SUGISAKAMASANORI

Journal of Robotics, Networking and Artificial Life
Vol. 9(3); December (2022), pp. 233–239
ISSN (Online):2352-6386; ISSN (Print):2405-9021
https://alife-robotics.org/jrnal.html

Research Article

Evaluation of Two Rollers Arrangement on a Hemisphere by
Kinetic Energy
Kenji Kimura1, Yusuke Abematsu2, Hiroyasu Hirai3, Kazuo Ishii3

Department of Control Engineering, National Institute of Technology, Matsue College, 14-4 Nishi-ikuma-cho, Matsue-shi, Shimane, 690-8518, Japan
Mathematics in General Education, National Institute of Technology, Kagoshima College, 1460-1 Shinko, Hayato-cho, Kirishima-shi, Kagoshima 8995193, Japan
3
Graduate School of Life Science and Engineering, Kyushu Institute of Technology, 2-4 Hibikino, Wakamatsu-ku, Kitakyushu-shi 808-0196, Fukuoka,
Japan
1
2

ARTICLE INFO
Article History
Received 10 November, 2021
Accepted 16 August 2022
Keywords
Omnidirectional movement
Kinetic energy
Angular velocity vector of the sphere

ABSTRACT

A driving roller arrangement of hemisphere is one of the important problems by omnidirectional
sphere conveyance. In this research, the roller arrangement problem, viewed as an evaluation
function, is thought of as mean of roller’s kinetic energy with respect to the sphere direction.
Furthermore, theoretically, we calculate the evaluation function, and find the contact point suchthat the evaluated value is minimal.

© 2022 The Author. Published by Sugisaka Masanori at ALife Robotics Corporation Ltd
This is an open access article distributed under the CC BY-NC 4.0 license(http://creativecommons.org/licenses/by-nc/4.0/)

1. Introduction
A sphere, one of the basic shapes of robots, is used
not only as a multi-fingered fingertip mechanism forhand robots but also as an actuator transmission
mechanism for omnidirectional movement and drivein mobile robots. Spheres are also used as drivingrollers for omnidirectional movement mechanisms,
and there are various arrangements, depending on theapplication of the movement mechanism. Figure1
shows the roller contact type for the number ofactuators (𝑁𝑁𝑤𝑤 ) per sphere.
Examples of mechanisms driven by two rollers
include a power transmission mechanism by Wada etal. [1] (see Figure 1(a)), a mobile device using Ishida's
Figure 1(b), and The abovementioned

Figure 1 Type of roller arrangement for sphere mobile robot

Corresponding author’s E-mail: k-kimura@matsue-ct.jp, y-abematsu@kagoshima-ct.ac.jp, multipletitan@gmail.com, ishii@brain.kyutech.ac.jp233

mechanisms can be used for the roller of a wheelchair.
The rollers are arranged on the equator, generate anangular velocity vector of sphere on the horizontal plane,
and can move in all direction [3]. The angular velocity
vector of the sphere has two- dimensional freedom. Thissituation is theoreticallyconsidered in [4].
The ball holding mechanism [5] (see Figure 1(c)) is
intended to transport the ball, and the roller is placedin the upper hemisphere to hold the ball by friction.
We conducted roller arrangement problem ofsphere conveyance by driving rollers as previous
study[6].
In this research, in the case of omnidirectional
movement, we define an evaluation function as meanof roller’s kinetic energy with respect to spheredirection angle, and we also derive the exact formula.
Furthermore, theoretically, we find the contact pointsuch that the evaluated value (mean of roller’s kinetic
energy) is minimal. Additionally, we perform
simulation and present energy distribution of several
contact points on a sphere.
The remainder of this research is organized as follows.
In Chapter 2, we calculate the exact evaluation value
given by the integral of the kinetic energy in spheredirection. In Chapter 3, we conduct simulation on theevaluation value on the sphere. In Chapter 4, we give asummary and future issues.

𝐸𝐸 = 𝐼𝐼(𝜔𝜔12 + 𝜔𝜔22 )

Where

(1)

Figure 2 The sphere rotational motion by driving rollers at 𝑷𝑷𝒊𝒊and omnidirectional condition is 𝜼𝜼𝟏𝟏 , 𝜼𝜼𝟐𝟐 ∈ 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬{𝑷𝑷𝟏𝟏 , 𝑷𝑷𝟐𝟐 }.

𝐼𝐼 =

1
𝑀𝑀𝑅𝑅2
2

(2)

Because that the sphere and roller engage with eachother at 𝑷𝑷𝑖𝑖 .
‖𝝎𝝎 × 𝑷𝑷𝑖𝑖 ‖ = 𝑅𝑅𝜔𝜔𝑖𝑖

(3)

Thus. 𝐸𝐸 is proportional with respect to sum of square tworoller’s speed.
𝑀𝑀
𝐸𝐸 = (‖𝝎𝝎 × 𝑷𝑷1 ‖2 + ‖𝝎𝝎 × 𝑷𝑷2 ‖2 )(4)
2
2.2 Mean of kinetic energy of rollers
To evaluate the value for roller arrangement, we
define the follows expressions. Eq. (5) presents themean of kinetic energy by integrating the total kinetic
energy of the rollers with respect to the direction 𝜑𝜑

