step-by-step motor dimensioning for linear arm implementation

[luigi0593]

Good evening to all,

I would kindly need help to size a mechanism that includes the following components:
-nema 8 dual shaft bipolar (holding torque = 0.03 nm, rotor inertia 0.36*10^-6 kg*m^2)
- two screws without end trapezoidal (single component characteristics: tr 10 x 2, with medium diameter = 9 mm , screw length = 25 mm, steel density= 7800 kg/m^3, screw weight = 0.0153 g ) the screw is inserted inside the joint with a diameter of 6 mm
- two shaft joints (one component carat: weight 13,60 g, inertia moment = 1,053*10^-6 kg*m^2, shaft hole diameter 6 mm, screw hole diameter 6 mm)
- two trapezoidal nails (one component carat: tr 10 x 2 , weight=0.029 kg)
-two arms (one component carat: weight=0,010 kg, aluminium density 2810 kg/m^3)

the project is to use a nema 8 step-by-step motor to build a linear actuator "symmetric", to allow the extension of two arms horizontally (using the two ends of the nema 8 shaft). the ring (vinculated to one arm) must travel a stroke of 10,5 mm.
the race must be carried out by every single arm in a maximum time between 0.5 s and 1 s. unfortunately the constraints are very stringent, starting from the engine nema 8 because of the reduced available space and especially the time to carry out the race.
what is possible to vary is the diameter/pass of the screw; consequently diameter/pass of the nail. I would be immensely grateful if someone could give me a hand through some advice, form, to understand how to solve the problem.

 

[luigi0593]

place the engine datasheet https://en.nanotec.com/fileadmin/fi..._schrittmotoren/sca2018m0804-l.pdf?1656012557torque curves/speed hollow-shaft motor – nema 8 – stepper motor | nanotec

 

[meccanicamg]

I read several times but it seems written in ostrogoth.
I think I'm getting less elastic than my mind.
but can you post a pattern? or do we have to imagine?
you have a motor with bisporing tree in the center on which, for each tree the joint and a trapezoid screw with snail. a sleigh will be attached.black: motor
red: joint
green: screw
blue: pigeon
orange: finger holdingI think that's what you were trying to write, but it's hard.

but what do you mean the screw is inserted into the joint? the screw is cast on the joint with a cylindrical diameter 6mm.

What does tree hole and screw hole mean? the joint is hollow and can accommodate two cylindrical trees from 6mm?

the first thing to do is to be clear, synthetic and schematic.... otherwise you start badly. engineers are not philosophers but technical.

then when you have the scheme you mark the inertias that eventually are cylinders and as such it is easy to deter the value.
[math]j=\frac{m•r^2}{2}[/math]you have to make the energy budget or powers or pairs or jobs.

you will have to deal with linear accelerations of the arms and therefore curves speed/time trapezoidal.If you don't have enough time, you're obliged to make triangular curves with zero or very small constant speed strokes.

the toe will suffer a force of inertia due to the linear acceleration necessary to respect the law of motion.
will he also have a force to generate? You know that.

You have to take into account the frictions. we have several discussions about screw/chip and friction.

then turn the linear motion into rotation through screw/chip. speed and acceleration.
here are the formulas related to the vines/kid taken by others post on the subject.
I bring everything to the engine, inertias and couples.

Here you see what and how much you need. If you put everything in excel you can change the various geometric values and get real-time responses to the system in power/rides.

overlap the cycle requirements with the engine curve and depending on where you are, you will need to increase or decrease the laps to stay in the optimal field or even change the engine.

all this, it is necessary to put in order ideas, understand what is being analyzed mechanically and then have a basis for optimization.

What can't you do? did the courses of rational mechanics and applied them? If you miss the bases we go uphill.

