Need Help Mechanical double lid lifter

[nvn]

The torsional spring is needed only if a motor drive mechanism is used. In general, you would probably purchase the torsional spring and hinge separately, remove the hinge pin, cut a notch for the torsional spring, then reassemble the parts. But how you do it is up to you. You can use any approach you prefer.

The cable can be attached in any way you prefer, and is relatively easy to attach. How you do it is up to you. The cable does not need to be an exact length. You want the cable to be longer than necessary, so that the cable makes approximately one and a half extra loops around the empty side of the spool when the mechanism is fully extended or contracted.

For the axial force in linkage bar BC, use the maximum linkage bar axial forces listed in post 104, paragraph 4.

 

[cybertron]

The torsional spring is needed only if a motor drive mechanism is used. In general, you would probably purchase the torsional spring and hinge separately, remove the hinge pin, cut a notch for the torsional spring, then reassemble the parts. But how you do it is up to you. You can use any approach you prefer.

The cable can be attached in any way you prefer, and is relatively easy to attach. How you do it is up to you. The cable does not need to be an exact length. You want the cable to be longer than necessary, so that the cable makes approximately one and a half extra loops around each spool when the mechanism is fully extended.

For the axial force in linkage bar BC, use the maximum linkage bar axial forces listed in post 104, paragraph 4.

nvn.

The axial force in linkage bar BC as stated in 104, paragraph 4. Dose this force have anything to with the tension and compression of the Torsional Spring and what the spring needs to conform to in holding the force on linkage bar BC. If not, Please explain!


Also in post 104 when the doors are at 0dgs (closed) coordinates A points in mm 10.00 and 29.60 these are two points where do these two points get attach to on the short tube at (A).

 

[nvn]

No. The torsional spring applies a small force only when the doors are opened greater than 85 deg. This torsional spring force is much smaller than the linkage bar forces when the doors are at 0 deg. Therefore, the torsional spring does not govern the design.

The location of a point is given as two coordinates (an x and y coordinate), as shown here[/color].

 

[cybertron]

nvn.

Can't open the file in post 124 for some reason it dose not open. its a svg type of file, do you know what software that will open this program.Because i cannot open it.

 

[nvn]

The location of a point is given as two coordinates (an x and y coordinate), as shown here[/color], or here[/color].

 

[cybertron]

nvn


In reading the topic on two points as two coordinates in post 126, assumming that 10.00mm and 29.60mm are the two coordinates for point (A). In that, the ending edge of the the shorter tube to point A is 29.60mm and from the bottom edge of the shorter tube to point A is 10.00mm which gives you coordinates x and y for point (A).
If this makes any sense at all and my understanding from what i read from the site you posted in 126. Please if I'm wrong correct me. explain!

 

[tyroman]

cybertron,

Attached is a sketch which is my understanding from the figures nvn gives in post #104. I have used a screen capture from nvn's latest animation and added the coordinates and calculated lengths from post #104.

Hope this helps...

.
(PS; edits to correct post reference # and added detail to sketch)

 

[cybertron]

tyroman.

Thanks for clarifying this up for me.


OK. So, I was about (80%) correct on the coordinates of x and y for point (A) from reading the topic in post 126. I had the dimensions backwards, and had based the origin point from the ending edge of the shorter tube instead of using the center point of the longer tube. As you indicated in your edited sketch of nvn animated design.

So the origin point X is the center point of the longer tube. correct! And all coordinated points are based off this center point of the longer Sq tube. correct!

Points J and K are these coordinates for the hinges.

Which would be the best and simplest way to construct the mechnism to my sq tube lifter inside or outside the sq botom tube.

Also these questions might need to be answered by nvn, if you can't answer tyroman.
1. If i was to construct this on the inside of the bottom tube how long would i need to cut a narrow slit where points B and C Slide right to left when the doors open and close from 0degs to 90degs and also on the shorter tube on the left side where linkage AB will need to slide in the top of the longer tube what would be the dimensions for all these cut slots.
2. Will point D's brown plate pass through the end of the longer tube or would i need to cut a slot at the longer tubes end side.

 

[tyroman]

cybertron,

Glad my sketch was helpful... but I need to be sure you understand how a coordinate system is used.

Where you say;
"the origin point X is the center point of the longer tube"

I would say;
"the origin point (X = 0.00, Y = 0.00) is at the LEFT END of the longer tube, at its centerline"

nvn said in post #104 [with edits by me];
"The coordinate system origin is ... located on the left-hand end [X = 0.00] of the long square tube, at the tube longitudinal centerline [Y = 0.00]."

The location of any point on the drawing is described with TWO numbers... (X, Y).

Your inside / outside question has been around for a while... Is there a reason you would want to put the mechanism inside the tube? I can think of reasons to put it outside the tube (aside from the motor-drive issue) such as; ease of access for lubrication, repair or adjustment. Additionally, from my understanding of the off-limits restrictions, most of the mechanism (link BC with its bushings and connections to links AB and CD) will have to be assembled inside the tube BEFORE the long tube is mounted to its base. Would this present a problem?

Your final questions;
"Also these questions might need to be answered by nvn, if you can't answer tyroman."
will most easily be answered by YOU when you build the device (or better yet, when you make a simple cardboard 2D model of the mechanism before you build anything).

.