2. Derivation of theoretical evaluation function
In this Chapter, we calculate the omnidirectionalenergy integral of the driving rollers.
As shown in Figure 2, The center 𝑶𝑶 of a sphere withradius r is fixed as the origin of the coordinate system𝛴𝛴 − 𝑥𝑥𝑥𝑥𝑥𝑥. The 𝑖𝑖 th constraint roller (i = 1 or 2) is in point
contact with the sphere at a position vector 𝑷𝑷𝒊𝒊 (𝑷𝑷𝟏𝟏 ≠ 𝑷𝑷𝟐𝟐 ).
𝝎𝝎 denotes the angular velocity vector of the sphere.
Because of 𝜼𝜼𝟏𝟏 , 𝜼𝜼𝟐𝟐 ∈ 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬{𝑷𝑷𝟏𝟏 , 𝑷𝑷𝟐𝟐 } (omnidirectional
condition), 𝝎𝝎 is on span{𝑷𝑷𝟏𝟏 , 𝑷𝑷𝟐𝟐 }. Sphere direction 𝜑𝜑(0° ≤ 𝜑𝜑 < 360°) is the angle from 𝑥𝑥-axis and 𝜌𝜌 is theangle from 𝑥𝑥𝑥𝑥 -plane to 𝝎𝝎 . Now, given the spheremobile velocity 𝑽𝑽 (the center velocity of sphere).

(0° ≤ 𝜑𝜑 ≤ 360°).

𝐸𝐸𝑀𝑀 =1 2𝜋𝜋
� 𝐸𝐸 𝑑𝑑𝜑𝜑
2𝜋𝜋 0

(5)

(i) Case of arbitrary roller arrangement
Quoting Equation (12) of Paper [6] (Kimura) as
follows:‖𝝎𝝎 × 𝑷𝑷1 ‖2 + ‖𝝎𝝎 × 𝑷𝑷2 ‖2(6)= (‖𝒆𝒆𝟑𝟑 × 𝑷𝑷1 ‖2 + ‖𝒆𝒆𝟑𝟑 × 𝑷𝑷2 ‖2 )𝜔𝜔𝑧𝑧2+ 2(〈𝝎𝝎́ × 𝑷𝑷1 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷1 〉 + 〈𝝎𝝎́ ×

2.1 Kinetic energy of the roller
Consider two rollers (right cylinder) with radius 𝑅𝑅, mass
𝑀𝑀, moment of inertia 𝐼𝐼, and roller’s angular velocity 𝜔𝜔𝑖𝑖
The total kinetic energy of the rollers is given by Eq. (1).

𝑷𝑷2 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷2 〉)𝜔𝜔𝑧𝑧
234+ ‖𝝎𝝎́ × 𝑷𝑷1 ‖2 + ‖𝝎𝝎́ × 𝑷𝑷2 ‖2whereThus. By substituting Eq. (13), Eq. (15) and Eq. (17) for𝑇𝑇

𝑷𝑷𝒊𝒊 = 𝑟𝑟�cos𝜃𝜃𝑖𝑖,1 cos 𝜃𝜃𝑖𝑖,2 , sin 𝜃𝜃𝑖𝑖,1 cos 𝜃𝜃𝑖𝑖,2 , sin 𝜃𝜃𝑖𝑖,2 �𝒆𝒆𝟑𝟑 = [0, 0, 1]𝑇𝑇 , 𝝎𝝎́ = �𝜔𝜔𝑥𝑥 , 𝜔𝜔𝑦𝑦 , 0�𝜔𝜔𝑧𝑧 =

‖𝑉𝑉‖
𝑟𝑟

𝑇𝑇

tan 𝜌𝜌

E q . (6), 𝐸𝐸𝑀𝑀 can be represented i n t eam s o f 𝑃𝑃𝑖𝑖,𝑥𝑥 ,(7)𝑃𝑃𝑖𝑖,𝑦𝑦 , 𝑃𝑃𝑖𝑖,𝑧𝑧 .
Here, quoting Eq. (5) and Eq. (6) in [6] (Kimura),
we have(8)(9)

tan 𝜌𝜌 =𝑇𝑇

Using 𝑷𝑷𝑖𝑖 = �𝑃𝑃𝑖𝑖,𝑥𝑥 , 𝑃𝑃𝑖𝑖,𝑦𝑦 , 𝑃𝑃𝑖𝑖,𝑧𝑧 � , 𝒆𝒆𝟑𝟑 × 𝑷𝑷𝑖𝑖 and 𝝎𝝎́ × 𝑷𝑷𝑖𝑖 are
represented as follow.
𝒆𝒆𝟑𝟑 × 𝑷𝑷𝑖𝑖 = �−𝑃𝑃𝑖𝑖,𝑦𝑦 , 𝑃𝑃𝑖𝑖,𝑥𝑥 ,0 �
𝑇𝑇

𝑇𝑇

=

(10)

