Читать книгу Robot Modeling and Control - Mark W. Spong - Страница 64

A common method of specifying a rotation matrix in terms of three independent quantities is to use the so-called Euler angles. Consider the fixed coordinate frame o₀x₀y₀z₀ and the rotated frame o₁x₁y₁z₁ shown in Figure 2.10. We can specify the orientation of the frame o₁x₁y₁z₁ relative to the frame o₀x₀y₀z₀ by three angles (ϕ, θ, ψ), known as Euler angles, and obtained by three successive rotations as follows. First rotate about the z-axis by the angle ϕ. Next rotate about the current y-axis by the angle θ. Finally rotate about the current z-axis by the angle ψ. In Figure 2.10, frame oaxayaza represents the new coordinate frame after the rotation by ϕ, frame obxbybzb represents the new coordinate frame after the rotation by θ, and frame o₁x₁y₁z₁ represents the final frame, after the rotation by ψ. Frames oaxayaza and obxbybzb are shown in the figure only to help visualize the rotations.

Figure 2.10 Euler angle representation.

In terms of the basic rotation matrices the resulting rotational transformation can be generated as the product

(2.27)

The matrix RZYZ in Equation (2.27) is called the ZYZ–Euler angle transformation.

The more important and more difficult problem is to determine for a particular R = (rij) the set of Euler angles ϕ, θ, and ψ, that satisfy

(2.28)

for a matrix R ∈ SO(3). This problem will be important later when we address the inverse kinematics problem for manipulators in Chapter 5.

To find a solution for this problem we break it down into two cases. First, suppose that not both of r₁₃, r₂₃ are zero. Then from Equation (2.27) we deduce that s_θ ≠ 0, and hence that not both of r₃₁, r₃₂ are zero. If not both r₁₃ and r₂₃ are zero, then r₃₃ ≠ ±1, and we have c_θ = r₃₃, so

(2.29)

or

(2.30)

where the function Atan2 is the two-argument arctangent function defined in Appendix A.

If we choose the value for θ given by Equation (2.29), then s_θ > 0, and

(2.31)

(2.32)

If we choose the value for θ given by Equation (2.30), then s_θ < 0, and

(2.33)

(2.34)

Thus, there are two solutions depending on the sign chosen for θ.

If r₁₃ = r₂₃ = 0, then the fact that is orthogonal implies that r₃₃ = ±1, and that r₃₁ = r₃₂ = 0. Thus, has the form

(2.35)

If r₃₃ = 1, then c_θ = 1 and s_θ = 0, so that θ = 0. In this case, Equation (2.27) becomes

Thus, the sum ϕ + ψ can be determined as

(2.36)

Since only the sum ϕ + ψ can be determined in this case, there are infinitely many solutions. In this case, we may take ϕ = 0 by convention. If r₃₃ = −1, then c_θ = −1 and s_θ = 0, so that θ = π. In this case Equation (2.27) becomes

(2.37)

The solution is thus

(2.38)

As before there are infinitely many solutions.