圆上清支大学 Example solution ▣Boundary conditions >When q=b,s=h,Thus h=C,+C+Cs cxxjcc > When 0=B,v=0,Thus 0=1/B[3C+4C+5Cg] .a.ceae)olcaa When 0=B,a=0,Thus 0=1/B2[6C3+12C+20Cg] cc)c) 圆上泽充大学 Example solution Solving gives:(Called the 3-4-5 polynomial) =-j Similar in shape to cycloidal ▣Discontinuous jerk 0 B cam angle 0 2525  Boundary conditions  When q=b, s=h, Thus h = C3+ C4+ C5 5 5 4 4 3 3 2 0 1 2                                                        s C C C C C C  When =, v=0, Thus 0 = 1/[3C3+ 4C4+ 5C5]                                             4 5 3 4 2 1 2 2 3 3 4 5 1          v C C C C C  When =, a=0, Thus 0 = 1/[6C3+ 12C4+ 20C5]                                    3 5 2 2 2 3 4 2 6 12 20 1        a C C C C Example solution  Solving gives: (Called the 3-4-5 polynomial)                                    3 4 5 10 15 6       s h Example solution  Similar in shape to cycloidal  Discontinuous jerk