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# luca-deltodesco

Member Since 10 May 2006
Offline Last Active Sep 07 2013 03:50 PM

### In Topic: Confused on Big Oh

14 March 2013 - 07:11 AM

I think the best example of that is triangulation:

Triangulating a polygon 'can' be done in O(n) time, but the algorithm to do this is so incredibly complex that unless you have a polygon with 100000000000000 (*number pulled out of my arse) vertices, an O(n.log(n)) or O(n^2) algorithm is going to do it faster.

### In Topic: the right way to represent inverse inertia tensor

11 February 2013 - 12:51 PM

inverse_inertia = inertia.inverse();

Anyways, if it aids:

By choosing the correct coordinate system for the object (For basic objects, the obvious one), you can always diagonalise the inertia tensor, and so it, and its inverse can be stored in a vector instead of a matrix, and then the inverse is just the component wise inverse:

(Ix, Iy, Iz)^-1 = (1/Ix, 1/Iy, 1/Iz)

### In Topic: Collision Physics Question

30 January 2013 - 04:21 PM

It is not that there is less trig involved when using x-y vectors, but that there is NO trig involved.

I already showed you earlier how to do bouncing.

That you say you have done bouncing, but cannot figure out this, plus that you talk about the 'angle' of the slope tells me you are using trigonometry to do the bouncing... *sigh*

The solution is incredibly simple, using simple vector operations.

You have the slope normal vector (As opposed to its angle)
You have the incident velocity.

The outgoing velocity is then: newVelocity = velocity - normal*(1+e)*dot(velocity, normal)
where e is the coefficient of restitution. For a perfectly elastic bounce e = 1, for a perfectly inelastic bounce (your 'slide') e = 0.

If you're struggling to compute the normal vector. Assuming you have two positions for the slope 'a' and 'b' that you're using to compute your angle like atan2(b.y-a.y, b.x-a.x), then the normal for this slope is instead: unit(a.y-b.y, b.x-a.x). The normal is a unit-length vector that points 'out' (or 'in' for these purposes, makes no difference to the result) of the surface, so the left side of a box will have normal (-1, 0) etc.

### In Topic: Collision Physics Question

30 January 2013 - 02:00 PM

What reason do you have to say that using x/y vectors would not permit you to have anything but right-angle ramps? Using angle/magnitude for vectors is almost NEVER the good way of doing things, it quite simply makes everything more complicated.

### In Topic: Collision Physics Question

27 January 2013 - 02:17 PM

That you say you have done bouncing, but cannot figure out this, plus that you talk about the 'angle' of the slope tells me you are using trigonometry to do the bouncing... *sigh*

The solution is incredibly simple, using simple vector operations.

You have the slope normal vector (As opposed to its angle)
You have the incident velocity.

The outgoing velocity is then: newVelocity = velocity - normal*(1+e)*dot(velocity, normal)
where e is the coefficient of restitution. For a perfectly elastic bounce e = 1, for a perfectly inelastic bounce (your 'slide') e = 0.

If you're struggling to compute the normal vector. Assuming you have two positions for the slope 'a' and 'b' that you're using to compute your angle like atan2(b.y-a.y, b.x-a.x), then the normal for this slope is instead: unit(a.y-b.y, b.x-a.x). The normal is a unit-length vector that points 'out' (or 'in' for these purposes, makes no difference to the result) of the surface, so the left side of a box will have normal (-1, 0) etc.

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