𝝎𝝎́ × 𝑷𝑷𝑖𝑖 = �𝜔𝜔𝑥𝑥 , 𝜔𝜔𝑦𝑦 ,0 � × �𝑃𝑃𝑖𝑖,𝑥𝑥 , 𝑃𝑃𝑖𝑖,𝑦𝑦 , 𝑃𝑃𝑖𝑖,𝑧𝑧 �𝑇𝑇= �𝜔𝜔𝑦𝑦 𝑃𝑃𝑖𝑖,𝑧𝑧 , −𝜔𝜔𝑥𝑥 𝑃𝑃𝑖𝑖,𝑧𝑧 , 𝜔𝜔𝑥𝑥 𝑃𝑃𝑖𝑖,𝑦𝑦 − 𝜔𝜔𝑦𝑦 𝑃𝑃𝑖𝑖,𝑥𝑥 �

(11)

1 2𝜋𝜋
� sin 𝜑𝜑 cos 𝜑𝜑 𝑑𝑑𝑑𝑑 = 02𝜋𝜋 0Using Eq. (18) and Eq. (19) ,Using Eq. (10), ‖𝒆𝒆𝟑𝟑 × 𝑷𝑷𝑖𝑖 ‖2 is calculated in teams of𝑇𝑇22‖𝒆𝒆𝟑𝟑 × 𝑷𝑷𝑖𝑖 ‖2 = 𝑃𝑃𝑖𝑖,𝑥𝑥+ 𝑃𝑃𝑖𝑖,𝑦𝑦(12)‖𝒆𝒆𝟑𝟑 × 𝑷𝑷1 ‖2 + ‖𝒆𝒆𝟑𝟑 × 𝑷𝑷2 ‖22𝜋𝜋

� tan2 𝜌𝜌 𝑑𝑑𝑑𝑑
0

(13)

2
2
2
2
2
2
= 𝑃𝑃1,𝑥𝑥+ 𝑃𝑃1,𝑦𝑦+ 𝑃𝑃2,𝑥𝑥+ 𝑃𝑃2,𝑦𝑦= 2𝑟𝑟 2 − 𝑃𝑃1,𝑧𝑧− 𝑃𝑃2,𝑧𝑧=

Using Eq. (10) and Eq. (11),〈𝝎𝝎́ × 𝑷𝑷𝑖𝑖 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷𝑖𝑖 〉(14)= −𝑃𝑃𝑖𝑖,𝑦𝑦 𝑃𝑃𝑖𝑖,𝑧𝑧 𝜔𝜔𝑦𝑦 − 𝑃𝑃𝑖𝑖,𝑥𝑥 𝑃𝑃𝑖𝑖,𝑧𝑧 𝜔𝜔𝑥𝑥
‖𝑉𝑉‖=−
(𝑃𝑃𝑖𝑖,𝑦𝑦 𝑃𝑃𝑖𝑖,𝑧𝑧 cos 𝜑𝜑 − 𝑃𝑃𝑖𝑖,𝑥𝑥 𝑃𝑃𝑖𝑖,𝑧𝑧 sin 𝜑𝜑)𝑟𝑟=−〈𝝎𝝎́ × 𝑷𝑷1 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷1 〉 + 〈𝝎𝝎́ × 𝑷𝑷2 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷2 〉

=

(15)

= 𝜋𝜋

{

(𝑃𝑃1,𝑦𝑦 𝑃𝑃1,𝑧𝑧 + 𝑃𝑃2,𝑦𝑦 𝑃𝑃2,𝑧𝑧 )(𝑃𝑃1,𝑧𝑧 𝑃𝑃2,𝑥𝑥 − 𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑧𝑧 )}2𝜋𝜋� ‖𝝎𝝎́ × 𝑷𝑷1 ‖2 + ‖𝝎𝝎́ × 𝑷𝑷2 ‖2 𝑑𝑑𝑑𝑑
0

= 𝜋𝜋

= 𝜋𝜋

2
2
{𝑃𝑃1,𝑧𝑧+ 𝑃𝑃2,𝑧𝑧

22
22+(𝑃𝑃1,𝑦𝑦+ 𝑃𝑃2,𝑦𝑦) sin2 𝜑𝜑+(𝑃𝑃1,𝑥𝑥+ 𝑃𝑃2,𝑥𝑥) cos 2 𝜑𝜑+2(𝑃𝑃1,𝑥𝑥 𝑃𝑃1,𝑦𝑦 + 𝑃𝑃2,𝑥𝑥 𝑃𝑃2,𝑦𝑦 )sin 𝜑𝜑 cos 𝜑𝜑} (17)1

𝑟𝑟 2 𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑦𝑦 −𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑥𝑥

Using Eq. (17),+2𝑃𝑃𝑖𝑖,𝑥𝑥 𝑃𝑃𝑖𝑖,𝑦𝑦 sin 𝜑𝜑 cos 𝜑𝜑) (16)
𝑟𝑟 2

‖𝑉𝑉‖2

(21)

