Acceleration Angular Acc:a =atBAl lAa b"b'/AB Dire:CW aBA=J(aBA)2+(anBa)2=laB√a2+o4=业aab' acA=(atcA)2+(a"cA)2=ICA/a2+4=uaac' acB=(a'cB)2+(a"cB)2 =ICBa2+4=uab'c then:a'b/le=b'c/lec=a'c/lcA △ab'c'∽△ABC p' Denote p'a'b'c'as acc image b B b” C% Velocity Vector graphical analysis method The fourbar Crank-slider Given linkage configuration VB3=VB2+VB3B2 1,Find VB3,@3 Mag: ?√ Dire: √√ ∥BC Direction of VB3B2:b2b3 o3=μpb3/lcB b3 p EX.Using IC b2 1111 Angular Acc:α＝at BA/ lAB then：a’b’/ lAB＝b’c’/ lBC＝ a’ c’/ lCA Denote p’a’b’c’ as acc image ∴ △a’b’c’∽△ABC aBA＝ (at BA)2+ (an BA)2 aCA＝ (at CA)2+ (an CA)2 aCB＝ (at CB)2+ (an CB)2 ＝μa b”b’ /μl AB Dire：CW b’ b” a’ p’ c”’ c” B c’ A C ＝lCA α2 +ω 4 ＝lCB α2 +ω 4 ＝lAB α2 +ω 4＝μaa’ b’ ＝μa a’ c’ ＝μa b’ c’ α Acceleration Vector graphical analysis method Direction of VB3B2 : b2b3 ω3 = μvpb3 / lCB VB3＝VB2+VB3B2 Mag： Dire： ? √ √ √ ? ∥BC p b2 b3 B 1 3 2 A C ω3 ω1 The fourbar Crank-slider ? Velocity Given linkage configuration , Find VB3, 3 EX. Using IC