step-by-step engine size

[Maury]

Good morning, gentlemen.
I find myself in the grips with the sizing of a step-by-step motor that has to move a truncator from the weight of 150kg. This truncator is mounted on 4 ball recirculation skates and performs a horizontal race of 2000mm
the movement takes place with a screw recirculation of spheres and a spiral.
the screw has a diameter of 40mm, a step of 14mm and is at 2 beginnings.
the speed is 0.5m/s and acceleration and deceleration have not yet been given but I suppose they are low but not negligible.
the motor must be possible in direct socket otherwise if pairs
are high will add a reducer but this is relatively important.Question :What couple do I have to apply on the screw considering everything including inertia and frictional events? if you can give me a simple calculation line to follow, a simple method of calculation/work to follow and the formulas to be applied and above all give me the possibility to be autonomous if I happen another case like happy serei.
I apologize if I wasn't clear and for my Italian.

hello thanks and good work

 

[PaoloColombani]

the screw behaves like a sloped plane whose base is represented by the circumference and its height from the step. with a simple diagram of the forces you find that the relationship between the tangential force on the primitive ray and the force is pushed on the axis is equal to the ratio between the primitive circumference of the vine and the step less than the yield that for a ball screw is about 0.8 (pure number).

As much as I can guess of your application, the step-by-step engine seems unsuitable to me. if you do not need particularly precise positionings (centesims) the ball screw could be advantageously replaced with another transmission system.

 

[Maury]

thanks for your answer your suggestion is definitely helpful thanks.
what I cannot have is the passage to get the axial load along the screw considering as already explained that the load moves horizontally and is subject to inertia.
how to sow the two principles?

 

[MBT]

thanks for your answer your suggestion is definitely helpful thanks.
what I cannot have is the passage to get the axial load along the screw considering as already explained that the load moves horizontally and is subject to inertia.
how to sow the two principles?

take the thk catalog or other endless screw builders.. .
Unfortunately, here I have no subject, but I remember very well that on one (perhaps of the star) there was a rich part of information and theoretical calculations, which took into account inertias, montages, lengths screws etc.

 

[Maury]

thanks now recover the material and I try
Hi.

 

[PaoloColombani]

what I can't have is the passage to get the axial load along the screw...

It is enough to translate into mathematical expression what I had previously written in words, so:
" ... the relationship between the tangential force on the primitive ray and the force is pushed on the axis is equal to the relationship between the primitive circumference of the vine and the step less than the yield that for a ball screw is about 0.8 (pure number). »And translate cosè:

(ft) / fa = 2πr / p

(ft ∙ r) is the tangential force on the primitive ray of the screw that is equal to the couple or moment (m) applied to the screw.
fa is the force pushed in axial direction, and p is the step of the life.

extrapolating:
fa = ρ ∙ m ∙ p / 2πrforces in newton, lengths in meters and moments in nm.
ρ is the yield of the screw.

for the calculation of inertias you can apply the second principle of dynamics.

 

[Maury]

Thank you.

 

[stevie]

m=(f*p)/(2*p*return)

 

[PaoloColombani]

It is enough to translate into mathematical expression what I had previously written in words, so:
" ... the relationship between the tangential force on the primitive ray and the force is pushed on the axis is equal to the relationship between the primitive circumference of the vine and the step less than the yield that for a ball screw is about 0.8 (pure number). »And translate cosè:

(ft) / fa = 2πr / p

(ft ∙ r) is the tangential force on the primitive ray of the screw that is equal to the couple or moment (m) applied to the screw.
fa is the force pushed in axial direction, and p is the step of the life.

extrapolating:
fa = ρ ∙ m ∙ p / 2πrforces in newton, lengths in meters and moments in nm.
ρ is the yield of the screw.

for the calculation of inertias you can apply the second principle of dynamics.

Naturally, my vice of getting my formulas quickly to avoid going to discard between my cards led me to error. I'm rewriting everything!

 

[PaoloColombani]

In fact from the diagram of the forces you have:

ft / fa = p / 2πr

since ft = m/r you get:
fa = ρ ∙ 2π ∙ m / p as it has already written