Optimal Geodesic Curvature Constrained Dubins’ Paths on a Sphere

In this article, we consider the motion planning of a rigid object on the unit sphere with a unit speed. The motion of the object is constrained by the maximum absolute value, \(U_{max}\), of geodesic curvature of its path; this constrains the object to change the heading at the fastest rate only when traveling on a tight smaller circular arc of radius \(r <1\), where r depends on the bound, \(U_{max}\). We show in this article that if \(0<r\le \frac{1}{2}\), the shortest path between any two configurations of the rigid body on the sphere consists of a concatenation of at most three circular arcs. Specifically, if C is the smaller circular arc and G is the great circular arc, then the optimal path can only be CCC, CGC, CC, CG, GC, C or G. If \(r>\frac{1}{2}\), while paths of the above type may cease to exist depending on the boundary conditions and the value of r, optimal paths may be concatenations of more than three circular arcs.

Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

1. Reference [5] considers maximum curvature to be 1, and therefore radius of small circle is \(\frac{1}{\sqrt{2}}\); reader is referred to the top of page 251, 253 of [5].

2. There is no loss in generality in considering a point traveling with unit speed on a unit sphere; one can scale distance and time appropriately to arrive at this conclusion.

This work has been supported in part by AFOSR Grant 21RQCOR084.

Authors and Affiliations

1. Texas A & M University, College Station, TX, 77843, USA

   Swaroop Darbha, Athindra Pavan, Rajagopal Kumbakonam & Sivakumar Rathinam

2. Air Force Research Laboratories, WPAFB, OH, 45433, USA

   David W. Casbeer

3. Infoscitex Corp., Dayton, OH, 45431, USA

   Satyanarayana G. Manyam

Corresponding author

Correspondence to Satyanarayana G. Manyam.

Communicated by Mauro Pontani.

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Appendix

1.1 Proof of Lemma 3.2

Proof of Lemma 3.2

1. (i)

   This can be seen from the governing equations:

   $$\begin{aligned} \frac{dX}{ds} = T(s), \quad \quad \frac{dT}{ds} = -X(s), \quad \quad \frac{dN}{ds} = 0. \end{aligned}$$

   Hence, \(N(s) = N_0\) is the normal to the plane containing the great circle, and

   $$\begin{aligned} X(s) = \begin{pmatrix} \cos (s) \\ - \sin (s) \\ 0 \end{pmatrix}, \quad \quad T(s) = \begin{pmatrix} -\sin (s) \\ - \cos (s) \\ 0 \end{pmatrix}. \end{aligned}$$

   (48)
2. (ii)

   Define \({{{\tilde{X}}}}(s):= u(s) X(s) + N(s)\). For \(s \in [a,b)\), we note that \(\frac{d{{{\tilde{X}}}}(s)}{ds} = U \frac{dX}{ds} + \frac{dN}{ds} = 0\). Hence, \({{{\tilde{X}}}}(s)\) remains constant on [a, b). Note that \(<{{{\tilde{X}}}}(s), T(s)> = 0\); define \({{{\tilde{N}}}}(s) = {{{\tilde{X}}}}(s) \times T(s) = UN(s) - X(s)\). Then:

   $$\begin{aligned} \frac{d{{{\tilde{X}}}}}{ds} = 0, \quad \quad \frac{dT(s)}{ds} = {\tilde{N}}(s), \quad \quad \frac{d{{{\tilde{N}}}}(s)}{ds} = -(1+U^2)T(s). \end{aligned}$$

   Clearly, then

   $$\begin{aligned} {{{\tilde{X}}}}(s)= & {} {{{\tilde{X}}}}(a), \end{aligned}$$

   (49)
   $$\begin{aligned} T(s)= & {} T(a) \cos (s \sqrt{1+U^2}) + {{{\tilde{N}}}}(a) \sin (s \sqrt{1+U^2}) \end{aligned}$$

