伺服电机的选型计算
丝杆螺距 p=10mm,输送质量 W=20kg,推力0kg(无),试选择伺服电机的容量.
(1）最大移动速度v 由于采用直接传动，减速比i=1,当选择电机的额定转速为3000r/min时，工作台 的最大移动速度为: 1 p ≔ 10 i≔1 n ≔ 3000 ⋅ ―― min v ≔ n ⋅ p ⋅ i = 500 ――
(2）负载惯性矩Jl 首先计算丝杆到电动机轴的惯性矩J1： 3 l ≔ 500 γ ≔ 7.85 ⋅ 10 ⋅ d ≔ 20
2 −5 ⋅γ⋅l⋅d J1 ≔ ―――― ⋅i =⎛ ⎝6.165 ⋅ 10 ⎞ ⎠ 32 ⋅ 4
―― 3
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2
移动部分折算到到电动机轴的惯性矩J2： W ≔ 20 ⎛ p ⎞ 2 −5 J2 ≔ W ⎜―― ⋅i =⎛ ⎝5.066 ⋅ 10 ⎞ ⎠ ⎟ ⎝2 ⎠ 总惯性矩J： −4 JL ≔ J1 + J2 = ⎛ ⎝1.123 ⋅ 10 ⎞ ⎠ ⋅
2 2
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2
(3）负载转矩Tl 工作台导轨移动摩擦系数=0.1，机械效率=0.9 μ ≔ 0.1 η ≔ 0.9 F≔0 ⋅
μ⋅W⋅ +F TL ≔ ―――― ⋅ p ⋅ i = 0.035 2 ⋅η (4）电机容量预选 TL Te ≔ ―― = 0.039 ⋅ 0.9 JL ⎛ −5 JM ≔ ― = ⎝3.744 ⋅ 10 ⎞ ⎠ 3
⋅
2
电机转速为3000转 TL = 0.035 TL ⋅ 2 ⋅ n PZ ≔ ―――― = 12.107 η 查表得MQMA-02 Pe ≔ 0.2 JM ≔ 0.42 ⋅ 10
−4
⋅
3 1 n=⎛ ⎝3 ⋅ 10 ⎞ ⎠ ――
⋅
⋅
2
Te ≔ 0.64
⋅
TAC ≔ 1.91
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(5）最短加减速时间 ⎛ ⎝JM + JL⎞ ⎠ ⋅ (n − 0) ⋅ 2 tAC ≔ ――――――― = 0.026 TAC − TL 工作时加减速时间为0.08秒，则加减速转矩为 tAC ≔ 0.08 ⎛ ⎝JM + JL⎞ ⎠ ⋅ (n − 0) ⋅ 2 TAC ≔ ―――――――+ TL = 0.641 tAC (6）运行模式及热校核 t1 ≔ 0.1 t2 ≔ 0.1 t3 ≔ 0.1 t0 ≔ 0.2 ⋅
T1 ≔ TAC = 0.641
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T3 ≔ T1
T2 ≔ TL = 0.035
⋅
2 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ T1 ⋅ t1 + T2 ⋅ t2 + T3 ⋅ t3 Trms ≔ ―――――――― = 0.405 t0 + t1 + t2 + t3
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