Solve angle issues for opening a door with a linear actuator

[marciokoko]

ok, setting the force issues aside (for now), Id like to use what I know about triangles to figure how to position my linear actuator in order to open and close this door correctly:

The vertical side is 84cm, the door is 92cm. The measured stroke of the AL is 25cm.

I would need to lower the attachment point which is currently at 26cm from the bottom edge:

If I were to randomy fix the attachment point to the door at 13cm from the bottom edge, how do I measure the angle now of the door with the bottom corner? I boiled it down to this:

So how do i solve Sin ⍬ = 0.325 on calc without arcsin (iphone)? 18.9 deg on website, but I am still missing an angle or two...

Please help.

 

[Tom.G]

https://www.rapidtables.com/calc/math/Arcsin_Calculator.html

(above and others found with:
https://www.google.com/search?&q=arcsin+.325+degrees)

Cheers,
Tom

 

[marciokoko]

Yes I got the arcsin but if you look at my last image I still need to figure out more angles in order to solve the problems.

 

[Tom.G]

Then enter them in Google like I did and let IT find the solution you like.
Or get a book with trig tables in it.
Or buy a $15 pocket calculator.

 

[marciokoko]

No, I don't mean i need to figure out other arcsines. I mean I need to figure out other angles in the figure. I need the blue angle.

 

[Tom.G]

Ahh, OK.
referring to the last sketch in your first post:

• Are the sketches and photo left-right reversed?
• There are some dimensions given, what part(s) of the structure do they represent?
• What is that right-most almost-vertical line, is it really vertical, is it part of the solution?
• It looks like you got the 108.9° correct.
• Please mark on the sketch which angles you need?
• Door pivoted at the bottom?

Thanks.

p.s. There was a similar thread about 6 months ago that may give you some ideas and insight. @JBA came up with a solution on that thread, hopefully he can jump in here and work his magic!
https://www.physicsforums.com/threads/how-to-put-solar-panels-on-top-of-my-rv.977763/#post-6241131

 

[marciokoko]

Ahh, OK.
referring to the last sketch in your first post:

• Are the sketches and photo left-right reversed?
• There are some dimensions given, what part(s) of the structure do they represent?
• What is that right-most almost-vertical line, is it really vertical, is it part of the solution?
• It looks like you got the 108.9° correct.
• Please mark on the sketch which angles you need?
• Door pivoted at the bottom?

Thanks.

p.s. There was a similar thread about 6 months ago that may give you some ideas and insight. @JBA came up with a solution on that thread, hopefully he can jump in here and work his magic!
https://www.physicsforums.com/threads/how-to-put-solar-panels-on-top-of-my-rv.977763/#post-6241131

Purple is the box frame.
Green is the acrylic door.
Blue is the body of the actuator.
Yellow is the extended actuator.

I'm trying to find out the best position to fully open the door with the given actuator.

 

[Baluncore]

@marciokoko
You need to specify three things.
1. What is the retracted length of the actuator. 40 cm ?
2. What is the extended length of the actuator. 65 cm ?
3. What angle must the door swing through to be called “fully open”.

 

[marciokoko]

1. Yes 40 cm
2. Yes 65 cm
3. ok so that's it really, if I determine the door should be open at 70deg, then I can position the actuator properly.

Ok so it was more a question of me deciding what I want fully open to be. Ok so no it turns into a forces problem, because as i see it I have 3 options:

 

[marciokoko]

top left is my current setup but the actuator is so far from the hinge or pivot, that the 25cm stroke is not enough to make the cover reach 70 deg.

top right & bottom right are basically the same, but the box is flipped.

what other options am I missing? I would try the bottom right today, I just need to move the actuator and attachment point.

 

[Baluncore]

what other options am I missing?

You have too many undefined degrees of freedom now.

You show a pin where the far end of the actuator body contacts the box.
You also show a pin where the actuator exits the box, why is that point relevant?
Does the actuator rod attach to the door with a pin, or just press on it with a roller?
What stops the actuator rod from rotating, or bending?

Some actuators mount between end pins, others mount the body near the rod seal.
How does the actuator mount? Which points are used?
Is there a data sheet on the actuator?

If you open more than 70°, the actuator may pass too close to the hinge, so the forces will be too high. That requires a different attachment point in the box.

 

[marciokoko]

You have too many undefined degrees of freedom now.

You show a pin where the far end of the actuator body contacts the box.
You also show a pin where the actuator exits the box, why is that point relevant?
Does the actuator rod attach to the door with a pin, or just press on it with a roller?
What stops the actuator rod from rotating, or bending?

