# Wedge and morpher

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June 12, 2012, 16:26
Wedge and morpher
#1
Senior Member

Henry Arrigo
Join Date: Jun 2010
Location: Italy
Posts: 100
Rep Power: 9
Hello guys,

I have a strange problem with morhper model.here is the problem, I run a simple model containing a heaving wedge with two symmetry planes on each side of the domain (I attached a photo of that) the wedge placed in the domain so that the center of area of the wedge is on the free surface level and the total height of the wedge is 0.15 m, means that the topmost part of the wedge is 0.05 m above the water level and bottommost is 0.1 m below the water level. Then I made the wedge to move with the function of 0.02*sin(25*\$Time) using the morpher model. First time I left the “Morph From Zero” unchecked, and the second time I checked this option to have the wedge to morph from zero at every time step and then I plot the time history location of the bottommost point of the wedge. Red dots plot indicates morph from zero, and the blue squares not from zero. As you can see there is no difference in these two conditions. Does anyone know why?
Attached Images
 1.jpg (49.4 KB, 37 views) 2.jpg (60.2 KB, 37 views) 3.jpg (46.5 KB, 36 views)

 June 13, 2012, 08:42 #2 Member   Ryan Coe Join Date: Jun 2010 Location: Albuquerque, NM Posts: 98 Rep Power: 9 I'm not an expert on deformable meshes, but I believe this option controls how the mesh at each timestep is produced. Without 'morph from zero' the grid points are deformed from one timestep to the next, but with it they are created based on the original mesh @ t = 0. The help manual seems to confirm this concept, and mentions that for a harmonic problem such as this, using the 'morph from zero' option is advisable to insure a harmonic mesh throughout the time of the simulation. Just curious, but is there a reason why you chose to use the morpher instead of a translating domain or overset mesh technique? __________________ Ryan

 June 13, 2012, 15:18 #3 Senior Member   Henry Arrigo Join Date: Jun 2010 Location: Italy Posts: 100 Rep Power: 9 Thanks ryancoe Because i use starccm+ v4.04

 June 15, 2012, 16:16 #4 Senior Member   Join Date: Oct 2009 Location: Germany Posts: 637 Rep Power: 15 Why the f... are you still using 4.04?!?!? Newer versions give a lot of advantages! Don't worry, you don't need to answer, I have my own thoughts Ryan is absolutely right. That's why it should not make any difference in theory. But just in theory, practically there are some differences. When you run a periodic motion for a long time without morph from zero option, it might happen that small differences are adding up to a bigger inaccuracy. Just assume, there is a vertex position error of 10^-5 in every cycle, and you run for 10 000 cycles. So you would add up a not negligible error... This is where the morph from zero option should give you an advantage: Repeating motions with relative small amplitudes. Now let's assume, you want to simulate a motion with a very large displacement within a time step. The morpher will destroy the mesh and the simulation crashes. But dividing this time step in several smaller ones, therefore morphing the mesh for a small portion in every time step will work. Now imagine, your mesh deformation should increase for 1mm in every second. You're running a time step of 1 second. With the morph from zero option unticked, you will morph the mesh for 1mm in every time step, no matter how many time steps you're doing. With the morph from zero option ticked, you will morph the mesh for: 1mm within the first time step 2mm in the second time step 3mm in the third time step ... ... 1483mm in the 1483rd time step... This is because this option will always take the initial volume mesh, calculate the displacement and try to morph the mesh from the initial position to the new one. After some time steps, the displacement gets too high and the mesh will be destroyed (or cell quality will be too low). This is where the morph from zero option should stay unticked: Motions with a very high deformation when running for a long time. Hope this gives you a better idea why there are two different options although they seem to do the same in your case... Henry Arrigo likes this. __________________ We do three types of jobs here: GOOD, FAST AND CHEAP You may choose any two!

 June 25, 2012, 10:31 #5 Senior Member   Henry Arrigo Join Date: Jun 2010 Location: Italy Posts: 100 Rep Power: 9 Thanks God i don t have to answer that f... question. And thank you abdul for you insightful reply. But as I said I got the error while I ticked the "morph from zero" option.

 July 6, 2012, 18:34 #6 Senior Member   Join Date: Oct 2009 Location: Germany Posts: 637 Rep Power: 15 What error? In your first post, you complain the two methods do give the same result, although that are two different settings, right? But that's no error, that's a perfect behaviour! As I wrote before, the only difference is the starting point for calculating the mesh deformation. When you've got a current deformation of 20mm and want to add 1mm deformation, an unticked "morph from zero" option would morph the mesh for 1mm, ending up in a total deformation of 21mm. A ticked "morph from zero" option would use the initial (undeformed) mesh and morph it for 21mm, ending up in a total deformation of 21mm. So the result is pretty much the same. The only difference is: The "morph from zero" option doesn't sum up all small errors from previous time steps while the unticked option doesn't need to deform the initial mesh for a huge displacement in one time step (and therefore is more stable for big deformations). __________________ We do three types of jobs here: GOOD, FAST AND CHEAP You may choose any two!

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