Polplan in STATICS (instantaneous center of rotation?)

[arestes]

TL;DR Summary

I need to find the ICR (hauptpol in german) in certain configurations.

Hello. I understand that plotting the hauptpole (I believe it's called "instantaneous center of rotation" in English) in STATICS is useful for virtual displacements.
I know the rules: Hauptpol (ICR, which is denoted (0, #member) ) are located where a support of a member is fixed. For a member with a perpendicular support (only 1 force) it is along the perpendicular line, etc.

However, even though this image (see figure, "totpunkt") is obviously a movable system, I can't reconcile the rules. It appears as if member III has not hauptpole when obviously it's at infinity (since the member III can only displace vertically, no rotation). But one of the rules needs that the Nebenpole (the articulation joining III and II (2,3) and the other articulation joining I and II (1,2)) should lie on a straight line. This means that the ICR of III should be where I marked it... which is a contradiction.

What am I missing?
Also, how do I find hauptpol (ICR) for member II ? it only has articulation for which I have no rule to try and determine (0,2)

 

[Lnewqban]

However, even though this image (see figure, "totpunkt") is obviously a movable system, ...

Sorry, I fail to see a movable mechanism as represented.
Could you explain the possible movements of each link a little further, please?

 

[arestes]

Sorry, I fail to see a movable mechanism as represented.
Could you explain the possible movements of each link a little further, please?

Sure, I am attaching two images (the one on the left is the one I was talking about). I drew the infinitesimal movements as a displacement du and angular displacements dphi1 dphi2. I guess it's right but only intuitively... but what I'm trying to figure out is the answers using Inst. Cent. of Rotat. (Hauptpol) by drawing a "Polplan". And for that, I'm having trouble applying the rules to find the Hauptpole (ICR) and Nebenpole.

 

[Lnewqban]

You may find some good ideas here:
https://www.real-world-physics-problems.com/instant-center.html

https://en.m.wikipedia.org/wiki/Instant_centre_of_rotation

Link II is simultaneously rotating and translating.
The top node follows a vertical line.
The bottom node follows a circular trajectory.
The geometrical center of link II must follow a trajectory that is a mix of both; therefore, its instantaneous center of zero velocity constantly changes positions during the movement of any slider-crank mechanism.

 

[berkeman]

I am attaching two images

Sorry, I'm no mechanical expert, but aren't the diagonal hashed lines supposed to represent a fixed wall or other anchor? How can they move vertically? Did you meant to instead draw some sort of a vertical slip joint?

 

[arestes]

Sorry, I'm no mechanical expert, but aren't the diagonal hashed lines supposed to represent a fixed wall or other anchor? How can they move vertically? Did you meant to instead draw some sort of a vertical slip joint?

View attachment 356490

Hi, I'm attaching the convention for figures used in my course and I believe it's common in many countries for Engineering: They only have one perpendicular reaction, allowing displacement parallel to the wall

 

[arestes]

You may find some good ideas here:
https://www.real-world-physics-problems.com/instant-center.html

https://en.m.wikipedia.org/wiki/Instant_centre_of_rotation

Link II is simultaneously rotating and translating.
The top node follows a vertical line.
The bottom node follows a circular trajectory.
The geometrical center of link II must follow a trajectory that is a mix of both; therefore, its instantaneous center of zero velocity constantly changes positions during the movement of any slider-crank mechanism.


View attachment 356488

Hi! First, thanks for the links and animation (may I ask where you found the animated gif? They might come in handy for me).
Also, unfortunately, I still can't for the life of me find the ICR for member II and member III. I am supposed to follow the rules to find them but I get a contradiction: On one hand I'm supposed to intersect to parallel lines perpendicular to the normal supports on the wall, forcing the ICR (Hauptpol) of II to be at infinity. That makes sense. But also, there's this rule that says that a Nebenpol and the Hauptpol should lie on a single line (possible counting infinity points of lines): (0,1) - (1,2) [Nebenpol, connecting I and II] - (2,0) but (0,1) - (1,2) is definitely not a line parallel to the previous ones.
Also, no idea as to how to find the Hauptpol (ICR) for member II. Still stuck.
Note: I understand these "rules" for Hauptpol and Nebenpol are a bit "niche" (I've only seen it in german schools and they're not explained in Statics book of the likes of Hibbeler). Here's a video explaining those rules, to get a sense of what I want to be able to do:

 

[erobz]

The velocity of the end attached to the shorter arm is perpendicular to the short arm. The velocity of the slide is always vertical. The lines for the IC come off perpendicular to the velocities of each endpoint.

 

[berkeman]

Hi, I'm attaching the convention for figures used in my course and I believe it's common in many countries for Engineering:

Maybe you should discuss this convention with your professor or Teaching Assistant...

https://www.dtytutoring.com/EngineeringStatics.html

 

[arestes]

Maybe you should discuss this convention with your professor or Teaching Assistant...

View attachment 356519

https://www.dtytutoring.com/EngineeringStatics.html

haha, yeah, I'm aware of the different conventions for the figures, but it seems to be something of a regional fashion. The notation I gave is found almost everywhere in german universities, probably all across Europe, idk. It's on Wikipedia and books on Statics by german authors, such as Dietmar Gross, even translated into English by Springer. But anyway, it's not what I was asking for... :(

 

[arestes]

The velocity of the end attached to the shorter arm is perpendicular to the short arm. The velocity of the slide is always vertical. The lines for the IC come off perpendicular to the velocities of each endpoint.

View attachment 356518

Hi! thanks. I was thinking of something like that but what I need is the IC when the whole mechanism is vertical. By analyzing it as a limit case I believe the IC of the middle member would be exactly at the joint at the upper end of it. But I can't make sense of it since it would have zero radius. Also, I was supposed to do this following the Hauptpole and Nebenpol rules, but at least I would like to know what's going on at that position.

It certaily baffles me that the upper end of the middle member would have zero radius

 

[Lnewqban]

Hi! First, thanks for the links and animation (may I ask where you found the animated gif? They might come in handy for me).

https://www.purdue.edu/freeform/me274/course-material/animations/

Also, unfortunately, I still can't for the life of me find the ICR for member II and member III.

Exactly when the three links are aligned, as shown in the original post, the IC of member II must be located at the top node.

The radius of rotation of link III is always infinite.
The radius of rotation of link II varies with time, but its value is 3/2l at top dead center (TDC).
The radius of rotation of link I is always l/2.

That position is normally called "top dead center" in slider-crank mechanisms because there is no vertical movement of the top link III or top node when the tangential instantaneous velocity of the lower node is perpendicular to that potentially sliding vertical movement (lowest link I or crank is in alignment with link II).

Please, see:
https://en.wikipedia.org/wiki/Slider-crank_linkage

 

[Lnewqban]

Additional information:
https://alistairstutorials.co.uk/mechanics_of_machines_content.html

https://alistairstutorials.co.uk/tutorial14.html

https://www.roymech.co.uk/Useful_Tables/Mechanics/Linkages.html