7

I found these images. Looks like they are driven by a motor. Belt Driven Model Screw Driven Model I suggest reviewing this document


6

Does anyone know if this would be a good idea? The proposed design has poor rigidity towards torsional loads. It doesn't matter whether the actuator is off-center or if the load is off-center: if load's center of gravity and actuator are not at the same position, the torsional load will likely damage the bearings. In your example with two guide rails, ...


6

Your mechanism is going to lock up like the old car jacks immediately, due to differential raising of the platform. I am sorry but it looks like a textbook example of what not to do. in most real-life cases where the lifting actuator has to be off-center, they use a linked pair of scissors mechanism, like what they do in trucks' gate lifter, or the hydraulic ...


5

It looks like it might be a air-powered propeller. You pump up the bottle with air, and the high pressure is used to run a air motor that spins the propeller. This might have been part of a toy airplane, with the wings and tail now gone.


4

Screw: M6 M6 coarse has a 1 mm pitch. M6 fine has a 0.75 mm pitch. Unless otherwise stated assume coarse, so 1 mm pitch. Your motor is 1.8° per step. This is 360/1.8 = 200 steps/rev. The screw will advance 1 mm every 200 steps. As usual with Amazon, there are no datasheets and poor specifications in the ad which is what you linked to. There's a question in ...


4

Paradoxically, maybe the home-sized magnetic levitation could be a narrow area, where superconducting solutions could be cheap. I explain, why. Although room-temperature superconductivity is a dream, room superconductivity is not. :-) There are soon superconducting materials which can be cooled by liquid air, and they aren't even costly. Liquid air is ...


4

You are considering it from an electrical engineering perspective. Speed vs torque. It's more a digital motor with a rotor position that can be easily controlled (read head on floppy drives). It is a brushless motor and its strongest torque (traction or holding) occurs when the coils are energized. They have fractional hp ratings (>34.3 mN·m). To do ...


3

Some linear actuators are self-locking — they can't be back-driven by force on the output connection. A leadscrew with a fine enough thread pitch is one example, and since you don't need a lot of speed for your application, wold probably be a good choice.


3

@StainlessSteelRat linked to a very good resource for stepper motors at All About Circuits, but I fear he didn't address your questions. I'll go through your question line by line. I'll probably have a little spool out there to hold the string for the weight so that probably changes how much torque it can deliver too. First, the sentence above is wrong. ...


3

There is nothing inherently wrong with running a motor in stall. The difficulty is that the power dissipation may be higher than the motor can handle long term. Every motor can dissipate some amount of power safely indefinitely. In stall, the power being dissipated as heat is the full voltage times current. As long as that doesn't exceed what the motor ...


2

For a minimum of moving parts, you can get linear stepper motors. With the right kind of controller (microstepping), the motion can be very smooth and quiet. For example, I have in the past worked with linear actuators from Nippon Pulse, driven by a motor controller from Technosoft. Very precise and powerful, but also very expensive!


2

IMHO (and I guess you already know that), from a structural point of view putting 100 kg at 1/4[m] will create a huge amount of bending moment on a actuator that it is rated for 120[kg]. An although you don't clarify whether the actuator electric/hydraulic, if bearings are involved you are bound to get failure sooner rather than later. Normally this type of ...


2

Your primary question, if a linear actuator can be moved with the power off, will be no, unless you find a mechanism specifically created for that purpose. These actuators are effectively a pinion gear design with a very high reduction ratio. You'll note in the link you provided that heavy lift mechanisms such as a motorcycle lift and a medical bed are ...


2

As a quick basic estimation approach we annotate the following: The length of bar 3 = L Top and bottom bearing force on each box $F_b$ The height and width of boxes 4 and its counterpart on channel 2 H and W. $$ \Sigma M=0 ,\quad 220*L/4 = H/2*4F_b \rightarrow \quad F_b=220*L/8H $$ Now you have to figure if the bearings and their connections are okay to ...


1

Let's convert 7 inches to meters, 7x0.0254 = 0.1778m You need a motor that can produce an average speed of 0.1778m/s so it must have an acceleration of 0.1778*2= 0.3556m/s^2 Now we can calculate the force your winch or motor needs. it must lift the 10kg against gravity and accelerate it $$F = m \alpha+ mg= 10kg*0.3556m/s^2+10kg*9.8m/s^2 =10*10.155m/s^2$$ $$...


1

In general the ideal travel per 1 revolution of the motor, will be depended on two things: screw dimensions the screw dimensions (in your case M6) and more specifically the pitch (see image below) Although its usually forgotten ISO 261 defines a coarse and a fine pitch. See wikipedia. The pitch should be written on the designation of the bolt M6-1.00 (...


1

What happens in the arrangement is that you are applying a force away from the center of gravity. Imagine you have a pencil on a flat surface and you try to push. If you don't push it around the center of gravity you will cause it to rotate. Although using linear bearings (especially longer ones) will mitigate the problem, you will be creating more normal ...


1

I suggest: joystick->arduino ->motor driver for arduino-> micro peristaltic pump 1) I need a failsafe. no prob as long as joystick returns to center position( joysticks usually have springs for that) 2) Its not a binary forward/backward so some voltage control arduino has pwm output, if you put a capacitor you get analog output, but maybe that's not ...


1

I might look into medical equipment, I was looking at actuators the other day, and they weren't too expensive.


1

You determined the torque at A correctly. You can get the force at B using the same equations. 650 Nm / .1 m = 6500 N. Yes, it's a lot, 5 times the load, but not close to where you ended up. The force will go down with the angle, meaning you should be dividing the force by the cosine, not multiplying by the sine (.5), but then you actually used .3 ...


1

I would say that the information on the datasheet it incomplete, you need at least a torque vs. speed graph for the rated voltage, and whether it is being driven at half steps, quarter steps, etc... You can then scale the torque based on the voltage applied. See for example this datasheet which provides that sort of data.


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