Since Maple Flow is built on Maple I expect the same interrupt behaviour. Sometimes it can pay of to wait. See here.

Alternatively, you could try to kill the mserver.exe process and reload the document. Optionally saving your work before reloading.

@Rouben Rostamian @Christopher2222

Also fixed step solvers reproduce two physical effects when the step size is reduced:

• reflection
• double rebounce

Reflection:

When the contact force is maximal (A), the velocity of the center of the disk is zero (B and C) and the penetration into the contact is max (F).

Since the contact force does not point to the center of mass of the disk (D), a torque is generated that lets the disk roll back. The torque is alo responsible for the deceleration of the horizontal velocity of the tiny mass.

Double rebounce:

After lift-off we see a second contact (E). This is caused by the backward rotation generated by the torque. The disk rotates fast ennough about the center of mass of the disk that the outer contour of the disk hits the flat surface a second time.

Does this make sense? (An extended explanation will take more work and is maybe worth a post visualising the contact force and the contact location. Both are not provided by MapleSim).

simplified_disk_pendulum.msim

Concerning the dislocated mass: This did not manifest with the above model and still might be related to solvers. (edit:) because the integration time ways shorter and more plot events where generated. -> It's a plot artefact.

Exporting the DAEs to Maple looks like this (just to give you an idea)

simplified_disk_pendulum_Equation_extraction.mw

@Christopher2222

I use MapleSim 2024.

I could not fix warnings and the initialization problem but I know where they are comming from. It is a connection between two scalar ports which is interpreted as an array to scalar connection. I can work around that but the effect at 12.5 s is still there.

I had a closer look at a "reflected" bounce (0.01 kg, d=0 (no damping), c=100000 N/m stiffness) similar to the bounce at 12.5 s. It's also not physical what we see there. The horizontal speed of the tiny mass changes (green) abruptly when the fat mass bounces.

Also the fat mass dislocates during the bounce while for m=0.1 kg (to the left) it stays in place.

This looks like a solver finding other solutions than the correct one.

I "bounce" back to my orginal surprise that I did not expect this for a mass ratio of moderate 1:100.

The solvers somehow have difficulties in keeping the horizontal speed contineous when the tiny mass crosses over the fat mass. I will see whether I can find setting that work better. If not I will try to export the DAE system for analysis in Maple.

@sursumCorda 

Your solution is better than nothing. Thank you

@C_R 

With a stiffer contact without damping it is possible to let the pendulum swing without angular reflection (which I called angular lock) and some bouncing. The bouncing depends on mass distribution to the contact point. The last bounce below looks strong because the two masses are vertically alligned (i.e. the 1 kg mass hits the contact point head on).

The bouncing is also reflected in ramps of the angular orientation where we have lift-off. We see a slight dependence on the solver type. In the beginning the disk rocks smoothly. There is a high number of events indicating that the numerical integration with the variable step solvers needs reconfiguration. Fixed step solvers do not work.

Overall the model is too simplified to work well together with the sophisticated contact elements. The contact element cannot be parametrised in a way to be infinite stiff, sticky and without friction. What we see here is what can be expected as a consequence of the low fidely of the modeling. Its neither a bug nor an insufficient implementation.

@sursumCorda 

That works if I paste from your answer or from https://en.wikipedia.org/wiki/Greek_and_Coptic

Its a bit complicated. Do I get unicode within Maple?

@Rouben Rostamian  

I thought I had excluded this possibitly by adding a mass of 0.001 kg at the center of the spherical contact element that does the rolling.

What you assumed was no mass at the center:

Disk_pendulum-_strip_down.msim

Without friction it should behave now like a pendulum with a horizontal prismatic joint at the pivot point.

Kinematically it does: the mass is falling straigt down.

Energetically after the first swing, energy is lost. This should not be for a conservative system.

However, in the elastic contact I added a bit of damping.

One could argue that this system is unphysical because no lateral forces act.

Effectively we see only a mass bouncing vertically. The erratic swing of the massless ridig body frame to the center of the contact element is only the solver searching for a solution that satisfies a constraint condition. In a vertical postition there are two possibilites left and right to the vertical.

