doubt tension strap lifting vertical

[davide75]

Hey, guys.
I am studying a new application for us and I have a doubt.
I attach the section of my scheme as I can't do more because my company doesn't want to disclose the project.
In any case, I have two open belted pulleys where two plates are connected, which covers the profile of the teeth, and which are attached to my mass to move vertically. the mass is 250 kg and must go up and down in a second.given the fast cycle not to use a big motor it is thought to use springs that give me the push at the beginning to lift my load.the springs I calculated so that when my cairco is down these are compressed, then the load rises with acceleration commanded by the motor and helped by the springs and then decelera when it comes to the top
Here's how I interpreted the cases by keeping the signs contyo:
t = voltage
as= acc. climb
ds= decel. climb up
a= acc. descent
dd= decel. descent

1) climb+acc.
t=-mg-mas+kx

2) climb+decel.
t=-mg+mds+k

3)descent+accc.
t=-mg+mad+kx

4)descent+decel.
t=-mg-mdd+kx

the poggi did not answer me unfortunately.
Is my reasoning okay? ?

 

[cacciatorino]

Now I don't know your application, but to compensate for a vertical load you usually use a pneumatic cylinder with a servovalve or a large air accumulator. you have the advantage of the strength about constant and more adjustable.

Bye.

 

[davide75]

hello, thanks for the answer but the option hydraulic cylinder or penumatic do not want it. cmq to me it would be enough to know if the reasoning for the calculation of the belt is right.

 

[cacciatorino]

hello, thanks for the answer but the option hydraulic cylinder or penumatic do not want it. cmq to me it would be enough to know if the reasoning for the calculation of the belt is right.

I think it's okay. The only problem is that you consider the accelerations to rise and descend constant, while they are probably polynomials (if you have a brushless motor).

Besides, you should find a relationship that, given the cycle time, calculates your ramps and accelerations, is not immediate but is feasible.

I'll attach you a spreadsheet that I did in time for a transferr, considering leaks.

Bye.

p.s. on these very fast applications it is important to take into account also of the inertias of the organs in rotation (swimmings, motor rotor, etc.).

 

[davide75]

then the ramps with accelerations and decelerations with the graph of the cycle I have because I gave it to the supplier of the engines for which I have replaced the various values (I have different values of accelerations and decelerations different depending on that I rise or that descend ) and I found the equivalent tensions in the 4 cases and I made the verification of the belt according to the greater voltage.
thanks for the spreadsheet but how do I open it? ? ?

 

[cacciatorino]

thanks for the spreadsheet but how do I open it? ? ?

downloaded openoffice, avoid converting it into xls format because I don't trust the translation of formulas is accurate.

 

[meccanicamg]

Hey, guys.
I am studying a new application for us and I have a doubt.
I attach the section of my scheme as I can't do more because my company doesn't want to disclose the project.
In any case, I have two open belted pulleys where two plates are connected, which covers the profile of the teeth, and which are attached to my mass to move vertically. the mass is 250 kg and must go up and down in a second.given the fast cycle not to use a big motor it is thought to use springs that give me the push at the beginning to lift my load.the springs I calculated so that when my cairco is down these are compressed, then the load rises with acceleration commanded by the motor and helped by the springs and then decelera when it comes to the top
Here's how I interpreted the cases by keeping the signs contyo:
t = voltage
as= acc. climb
ds= decel. climb up
a= acc. descent
dd= decel. descent

1) climb+acc.
t=-mg-mas+kx

2) climb+decel.
t=-mg+mds+k

3)descent+accc.
t=-mg+mad+kx

4)descent+decel.
t=-mg-mdd+kx

the poggi did not answer me unfortunately.
Is my reasoning okay? ?

clearly consider the accelerations in form and verses + - in the formula.
I think they're right.

a question that I ask you: in addition to the tensions generated by the masses, the inertias and the springs... don't you take into account a belt pretension system? Do you need positioning accuracy? Do you have a linear ecnoder or resolver?

However if you have to move 250 kg vertically if you are granted in the project you think you use a counterweight so reduce the pairs to the engines (see that the pneumatic or hydraulic cylinder does not want it).

Do you have to stop right? with high accelerations and high dynamics think of a brake if you do not want to overheat and reduce the life of the engines (see nexen brakes for linear guides).

do vobis:biggrin:

 

[davide75]

thanks for the answers.
I had also thought of a counterweight only that the solution of the springs pleases more to the customer... it would have been much easier for me but instead... .
As for tensioning I studied a system that allows me to tension the belt that consists of using the plates that covers the profile of the belt so as to anchor me to my mass to marry with plates connected to each other through a threaded bar so as to tension the belt.
Thanks again for the answers.

