topic Re: Dynamic Simulation - Output Grapher question in Inventor Forum

Dynamic Simulation - Output Grapher question

Wapster — Sat, 17 Mar 2018 12:23:06 GMT

Hi there,

I’m having trouble understanding the Output Grapher in Dynamic Simulation.

I have a door, which I’m going to open and close using an actuator.

I set it up in Simulation and use an imposed motion on the prismatic joint in the actuator.

Then I use the output grapher to look at the force (Z) of the joint, needed for the actuator.

When I look at the force, there is a negative force showing.

How is that possible? I’m stuck on this one.

If there is anyone who can tell me what I’m doing wrong I would be very thankful.

Cheers,

Wap.

Re: Dynamic Simulation - Output Grapher question

admaiora — Sat, 17 Mar 2018 14:12:07 GMT

Hi Wap,

it's correct.

Each Joint has its reference (created at the joint creation)

1 arrow x

2 arrows y

3 arrows z

Re: Dynamic Simulation - Output Grapher question

Wapster — Sat, 17 Mar 2018 15:07:06 GMT

Hi admaiora,

Thank yo for looking in to this.

I don't know if I understand your answer tough.

I know the joint is OK, but why is the force negative in the Grapher..?

The model is a simple model now, but I'm gonna add glass, (wind) pressure, and friction later.

The goal here is to find out if the actuator is strong enough but I first wanna know if I'm "reading" the Grapher the right way.

A negative force suggests that the door would open by itself (the first few seconds) only using gravity

I just don't believe that (and tested it removing the imposed motion), so I must be doing something wrong.

Thanks again,

Wap.

Re: Dynamic Simulation - Output Grapher question

admaiora — Sat, 17 Mar 2018 15:42:43 GMT

Hi Wap,

you have set an imposed motion (velocity, 30 mm/s).

To ensure that, it's necessary a force (mainly) oriented in the opposite direction of the three joint blue arrows.

Opposite direction means negative.

If i reverse the prismatic joint orientation and velocity

Re: Dynamic Simulation - Output Grapher question

Wapster — Sat, 17 Mar 2018 20:05:36 GMT

Hi,

It seems that when it comes to DS you’re on a much higher level then me.

So.... could you perhaps tell me how much force needs to be applied on (the back off) the actuator to open the door in the current setup (no other loads applied).

Many thanks in advance.

Wap.

Re: Dynamic Simulation - Output Grapher question

admaiora — Mon, 19 Mar 2018 09:53:40 GMT

Hi Wap,

you have done a globally good dyn sim work.

Just some consideration:

- there is not actual opposing forces (no forces, no frictions)

- conseguence there are only inertial forces, (not always so easy to be interpreted)

- as consequence, you have a velocity of 30 mm/s, if you reduce it you will need a reduced cylinder axial force (it will tend to 0)

I suggest to suppress contact joint in this fase, because they can push result in an system like that.

Re: Dynamic Simulation - Output Grapher question

jan_priban — Tue, 20 Mar 2018 13:30:39 GMT

Hi Wap,

basically when moving part (you door) is decreasing speed - decelerates, then force can be negative value. I think it could be your case, doors at initial position move faster than in the end position while piston speed/imposed motion is constant. See please the attached image what I am trying to explain. Speed in position 1 is faster than speed in position 2, due to angle change.

Regards

Jan Priban, Autodesk Inventor team

Re: Dynamic Simulation - Output Grapher question

jan_priban — Wed, 21 Mar 2018 11:05:14 GMT

Hi Wap,

see please my better answer/update

I created similar model,

And define imposed motion like you. Graph shows angular velocity of rotating door, see it is decreasing, means deceleration from start to time 2.8 sec. Than angular speed is growing, door accelerates. That's why Imposed motion force or FZ in prismatic joint has negative value to t= 2.8 s, after that positive.

and

Regards

Jan Priban