds to step 115.

In step S103, the direction of the impulse Ft coincides with the collision direction N since the contact point is smooth and there is no friction from the friction coefficient μ = 0. No friction and coefficient of restitution ε = 1
From the condition that the kinetic energy is preserved before and after the collision, the equation (18) is substituted into the energy equation of the equation (16), k is calculated from the equation (19), and the impulse at this time is calculated. Calculate F01.

In step S104, the impulse F01 is calculated by the following equation (8).
Then, the velocities V′1 and V′2 immediately after the collision are calculated by substituting into Equation 9.

In step S105, the impulse F01 is calculated by the equation (1).
Substituting 0 into Equation 11 to calculate the velocities W′1 and W′2 immediately after the collision.

In step S106, since the friction coefficient μ = 1, rolling occurs without slipping at the contact point, and the velocities of the two objects on the collision tangent plane after the collision become equal. And from the coefficient of restitution ε = 1,
Under the condition that the kinetic energy is preserved before and after the collision, Equation 22 is substituted into the energy equation of Equation 16 to obtain Equation 21
Then, k is calculated, and the impulse F11 at this time is calculated.

In step S107, the impulse F11 is calculated by the following equation (8).
Then, the velocities V′1 and V′2 immediately after the collision are calculated by substituting into Equation 9.

In step S108, the impulse F11 is calculated by the equation (1).
Substituting 0 into Equation 11 to calculate the velocities W′1 and W′2 immediately after the collision.

In step S109, since the contact point is smooth and there is no friction from the friction coefficient μ = 0, the direction of the impulse Ft coincides with the collision direction N, and since the coefficient of restitution ε = 0, the collision after the collision From the condition that the speeds in the directions match and the direction of the relative speed is perpendicular to the collision direction, k is calculated from Equation 25, and the impulse F00 at this time is calculated.

In step S110, the impulse F00 is calculated by the following equation (8).
Then, the velocities V′1 and V′2 immediately after the collision are calculated by substituting into Equation 9.

In step S111, the impulse F00 is calculated by the following equation (1).
Substituting 0 into Equation 11 to calculate the velocities W′1 and W′2 immediately after the collision.

In step S112, since the friction coefficient μ = 1, rolling occurs without slipping at the contact point, the velocities of the two objects on the collision tangent plane after the collision become equal, and the coefficient of restitution ε =
From 0, the impulse F10 is calculated by substituting Expression (26) into Expression (17) and substituting the relative speed of Expression (17) into the expression under the condition that the speeds in the collision direction match after the collision.

In step S113, the impulse F10 is calculated by the following equation (8).
Then, the velocities V′1 and V′2 immediately after the collision are calculated by substituting into Equation 9.

In step S114, the impulse F10 is calculated by the equation (1).
Substituting 0 into Equation 11 to calculate the velocities W′1 and W′2 immediately after the collision.

In step S115, first, the impulse F00,
Find F01, F10, F11. The way to find each is
Step S109, step S104, step S11
2. This is the same as the method of obtaining each impulse in step S106. Then, the impulse values F00, F01, F10, and F11 obtained are calculated according to Equation 2.
The impulse Ft is calculated by substituting into the interpolation formula of FIG.

In step S116, the impulse Ft is calculated by the following equation (8).
Then, the velocities V′1 and V′2 immediately after the collision are calculated by substituting into Equation 9.

In step S114, the impulse Ft is calculated by the following equation (1).
Substituting 0 into Equation 11 to calculate the velocities W′1 and W′2 immediately after the collision.

By executing the above-described procedure by the CPU 200, it is possible to easily obtain the motion state after the collision of the two objects. Then, the obtained motion state after the collision of the two objects is displayed on the display monitor.

The present invention may be applied to a system constituted by a plurality of devices or an apparatus constituted by a single device. Needless to say, the present invention can be applied to a case where the present invention is achieved by supplying a program to a system or an apparatus.

As described above, when simulating a collision between virtual objects in the virtual three-dimensional space constructed inside the computer, the state of the virtual object at the time of collision, the collision position, and the collision direction are determined based on the state of the collision. By calculating the state of the virtual object, the calculation can be performed for an arbitrary object shape. Also, by calculating the state calculation of the virtual object after the collision by calculating the impulse generated by the collision taking into account the coefficient of restitution and the coefficient of friction between the two objects, the calculation corresponding to various collision phenomena can be performed. It can be carried out. As a result, even when the coefficient of restitution is 1, when the friction coefficient is between 0 and 1 (when rolling while sliding), actual natural phenomena are well represented, for example, kinetic energy is not preserved.

[0071]

As described above, according to the present invention, the motion of an object at the time of collision between objects can be approximated with high accuracy.

[0072]

[Brief description of the drawings]

FIG. 1 is a diagram illustrating a state at the moment when two objects collide.

FIG. 2 is a diagram illustrating movement of two objects after a collision.

FIG. 3 is a diagram showing a state at the moment when balls collide with each other.

FIG. 4 is a diagram illustrating a state after collision between balls under the condition of friction coefficient = 0 and restitution coefficient = 1.

FIG. 5 is a diagram for explaining a state after balls collide with each other under the conditions of friction coefficient = 1 and restitution coefficient = 1.