+ 〈𝝎𝝎́ × 𝑷𝑷2 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷2 〉)𝜔𝜔𝑧𝑧 𝑑𝑑𝑑𝑑

(𝑃𝑃1,𝑥𝑥 𝑃𝑃1,𝑧𝑧 + 𝑃𝑃2,𝑥𝑥 𝑃𝑃2,𝑧𝑧 )( 𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑧𝑧 − 𝑃𝑃2,𝑦𝑦 𝑃𝑃1,𝑧𝑧 ) +22
22(𝑃𝑃𝑖𝑖,𝑧𝑧 + 𝑃𝑃𝑖𝑖,𝑦𝑦 sin 𝜑𝜑 +𝑃𝑃𝑖𝑖,𝑥𝑥
cos 2 𝜑𝜑‖𝑉𝑉‖2

(𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑧𝑧 − 𝑃𝑃2,𝑦𝑦 𝑃𝑃1,𝑧𝑧 )2 + (𝑃𝑃1,𝑧𝑧 𝑃𝑃2,𝑥𝑥 − 𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑧𝑧 )(𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑦𝑦 − 𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑥𝑥 )2

0

2
2
2
= (𝜔𝜔𝑥𝑥2 + 𝜔𝜔𝑦𝑦2 )𝑃𝑃𝑖𝑖,𝑧𝑧+ 𝑃𝑃𝑖𝑖,𝑦𝑦𝜔𝜔𝑥𝑥2 +𝑃𝑃𝑖𝑖,𝑥𝑥𝜔𝜔𝑦𝑦2 −2𝑃𝑃𝑖𝑖,𝑥𝑥 𝑃𝑃𝑖𝑖,𝑦𝑦 𝜔𝜔𝑥𝑥 𝜔𝜔𝑦𝑦‖𝝎𝝎́ × 𝑷𝑷1 ‖2 + ‖𝝎𝝎́ × 𝑷𝑷2 ‖2 =((𝑷𝑷𝟏𝟏 × 𝑷𝑷𝟐𝟐 )𝑥𝑥 )2 + ((𝑷𝑷𝟏𝟏 × 𝑷𝑷𝟐𝟐 )𝑦𝑦 )2𝜋𝜋((𝑷𝑷𝟏𝟏 × 𝑷𝑷𝟐𝟐 )𝑧𝑧 )22𝜋𝜋2‖𝝎𝝎́ × 𝑷𝑷𝑖𝑖 ‖2 = �𝜔𝜔𝑥𝑥2 + 𝜔𝜔𝑦𝑦2 �𝑃𝑃𝑖𝑖,𝑧𝑧+ (𝜔𝜔𝑥𝑥 𝑃𝑃𝑖𝑖,𝑦𝑦 − 𝜔𝜔𝑦𝑦 𝑃𝑃𝑖𝑖,𝑥𝑥 )2

𝑟𝑟 2

(20)

� (〈𝝎𝝎́ × 𝑷𝑷1 , 𝒆𝒆𝟑𝟑 × 𝑷𝑷1 〉Using Eq. (11),

=

(19)

Using Eq. (15) and Eq. (18),‖𝑉𝑉‖{�𝑃𝑃1,𝑥𝑥 𝑃𝑃1,𝑧𝑧 + 𝑃𝑃2,𝑥𝑥 𝑃𝑃2,𝑧𝑧 � sin 𝜑𝜑
𝑟𝑟+(𝑃𝑃1,𝑦𝑦 𝑃𝑃1,𝑧𝑧 + 𝑃𝑃2,𝑦𝑦 𝑃𝑃2,𝑧𝑧 ) cos 𝜑𝜑}

‖𝑉𝑉‖2

(18)

�𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑧𝑧 − 𝑃𝑃2,𝑦𝑦 𝑃𝑃1,𝑧𝑧 � sin 𝜑𝜑 − �𝑃𝑃1,𝑧𝑧 𝑃𝑃2,𝑥𝑥 − 𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑧𝑧 � cos 𝜑𝜑
𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑦𝑦 − 𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑥𝑥2𝜋𝜋
1 2𝜋𝜋 21
� sin 𝜑𝜑 𝑑𝑑𝑑𝑑 = � cos 2 𝜑𝜑 𝑑𝑑𝑑𝑑 =
2𝜋𝜋 0
2
0

𝑇𝑇

𝑷𝑷𝑖𝑖 = �𝑃𝑃𝑖𝑖,𝑥𝑥 , 𝑃𝑃𝑖𝑖,𝑦𝑦 , 𝑃𝑃𝑖𝑖,𝑧𝑧 � .

(𝑷𝑷𝟏𝟏 × 𝑷𝑷𝟐𝟐 )𝑥𝑥 sin 𝜑𝜑 − (𝑷𝑷𝟏𝟏 × 𝑷𝑷𝟐𝟐 )𝑦𝑦 cos 𝜑𝜑(𝑷𝑷𝟏𝟏 × 𝑷𝑷𝟐𝟐 )𝑧𝑧

235

‖𝑉𝑉‖2
𝑟𝑟 2

‖𝑉𝑉‖2
𝑟𝑟 2

(22)

2
2
2
2
2
2
(2𝑃𝑃1,𝑧𝑧+ 2𝑃𝑃2,𝑧𝑧+ 𝑃𝑃1,𝑥𝑥+𝑃𝑃1,𝑦𝑦+ 𝑃𝑃2,𝑥𝑥+𝑃𝑃2,𝑦𝑦)22(2𝑟𝑟 2 + 𝑃𝑃1,𝑧𝑧+ 𝑃𝑃2,𝑧𝑧)

