TM 5-3895-374-24-1OptionsResistance calculation exampleA motor with the following characteristics: - power4 kW,-rated speed 1450 rpm,- moment of inertia 0,0135 kg m2drives a machine with: - inertia 10 times that of the motor,- resistive torque one tenth of the rated motor torqueThe requirement is to stop in 5 seconds from rated speed at a rate of 2 cycles per minuteRated angular speed:wn = 2pN = 2p1450= 152 rad/s6060Rated motor torqueCnPn= 4000= 26,3 N mwn152*/Resistive torqueCr = 26,3= 2,63 N m10Total inertiaJ = 0,0135 + 10 x 0,0135 = 0,1485 kg m2Braking torqueC= J Dw = 0,1485 x 152=4,52 N mDt5Motor braking torqueCb = C - Cr = 4,52 - 2,63 = 1,89 N mInstantaneousbraking powerPb = Cb w = 1,89 x 152 = 287 WAverage brakingpower duringdecelerationPbd = 0,5 Cb Dw = 0,5 x 1,89 x 152 = 144 WCycle timeT =60= 30 s2Average braking powerduring one cyclePbm = Pbd t = 144 x 5 = 24 WT30Standardized resistance VY1-ADR100W072 is suitable- rated power Pn = 72 W, donc> Pfm,- maximum power possible for 5 seconds (curve p 65)Pmax = 10 x 72 = 720 W, thus > PfCAUTIONA Precise calculation of the resistance, as shown in the example above, is absolutely essentialfor severe applications requiring high braking powers hoisting (vertical movements), machines with very high inertia, driving loads,....If the required braking torque is high, choose a resistance with an ohmic value equal to, or slight’greater than the minimum value given in the table on table page 64(page 3-218)