Some actuators mount between end pins, others mount the body near the rod seal.
How does the actuator mount? Which points are used?
Is there a data sheet on the actuator?

If you open more than 70°, the actuator may pass too close to the hinge, so the forces will be too high. That requires a different attachment point in the box.

Here are more pictures:

 

[Lnewqban]

Rather than going into calculations, and since you have everything right there, I would use a marker to determine the best you can do with that actuator in tgat arrangement.

With the door closed and the actuator fully retracted, I would mark several pairs of points of contact on surfaces of door and box, giving sequential numbers to each pair.
Next, I would open the door to desired angle and fully extend the actuator, trying all pairs of marked points.
One combination should be best.

In case full extension of actuator results to be not long enough, you will need to use some levers or pulleys and cable to multiply the reach at expense of the force.

 

[berkeman]

• Are the sketches and photo left-right reversed?
• There are some dimensions given, what part(s) of the structure do they represent?
• What is that right-most almost-vertical line, is it really vertical, is it part of the solution?
• It looks like you got the 108.9° correct.
• Please mark on the sketch which angles you need?
• Door pivoted at the bottom?

Yeah, I'm completely confused by the sketches. Apparently there is also a ramp that opens downward in some of the sketches... Not sure what that it all about. It's hard for me to help when I can't even decode the sketches...

 

[marciokoko]

Ok here is the new setup. I moved the actuator to the top part of the box. The problem now is that I need to fix the body of the actuator so it doesn't "drop" down.

 

[Baluncore]

Ok here is the new setup. I moved the actuator to the top part of the box. The problem now is that I need to fix the body of the actuator so it doesn't "drop" down.

That is a worse arrangement. The actuator should only be attached with pins at the ends. That way it will remain in a straight line and not be bent by holding the actuator body.

As the door swings out, the force needed to move the door weight increases from zero to a maximum.
At the same time, the actuator line gets closer to the hinge, so the mechanical advantage gets worse.
It would be better to attach the actuator lower down on the box. Then the two effects could cancel, and you would have them working together.

 

[Tom.G]

Try this:

Cheers,
Tom

 

[berkeman]

The problem now is that I need to fix the body of the actuator so it doesn't "drop" down.

One way that I deal with such strengthening problems is to make I-Beams out of the edges that are curving. In this case, I'd probably start with 1x4" finished wood, cut it down to about 1x3" finished dimensions, and then rip a dado groove lengthwise down the length of the beam. Make two such I-Beam outside pieces, and glue them to the curved edges of the plexiglass (the vertical outside edges). If it's going to be a long-term setup, be sure to weatherize/finish the wood before gluing it onto the plexiglass.

The wood edge pieces will looks something like those in this picture:

http://www.hobbithouseinc.com/personal/woodpics/_g_IJK.htm

 

[marciokoko]

So you mean On the side of the wall? The door is acrylic, it's too wobbly. I would need a second actuator. Well i fixed a 2x2 on the wall and mounted the actuator on it. It works ok. I didnt measure the angle but its at least half way I think. The acrylic door is really have actually because its thick, about 1/4" i think.

Well I did this and Ill try to reinforce the door.

I would need to modify it to give me more opening then because I need to fit an rc car or from there.

Try this:

View attachment 260159

Cheers,
Tom

 

[marciokoko]

This is a better take:

 

[pandem]

ok, setting the force issues aside (for now), Id like to use what I know about triangles to figure how to position my linear actuator in order to open and close this door correctly:
View attachment 260064

The vertical side is 84cm, the door is 92cm. The measured stroke of the AL is 25cm.

I would need to lower the attachment point which is currently at 26cm from the bottom edge:
View attachment 260066

If I were to randomy fix the attachment point to the door at 13cm from the bottom edge, how do I measure the angle now of the door with the bottom corner? I boiled it down to this:

View attachment 260069
So how do i solve Sin ⍬ = 0.325 on calc without arcsin (iphone)? 18.9 deg on website, but I am still missing an angle or two...

Please help.

You can activate arcsin (sin exp-1) or any inverse trig functions on the iPhone scientific calculator (rotate to landscape) with the 2nd key( far right )

 

[Nik_2213]

Having recently disconnected a fused linear actuator, a screw-thread type, from our 'up&over', I'm reminded the door had a big stand-off bracket to mitigate the effects of adverse Sin/Cos etc, provide adequate 'starting' force...