Integral by 𝜑𝜑(0° ≤ 𝜑𝜑 ≤ 360°) is represented as follow.
By substituting Eq. (13), Eq. (20), Eq. (21), and Eq. (22)
into Eq. (4), 𝐸𝐸𝑀𝑀 can be represented as

As shown in Table 1, the value increases from thelower left of the table to the right and upward
correspondingly (see [Property 2]. 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) diverges
infinitely as it approaches (𝜃𝜃1 , 𝜃𝜃2 ) = (90°, 0°) . In
particular, when 𝜃𝜃2 = 0 , 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) is constantregardless of the contact position.
As shown in [1] and [2], when two constraintrollers are placed on the equator, the evaluation value
is constant regardless of the angle of the two position
vectors (see [Property 1]).

2
2
− 𝑃𝑃2,𝑧𝑧
2𝑟𝑟 2 − 𝑃𝑃1,𝑧𝑧
4𝑟𝑟2
2
𝐸𝐸
=
𝑀𝑀
2 {�𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑧𝑧 − 𝑃𝑃2,𝑦𝑦 𝑃𝑃1,𝑧𝑧 �
2
‖
‖
𝑀𝑀 𝑉𝑉
�𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑦𝑦 − 𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑥𝑥 �2+�𝑃𝑃1,𝑧𝑧 𝑃𝑃2,𝑥𝑥 − 𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑧𝑧 � } +2{
𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑦𝑦 − 𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑥𝑥(𝑃𝑃1,𝑥𝑥 𝑃𝑃1,𝑧𝑧 + 𝑃𝑃2,𝑥𝑥 𝑃𝑃2,𝑧𝑧 )(𝑃𝑃1,𝑦𝑦 𝑃𝑃2,𝑧𝑧 − 𝑃𝑃2,𝑦𝑦 𝑃𝑃1,𝑧𝑧 ) + (𝑃𝑃1,𝑦𝑦 𝑃𝑃1,𝑧𝑧 +2
2+𝑃𝑃2,𝑧𝑧
𝑃𝑃2,𝑦𝑦 𝑃𝑃2,𝑧𝑧 )(𝑃𝑃1,𝑧𝑧 𝑃𝑃2,𝑥𝑥 − 𝑃𝑃1,𝑥𝑥 𝑃𝑃2,𝑧𝑧 )} + 2𝑟𝑟 2 + 𝑃𝑃1,𝑧𝑧(23)

By theoretical calculation, we get the following
properties.
[Property 1]: Optimality of the evaluated value
If �𝜃𝜃1,2 , 𝜃𝜃2,2 � = (0, 0) (𝑷𝑷1 and 𝑷𝑷2 are on the equator),
𝐸𝐸𝑀𝑀 takes the minimal value 𝑀𝑀‖𝑉𝑉‖2 /2 (seeAppendix(A)).
(ii) Case of symmetry roller arrangement

Figure 3 The distribution of contact points on the upper
hemisphere. (a) Isometric view. (b) Right overhead view.

Especially, in case of symmetry arrangement ( 𝑃𝑃1,𝑥𝑥 =−𝑃𝑃2,𝑥𝑥 , 𝑃𝑃1,𝑦𝑦 = 𝑃𝑃2,𝑦𝑦 , 𝑃𝑃1,𝑧𝑧 = 𝑃𝑃2,𝑧𝑧

),

using

(𝜃𝜃1 , 𝜃𝜃2 ) =(𝜃𝜃1,1 , 𝜃𝜃1,2 ). Eq. (23) is represented as follow.

Table 1 The distribution of energy function 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) in
the upper hemisphere22(𝑟𝑟 2 − 𝑃𝑃1,𝑧𝑧)
2𝑃𝑃1,𝑧𝑧
𝑀𝑀‖𝑉𝑉‖2
2
𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) ={2𝑟𝑟 2 − 2𝑃𝑃1,𝑧𝑧
+
}
2
2
4𝑟𝑟
𝑃𝑃1,𝑦𝑦=𝑀𝑀‖𝑉𝑉‖2 (1 − cos 2 𝜃𝜃1 cos 2 𝜃𝜃2 )2

sin2 𝜃𝜃1(24)(0° < 𝜃𝜃1 < 90°, 0° ≤ 𝜃𝜃2 < 90°)

By theoretical calculation, we prove thefollowing fact.

32.19

29.40

25.12

19.87

14.29

9.04

4.76

1.97

70°
60°

8.32
3.91

7.67
3.65

6.66
3.25

5.43
2.76

4.12
2.24

2.89
1.75

1.88
1.35

1.23
1.09

50°

2.38

2.25

2.07

1.83

1.59

1.36

1.17

1.04

40°

1.68

1.62

1.53

1.41

1.29

1.18

1.08

1.02

30°

1.32

1.29

1.25

1.20

1.14

1.08

1.04

1.01

20°

1.13

1.12

1.10

1.08

1.05

1.03

1.02

1.00

10°

1.03
1.00

1.03
1.00

1.02
1.00

1.02
1.00

1.01
1.00

1.01
1.00

1.00
1.00

1.00
1.00

10°

20°

30°

40°

50°

60°

70°

80°

0°

[Property 2]:Monotonicity of the evaluation function𝜃𝜃2 ,𝜃𝜃1(i) When 𝜃𝜃1 increases, 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) also
decrease.
(ii) When 𝜃𝜃2 increases, 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) also
increase.

