本論文提出兩指抓取圓球之逆向動力學的力量分析程序，亦即在物體的線性加速度及角加速度皆已知的情況，求出手指的挾持力。由於剛體模型會造成不確定性，本研究建議加入彈性變形，利用赫式接觸之力與位移關係式、相容方程式、以及運動方程式合併解題。本文針對二指抓取彈性球體時的情況，從事以下兩種情況的逆向動力分析:第一，針對物體的線性加速度及角加速度為已知，且兩指間的相對位移向量亦為已知的情況，求得挾持力的數值解。由於運動方程式具一階不確定性，這表示可選取一個抓取力當作主要變數，而其他的抓取力都表示成這個力的函數，因此數值程序僅包含單一未知數，可快速求出接觸力。第二，針對特定的線性加速度及角加速度，若兩指間的夾緊方向亦為已知，本文可求出手指不會滑動所需的最低夾緊量、以及這時的接觸力，而且是以閉合解呈現。

In this thesis a procedure for inverse dynamics analysis of two-fingered grasping of a sphere is proposed. Contact forces may be found for given linear and angular accelerations of a spherical body. Elastic force-displacement relations predicted by Hertz contact theory are used to remove the indeterminancy produced by rigid body model. Two types of inverse dynamics analysis are performed for two-fingered grasping of a sphere. Firstly, as linear and angular accelerations, as well as the relative displacement vector of the finger tips, are given, grasping forces may be obtained by a numerical procedure. In this procedure one degree of indeterminancy produced by the equations of motion are utilized. Specifically, one particular contact force may be chosen as the principal unknown, and all other contact forces are expressed in terms of this force. The numerical procedure is hence very efficient since it contains only one principal unknown. Secondly, for given linear and angular accelerations, if the relative grasping direction of the two fingers is also known, then the closed form solution for the minimum tightening displacement for sliding not to occur can be obtained.

目錄
中文摘要.............................................I
英文摘要............................................II
目錄....................................................III
圖目錄...............................................IV
第一章 緒論.......................................1
1.1 前言..............................................1
1.2 文獻回顧.....................................1
1.3 研究目的.....................................5
第二章 運動方程式..........................6
2.1 簡介..............................................6
2.2 運動方程式.................................7
第三章 研究方法	............................11
3.1 研究方法....................................11
3.2 解題方法....................................14
第四章 結果與討論.........................17
第五章 結論及未來展望................21
參考文獻...........................................22
附錄...................................................26

圖目錄
圖1     半徑為r的球體，A點受力為N1，T1，T2，B點受力為N2，T3， T4...................................................................................................................................45
圖2     N1在x軸挾角及z軸挾角皆為π/2，只有z方向質心加速度且沒有此球體角加速度，y方向挾緊量為0.09mm時與x z方向挾緊位移的關係......46
圖3    當z軸挾角為π/2，在不同x軸挾角情況下且只有z方向質心加速度，角加速度向量α=0時N1 與y方向夾緊位移的關係.........................................47
圖4    當z軸挾角為π/2，在不同x軸挾角情況下且只有z方向質心加速度，角加速度向量α=0時N1 與AB方向挾緊位移的關係......................................48
圖5    當z軸挾角為π/2，在不同x軸挾角情況下且只有y方向質心加速度，角加速度向量α=0時N1 與AB方向挾緊位移的關係......................................49
圖6    當z軸挾角為π/2，在不同x軸挾角情況下且只有z方向質心加速度，質心加速度向量a=0時N1 與AB方向挾緊位移的關係...................................50
圖7    當x軸挾角為π/2，在不同z軸挾角情況下且只有z方向質心加速度，角加速度向量α=0時N1 與AB方向挾緊位移的關係.......................................51
圖8    當x軸挾角為π/2，在不同z軸挾角情況下且只有z方向質心加速度，角加速度向量α=0時N1 與AB方向挾緊位移的關係.......................................52
圖9    至少所需夾緊量d與x y方向質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/2，且沒有z方向質心速度及角加速度向量α=0....................53
圖10  至少所需夾緊量d與此球體質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/2，且沒有z方向質心速度及角加速度向量α=0....................54
圖11  至少所需夾緊量d與y z方向質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/2，且沒有x方向質心速度及角加速度向量α=0...................55
圖12  至少所需夾緊量d與z x方向質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/2，且沒有y方向質心速度及角加速度向量α=0...................56
圖13  至少所需夾緊量d與z x方向角加速度的關係此時B點與x軸挾角及z軸挾角皆為π/2，且沒有y方向角速度及質心加速度向量a=0.........................57
圖14  至少所需夾緊量d與x y方向質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/4，且沒有z方向質心速度及此角加速度向量α=0................58
圖15  至少所需夾緊量d與y z方向質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/4，且沒有x方向質心速度及此角加速度向量α=0................59
圖16  至少所需夾緊量d與z x方向質心加速度的關係此時B點與x軸挾角及z軸挾角皆為π/4，且沒有y方向質心速度及此角加速度向量α=0................60
圖17  至少所需夾緊量d與x y方向角加速度的關係此時B點與x軸挾角及z軸挾角皆為π/4，且沒有z方向角速度及質心加速度向量a=0.........................61
圖18  至少所需夾緊量d與y z方向角加速度的關係此時B點與x軸挾角及z軸挾角皆為π/4，且沒有x方向角速度及質心加速度向量a=0.........................62
圖19  至少所需夾緊量d與z x方向角加速度的關係此時B點與x軸挾角及z軸挾角皆為π/4，且沒有y方向角速度及質心加速度向量a=0..........................63
圖20  至少所需夾緊量d與x y方向質心加速度的關係此時B點與x軸挾角為π/2、z軸挾角為π/4，且沒有z方向質心速度及此角加速度向量α=0.........64
圖21  至少所需夾緊量d與y z方向質心加速度的關係此時B點與x軸挾角為π/2、z軸挾角為π/4，且沒有x方向質心速度及此角加速度向量α=0.........65
圖22  至少所需夾緊量d與z x方向質心加速度的關係此時B點與x軸挾角為π/2、z軸挾角為π/4，且沒有y方向質心速度及此角加速度向量α=0.........66
圖23  至少所需夾緊量d與y z方向角加速度的關係此時B點與x軸挾角為π/2、z軸挾角為π/4，且沒有x方向角速度及質心加速度向量a=0...................67
圖24  至少所需夾緊量d與x y方向質心加速度的關係此時B點與x軸挾角為π/4、z軸挾角為π/2，且沒有z方向質心速度及此角加速度向量α=0..........68
圖25  至少所需夾緊量d與y z方向質心加速度的關係此時B點與x軸挾角為π/4、z軸挾角為π/2，且沒有x方向質心速度及此角加速度向量α=0..........69
圖26  至少所需夾緊量d與z x方向質心加速度的關係此時B點與x軸挾角為π/4、z軸挾角為π/2，且沒有y方向質心速度及此角加速度向量α=0..........70
圖27  至少所需夾緊量d與x y方向角加速度的關係此時B點與x軸挾角為π/4、z軸挾角為π/2，且沒有z方向角速度及質心加速度向量a=0....................71
圖28  至少所需夾緊量d與y z方向角加速度的關係此時B點與x軸挾角為π/4、z軸挾角為π/2，且沒有x方向角速度及質心加速度向量a=0....................72
圖29  至少所需夾緊量d與z x方向角加速度的關係此時B點與x軸挾角為π/4、z軸挾角為π/2，且沒有y方向角速度及質心加速度向量a=0....................73

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