Chapter Instant Centers of Velocity KUAS ME, C. F. Chang B C Definition of Instant Center (ref pp. 5-7 7 If points C and B are fixed on different bodies, but they are coincident and hae the same elocity at an instant, the location of the coincident point is called the instant center of elocity for the two bodies. That is, or B / / C C B 0 B, C 簡言之, 瞬心是兩桿件之一個重合點, 兩桿件在該點有相同之速度 ( 即無相對速度 In general, there is only one instant center for two bodies. If more than one location is found to be an instant center for two bodies, then all locations are instant centers and the two bodies can be considered to be instantaneously fixed to each other. If no finite location can be found as an instant center of relatie motion of two bodies, then the two bodies are translating w.r.t each other. ermanent instant center: points A and B are permanently coincident Instantaneous instant center: points A and B are momentarily coincident KUAS ME, C. F. Chang 國立高雄應用科大機械系, C. F. Chang
Location of an Instant Center from the Directions of Two Velocities (ref pp. 7-8 8 oints and Q are fixed in body C Gien: the directions of relatie elocities and Q (relatie to body B Find: the instant center, I BC, for bodies B and C 桿 C 上之點 相對桿 B 的速度 桿 C 上之點 Q 相對桿 B 的速度 KUAS ME, C. F. Chang rimary Instant Centers ( 主要瞬心 All instant centers which can be found merely by inspection are called primary instant centers Case : for the two bodies connected with reolute pair IC is located at the reolute joint Case : for the two bodies connected with sliding pair IC is located at the center of curature KUAS ME, C. F. Chang 國立高雄應用科大機械系, C. F. Chang
rimary Instant Centers- Case : for the two bodies connected with rolling pair IC is located at the point of contact Case : for the two bodies connected with cam pair 桿 C 上之 點在桿 B 上之速度 IC is located on the common normal, O B O C, through the contact point KUAS ME, C. F. Chang 5 Example: The Instant Center Location Between Two Frame-Mounted Cams Q is the only point where both elocities Q and Q hae the same direction KUAS ME, C. F. Chang 6 國立高雄應用科大機械系, C. F. Chang
The Kennedy-Aronholdt Theorem ( 三心定理 The total number of instant centers for n bodies is gien by N=n(n-/ 類比 : 兩桿之相對運動有一瞬心 兩點間有一連線 Ex: For four-bar linkage, we hae N=(-/=6 If three bodies are in relatie planar motion, there are three instant centers pertaining to the relatie motion of pairs of those bodies. Those three instant centers are collinear. 進行平面相對運動之三支桿件的三個瞬心恒在一直線上 f: 設若 I 不在 I I 之連線上, 例如在點 A 處, 則桿 和桿 在瞬心處之速度方向必不同, 即 A A, 亦即違反瞬心之定義, 故該三瞬心必在一直線上 KUAS ME, C. F. Chang 7 Find all of the instant centers Example (p. 55 Step : determine primary instant centers by inspection I I I I KUAS ME, C. F. Chang 8 國立高雄應用科大機械系, C. F. Chang
Example (cont. Step : determine the other instant centers by using the Kennedy- Aronholdt theorem - - - - I I I I I I KUAS ME, C. F. Chang 9 Find all of the instant centers Example (p. 56 Step : determine primary instant centers by inspection I I at I I KUAS ME, C. F. Chang 0 國立高雄應用科大機械系, C. F. Chang 5