In the ball holding mechanism (evaluation of theplacement of the world team) [5], the roller arrangement
is on the upper hemisphere for ball transportation, but it
is less-energy efficient than on the equator. Since the ballis not fixed by a pole caster, it is required to be placed on
the upper hemisphere.

(See Appendix(B)).

3. simulation of Evaluation value on sphereThis Chapter presents the simulation results 𝐸𝐸𝑀𝑀 (Eq.
(24)), with 0° < 𝜃𝜃1 < 90°, 0° ≤ 𝜃𝜃2 < 90°, ‖𝑽𝑽‖ = 1[m/s], 𝑀𝑀 = 2.
Figure 3 shows the contact points on the upper
hemisphere. Table 1 shows the distribution of 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 )at the contact points on the upper hemisphere in steps of𝜃𝜃1 (0° < 𝜃𝜃1 < 90°) and 𝜃𝜃2 (0° ≤ 𝜃𝜃2 < 90°).

80°

4. Conclusion
In this research, we defined an evaluation function
as mean of roller’s kinetic energy with respect tosphere direction angle and derived the exact formula.
Furthermore, theoretically, we proved that points onequator are minimal.
236

Future issues include consideration of motion related tovariable mechanisms with offset.

𝑅𝑅 −

−

Appendix

Proof of [Property 1](A)

𝑃𝑃2,𝑥𝑥 = 𝛼𝛼, 𝑃𝑃2,𝑦𝑦 = 𝛽𝛽, 𝑃𝑃2,𝑧𝑧 = 𝑦𝑦

4𝑟𝑟 2
𝐸𝐸� =
𝐸𝐸
𝑀𝑀‖𝑉𝑉‖2 𝑀𝑀
𝐸𝐸� =+(2𝑟𝑟 2(A,1)+2𝑟𝑟 2 + 𝑥𝑥2 + 𝑦𝑦22− 𝛽𝛽𝛽𝛽) + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝) }𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞

(𝑟𝑟 2 𝑥𝑥 2 + 𝑟𝑟 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 )2 ≥ 4(𝑟𝑟 2 − 𝑥𝑥 2 )(𝑟𝑟 2 − 𝑦𝑦 2 )𝑥𝑥 2 𝑦𝑦 2

(A,12)

𝑟𝑟 2 𝑥𝑥 2 + 𝑟𝑟 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 ≥ 2(𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞)𝑥𝑥𝑥𝑥

𝐸𝐸𝑀𝑀 = 𝑃𝑃𝑋𝑋 2 + 2𝑄𝑄𝑄𝑄 + 𝑅𝑅

𝑋𝑋 = 1/(𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞), (𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞 ≠ 0)

(A,3)

(A,14)(A,4)

From 𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞 ≠ 0 (𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞 = 0 is equality

condition of Cauchy-Schwarz inequality),

(A,5)

2

(𝑝𝑝2 + 𝑞𝑞 2 )(𝛼𝛼2 + 𝛽𝛽 ) > (𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞)2

𝑃𝑃 = (2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 ){(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }2 24(𝑝𝑝2 + 𝑞𝑞 2 )(𝛼𝛼 2 + 𝛽𝛽 2 )𝑥𝑥 2 𝑦𝑦 2 ≥ 4(𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞)2 𝑥𝑥 2 𝑦𝑦 2

𝑄𝑄 = (𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽)(𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)(A,16)(A,7)

Using Eq. (A,14) and Eq. (A,16), it is given.

(A,8)

4(𝑟𝑟 2 𝑥𝑥 2 + 𝑟𝑟 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 )2 ≥ 4(𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞)2 𝑥𝑥 2 𝑦𝑦 2

(A,17)

From 2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 > 0 ⇒ 𝑃𝑃 ≥ 0, we consider caseof 𝑃𝑃 > 0 and 𝑃𝑃 = 0.

Focus on 𝑟𝑟 𝑥𝑥 + 𝑟𝑟 𝑦𝑦 − 2𝑥𝑥 𝑦𝑦 and AM-GM inequality.
2 2

Case of 𝑃𝑃 > 0

From completing the square of Eq. (A,4),
𝐸𝐸� = 𝑃𝑃𝑋𝑋 2 + 2𝑄𝑄𝑄𝑄 + 𝑅𝑅

(A,15)

Using Cauchy-Schwarz inequality and 𝑥𝑥 𝑦𝑦 ≥ 0,

(A,6)

𝑅𝑅 = 2𝑟𝑟 2 + 𝑥𝑥 2 + 𝑦𝑦 2

(A,13)PROOF:
Eq.(A,3) is substituted in right side of Eq.(A,12). Thus,
it is given.
2(𝑟𝑟 2 𝑥𝑥 2 + 𝑟𝑟 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 )2 ≥ 4(𝑝𝑝2 + 𝑞𝑞 2 )(𝛼𝛼2 + 𝛽𝛽 )𝑥𝑥 2 𝑦𝑦 2(A,2)

Furthermore.