 

[meccanicamg]

attention for example to system returns. as it is symmetrical you could calculate everything and multiply by 2, considering that the two screws have identical behavior.
the two fingers or the two pieces of final rods if they have a weight and run you will deal with friction forces.
to move the system you will find that you will consume very little power provided the frictions are discreetly low.
also for the curved speech of the motor see that the more you lift the tension and the more you have constant torque at higher regimes.
vaguely up to 400rpm you have maximum torque and therefore if you need strength work in this field. then instead if you want to go fast you can raise the rpm regimen as long as your system asks less torque according to the curves....without exceeding.

 

[luigi0593]

All right, thank you for the disappointments. I'm trying to get rid of the accounts.
these were some considerations I had already made: the motor provides the same pair to both two ledges. as a mass to move I had considered both arms with their respective snails. in fact in the count je turns out to be double as I make this consideration. Also jv turns out to be double as I consider the inertias of the two screws.
correct me if I'm wrong?

 

[meccanicamg]

All right, thank you for the disappointments. I'm trying to get rid of the accounts.
these were some considerations I had already made: the motor provides the same pair to both two ledges. as a mass to move I had considered both arms with their respective snails. in fact in the count je turns out to be double as I make this consideration. Also jv turns out to be double as I consider the inertias of the two screws.
correct me if I'm wrong?

for now reasoning is correct.

 

[luigi0593]

in the count of the torque necessary to win the linear inertia, I also considered the inertia of the motor, besides the two screws, two arms, two joints. (correct ornamentation? )

from the count back to me that using a screw with 8mm diameter and 1.5mm pitch, the speed of rotation of the motor is 860 rpm. the torque that gives me the motor regardless of the voltage, for a rotation of 860 rpm is two orders of magnitude greater than the torque needed to beat friction and inertia. (I expect to use a 24v voltage)

I used a triangular profile, because as you had predicted with a trapezoidal profile 1/3 1/3 1/3 1/3 I would have pushed over 1200 rpm due to the screw step. (I prefer to keep me in the almost linear behavior of the torque chart) (what do you think? )

I know that I have a fairly low screw efficiency 37 %, but it was necessary to be less than 50% so as not to reverse in the case of a contraccolpo
I look forward to your comment
p.s some data are varied as screw diameter, step and load system.

 

[meccanicamg]

Surely in the budget there is also the inertia of the engine as you calculated.
the yield of the screws is never great that, especially if the step is short and therefore the narrow helix angle.
If you embellish higher accelerations you can definitely do the trapeze instead of the triangle. because in the text you said that you could use us a time between 0.5 and 1 second....beh with 1 second the trapeze is there.
orders of magnitude seem to be there.
I had done two quick accounts, leaving a few things, and it didn't seem so scary that engine.

 

[luigi0593]

Yes, if I increase the time with the trapeze I'm in. If not I should simply increase the number of laps. for now I prefer to respect the request of 0.5 s.

with the trapeze I return around 1200 rpm in 0.5s. therefore independently of the voltage the pairs provided by the motor are always two orders of magnitude greater than the necessary torque.
But as I told you earlier, I am in the descending part of the pair chart.
what triangular profile tips with lower rpm or trapeze profile though with higher spins?

thank you for the advice and support provided to me

 

[meccanicamg]

Yes, if I increase the time with the trapeze I'm in. If not I should simply increase the number of laps. for now I prefer to respect the request of 0.5 s.

with the trapeze I return around 1200 rpm in 0.5s. therefore independently of the voltage the pairs provided by the motor are always two orders of magnitude greater than the necessary torque.
But as I told you earlier, I am in the descending part of the pair chart.
what triangular profile tips with lower rpm or trapeze profile though with higher spins?

thank you for the advice and support provided to me

If you want a softer application you should use the trapeze, climb turns and then modulate the accelerations and decelerations.

if instead you have the risk of having to win accidental external forces, better stay in the area with extra couple available.

as it seems to me the application would go for the first solution, guaranteeing the margin anyway with the couple.

 

[luigi0593]

All right, thank you very much for the advice