View Tests with a Line Segment

In this section, we introduce performing view tests with a view-line.

Figure 3.5: Triangle orientation determination with a view-line at a certain time.

At a certain time

At first, we perform the view test of a triangle at a certain time. Assume that there is a triangle T (v₀, v₁, v₂), and N = (v₁− v0)× (v2− v0) is the normal vector of the triangle.

In addition, q₀q₁is the view-line, as shown in Figure 3.5. We can choose a vertex v of the triangle T and obtain a vector v− vp, where vp is the check point on the view-line q0q1. Then, the view test is performed as follow.

V T_l = (v− vp)· N (3.3)

v_p has three types according to the location of v.

1. If v ∈ R0, then vp = q0. 2. If v ∈ R1, then v_p = q₁.

3. If v ∈ R2, then v_p = the projective point of v on the view-line q₀q₁.

Therefore, view tests based on the view-point model and the view-line model are different. For the view-line model, we divide space and deformable objects into three regions, and the check points are different with respect to different vertices. Hence, before performing view tests of a triangle, we determine the regions for three vertices of a triangle.

For a vertex v of the triangle T , the projective point of v on the view-line q₀q₁ should be v_p = q₀+ u× (q1− q0).

Then, we can determine the regions for the vertex v based on the value of u.

• If u < 0, then v ∈ R0.

• If u > 1, then v ∈ R1.

• If u ∈ [0, 1], then v ∈ R2.

Theorem 3. Suppose that there is a triangle T (v₀, v₁, v₂), and N is the normal vector of the triangle. q₀q₁ is a fixed view-line without intersecting with the triangle. Then, for all vertices of the triangle T , the results of view tests are the same.

Proof. We prove it as follows.

• Case 1: For any two vertices, if they are in the same region, R₀ or R₁, then it is proved by Corollary 1.

• Case 2: For any two vertices, if they are in the same region, R₂, then suppose that v₀ and v₁ are chosen, and v_0p and v_1pare the projective points of v₀ and v₁ on the By Corollary 1, we have the following results.

The results of A and D are the same according to the fixed point v_0p. The results of B and C are the same according to the fixed point v_1p. The results of A and C are the same according to the fixed point v₀. The results of B and D are the same according to the fixed point v₁.

Therefore, the results of V T_v0 = (v₀− v0p)· N and V Tv1 = (v₁− v1p)· N are the same.

• Case 3: For any two vertices, if one vertex belongs to R₀ and the other belongs to R₂, then suppose that v₀ and v₁ are chosen that v₀ ∈ R0 and v₁ ∈ R2, as shown in Figure 3.6. We can find a point m of the triangle that the projective point of m on the view-line q₀q₁ is q₀. The results of V T_v0 and V T_m are the same by case1. The results of V T_m and V T_v1 are the same by case2. So, the results of V T_v0 and V T_v1 are the same.

• case4: For any two vertices, if one vertex belongs to R₁, and the other belongs to R₂, then the result is similar to case3.

• case5: For any two vertices, if one vertex belongs to R0 and the other belongs to R₁, then suppose that v₀ and v₁ are chosen that v₀ ∈ R0 and v₁ ∈ R1, as shown in Figure 3.7. Similar to case3, we can find two points m₀ and m₁of the triangle that

Figure 3.6: Triangle orientation determination with a view-line at a certain time when the triangle lies in multiple regions.

Figure 3.7: Triangle orientation determination with a view-line at a certain time when the triangle lies in multiple regions.

the projective points of m₀ and m₁on the view-line q₀q₁are q₀ and q₁. The results of V T_v0 and V T_m0 are the same, and the results of V T_m1 and V T_v1 are the same by case1. The results of V T_m0 and V T_m1 are the same by case2. So, the results of V T_v0 and V T_v1 are the same.

According to the above results, it is proved. Hence, for all vertices of the triangle T , the results of view tests are the same.

Therefore, we can classify the type of a triangle by performing view tests with any point of the triangle according to the view-line. In addition, the chosen vertex belongs to only one region at a certain time, so view tests are performed for just one time. By Theorem 3 and Equation 3.3, we have three conclusions. For any point of a triangle T ,

• If V T_l > 0, then the triangle faces the view-line.

• If V T_l < 0, then the triangle is back to the view-line.

• If V T_l = 0, then the triangle is coplanar to the view-line.

Over a time interval

Next, we perform view tests of a triangle over a time interval [0, ∆t], where ∆t is the time step of the simulation. During the simulation, all vertices move with linear velocities within a frame. Therefore, the position of a vertex v at a certain time in [0, ∆t] should be v^t= v_bgn+ V t, where v_bgnis the initial position of v, v^tis the new position of v, V is the linear velocity of v, and t∈ [0, ∆t].

Assume that there is a triangle T (v₀, v₁, v₂) and N = (v₁ − v0)× (v2 − v0) is the normal vector of the triangle in the beginning. After the time step ∆t, the triangle moves to T^′(v₀^′, v₁^′, v₂^′), and N^′ is the normal vector of the triangle T^′. In addition, q₀q₁ is the view-line, as shown in Figure 3.8.

Before performing view tests of a triangle, we determine the regions that the triangle belongs to because the triangle can spread and move across multiple regions in [0, ∆t].

The regions for a triangle are determined by its three vertices. We compute the regions

Figure 3.8: Triangle orientation determination with a view-line over a time interval when the triangle moves across multiple regions.

for three vertices, then the regions for the triangle are the union of the regions for its three vertices.