Example (cont. Step : determine the other instant centers by using the Kennedy- Aronholdt theorem - - - - I I I I at I I KUAS ME, C. F. Chang Example. Find: all instant centers for the Quick-Return Mechanism rocedure: Determine primary instant centers: I, I, I, I 5 and I 5 Applying Kennedy- Aronholdt theorem to determine the other instant centers I I 5 I I I 5 at I KUAS ME, C. F. Chang 國立高雄應用科大機械系, C. F. Chang 6
Example. (I I I 5 - I I 5 at I I KUAS ME, C. F. Chang Example. (I - - 5-5 I 5 at KUAS ME, C. F. Chang 國立高雄應用科大機械系, C. F. Chang 7
Example. (I - - 5-5 I 5 at - - KUAS ME, C. F. Chang 5 Example. (I 5 - - 5-5 I 5 at - - -5 5-5 KUAS ME, C. F. Chang 6 國立高雄應用科大機械系, C. F. Chang 8
Example. (I 5 - - 5-5 I 5 at - - -5 5-5 -5 5-5 KUAS ME, C. F. Chang 7 Example. (whole procedure - - 5-5 I 5 at - - -5 5-5 -5 5-5 KUAS ME, C. F. Chang 8 國立高雄應用科大機械系, C. F. Chang 9
Rotating-Radius Radius Method (pp. 56-57 57 Gien: the elocity of a point on link j ( the instantaneous rotating center of link j (I( j Find: the elocity of any other point on link j ( S j I I S S S ' I js' j I I I j j j j S' S S I j j S ' I I j j S' KUAS ME, C. F. Chang 9 Rotating-Radius Radius Method (cont. Gien: Gien: the theelocity of ofa point pointon onlink link ( ( the theinstantaneous rotating rotatingcenter centerof oflink link (I( (I Find: Find: the theelocity elocityof ofany anyother otherpoint pointon onlink link ( ( S S I I S S ' IS' I I I S' S ' I I S' KUAS ME, C. F. Chang 0 國立高雄應用科大機械系, C. F. Chang 0
Example. (Rotating-Radius Radius Method, p.58 Gien: A Find: B Analysis: oint A lies on link and link oint B lies on link 5 Method (use I 5 :. Find I 5, I and I 5. Use I5 / A = (I I 5 /(I A to determine I5 [I 5 and A are both on link ]. Use B / I5 = (I 5 B/(I 5 I 5 to determine B [I 5 and B are both on link 5] Method (use I 5. Find I 5, I and I 5. Use I5 / A = (I I 5 /(I A to determine I5 [I 5 and A are both on link ]. Use B / I5 = (I 5 B/(I 5 I 5 to determine B [I 5 and B are both on link 5] KUAS ME, C. F. Chang Example. Method - - I 5 I 6 at I 5-5 5 6-56 5 B I56 Method (use I 5. Find I 5, I and I 5. Use I5 / A = (I I 5 /(I A to determine I5 [I 5 and A are both on link ] I 5. Use B / I5 = (I 5 B/(I 5 I 5 to determine B [I 5 and B are both on link 5] 國立高雄應用科大機械系, C. F. Chang
Example. Method KUAS ME, C. F. Chang Example. (pp.59-60 60 Gien: Find: 5 Analysis: I 5 lies on link and link 5. considering I 5 as a point in link yields I5 = (I I 5. considering I 5 as a point in link 5 yields I5 = (I 5 I 5 5 rocedures:. Find I 5, I and I 5. Use I5 = (I I 5 to determine I5. 5 = I5 /(I 5 I 5 KUAS ME, C. F. Chang 國立高雄應用科大機械系, C. F. Chang
Example. (cont. -5-5 5 5 I5 = (I I 5-5 5-56 5 = I5 /(I 5 I 5 KUAS ME, C. F. Chang 5 Example.6 (pp.6-6 6 Gien: =0 rad/s, CW Find: C rocedures:. Find I 5, I and I 5. Use I5 = (I I 5 =(I 5 I 5 5 to determine 5. Vc= (I 5 C 5 KUAS ME, C. F. Chang 6 國立高雄應用科大機械系, C. F. Chang
Example.6 (cont. -5 5 5-56 - - - -5 5-5 5 = (I I 5 /(I 5 I 5 Vc= (I 5 C 5 KUAS ME, C. F. Chang 7 End of chapter KUAS ME, C. F. Chang 8 國立高雄應用科大機械系, C. F. Chang