Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How can I simulate a rigid sword bounced from a wall and hit the ground(like in physical world)?

I want to simulate a simple animation.

The sword is controlled by a center/mass point.(Actually finally it't in a game engine).

I can detect the figure and the size of the sword(maybe needed in doing bounce).

And the position is $\{x,y,z\}$

rotation can be controled by quaternions/matrix/euler angles.[In mathematica, by default is matrix]

Step1


point1 = {0, 0, 500};point2 = {500, 0, 500};

Fly 500 miles. (Here I can add an acceleration)

Graphics3D[{Red, Cuboid[{600, -100, 400}, {510.1, 200, 600}],   GeometricTransformation[swordX[[1]],    TranslationTransform /@ {point1, point2}]}, ViewPoint -> Front]

enter image description here

As you see in the picture, my sword inserted into the Red wall.

step2


sword bounced from the wall, It should turn the head and do rotations and fly to the ground.

Something like a projectile

How can I simulate this physical process?

Maybe what I need is an equation and some parameters?

Any references?

I've played with this Mathematica Demonstration model, and I've made a video before.

And I need make it more physical, especially how the sword hit and bounced from the wall and fly like a hammer in this model...

video[03:07]

step3


hit the ground, (maybe can bounce from the ground and hit again...)

Some model maybe the same as that in step2.


I have no physical engines in Mathematica, if you know, welcome to tell me something related.

share|improve this question
    
Have you considered momentum conservation? –  Kyle Kanos Oct 15 '13 at 13:31
    
@KyleKanos ?what do you mean? When sword hit on the ground for several times, it'll stop, of course. –  HyperGroups Oct 15 '13 at 13:32
    
Obviously, but it is a starting point, isn't it? –  Kyle Kanos Oct 15 '13 at 13:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.