Suppose that there is a vertex of the triangle, which moves from v to v^′ in [0, ∆t].

Then, the projective points of v and v^′ on the view-line q₀q₁ should be v_p = q₀ + u_bgn× (q₁−q0) and v_p^′ = q₀+ u_end×(q1−q0). Because the vertices move linearly, [u_bgn, u_end] is also linear. We can determine the regions that the vertices move across in [0, ∆t] according to the values of u_bgnand u_end.

Let R_v0, R_v1, and R_v2 are the regions that the three vertices v₀, v₁, and v₂move across in [0, ∆t]. So, the regions for the triangle are R_v0∪ Rv1 ∪ Rv2. Note that the check points are different based on different vertices of the triangle. Actually, view tests are performed based on the vertices. Suppose that v is one of the vertices of the triangle T . Then, the view test is performed as follow.

V T_l(t) = v(t)· N(t) (3.4)

V T_l(t) is a cubic function in the time domain, N (t) is the time normal vector of triangle T with quadratic form in the time domain, and v(t) is a linear function in the time domain for the vector variation from the vertex to the check points in [0, ∆t]. Note that we compute v(t) separately according to the region for the vertex. For example, if a vertex v

moves from R₀to R₂in [0, ∆t], then the check point is variable. So, the function of vector variation from the vertex to the check point is not linear. We should compute n kinds of v(t), where n is the number of regions for the vertex in [0, ∆t].

Suppose that the velocities of vertices v₀, v₁, and v₂ in [0, ∆t] are V₀, V₁, and V₂. As mentioned above, the positions of v₀, v₁, and v₂at a certain time in [0, ∆t] are

v^t₀ = v₀+ V₀· t, t ∈ [0, ∆t]

v^t₁ = v₁+ V₁· t, t ∈ [0, ∆t]

v^t₂ = v₂+ V₂· t, t ∈ [0, ∆t]

Then, v(t) and N (t) are computed as follows.

• For any vertex v, v moves across R₀ or R₁in [0, ∆t].

v(t) = (v− vq) + ((v^t− vq)− (v − vq))

= (v− vq) + ((v + V · t − vq)− (v − vq))

= (v− vq) + ((v− vq) + V · t − (v − vq))

= (v− vq) + V · t v_qis q₀or q₁ according to the location of v.

• For any vertex v, v moves across R₂ in [0, ∆t].

v(t) = (v− vq0) + ((v^t− vqt)− (v − vq0))

= (v− vq0) + ((v + V · t − vqt)− (v − vq0))

= (v− vq0) + (v− vqt) + V · t − (v − vq0)

= (v− vqt) + V · t

v_q0 is the projective point of v in the beginning, and v_qtis the projective point of v at a certain time in [0, ∆t].

• Let B_s = v₁− v0, B_t= V₁− V0, C_s= v₂− v0, C_t= V₂− V0 and N (t) is the time normal vector of the triangle T . q₀q₁is a fixed view-line without in-tersecting with and penetrating the triangle in [0, ∆t]. Then, for all vertices of the triangle T , the results of view tests are the same in [0, ∆t].

Proof. We prove it as follows.

• Case 1: For any two vertices, if they move in the same region in [0, ∆t], then the results of view tests are always the same by Theorem 3.

• Case 2: For any two vertices, if they move across multiple regions in [0, ∆t], the proof is described as follows. For example, in Figure 3.9, v₀moves from R₂to R₀, and v₁ and v₂ move inside R₂ in [0, ∆t]. Suppose that the triangle T moves to T_t at time t, where t ∈ [0, ∆t]. In [0, t], all vertices move in the same region and the results of view tests are the same by Theorem 3. On the other hand, in (t, ∆t], We can always find a point m_tthat the projective point of m_ton the view-line q₀q₁ is q₀. So, the results of view tests of v₀ are equal to the point m_tin (t, ∆t]. And the results of view tests of v₁and v₂are equal to the point m_tin (t, ∆t]. Therefore, the results of view tests of v₀, v₁, and v₂ are the same.

By Case 1 and Case 2, it is proved. Hence, for all vertices of the triangle T , the results of view tests are the same in [0, ∆t].

Therefore, we can classify the type of a triangle by performing view tests with any point of the triangle over a time interval according to the view-line. View tests are per-formed for nr times, where nr is the number of regions that the chosen vertex moves

Figure 3.9: Triangle orientation determination with a view-line over a time interval when the triangle moves across multiple regions.

across in [0, ∆t]. By Theorem 4 and Equation 3.4, we have three conclusions. Suppose that the result of view tests for a triangle T is V T_l(t) = ∪V Tli(t), where V T_li(t) are the results of view tests according to region R_i, i = 0, ..., n− 1, and 1 ≤ n ≤ 3. For any point of a triangle T in [0, ∆t],

• If ∀V Tli(t) > 0, then V T_l(t) > 0. So, the triangle faces the view-line, and it is assigned to the view set V_l⁺.

• If∀V Tli(t) < 0, then V T_l(t) < 0. So, the triangle is back to the view-line, and it is assigned to the view set V_l⁻.

• If V T_li(t) contain both positive and negative values, then the triangle is coplanar to the view-line at a certain time in [0, ∆t], and it is assigned to the view set V_l⁰.