How To Build Your Own Scissor Lift?

Motor choosing for screw lift scissor lift?

• For a scissor lift, I need around +-102Watt to lift it, and while at lowest position before lifitng, it is around 9.08Nm torque required, and at highest position the speed is fastest where,447.8RPM. Now i do not know how to choose a DC motor and gear reduction ratio. because i cannot search any 120 watt catalogue have 10Nm stall torque. Can anyoe have any idea? or can we gear the motor torque over the motor stall torque at gearbox output? Thanks ...please help, and also the torque at output of gear box is gradually reducing,so RPM increasing? So what should be the ratio reduction of gear? From input motor speed to gear output speed..Which output speed location should i choose? Thanks ADDED, ThanksEcko, Now i am learning to design a project.So no motor buy yet. need to choose from calculation and get the budget to buy. please help ...thnaks My scissor lift is to lift up a person sitting on it. I assume 100kg capacity with 0.5 safety factor. Then become 150kg. I did the static analysis on it, without include the weight of material and friction in my static calculation..Because from static analysis only i will find stress analysis and find the geometry minimum thickness of material required. Then from static, i get the horizontal force act on ACME screw. find the torque rise with friction factor included. From mechanical advantage i assume vertical speed always constant along lifting then i find the horizontal speed required at few position. From lowest, intermediate and to highest position. Then i divided it with thread lead. now i got RPS and multiply with 60 get RPM. So for every location, the torque from static analysis and speed from mechanical advantage i got

• Answer:

  You have already specified the system as: Power 102 watts Torque 9.8Nm (maximum load) Speed 447.8RPM (lightly loaded). For a quick check: Power_W = (torque_N.m x 2pi x RPM) / 60 = (9.8Nm x 6.2832 x 448) / 60 = 459W. I realise you don't need the high torque and the high rpm at the same time, but the motor is more or less constant maximum torque, and if it has to go this speed, that is the size. The motor goes at a speed according to the voltage applied, and the torque according to the load. If it is not allowed to run at the correct speed for the voltage (due to overload) it will draw too much current (and burn out). If you want slower speed at the start for example you would have a speed controller which varies the voltage. If your test example varied speed dramatically it is because the motor is overloaded part of the time or all of it. Measure the voltage and current. The electrical power is V x I. The electrical power and mechanical power are related. Multiply the electrical power by 0.8 (allowing for losses) to get the mechanical power. Then if the RPM is determined, calculate the torque using the formula above.. Voltage is for RPM, current is for torque. DC Motors of this size might be 2000RPM. They should be 24V as the current is high for a 12V motor this size. This is about starter motor size, but they are only rated for a few seconds. An AC motor is probably cheaper especially if you need a power supply for the DC motor.. With reduction gears, the spec above is the output of the gears. Lets say the available motor speed with rated voltage and load is 2000 RPM so the gears are about 2000/448 or 4.46:1 reduction. The losses in the gears may be 10-30% (find the data sheet). Thus the motor power is 10-30% more than at the gear output. The RPM is fixed by the voltage. This is 459W / 0.8 = 574W. The motor will probably be a 3/4hp size for this, but even a 1hp will do, and it may be more available. Note that the torque is less for the motor, and multiplied by the gears ratio, but with losses through the gears. The RPM is divided by the gears ratio. A suggestion, a larger size battery operated electric drill may be exactly what you want for this. It can also be used to measure the voltage, and current. This gives the power, and with the RPM and gear ratio of the drill you can calculate the torque (for 80% efficiency). It may not be continuously rated, but I think the operation is intermittent anyway. How did you get the figures for the scissor lift? You might have to leave a margin for friction and load, also changes as it ages. Do check them again. The link shows a suitable gear motor (if the figures are correct), but these are much more expensive than you want I think. Go for an electric drill.