   (50)
   $$\begin{aligned} {{{\tilde{N}}}}(s)= & {} \sqrt{1+U^2}[-T(a) \sin (s\sqrt{1+U^2}) + {\tilde{N}}(a) \cos (s \sqrt{1+U^2})]. \end{aligned}$$

   (51)

   Clearly, the motion is periodic and the length of the period (circumference of the smaller circle) is \(\frac{2\pi }{\sqrt{1+U^2}}\). The motion of the object corresponding to \(s \in [a,b)\) is a circular arc of radius \(r=\frac{1}{\sqrt{1+U^2}}\) and is in the plane with a normal \({{{\tilde{X}}}}(a)\).

\(\square \)

1.2 Identities Needed for Proof of Lemma 3.8

We will encounter the following quantities frequently and hence, we list them for easy reference:

As we have seen before, the axial vectors of \({{\hat{\Omega }}}_L\) and \({{\hat{\Omega }}}_R\) are respectively given by:

Furthermore, the corresponding rotation matrices are:

Let

The following relationships will be useful in solving for \(a_1, a_2, a_3, b_1, b_2\) and \(b_3\):

Claim 1:

• If \({{\hat{\Omega }}} \in \{{{\hat{\Omega }}}_L, {{\hat{\Omega }}}_R\}\), then \(e_2^T{{\hat{\Omega }}}^2e_2 = -1\).

• \(e_2^T{{\hat{\Omega }}}_R{{\hat{\Omega }}}_Le_2 = 1-2r^2 = e_2^T{\hat{\Omega }}_L{{\hat{\Omega }}}_Re_2\).

• \(e_2^T{{\hat{\Omega }}}_R^2{{\hat{\Omega }}}_L e_2 = 0\).

Proof

It is easy to see that \(e_2^T{{\hat{\Omega }}}_L{{\hat{\Omega }}}_R e_2 = 1-2r^2\) and \(e_2^T{{\hat{\Omega }}}_L^2{{\hat{\Omega }}}_Re_2 = 0\). \(\square \)

Claim 2:

1. 1.

   \(A_{100LR} = {{\textbf{u}}}_L^T{{\hat{\Omega }}}_R R_R(\pi + \phi ) R_L(\pi + \phi ) {{\textbf{u}}}_R =4r^2(1-r^2)a_1\sin \phi \).

2. 2.

   \( A_{010LR} = {{\textbf{u}}}_L^T R_R(\pi + \phi ) R_L(\pi + \phi ){{\hat{\Omega }}}_L {{\textbf{u}}}_R=4r^2(1-r^2)a_1\sin \phi .\)

Proof

Recognizing that

we obtain using Euler–Rodrigues formula that

Since \(e_2\) is perpendicular to both \({{\textbf{u}}}_R\) and \({\textbf{u}}_L\), and \({{\hat{\Omega }}}_R{{\textbf{u}}}_R = 0\), we get

Using a similar approach:

\(\square \)

Claim 3:

1. 1.

   \(A_{000RL}:= {{\textbf{u}}}_R^T{{\hat{\Omega }}}_L R_R(\pi + \phi ) {{\textbf{u}}}_L = -4r^2(1-r^2) \sin \phi .\)

2. 2.

   \(A_{001RL}:= {{\textbf{u}}}_R^TR_L(\pi + \phi ) {\hat{\Omega }}_R{{\textbf{u}}}_L = -4r^2(1-r^2) \sin \phi .\)

Proof

Similarly,

\(\square \)

Claim 4:

1. 1.

   \(A_{000RR}:= {{\textbf{u}}}_R^T{\hat{\Omega }}_LR_R(\pi +\phi )R_L(\pi +\phi ){{\textbf{u}}}_R = 4r^2(1-r^2)\sin \phi [1-2r^2(1+\cos \phi )].\)

2. 2.

   \(A_{010RR}:= {{\textbf{u}}}_R^TR_L(\pi + \phi ) {\hat{\Omega }}_L(\pi + \phi ){{\textbf{u}}}_R = 4r^2(1-r^2) \sin \phi .\)

Proof

Similarly,

\(\square \)

Claim 5:

1. 1.