𝐸𝐸� takes minimal value(A,11)
2

We make preparations several Lemma for prove
Optimality.
[Lemma 1]:(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)}

𝑄𝑄
𝑄𝑄2= 𝑃𝑃(𝑋𝑋 + )2 + 𝑅𝑅 −𝑃𝑃
𝑃𝑃− 𝑥𝑥 − 𝑦𝑦⟺2 ){(𝑞𝑞𝑞𝑞[Lemma 2]:

𝑝𝑝2 + 𝑞𝑞2 + 𝑥𝑥 2 = 𝑟𝑟 2 , 𝛼𝛼 2 + 𝛽𝛽2 + 𝑦𝑦 2 = 𝑟𝑟 2(i)2

= −(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) − (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)

2{
𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞

Where{(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)}2(2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 ){(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }𝑃𝑃𝑃𝑃 = −𝑄𝑄

2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2{(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }(𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞)2

Where(A,10)

And. Equality condition is

We replace as follow in Eq. (23).

𝑃𝑃1,𝑥𝑥 = 𝑝𝑝, 𝑃𝑃1,𝑦𝑦 = 𝑞𝑞, 𝑃𝑃1,𝑧𝑧 = 𝑥𝑥

𝑄𝑄2
= 2𝑟𝑟 2 + 𝑥𝑥 2 + 𝑦𝑦 2
𝑃𝑃

2 2

2 2

𝑟𝑟 2 𝑥𝑥 2 + 𝑟𝑟 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 = 𝑟𝑟 2 (𝑥𝑥 2 + 𝑦𝑦 2 ) − 2𝑥𝑥 2 𝑦𝑦 2

≥ 𝑟𝑟 2 × 2�𝑥𝑥 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 = 2𝑥𝑥𝑥𝑥(𝑟𝑟 2 − 𝑥𝑥𝑥𝑥) ≥ 0

From |𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞| ≥ 0, |𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞| ≥ 𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞,

(A,9)
237

𝑟𝑟 2 𝑥𝑥 2 + 𝑟𝑟 2 𝑦𝑦 2 − 2𝑥𝑥 2 𝑦𝑦 2 ≥ 2|𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞|𝑥𝑥𝑥𝑥(A,18)

≥ 2(𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞 )𝑥𝑥𝑥𝑥

From Eq. (A,6) and 2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 > 0,

(A,19)

𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽 = 𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝 = 0

Equality conditions are 𝑥𝑥 = 𝑦𝑦 , 𝑥𝑥 2 𝑦𝑦 2 = 0 and{ = 0 or 𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞 ≥ 0 }.
Thus. 𝑥𝑥 = 𝑦𝑦 = 0. [END]

From Eq. (A,6) and Eq. (A,24),[Lemma 3]:
2

2

2

2

𝐸𝐸� = 𝑃𝑃𝑋𝑋 2 − 2𝑄𝑄𝑄𝑄 + 𝑅𝑅

(𝑥𝑥 + 𝑦𝑦 )(2𝑟𝑟 − 𝑥𝑥 − 𝑦𝑦 )−(𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) = 2𝑟𝑟
2−(𝑝𝑝 + 𝑞𝑞

2 2

2

2 2

2 (𝑥𝑥 2

)𝑥𝑥 2

+ 𝑦𝑦
2

2)

− (𝛼𝛼 + 𝛽𝛽
2 2

−

2

(𝑥𝑥 2

)𝑦𝑦 2

+ 𝑦𝑦

Proof of [Property 2](B)

We put 𝑋𝑋 = sin2 𝜃𝜃1 and 𝑌𝑌 = sin2 𝜃𝜃2 in Eq. (23).

2 )𝟐𝟐𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 )− 2(𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞)𝑥𝑥𝑥𝑥=

Equality conditions are 𝑥𝑥 = 𝑦𝑦 = 0. [END]𝑀𝑀‖𝑉𝑉‖2 1 − (1 − 𝑋𝑋)(1 − 𝑌𝑌)
2Using [Lemma 3] , Cauchy-Schwarz inequality and
𝑃𝑃 > 0,≥ {(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)}2

References[1]

2(A,22)[2]

From Eq. (A,10) and Eq. (A,22),−

𝑄𝑄2= 2𝑟𝑟 2 + 𝑥𝑥 2 + 𝑦𝑦 2
𝑃𝑃

(A,23){(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)}(2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 ){(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }

[3]

2

[4]

≥ 2𝑟𝑟 2And. equality condition is Eq. (A,11) , 𝑥𝑥 = 𝑦𝑦 = 0 and(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝) − (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) = 0.
Thus. Minimal value is 𝑀𝑀‖𝑉𝑉‖2 /2 when 𝑥𝑥 = 𝑦𝑦 = 0(The contact points are on the equator).
[END]

(ii)

Case of 𝑃𝑃 = 0

𝑋𝑋𝑀𝑀‖𝑉𝑉‖ 𝑋𝑋 + 𝑌𝑌 − 𝑋𝑋𝑋𝑋
2

=

2

𝑋𝑋

𝑀𝑀‖𝑉𝑉‖2
2(A,27)

1

�1 + ( − 1)𝑌𝑌�
𝑋𝑋

From 0 < − 1, 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) is an decreasing function
𝑋𝑋with respect to 𝜃𝜃1 . and 𝐸𝐸𝑀𝑀 (𝜃𝜃1 , 𝜃𝜃2 ) is an increasing
function with respect to 𝜃𝜃2 .

