Ultrasound-guided prostate percutaneous intervention robot system and calibration by informative particle swarm optimization

doi:10.1007/s11465-021-0659-x

Frontiers of Mechanical Engineering

2022, Vol. 17

Issue (1): 3 https://doi.org/10.1007/s11465-021-0659-x

本期目录

Ultrasound-guided prostate percutaneous intervention robot system and calibration by informative particle swarm optimization

Jiawen YAN, Bo PAN(

), Yili FU

State Key Laboratory of Robotics and System, Harbin Institute of Technology, Harbin 150001, China

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Abstract：

Applying a robot system in ultrasound-guided percutaneous intervention is an effective approach for prostate cancer diagnosis and treatment. The limited space for robot manipulation restricts structure volume and motion. In this paper, an 8-degree-of-freedom robot system is proposed for ultrasound probe manipulation, needle positioning, and needle insertion. A novel parallel structure is employed in the robot system for space saving, structural rigidity, and collision avoidance. The particle swarm optimization method based on informative value is proposed for kinematic parameter identification to calibrate the parallel structure accurately. The method identifies parameters in the modified kinematic model stepwise according to parameter discernibility. Verification experiments prove that the robot system can realize motions needed in targeting. By applying the calibration method, a reasonable, reliable forward kinematic model is built, and the average errors can be limited to 0.963 and 1.846 mm for insertion point and target point, respectively.

Key words： ultrasound image guidance prostate percutaneous intervention parallel robot kinematics identification particle swarm optimization informative value

收稿日期: 2021-05-16 出版日期: 2022-01-28

Corresponding Author(s): Bo PAN

引用本文:

. [J]. Frontiers of Mechanical Engineering, 2022, 17(1): 3.
Jiawen YAN, Bo PAN, Yili FU. Ultrasound-guided prostate percutaneous intervention robot system and calibration by informative particle swarm optimization. Front. Mech. Eng., 2022, 17(1): 3.

链接本文:

https://academic.hep.com.cn/fme/CN/10.1007/s11465-021-0659-x
https://academic.hep.com.cn/fme/CN/Y2022/V17/I1/3

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Tab.1

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Tab.4

Stage information	δθ₁/rad	δθ₂/rad	$r ∗$ /mm	$d 1 ∗$ /mm	$d 2 ∗$ /mm	δx/mm	δy/mm	$z ∗$ /mm
Top stage	−0.012	−0.015	66+1.466	30+0.966	45−0.009	0.4381	−1.6791	−27.150
Bottom stage	0.002	−0.007	66+2.193	30+1.395	45+0.001	0.4535	0.1635	−134.850

Tab.5

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Fig.19

Parameter	Component	Informative value
δα₁	x	$[16 (d 1 + 2 d 2) (2 (d 12 − r 2) c o s θ 1 + r 2 c o s (2 θ 1 − θ 2)) − (2 d 13 + 4 d 12 d 2 − d 1 r 2 − 2 d 2 r 2) c o s θ 2 − 2 d 12 r 1 − c o s (θ 1 − θ 2) 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12 s i n θ 1 \Bigggr] 16 s i n 2 (θ 1 − θ 2 2) / (2 d 12 (1 − c o s (θ 1 − θ 2)) 32 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12)$
δα₁	y	$s i n 2 (θ 1 − θ 2 2) [152 d 12 r c o s θ 1 1 − c o s (θ 1 − θ 2) 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12 + (d 1 + 2 d 2) (2 (d 12 − r 2) s i n θ 1 + r 2 s i n (2 θ 1 − θ 2) + (r 2 − 2 d 12) s i n θ 2)] 15 / (2 d 12 (1 − c o s (θ 1 − θ 2)) 32 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12)$
δα₂	x	$− s i n 2 (θ 1 − θ 2 2) [15 (d 1 + 2 d 2) (− (2 d 12 − r 2) c o s θ 1 − r 2 c o s (θ 1 − 2 θ 2)) + (2 d 13 + 4 d 12 d 2 − 2 d 1 r 2 − 4 d 2 r 2) c o s θ 2 + 2 d 12 r 1 − c o s (θ 1 − θ 2) 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12 s i n θ 2] 15 / (2 d 12 (1 − c o s (θ 1 − θ 2)) 32 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12)$
δα₂	y	$s i n 2 (θ 1 − θ 2 2) [152 d 12 r c o s θ 2 1 − c o s (θ 1 − θ 2) · 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12 + (d 1 + 2 d 2) · ((2 d 12 − r 2) s i n θ 1 + r 2 s i n (θ 1 − 2 θ 2) + 2 (r 2 − d 12) s i n θ 2)] 15 / (2 d 12 (1 − c o s (θ 1 − θ 2)) 32 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12)$
δl₁	x	$(d 13 + d 2 r 2 − d 2 r 2 cos ? (θ 1 − θ 2)) (sin ? θ 1 − sin ? θ 2) / (d 13 1 − cos ? (θ 1 − θ 2) 2 + r 2 (cos ? (θ 1 − θ 2) − 1) d 12)$
δl₁	y	$− (d 13 + d 2 r 2 − d 2 r 2 c o s (θ 1 − θ 2)) (c o s θ 1 − c o s θ 2) / (d 13 1 − c o s (θ 1 − θ 2) 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12)$
δl₂	x	$2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12 (s i n θ 1 − s i n θ 2) / 1 − c o s (θ 1 − θ 2)$
δl₂	y	$2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12 (− c o s θ 1 + c o s θ 2) / 1 − c o s (θ 1 − θ 2)$
δr	x	$12 (c o s θ 1 + c o s θ 2 + (d 1 + 2 d 2) r 1 − c o s (θ 1 − θ 2) (− s i n θ 1 + s i n θ 2) d 12 2 + r 2 (c o s (θ 1 − θ 2) − 1) d 12)$
δr	y	$12 (s i n θ 1 + s i n θ 2 + (d 1 + 2 d 2) r 1 − c o s (θ 1 − θ 2) (c o s θ 1 − c o s θ 2) 2 d 14 + r 2 d 12 (c o s (θ 1 − θ 2) − 1))$
δx	x	1
δx	y	0
δy	x	0
δy	y	1

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