Linear screw actuator
Linear screw actuator
(OP)
Due to space restrictions I wish to design a linear actuator using trapezoidal screw, see attached sketch.
Travel required=730mm
Required time=25sec
Motor Speed=584rpm
Mass=45Kg
l=Screw pitch=3mm
dm=Mean pitch diameter=12.45mm
f=Friction coefficient=0.15
F=Load=441.45N
Calculated torque: Fdm/2*((l+pi*f*dm)/(pi*dm-f*l) = 0.7Nm
My question is, will the required motor torque be the same as the nut, i.e. 0.7Nm?
I cant figure out if the torque on the gear be more than the nut (as this has a large radius)
Travel required=730mm
Required time=25sec
Motor Speed=584rpm
Mass=45Kg
l=Screw pitch=3mm
dm=Mean pitch diameter=12.45mm
f=Friction coefficient=0.15
F=Load=441.45N
Calculated torque: Fdm/2*((l+pi*f*dm)/(pi*dm-f*l) = 0.7Nm
My question is, will the required motor torque be the same as the nut, i.e. 0.7Nm?
I cant figure out if the torque on the gear be more than the nut (as this has a large radius)





RE: Linear screw actuator
If I understand the sketch correctly then your driving the screw through a set of gears with one gear fixed to the screw and a ratio of 1:1, that being the case then your motor needs to produce that same torque plus losses.
RE: Linear screw actuator
RE: Linear screw actuator
Using this link I get the torque to be 0.63Nm so it looks good to me.
http://www.eng.auburn.edu/users/marghitu/Screws.pd...
RE: Linear screw actuator
Since you have a gear set between the motor and jackscrew nut, there will be a difference in the relative torque at each location due to mechanical losses in the supporting bearings and gear contact. When you calculate the total mechanical losses in your system for sizing the electrical motor, you should make sure to consider the combined worst case for all of the contact friction losses.
Good luck.
Terry
RE: Linear screw actuator