However, increasing to a mass at the center to 0.01 kg, i.e. 1%, is not sufficient. Only with 0.1 kg the effect vanishes. This is 10% of the excentric mass and unexpected high.

At the moment I am a little surprised that the problem is still illconditionned with a mass ratio of 1%.

Adding to @Rouben Rostamian suggestion: Once you have computed the normal force of the contact you could investigate for slippage. In this question about some numerical artefacts I have uploaded a MapleSim model that still shows slippage for mu=1.

To implement slippage, numerical integration has to switch between two configurations of odes: One for slippage and one for rolling without slippage (this one you have implemented already).

Maybe someone knows whether dsolves event control can be/has been used for this.

P.S.:

The red curve red below (with reduced slippage) shows a good agreement with your implementation 👍

@Gabriel Barcellos

Your interpretation is correct (based on one attempt using the procedure p_O2 in Médio_spin_7_2_-_Forum_optimize_03_b.mw) all results for varying d close to d=3 are the same.

Before zooming in I would now create a coarse d-T map. Finding solution far away from  d=3 seems to be difficult. So far I could not find anything.

Everything else next year.

@dharr @Gabriel Barcellos

To wrap up, I have added my findings in the document from dharr

Campo_Médio_spin_7_2_new5MP_C_R_2.mw

in case more computing resources or smart simplifications are available in the future.

I could run the parameter x up to a point where the problem becomes numerically illconditioned, which requires higher precision and therefore computation time goes up.

I have also notices in my original worksheet all the below fsolve calls produce the same result

It could be that numerically there is nothing to gain but the result is stable with the given Digits whereas in the xyz there is more to gain at the cost of computational effort.

Happy holidays

@acer 

I could not find such settings. I also tried Edge. What seems to work is selecting the text with the pasted font, changing it to another font and then changing it to default. Whatever font default is.

@acer 

Your worksheet gives

(is(abs(x) = max(x, -x)) assuming positive);
                              true

(is(abs(x) = -min(-x, x)) assuming positive);
                              true

(is(abs(x) = -min(-x, x)) assuming negative);
                              true

@dharr 

I have used the above worksheet to compute solutions for x, y and y2 for a given z with fsolve similar to the original call

A first result is about 10 times faster

Substituting this results into the original expression for eqm gives not really equality

I attribute this to the large values for y and y2. So maybe that is not a real solution to the origianal problem with exp().

On the other hand if I give a solution from my worksheet as inital values, the agreement is better

All this without procedures. I think that a further gain with codgen,makeproc and codegen,optimized is possible. We might see a gain in speed of about x25.

Before going further in this direction, the following should be discussed:

Technically we could run x as a parameter (which means varying T) and with that solve for y, y2 and z.
z will give us d. I have tried this with initial values starting from here:

This can only be done in tiny increments of x. At each increment the former solution is used as start value. Very soon the solution jumps to negative values for z. So either there is a bug in the xyz approach, which I do not think, or we are dealing with a singularity.

@dharr : Thank you for the background. I was realy thinking about something else.

@Gabriel Barcellos

I have tried modified equations with a reduced number of arithmetic operations (removing 1/T in the sums, see eqm_eqm2_eqTPO.mw). I did this by chanceling beta in m0. To my surprise fsolve took longer with the simplified equations. Now there is only one simplifcation left which is removing redundancies in the 3 equations (see update 3). This is in itself a new question where I have no good answer for and my time to work on it is limited. However, worksheets with a reduced number of terms in the sums are a big help to work on this problem.

Concerning physics: It is the first time that I see such expressions (big sums which are not series) in a physical context. Since it is new to me I share @one man doubts and I cannot interprete results. Any bockgound would be of help. I was surprise to see that only S0 has to be modifed for a lower spin number. I did expect Z0 also to have a reduced number of additions. Mathematically we can always plug in results in the set of equations that fsolve solves to verify Maples numerical results. This is less of a worry to me.

Concerning solving the problem: You have to tell us what you are finally looking for. Is it a T-d diagram, as we see it at the end of your last attachment for 3/2 spin? Depending on that answer a grid of data point might be ennough for you or is a restructured problem with less redundancies (where we do not know how much fsolve will get faster) needed.