Diogo de Andrade | 1 Apr 01:17 2003

Re: [Algorithms] Triangulation of polygons with holes


Hey all!

Just to say that I've made the TrueType to polygon application working and
online (with source, of course) at
http://www.spellcasterstudios.com/~diogo.andrade/code.htm, if someone wants
to take a look/use it...

I've just ended using the GLUT tesselator (internally, it must be a VERY
nice piece of code, very stable and usable)...

Thanks all for the help

Diogo de Andrade
diogo.andrade <at> spellcasterstudios.com

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Jay Stelly | 1 Apr 04:04 2003

RE: [Algorithms] Field-Of-View maths trick


> I believe there's a maths solution something to do with 
> scaling by tan((FOV
> A)*0.5f)/tan((FOV B)*0.5f) but can't quite get it right. Can 
> anyone work
> this out?

If you scale the coordinate along the axes in the plane of the camera by
this factor, but not the coordinate along the depth axis, this method works
perfectly - unless I misunderstand the question.

Jay

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Research (GameBrains | 1 Apr 05:28 2003

Re: [Algorithms] game networking security

For the record, at least one console manufacturer won't let code be downloaded and run, only data.  I am still
checking on the other two console makers policies, but if one doesn't allow this, then program downloads
are not an option for us anyway.

As far as I can tell, there is no way to stop cheating because as Brian pointed out the cheater will modify the
code until they are just superhuman.  They can add enough error to the cheat to always be right on the line of
impossible but nothing you can do.  No amount of client voting or server authentication can fix that.

The solution we are pursuing at the moment is to assume the clients to be perfect and completely hacked.  The
server maintains the offical player stats and skills and will use those to determine the outcome of events
in multiplayer games and notify the clients.  If the client ever sends a message that exceeds the values
that are possible, even once, the account is suspended.  For our particular game this can work as we
modified the design to account for this, but it wouldn't work on the vast majority of FPS games as they are
now.  The single player and LAN games don't do this stat and skill calc and so the client code doesn't have any
idea what the limits are.  It would simply be too painful for a cheater to create and account and go through
trial and error to try and find what those limits are (every weapon is encoded completely differently
[different structs with elements in different order and different sizes] with different values and
different ranges and there are more than 100 weapons).

Brett

----- Original Message ----- 
From: "Lucas Ackerman" <ackerman7 <at> llnl.gov>
To: <gdalgorithms-list <at> lists.sourceforge.net>
Sent: Saturday, March 29, 2003 4:32 AM
Subject: Re: [Algorithms] game networking security

> > Ron Hay wrote:
> > Brian Hook wrote:
> > > Problem #1 is that if the local client is authoritative, you're 
(Continue reading)

Jim Offerman | 1 Apr 08:15 2003

Re: [Algorithms] transforming vectors...

Hi Pierre,

> There's some confusion in the thread (I'll let the "Maths Police" handle
> that) but your first post was right. It's worth recalling this opens the
> door to much much more confusion when scaling is out. For example :
> ...
> Which shortcuts all the maths and gives the correct answer, but uh ! Your
> poor reader now has no clue what the hell you're doing or what conventions
> you're using.

Very good to point this out.

There's a couple of things one can do to avoid such confusion, of course. I
enforce people to use vector = vector * matrix' and the inverse-world matrix
is calculated automatically in my scenegraph. So, to rotate a normal back to
local space, the code would read:

Vector3 localNormal = worldNormal * mInverseWorldMatrix;

(Note that in my engine Vector3 is treated as a true 3d vector, not 4d
homogenous, so any translations stored in mInverseWorldMatrix are simply
ignored)

Jim Offerman
Crevace Games
www.crevace.com

' I've actually been doing that since the last big discussion on matrix
conventions, before I had code where matrix * vector = vector * matrix,
which is obviously very mathematically incorrect...
(Continue reading)

Igor Kravtchenko | 1 Apr 09:14 2003

Re: [Algorithms] rigid body friction

 
We use something fairly simple but that has the effect to consequently reduce the "jittering" effect.
At each iteration, we cumulate over the time the dot product between the current velocity vector
and the previous. If that become too low, it means a "chaotic" displacement is being applied and
thus, all further velocities vector are blended with the previous. The rigid body then quickly stop to
jitter as a tuning fork loosing its energy.
 