@cacciatorino not yet had way to see your spreadsheet cmq thanks again

 

[cacciatorino]

I had also thought of a counterweight only that the solution of the springs pleases more to the customer... it would have been much easier for me but instead... .

mhhh, but then you'd have double the inertias to take around...

As for tensioning I studied a system that allows me to tension the belt that consists of using the plates that covers the profile of the belt so as to anchor me to my mass to marry with plates connected to each other through a threaded bar so as to tension the belt.

find everything commercial on the catalog clavals, both for the straps at and those htd

@cacciatorino not yet had way to see your spreadsheet cmq thanks again

have you installed openoffice?

 

[meccanicamg]

mhhh, but then you'd have double the inertias to take around...

I would say no, there are no inertias because with the counterweight they always subtract and is the solution used for the axis z of the working centers.

try to make two accounts....and you will see:finger:

 

[cacciatorino]

I would say no, there are no inertias because with the counterweight they always subtract and is the solution used for the axis z of the working centers.

try to make two accounts....and you will see:finger:

You didn't convince me....
We are talking about moving a mass of 250 kg in a second, which with the counterweight would become 500.

 

[Borgotech]

You didn't convince me....
We are talking about moving a mass of 250 kg in a second, which with the counterweight would become 500.

in fact it is not true because the counterweight on the other arm of the belt wipes the weight (250 kg rise and 250 drops), clearly the pulley, its support and the bearing have double the work to endure, but surely the engine benefits with a really minimal effort

 

[cacciatorino]

in fact it is not true because the counterweight on the other arm of the belt wipes the weight (250 kg rise and 250 drops), clearly the pulley, its support and the bearing have double the work to endure, but surely the engine benefits with a really minimal effort

we are prying two different loads: strength and strength of inertia. the counterweight cancels the effort due to the weight force, but doubles that due to the force of inertia. It is necessary to see which of the two terms draws more: if the system is slow, the importance of the weight (axis z of the working center) prevails if the dynamic is rapid, the contribution of the inertia force prevails.

 

[meccanicamg]

we are prying two different loads: strength and strength of inertia. the counterweight cancels the effort due to the weight force, but doubles that due to the force of inertia. It is necessary to see which of the two terms draws more: if the system is slow, the importance of the weight (axis z of the working center) prevails if the dynamic is rapid, the contribution of the inertia force prevails.

certainly that cancels the weight effect if m1=m2 does not have the m*g effect.
The term m*a is double therefore 2*m*a. for this reason they started on z axes with high dynamics to use nexen brakes on electric guides or linear actuators.

We need to know whether they need high accelerations or not as an application.

 

[davide75]

Then guys apologized if I answer only now but I am doing work shifts massacres. As for the counterweight I agree with hunter and in fact doing two accounts with a sheet of excel the inertia of the counterweight affects not little on the tension of the belt. I made two accounts with different counterweight values and at the end the tension on the belt does not vary much.
@cacciatorino excuse but I haven't had time to look at your calculation program yet because as I repeat I have to deliver to you as soon as possible and I have breath on the neck.I have to see it this weekend, cmq thanks again so much for the tips

 

[cacciatorino]

@cacciatorino excuse but I haven't had time to watch your calculation program yet

I converted it into xls format, so at least if there are errors I can always blame the conversion program :biggrin:

If I don't remember here, it didn't take into account the weight of the object because it was compensated by a pneumatic cylinder with a servovalve.

 

[stefanobruno]

What are you going to do?
Should the springs only give an initial help or work for the whole race?

Bye.

 

[davide75]

the springs help me only for the departure in fact to half run the springs balance my weight more or less.the stroke is 260 mm.
@cacciatorino thanks

 

[Esselle]

I attach myself to this discussion to post my work, for which I have weighed heavily from the sheet of @cacciatorino Thank you. My case is simpler because it is only about a vertical translation of a load but I have inserted in the calculation the mass to be lifted.
the system is identical to that schematized at the beginning of the post, except for the absence of the spring and for the presence of an additional transmission between the lifting pulley and the gearbox, realized with a pair of pulleys and belt.
would be interesting for me to put in the calculation the inertias to have some complete, but I did not succeed, do you have any indication to give me?

 

[Esselle]

I add an image to explain. the other day I hadn't defined the details so I could show something.