件加速度规律单位°方案3160010408401.350.39110001/301930653575正弦连杆机构的设计及运动分长度计算:已知:由条件可得:△等边三角形△△△△△在三角形△ADC和△∴AB=(AC-AC’)/2=69.015mmBC=(AC+AC’)/2=314.425mm∵BS2/BC=1/2,DS3/DE=1/2∴BS2=BC/2=314.46/2=157.2125mmDS3=DE/2=210/2=105mm由上可得:ABBCBS2CDDEDS3EF69.015mm314.425mm157.2125mm140mm210mm105mm52.5mm位置8机构运动速度分析:已知:n1=90r/min;=rad/s==9.425逆时针=·=9.425×0.069015=0.650m/s=+大小?0.65?方向⊥CD⊥AB⊥BC选取比例尺=0.01(m/s)/mm,作速度多边形=·=0.01x60=0.600m/s=·=0.01x18=0.180m/s=·=0.01x90=0.900m/s=·=0.01x88=0.880m/s=·=0.01x20=0.200m/s=·=0.01x62=0.620m/s=·=0.01x44=0.440m/s∴==0.18/0.314425=0.572rad/s(逆时针)ω==0.60/0.140=4.290rad/s(顺时针)ω==0.20/0.0525=3.809rad/s(顺时针)项目数值0.6500.6000.9000.8800.6200.449.4250.5724.2903.809单位m/sRad/s位置8机构运动加速度分析:==9.4252×0.069015=6.130m/s2==0.5722×0.314425=0.103m/s2==4.2902×0.14=2.577m/s2==3.8092×0.0525=0.762m/s2
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