Igor.
 
 
----- Original Message -----
From: Paul Firth
Sent: Monday, March 31, 2003 12:52 PM
Subject: [Algorithms] rigid body friction

Hi ppl,

I was wondering if anyone had any ideas about how to avoid the "jittering"
which the standard coulomb friction model introduces at very low velocities.
I have seen papers which talk about using two tangential friction directions
forming a basis with the normal of the force, but I'm not sure where these
two directions come from.

Also, does anyone know how to tell if the frictional force required to hold
the object static is within the friction cone?

Cheers, Paul.



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Willem H. de Boer | 1 Apr 10:48 2003

RE: [Algorithms] transforming vectors...

Yes, you're absolutely right. As my final post to this thread
before I get chased away for bringing math to the peaceful and
quiet community that is GDAlgorithms:

I claim, that if we only have uniform scales and rotations
then:
      M = SA = AS

and thus:

      S^{-1} A^{-1} = A^{-1} S^{-1}

Then:
	M^{-1} = (SA)^{-1}					
             = (AS)^{-1}
             = S^{-1} A^{-1}
             = S^{-1} A^t					(1)
And:
	M^-t   = (M^{-1})^t
             = (S^{-1} A^t)^t				
             = (A^t S^{-1})^t
             = (S^{-1})^t (A^t)^t
             = S^{-1} A						(2)

Using (1),(2) in the proof in my previous post shows that you
can just use M to transform normal vectors, rather than
its inverse transpose M^-t, if and only if the matrix M contains
rotations and _uniform_ scales. And _only_ if you _always_
normalise your normals afterwards do normals transform
exactly the same under either transformation.

Willem H. de Boer

> -----Original Message-----
> From: Jamie Fowlston [mailto:jamief <at> qubesoft.com]
> Sent: 31 March 2003 18:24
> To: gdalgorithms-list <at> lists.sourceforge.net
> Subject: RE: [Algorithms] transforming vectors...
> 
> 
> No.
> 
> 	(M^{-1})^t = (S^{-1} A^t)^t
> 		     = (S^{-1})^t (A^t)^t
> 		     = S^{-1} A
> 
> is not true. t(AB) = t(B)t(A), so 
> 
> t(inv(M)) = t(inv(S) t(A))
> 	= t(t(A)) t(inv(S))
> 	= A inv(S)
> 
> (_not_ inv(S) A)
> 
> 
> Regardless, my previous post showed that if
> 
> 	a = Sx
> 	b = S^{-1}x
> 
> then a and b are not in general colinear.
> 
> Jamie
> 
> 
> -----Original Message-----
> From: gdalgorithms-list-admin <at> lists.sourceforge.net
> [mailto:gdalgorithms-list-admin <at> lists.sourceforge.net]On Behalf Of
> Willem H. de Boer
> Sent: 31 March 2003 17:16
> To: gdalgorithms-list <at> lists.sourceforge.net
> Subject: RE: [Algorithms] transforming vectors...
> 
> 
> >And it wouldn't work for any scalings (apart from the trivial
> >one: all 1's) either. Because a matrix that contains a scaling
> >in whatever direction is not an orthogonal matrix. Before anyone
> >starts objecting: With orthogonal I mean M^T M = I, it's what
> >many people would call orthonormal.
> 
> Actually, it _would_ work for scalings, as I've just found out.
> 
> [People not interested in maths should click away, NOW]
> 
> Suppose M is a transformation (an NxN matrix), composed of two 
> other transformations M = SA, where both S and A are NxN matrices.
> S is a scaling matrix, and is thus diagional. And A is an orthogonal
> matrix (or orthonormal, if you will). Now we want to "prove"
> that for a given vector v:
> 
> 	Mv . Nv = |Mv| |Nv|				(1)
> 
> Or, the vector a=Mv, is colinear but not necessarily of same magnitude
> to the vector b=Nv, where N is the inverse transpose of M. Let's
> calculate the inverse transpose of M, firstly, its inverse:
> 
> 	M^{-1} = (S A)^{-1} = S^{-1} A^{-1} = S^{-1} A^t
> 
> The last bit comes from the fact that the inverse of an orthogonal
> matrix is just its transpose. Now let's take the transpose of M^{-1}
> 
> 	(M^{-1})^t = (S^{-1} A^t)^t
> 		     = (S^{-1})^t (A^t)^t
> 		     = S^{-1} A
> 
> The last bit coming from the fact that the transpose of a diagonal
> matrix (S^{-1}, in this case, the inverse of S) is the same
> matrix. And the transpose of the transpose of a matrix A, is
> just A.
> 
> So, now back to the original statement (1). We can rewrite the vector
> a as:
> 	a = Mv = (SA)v
> and b:
> 	b = M^-tv = (S^{-1} A)v
> 
> Now we now that both a and b transform similarily under A:
> so, let's call Av, x, both equations reduce to:	
> 
> 	a = Sx
> 	b = S^{-1}x
> 
> So, yes, you're assumption was right. a, and b will be colinear
> but of different magnitudes, if and only if M is a transformation
> that consists only of rotations, and scalings (even non-uniform ones).
> 
> Willem
> 
> 
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> 
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Christopher Phillips | 1 Apr 11:04 2003

