汽车门锁耐惯性力计算
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Door latch system subjected to a 30g deceleration
Average Push-Button Spring Output Force=0.5kgf
Pawl Spring Output Torque = 0.079kgfm
A = 30g(m/s2)
F= Ma = M*30g = M*294.2
M1=0.1kg, M2=0.046kg, M3=0.028kg, M4=0.013kg
d1=25.9mm, d2=24.2mm, d3=12.6mm, d4=15.8mm, d5=8.3mm, d6=17.6mm
F1 = M1*a - Average load on knob spring = (0.1kg*30g) - 0.5kgf = 2.5kgf
F2 = M2*a = 0.046kg*30g = 1.38kgf
F3 = M3/2*a = 0.028/2*30g = 0.42kgf
ΣMo = F1*d1+F2*d2-F3*d3 = 2.5*25.9+1.38*24.2-0.42*12.6 = 92.854 kgfmm
F5 = M o/d4 = 92.854/15.8 = 5.8768 kgf
F6 = M4*a = 0.013kg*30g = 0.39 kgf
ΣM p = Pawl spring output torque - (F5d5+F6d6)/1000 = 0.079-(5.8768*8.3+0.39*17.6)/1000 =0.0234 kgfm49
The result indicates ΣM p is more than zero shows that handle component can not transmit a thrust to the lock and can not bring a tendency for unlocking, when the handle and connecting rod are under the action of 30g inertia loading.
:16 Title: Calculation note for 30g - front door longitudinal Drawing no. : 10 30g门锁耐惯力计算-前门纵向（X向）Directive:-Type : XXXX Regulation:11
Door latch system subjected to a 30g deceleration
Average Push-Button Spring Output Force=0kgf
Pawl Spring Output Torque=0.079kgfm
A = 30g(m/s2)
F= Ma = M*30g = M*294.2
M1 = 0.1kg, M2 = 0.046kg, M3 = 0.028kg, M4 = 0.013kg
d1 = 27mm, d2 = 0mm, d3 = 0mm, d4 = 13.8mm, d5 = 13.3mm, d6 = 0mm
F1 = M1*a - Average load on knob spring = 0.1kg*30g - 0.5 kgf = 2.5kgf
F2 = M2*a = 0.046kg*30g = 1.38kgf
F3 = M3/2*a = 0.028/2*30g = 0.42kgf
ΣMo = F1*d1+F2*d2-F3*d3 = 2.5*27+1.38*0-0.42*0 = 67.5 kgfmm
F5 = M o/d4 = 67.5/13.8 = 4.891kgf
F6 = M4*a = 0.013kg*30g = 0.39 kgf
ΣM p = Pawl spring output torque - (F5d5+F6d6)/1000 = 0.079-(4.891*13.3+0.39*0)/1000 =0.0139 kgfm
The result indicates ΣM p is more than zero shows that handle component can not transmit a thrust to the lock and can not bring a tendency for unlocking,when the handle and connecting rod are under the action of 30g inertia loading.
:17 Title: Calculation note for 30g - rear door transversal Drawing no. : 10 30g门锁耐惯力计算-后门横向（Y向）Directive:-Type : XXXX Regulation:11
Door latch system subjected to a 30g deceleration
Average Push-Button Spring Output Force=0.5kgf
Pawl Spring Output Torque=0.079kgfm
A = 30g(m/s2)
F= Ma = M*30g = M*294.2
M1=0.1kg, M2=0.046kg, M3=0.012kg, M4=0.017kg
d1=25.9mm, d2=24.2mm, d3=11.2mm, d4=16.2mm, d5=10mm, d6=0.7mm
F1 = M1*a - Average load on knob spring = (0.1kg*30g) - 0.5kgf =2.5kgf
F2 = M2*a = 0.046kg*30g = 1.38kgf
F3 = M3/2*a = 0.012/2*30g = 0.18kgf
ΣMo = F1*d1+F2*d2-F3*d3 = 2.5*25.9+1.38*24.2-0.18*11.2 = 96.13 kgfmm
F5 = M o/d4 = 96.13/16.2 = 5.934 kgf
F6 = M4*a = 0.017kg*30g = 0.51 kgf
ΣM p = Pawl spring output torque - (F5d5+F6d6)/1000 = 0.079-(5.934*10+0.51*0.7)/1000 =0.0193 kgfm
The result indicates ΣM p is more than zero shows that handle component can not transmit a thrust to the lock and can not bring a tendency for unlocking, when the handle and connecting rod are under the action of 30g inertia loading.