≥ {(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)2 + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽)2 }{(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }𝐸𝐸� ≥ 𝑅𝑅 −=1

(𝑥𝑥 2 + 𝑦𝑦 2 )(2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 ){(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }{(𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽) + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽) (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)}(2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 ){(𝑞𝑞𝑞𝑞 − 𝛽𝛽𝛽𝛽)2 + (𝛼𝛼𝛼𝛼 − 𝑝𝑝𝑝𝑝)2 }(A,26)

Equality condition is 𝑥𝑥 = 𝑦𝑦 = 0.
From (i) and (ii), It is proved completely.

= 𝑟𝑟 𝑥𝑥 + 𝑟𝑟 𝑦𝑦 − 2𝑥𝑥 𝑦𝑦 − 2(𝑝𝑝𝑝𝑝 + 𝑞𝑞𝑞𝑞)𝑥𝑥𝑥𝑥 ≥ 0(A,21)⟺ 𝑥𝑥2 + 𝑦𝑦2 ≥(A,25)= 𝑅𝑅 = 2𝑟𝑟 2 + 𝑥𝑥 2 + 𝑦𝑦 2 ≥ 2𝑟𝑟 2≥ (𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)2 + (𝑞𝑞𝑞𝑞 + 𝛽𝛽𝛽𝛽)2 (A,20)

2

𝑄𝑄 = 0Thus

2

PROOF:
Using Eq. (A,3) and [Lemma 2],(𝑥𝑥 2 + 𝑦𝑦 2 )(2𝑟𝑟 2 − 𝑥𝑥 2 − 𝑦𝑦 2 ) − (𝑝𝑝𝑝𝑝 + 𝛼𝛼𝛼𝛼)2(A,24)

[5]

[6]

238

M.Wada, K.Kato, "Kinematic modeling and simulationof active-caster robotic drive with a ball transmission(ACROBAT-S)".
2016
IEEE/RSJInternational
Conference on Intelligent Robots and Systems. Daejeon,
2016-12-9/25, IEEE Robotics and Automation Society,S.
Ishida,H.
Miyamoto,“Holonomic
Omnidirectional Vehicle with Ball Wheel Drive
Mechanism, 2012 The Japan Society of MechanicalEngineers.Vol.78,No.790, pp.2162-2170, 2012.
K. Kimura, S. Chikushi, et al, Motion Analysis of aSphere Driven by Rollers, Journal of the Robotics
Society of Japan. Vol.38, No.5, pp.485-495, 2020.
K. Kimura, K. Ishii, Y. Takemura, M. Yamamoto,
Mathematical Modeling and motion analysis of the wheel
based ball retaining mechanism, SCIS & ISIS, pp.41064111, 2016.
K. Kimura, S. Chikushi, K. Ishii, Evaluation of the Roller
Arrangements for the Ball-Dribbling Mechanisms
adopted by RoboCup Teams, Journal of Robotics,
Nrtworking and Artificial Life, Vol.6, issue 3, pp. 183190, 2019.
K. Kimura, K. Ishii, The Spherical Robot Transfer
Problem With Minimal Total Kinetic Energy, The 2021

international Conference on Artificial Alife and Robotics(ICAROB2021), pp.226-270,2021.

Dr. Kazuo Ishii
He is a Professor in the Kyushu
Institute of Technology, where hehas been since 1996. He received
his Ph.D. degree in engineeringfrom University of Tokyo, Tokyo,
Japan, in 1996. His researchinterests span both ship marine
engineeringandIntelligent
Mechanics. He holds five patents derived from his
research.

Authors Introduction
Dr. Kenji Kimura

He is a Lecturer in Department ofControlEngineering,
National
Institute of Technology, Matsue
College, where he has been since2022. He
receivedtheME(mathematics)
fromKyushuUniversity in 2002 and received his
Ph.D. degree in engineering from
Kyushu Institute of in 2020. Then,
He was mathematical teacher of International
Baccalaureate Diploma Program (Mathematics) andengineering course chief in Fukuoka Daiichi HighSchool, His research interests are spherical mobile robot
kinematics, control for object manipulation.
Mr. Yusuke Abematsu
He is an Associate professor of
Mathematics in general education at
National Institute of Technology,
Kagoshima College, where he has
been since April in 2022. Hereceived the ME (mathematics)
from Kyushu University in 2002. Heworked as a public high school math teacher in
Kagoshima for 20 years. He is positively working on
establishing an educational method that connects closely
mathematics and engineering.
Mr. Hiroyasu Hirai
He received his B.E., M.E., in
Computer Science from Nippon
Bunri University, Japan, in 2013,2016, respectively. He is a 3nd year
student in the doctoral program of
the Kyushu Institute of Technology.
He is an engineer of Garuda
incorporated in 2016. His research interest includes
autonomous robotics and machine learning (deeplearning) .

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