FIG. 6 is a diagram illustrating a state after the balls collide with each other under the conditions of friction coefficient = 0 and restitution coefficient = 0.

FIG. 7 is a diagram illustrating a state after the balls collide with each other under the conditions of friction coefficient = 1 and restitution coefficient = 0.

FIG. 8 is a diagram illustrating a state after the balls collide with each other under the conditions of a coefficient of friction = 0.5 and a coefficient of restitution = 0.5.

FIG. 9 is a diagram showing a state at the moment when a cuboid and a ball collide.

FIG. 10 is a diagram illustrating a state after a rectangular parallelepiped and a ball collide under the conditions of friction coefficient = 0 and restitution coefficient = 1.

FIG. 11 is a diagram illustrating a state after a rectangular parallelepiped and a ball collide under the conditions of a friction coefficient = 1 and a restitution coefficient = 1.

FIG. 12 is a diagram illustrating a state after a rectangular parallelepiped and a ball collide under the conditions of friction coefficient = 0 and restitution coefficient = 0.

FIG. 13 is a diagram illustrating a state after a rectangular parallelepiped and a ball collide under the conditions of a friction coefficient = 1 and a restitution coefficient = 0.

FIG. 14 is a diagram illustrating a state after a rectangular parallelepiped and a ball collide under the conditions of a coefficient of friction = 0.5 and a coefficient of restitution = 0.5.

FIG. 15 is a diagram showing kinetic energy after spheres collide with each other.

FIG. 16 is a diagram showing kinetic energy after a rectangular parallelepiped and a ball collide.

FIG. 17 is a diagram illustrating a configuration of a data processing device according to the present embodiment.

FIG. 18 is a diagram illustrating a simulation processing flow according to the present embodiment.

FIG. 19 is a diagram illustrating a simulation processing flow according to the present embodiment.

FIG. 20 is a diagram illustrating a simulation processing flow according to the present embodiment.

FIG. 21 is a diagram illustrating a simulation processing flow according to the present embodiment.

FIG. 22 is a diagram illustrating a simulation processing flow according to the present embodiment.

FIG. 23 is a diagram illustrating a simulation processing flow according to the present embodiment.

[Explanation of symbols]

Reference Signs List 1 object 1 2 object 2 3 sphere 1 4 sphere 2 5 x axis 6 y axis 7 z axis 8 rectangular parallelepiped 9 sphere 10 basic orthogonal coordinate axis P collision position N collision direction F impulse generated at the collision O1, O2 The center of gravity of each object R1, R2 Collision position of each object with respect to the center of gravity M1, M2 Mass of each object [I1], [I2] Matrix [S1], [S2] representing moment of inertia of each object Matrix L1, L2 representing inclination of each object Radius of sphere, Lx, Ly, Lz Length of each side of rectangular parallelepiped, V1, V2 Velocity of each object before collision, W1, W2 Angular velocity of each object before collision F1, F2 Impulse for each object generated at collision V'1, V'2 Velocity of each object after collision W'1, W'2 Angular velocity of each object after collision μ Friction coefficient ε Restitution coefficient.

────────────────────────────────────────────────── ─── Continuation of the front page (56) References JP-A-2-61773 (JP, A) JP-A-5-35846 (JP, A) JP-A-3-11668 (JP, A) JP-A-4-167 181363 (JP, A) (58) Fields investigated (Int. Cl. ⁷ , DB name) G06T 13/00 G06T 15/70 G06F 19/00 G06F 17/50 G09B 23/10 CSDB (Japan Patent Office)

Claims (4)

(57) [Claims] 1. A method of simulating a collision between virtual objects in a virtual three-dimensional space, the method comprising: calculating state data of the virtual object after the collision based on characteristic data of the virtual object and state data of the virtual object at the time of the collision. a calculation step of calculating a Bei and a display step of the above based on the state data of the virtual object after the collision, to display the status of the virtual object after the collision
The coefficient of restitution and coefficient of friction between virtual objects in the calculation step
If at least one of is not 0 or 1, the coefficient of restitution
Using impulse when friction coefficient is 0 and 1
A virtual three-dimensional simulation method, wherein state data after a collision is calculated by the following method. 2. The state data of the virtual object at the time of the collision,
The virtual three-dimensional collision simulation method according to claim 1, further comprising a movement position, a tilt, a movement speed, a rotational movement angular velocity, or a collision direction. 3. The system according to claim 2, wherein the coefficient of restitution and the coefficient of friction are zero.
Claim 1 or Claim 2 characterized by being set in the range of
Is a virtual three-dimensional collision simulation method according to 2. 4. An apparatus for simulating a collision between virtual objects in a virtual three-dimensional space, wherein a state of the virtual object after the collision is determined based on characteristic data of the virtual object and state data of the virtual object at the time of the collision. Bei and calculating calculating means, based on the state data of the virtual object after the collision, and display means for displaying the status of the virtual object after the collision
The coefficient of restitution and coefficient of friction between virtual objects in the calculation means
If at least one of is not 0 or 1, the coefficient of restitution
Using impulse when friction coefficient is 0 and 1
A virtual three-dimensional simulation apparatus, which calculates post-collision state data according to: