How Do You Select a DC Motor for a Scissor Lift Using a Power Screw?

In summary, the conversation discusses the considerations when choosing a DC motor for a scissor lift. It is mentioned that the highest torque is at the lowest position and the speed should not be too fast when lifting a patient. The conversation also touches on the assumption of constant speed along the lifting process and the use of a motor with a control to supply a DC current. It is recommended to find the RPM and torque at different heights to better understand the speed-torque relationship. The use of mechanical advantage in calculating speed and how to compare manufacturer specifications is also mentioned. Finally, a picture of the scissor linkage is provided for reference.
  • #1
sporty
30
0
Can anyone tell me . if I am going to lift up a scissor lift using power screw drive by a motor, the highest torque is during lowest position before lifting so this is a maximum torque and the speed RPM is the speed i wish to lift scissor up with a turn equal to a lead of the screw?
So during choosing a DC motor, what should i consider? IF the torque required to lift from lowest possition is 10Nm and speed is 619RPM.. How to get a suitable motor from manufacturer catalogue?
On e more doubt is during lifting, the torque will gradually reduce is it means the speed will increase? So Due to power= Torque * speed in radian...So what i assume in beginning RPM required will change due to increasing speed ?
Finally from motor spec choose wth compared the load rpm and motor rpm to decide the gear ratio...am i correct?
Please explain to me...thanks.
 
Engineering news on Phys.org
  • #2
Some things seem a bit mixed up -

Yes, the highest torque is at the bottom.

Assuming the friction is constant (probably won't be):
- If you provide a constant RPM to the screw, then the lift will slow down as it gets higher.
- If you provide a constant torque to the screw then the lift will speed up as it gets higher.
- If you provide a constant power to the screw, then the lift will rise at a constant speed.
However a DC motor does neither, so it isn't clear if it'll speed up or slow down without further information.

Assuming you don't really care how fast it goes at the top, and just want to be sure it can lift all the way, then you can select a motor that can produce 10Nm at 619RPM.

To identify that, look at (or plot) the speed-torque curve (typically a straight line) for a candidate motor, and see if that point is on or below the curve. The operating point should normally be roughly around 75% of the no-load speed. That gives you reasonable compromise between efficiency and power. But it's easy enough to calculate the optimum to be sure.

The main thing to be careful of is if a motor is rated with 10Nm torque and 619RPM then it usually doesn't mean both at the same time, so it will be grossly underpowered. Unless it says 10Nm at 619RPM.
 
  • #3
Thanks a lot unrest,
I have no idea in choosing a motor in my product, because i do not well know the motor.
If I am goin to lift up my scissor lift, i found the highest torque is at lowest position, So the 10Nm is the torque required from my calculation.
At the same time, the scissor lift is to lift up a patient from a seat, so the speed should not be too fast, from the power screw thread i chose, i decide a speed 1.8cm/s as for eldery people to lift up. RPM can get from the screw lead or pitch from this speed. So i get 619rpm.
But i don't know this assumption correct or not. because i assume the speed constant along lifting process.
So for choosing a motor is it choose the torque base on lowest position, and the speed is assume constant in what i need?

I have a limited knowledge in motor, so hopefully u can share some link or knowledge in choosing a motor...thanks a lot
 
  • #4
If it was me I'd find the RPM and torque required at a range of heights, all at 1.8cm/s, then plot a graph to get a feel for how it behaves and how much error you can tolerate. It's definitely an ugly function, with aymptotes! You might be lucky and find that the speed-torque relationship is close to that of a motor for the range of extension you have.

Then just pick a motor and see how well it compares over the whole range of movement. There's plenty of "gear motor" makers on the internet, with spec sheets.

The speed-torque curve of a motor is the key tool in working out how it'll perform under different loads. You can plot that if you know the stall torque and the no-load speed. It's just a straight line.
 
  • #5
thanks, if assume 1.8 cm/s along lifting, this is correct?
Because torque reduce along lifting, but speed should be increase, then how to get the spped?
for rpm calculation it is base on 2 factor, one is lead or pitch of screw, and one is linear speed...but this 2 is constant then the rpm at lowest and top should be same?

Along the lifting, the DC motor will supply a constant torque, constant speed, or constant power?

sorry , i am really no idea in motor...please help... thanks
 
  • #6
I think you really should find the RPM and torque at several different heights, all with the same 1.8cm/s linear speed. Then it should be clearer what's going on. If the lift speed is constant then the RPM won't be the same at the top and bottom. That's because the mechanical advantage of a scissor mechanism changes with height.

DC motors are typically neither constant torque, speed or power. Tho you can use an electronic driver to get those effects if you really want. But I doubt that'll be necessary here.
 
  • #7
Unrest, thanks
is it what you means here, we do not need any electronic to do this, we just simply use a motor with a control to supply a dc current to it, reverse the current than it will turn other direction.

can u teach me how to use mechanical advantage to calculate speed? is it like lever or a velocity ratio(VR)?

And after calculation, base on what factor we use to compare with manufacturer spec? because different manufacturer different way they list their motor spec...


Thanks again
 
  • #8
Last edited by a moderator:
  • #9
sporty said:
This is my scissor linkage...can u teach me how to calculate the mechanical advantage for it...Just show me one for me to understanding..thanks.

http://filesup.net/R6P3SQJEKQ83/toy.bmp.html"

Do a free body diagram of each rigid member. It has to be in static equilibrium so all the forces on each member must add to zero, and all the moments about every point must add to zero.

Applying these restrictions allows you to form equations involving the variables weight (W) and the force applied by the leadscrew (Fscrew). Then you can get the ratio between the two, and this is the mechanical advantage. The inverse of this ratio is the ratio of linear speeds. If you plug in the known weight, you can find the required screw force. Then do the whole thing again for several other heights.

This diagram uses the dimension from the left-hand case in your diagram. I assumed no friction, and that the top ends of the arms are attached by rollers/sliders so they have no horizontal force applied to them. You system is symmetric so it's only necessary to analyse one half of it.

[PLAIN]http://dl.dropbox.com/u/21857463/scissor.png

Any motor should at least tell you one speed and torque point that it can operate at. Often they'll give you two points - (zero torque, maximum speed) and (maximum torque, zero speed). In that case you can interpolate betwen them to find how fast it runs when loaded with different torques.
 
Last edited by a moderator:
  • #10
http://www.engineersedge.com/mechanics_machines/scissor-lift.htm"

Thanks unrest, the static analysis i have already did it last month but currently i found that some other equation show in above forum... so i am doubt, which one is correct because, both will produce different answer.

Mechanical advantade is it =W/(2 F screw)?
Ratio of Linear speed is inverse of it? = 2 F screw / W?

What is the purpose to calculated a Mechanical advantage ? What can it use for analysis?

For Speed because the height is vary from lowest postion to highest, thus the speed along lifting also vary and too the torque...We only choose on motor base on max torque < stall torque of motor spec and at highest height the speed is highest so it could not excees the max speed of the motor sped...Am i right?

Than mechanical advantage is calculated for?
Thanks for your motor choosing step...It is very clear for me now..

Thanks
 
Last edited by a moderator:
  • #11
sporty said:
but currently i found that some other equation show in above forum... so i am doubt, which one is correct because, both will produce different answer.
Can you be more specific?


Mechanical advantade is it =W/(2 F screw)?
No, just W/Fscrew. Even tho there are 2 Fscrews in my diagram, one is just a reaction force.

Ratio of Linear speed is inverse of it?
Yes.

What is the purpose to calculated a Mechanical advantage ? What can it use for analysis?
To find the entire range of screw forces needed for every lift height. Also to find the range of speeds needed.

lifting also vary and too the torque...We only choose on motor base on max torque < stall torque of motor spec and at highest height the speed is highest so it could not excees the
max speed of the motor sped...Am i right?
Keep the continuous torque well below the stall torque or it will burn out/stop running. Otherwise that approach would guarantee that it lifts all the way up. But it may exceed your maximum allowable lift speed, or be impractically slow.
 
  • #12
What i means is in this link, the calculation didn't show the reaction force in the middle connection, and too the final answer for force from that forum is different to us which we using static analysis The link, http://www.engineersedge.com/mechani...issor-lift.htm

Mechanical advantage initially i thought is for 1 bar linkage because the weight is divided into 2 side so Wg=W/2, so for 1 bar i thought is ratio for it is (W/2)/F screw...ok...W/F is ratio now i know, thanks.

ya..i think so, so the stall torque of motor spec means when there is zero speed, so if our maximum match with this value it will be no speed but only torque..Thus we need higher stall torque than what we required right?
Thanks. There are a lot of variety of DC motor, brushless, gear, servo...really headache in choosing. any idea? Thanks
 
Last edited by a moderator:
  • #13
It looks like that site is just static calculations like mine. Can you show a specific case where they're different? It's a lot of numbers for me to crunch. I could have made a mistake.

Furthermore, how do the results compare to your existing data of 10Nm, 619RPM?

Whether you need a gearmotor or not just depends on the speed and torque requirements - once you nail those down the motor selection will be much easier, and from the lifting point of view other factors don't matter. Those other decisions depend on concerns like lifetime, cost, size, noise, weight, etc.
 
  • #14
[PLAIN]http://www.fileupyours.com/view/303536/tis%20is.bmp

This is what i means..thanks
and one more is mechnaical advantage in your picture for 1 of linkage the applied weight only W/2 so the Mechanical advantage is (W/2)/F screw or W/Fscrew?
thanks again
 
Last edited by a moderator:
  • #15
OK I see your point and I'm getting confused.

Suppose you just had one of the arms, and its center pivot was connected to a vertical slide. With a load of 40N it would require a force of 60N to hold it up. Right? Mech. Adv. = 40/60

Then if you add the other arm, which takes its force from the reaction force of the vertical slide, that can lift an additional 40N with the same 60N input force.

So for a total load of 80N it still requires 60N input force. Mechanical advantage = 80/60.

Well I can't quite understand how that website derives their equations, but it might be worth nutting it out. I did try using software to model this, and it told me that a 40N load on each arm of this X mechanism produces a 60N horizontal reaction force at the bottom of each arm.
 
  • #16
No,Actually this is not my design, i just use this as basic analysis on this 2 type of equation..Because i am doubt why the answer will be different. actually in the beginning the symmetric scissor linkage with middle pivot is supprted by 2 arm linkage in 'X' and the force..abd in overall free body diagram, we use equilibrium equation to calculated both reaction vertical from floor. Then we continue analysis by member to find out horizontal forces F screw required.

SO that means our calculation is right? Thanks a lot from your explanation...Thanks...
 
  • #17
unrest, may i know from my design, how to calculate the speed in choosing a motor can u teach me to calculate the speed using mechnical advantage??
how to calculated speed in varies position? especislly at higest speed position...can you show me the formula? 1 rev turn of screw will move 1.75mm pitch..total vertical distance lift is 14cm while horizontal is 18.06cm.
lowest angle is 24 degree and highest is 61.81 degree. Thanks..
 
Last edited:

Related to How Do You Select a DC Motor for a Scissor Lift Using a Power Screw?

1. What is a motor and how does it work?

A motor is an electromechanical device that converts electrical energy into mechanical energy. It works by using the interaction between a magnetic field and electric current to generate rotational motion.

2. What is a gear and how does it contribute to a motor's function?

A gear is a mechanical component that transmits power and motion from one part of a machine to another. In the case of a motor, gears are used to increase torque and decrease speed, allowing the motor to produce more force to perform a specific task.

3. What is a power screw and what is its purpose in a motor?

A power screw is a type of mechanical device that converts rotational motion into linear motion. In a motor, it is used to convert the rotational motion of the motor's shaft into the linear motion of the load being moved.

4. What factors should be considered when selecting a motor, gear, and power screw?

The factors to consider when selecting a motor, gear, and power screw include the power and speed requirements of the application, the load being moved, the available space, and the desired level of precision and control.

5. What are some common applications of motors, gears, and power screws?

Motors, gears, and power screws are commonly used in a wide range of applications, including robotics, manufacturing equipment, vehicles, household appliances, and medical devices. They are also used in many industrial and commercial settings to power conveyor systems, pumps, and other machinery.

Similar threads

  • Mechanical Engineering
Replies
1
Views
684
Replies
2
Views
1K
  • Mechanical Engineering
Replies
2
Views
3K
Replies
34
Views
2K
  • Mechanical Engineering
Replies
17
Views
3K
Replies
8
Views
2K
  • Mechanical Engineering
Replies
12
Views
2K
  • Mechanical Engineering
Replies
1
Views
3K
Replies
16
Views
2K
  • Mechanical Engineering
Replies
1
Views
1K
Back
Top