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opening doors

  • Thread starter Thread starter Tarkus
  • Start date Start date

Tarkus

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Hi, I don't know if it's right to sit down on this debate open by braunfish, so the question is,
I have a door that will have to be on hold from 2 gas springs, like those from car hatch, with a force of 500 n each.

and possible with inventor simulate the thing?
I mean, can you know if the blessed spring is able to hold on the door, and not get it up if it's brought down?
I'll do some things.
p.s.
for convenience the gas spring is one, so it will be considered by 1000 n...
 

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Tarkus said:
...that you know if the blessed spring is able to hold on the door, and not to raise it if it is brought down?
...
Click to expand...
inventor simulation apart (which I do not know), be careful that the gas spring must be mounted in a very precise way if you want the door not to rise.
the piston always pushes, both open and closed and of course you have to choose it so that it wins the weight (the couple if hinged).
"graphically" you have to place it so that after the closing rotation the result of the push is against the beat; when you open with your hand once you pass the "critical corner" then you can open it yourself. .

I know, I explained as a dog, but if you make a pattern, you know what I mean:

greetings
Mar
 
No, no, no.
You didn't explain yourself wrong, you're saying the geometry is kneeling, as they say in my parts, that the spring push direction goes the opposite side of the hinge axis when the door is open.
But I can't put the spring push point at the right point for various reasons of bulk, so I'd like a simulation to check:
1- that the springs are adequate to hold on the door.
2- that if I put the door in closing, the spring finds itself working with a very unfavorable push angle and therefore should not open anyway,if you start the opening by hand until the spring is working with an efficient angle.
However, it would be enough to check point 1....
 
I posted a set with the relative simulation. I have eliminated the two jacks because in sd there is the appropriate joint, through the unknown force command I have determined the force necessary to rotate the door of 90° (518 n) therefore to the two cylinder joints I have assigned a force equal to half of that obtained.
look at the file if you have any doubts ask.
Hi.
 

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diegus said:
I posted a set with the relative simulation. I have eliminated the two jacks because in sd there is the appropriate joint, through the unknown force command I have determined the force necessary to rotate the door of 90° (518 n) therefore to the two cylinder joints I have assigned a force equal to half of that obtained.
look at the file if you have any doubts ask.
Hi.
Click to expand...
I'll take a look at him and then I'll tell you.
for now I have a nice weekend in the north in the rain...
 
diegus said:
..... I determined the force necessary to rotate the door of 90° (518 n) then to the two cylinder joints I assigned a force equal to the half of that obtained.
look at the file if you have any doubts ask.
Hi.
Click to expand...
Good morning, I've seen and thank you again. I think I'm gonna have a little simultaneous experience because I didn't understand much.
So two questions:
How did you determine the unknown force?
I see that you have set (justly in my case) the rotation at 90°, but the simulation begins with the door open halfway.
How do I start the simulation with the closed door?
 

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