Polygons

Jay

Jay

Member
May 19, 2012
478
3
I just watched the long video on the Pinball Arcade home page that explains their process of bringing the tables to virtual life. One thing that struck me (at about 11:00 into the video) is the discussion of the physics engine related to the objects on the table. Apparently, the ball is "colliding" not with solid objects but with the wireframe mesh that the object is constructed from. If that's true, I wonder if the mesh itself has an effect on the ball's trajectory. It would be like the difference between kicking a soccer ball into a stiff chain link fence or kicking it into a solid wall.
 
brakel

brakel

New member
Apr 27, 2012
2,305
1
The "mesh" has the properties that the physics programming gives it. They can make it act like steel, wood, concrete, water, gravy, etc. It doesn't act like chain link fence unless that's what they want it to do. Describing it as a mesh is just easier for us to understand and probably easier for the people doing the coding. They model objects in virtual wire frames which can then probably be translated to the "mesh". At some point its all just numbers going through the processor. The processor doesn't know that it is "mesh" or water or dirt it just crunches the numbers. I don't know if TPA uses a physics enginge that they bought and are tweeking or if they designed the physics from the ground up. Some coding packages have a graphical representation that the coder can render quickly to see what is going on without having to fully render the scene. Even if that package uses a mesh to represent a physical object, that mesh has no physical properties like a metal cyclone fence than the words on this page have any properties of ink. Its all just code.
 
Jay

Jay

Member
May 19, 2012
478
3
Okay. I wasn't sure if the mesh surface was solid only on the connective lines or whether the faces between them were interpreted as solid. Even so, some of the curved structures seem to need a much higher polygon count to produce realistic rebounds.
 
M

Matt McIrvin

New member
Jun 5, 2012
801
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It'd make no sense to use a mesh without solid surfaces; it'd actually be harder than doing it the right way. I interpreted the images in the video to just mean that the objects were represented in the physics model by polyhedra with flat faces, not necessarily as numerous as the polygons in the visible representation we see on the screen (which makes sense; some simplification is necessary to make the computations tractable).

Incidentally, the other day when I was playing Black Hole on Android, I think I saw my first unambiguous instance of a ball flying straight through a flipper like it wasn't there. Sometimes it's the finite time step that can get you in trouble...
 
R

ridcully

New member
Jun 5, 2012
59
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The oposite of a infinite time step?

I think he means, that you have a limited time frame, to do calculations. If it would take a lot longer, the simulation would get jerky..
 
M

Mike Reitmeyer

FarSight Employee
Mar 13, 2012
1,735
1
The mesh represents the surface of the physics objects. The wireframe you see is just outlines of each polygon. We used triangles to represent meshes because they are the simplest geometric shape that can be used to represent any other shape.

We wrote our own physics engine built specifically for pinball.
 
brakel

brakel

New member
Apr 27, 2012
2,305
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Mike Reitmeyer said:
The mesh represents the surface of the physics objects. The wireframe you see is just outlines of each polygon. We used triangles to represent meshes because they are the simplest geometric shape that can be used to represent any other shape.

We wrote our own physics engine built specifically for pinball.
Click to expand...

Cool!
 

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