   \(A_{200LR}:= {{\textbf{u}}}_L^T{{\hat{\Omega }}}_R^2 R_R(\pi + \phi ) R_L(\pi +\phi ){{\textbf{u}}}_R = -4r^2(1-r^2)[1-\cos ^2 \phi + (1-2r^2) \cos \phi (1+ \cos \phi )].\)

2. 2.

   \(A_{110LR}:= {{\textbf{u}}}_L^T{{\hat{\Omega }}}_R R_R(\pi + \phi ) R_L(\pi +\phi ){{\hat{\Omega }}}_L {{\textbf{u}}}_R = 4r^2(1-r^2)[\cos ^2 \phi + (1-2r^2) (1- \cos ^2 \phi )].\)

3. 3.

   \(A_{020LR}:= {{\textbf{u}}}_L^T R_R(\pi + \phi ) R_L(\pi +\phi ){{\hat{\Omega }}}_L^2 {{\textbf{u}}}_R = -4r^2(1-r^2)[1-\cos ^2 \phi + (1-2r^2) \cos \phi (1+ \cos \phi )].\)

Proof

Similarly,

\(\square \)

Claim 6:

1. 1.

   \(B_{000RL}:= {{\textbf{u}}}_R^T{{\hat{\Omega }}}_L^2R_R(\pi + \phi )R_L(\pi + \phi ){{\textbf{u}}}_L = 4r^2(1-r^2)(1-2r^2)(1+\cos \phi )\).

2. 2.

   \(A_{011RL}:= {{\textbf{u}}}_R^TR_R(\pi + \phi )R_L(\pi +\phi ){{\hat{\Omega }}}_L{{\hat{\Omega }}}_R{{\textbf{u}}}_L = -4r^2(1-r^2) \cos \phi \).

3. 3.

   \(A_{002RL}:= {{\textbf{u}}}_R^T R_L(\pi + \phi ){\hat{\Omega }}_R^2{{\textbf{u}}}_L = 4r^2(1-r^2)(1-2r^2)(1+ \cos \phi ).\)

Proof

Finally,

Claim 7: Consider two configurations where the initial and final positions are same, and the initial and final headings are opposite to each other; the paths of type CCC are possible only for \(r\in \Big (0,\frac{\sqrt{3}}{2}\Big ]\).

Proof

It could be shown that the initial and final arc angles are equal, but, we do not present here for brevity. Let \(\alpha \) be the initial and final arc angle, and \(\beta \) be the second arc angle. Without loss of generality, we can assume that the initial and final configurations are I and \(diagonal([1, -1, -1])\), respectively. Therefore, we have the following:

Evaluating the above gives \(\cos {\beta }=\frac{1-2r^2}{2(1-r^2)}\). Considering the fact that the range of the cosine function is \([-1,1]\) gives \(r\le \frac{\sqrt{3}}{2}\). By definition, \(r>0\) since r is the radius of the tightest turn. \(\square \)

Claim 8: For the initial and final configurations as above, the paths of type CGC are possible only for \(r\in \Big (0,\frac{1}{\sqrt{2}}\Big ]\).

Proof

Let \(\alpha \) be the initial and final arc angle, and \(\beta \) be the second arc angle. Without loss of generality, we can assume that the initial and final configurations are I and \(diagonal([1, -1, -1])\), respectively. Therefore, we have the following:

Here \(R_G(\beta ) = I + \Omega _{G} \sin \beta + \Omega _G^2(1-\cos \beta )\), where \(\Omega _{G} = [0 ~-1 ~0; 1 ~0 ~0; 0 ~0 ~0]\). Evaluating the above gives \(\cos {\beta }=\frac{1-3r^2}{1-r^2}\), which implies that \(r\le \frac{1}{\sqrt{2}}\). \(\square \)

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