RE: [Algorithms] game networking security


> -----Original Message-----
> From: Research (GameBrains) [mailto:research <at> gamebrains.com]
> Sent: 01 April 2003 04:28
> To: gdalgorithms-list <at> lists.sourceforge.net
> Subject: Re: [Algorithms] game networking security
> 
> 
> For the record, at least one console manufacturer won't let 
> code be downloaded and run, only data.  I am still checking 
> on the other two console makers policies, but if one doesn't 
> allow this, then program downloads are not an option for us anyway.
> 

Has said manufacturer made any specific statements about what 
their definition of code is?   The post you responded to was 
basically advocating downloading tokenised scripts, which is 
on the codier end of grey, but given the amount of discussion 
about data-driven code we've had here,  even replacing level 
files that contain a few attributed entities can drastically 
affect the execution path of your engine..

-----------------------------
-Virus scanned and cleared ok

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Sin Chian | 1 Apr 11:39 2003

Re: [Algorithms] rigid body friction

Hi,
 
Miguel Gomez presents a very interesting solution in his article "Coping with Friction in Dynamic Simulations" (section 2.7, Game Programming Gems 3).
Basically, the object's forces are summed up to get the acceleration, a.
Then, the simulation time step is clipped to the moment when the velocity, v is reduced to 0 where h = -(v.v) / (a.v);
Once you have | v | = 0, the static friction should kick in, preventing the object from moving unless acted upon by a sufficiently large force.
The simulation is then resumed from the clipped moment.
 
The method that I currently use is to integrate with Euler Method enough times to get v to converge to zero, once | v | nears zero, it is set to zero.
 
Hope this information is useful.
SinChian  
----- Original Message -----
Sent: Tuesday, April 01, 2003 3:14 PM
Subject: Re: [Algorithms] rigid body friction

 
We use something fairly simple but that has the effect to consequently reduce the "jittering" effect.
At each iteration, we cumulate over the time the dot product between the current velocity vector
and the previous. If that become too low, it means a "chaotic" displacement is being applied and
thus, all further velocities vector are blended with the previous. The rigid body then quickly stop to
jitter as a tuning fork loosing its energy.
 
Igor.
 
 
----- Original Message -----
From: Paul Firth
Sent: Monday, March 31, 2003 12:52 PM
Subject: [Algorithms] rigid body friction

Hi ppl,

I was wondering if anyone had any ideas about how to avoid the "jittering"
which the standard coulomb friction model introduces at very low velocities.
I have seen papers which talk about using two tangential friction directions
forming a basis with the normal of the force, but I'm not sure where these
two directions come from.

Also, does anyone know how to tell if the frictional force required to hold
the object static is within the friction cone?

Cheers, Paul.



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Research (GameBrains | 1 Apr 12:28 2003

Re: [Algorithms] game networking security

Christopher,
The manufacturer has not stated explicitly what that means, but I think the intent is to include scripts. 
The rationale being that they don't want a third party (user?) to be able to download and execute anything
on the console.  Also they are worried about quality issues that might arise because the code hasn't been
officially tested and approved by them so what happens if there are bugs...
Brett

----- Original Message ----- 
From: "Christopher Phillips" <cphillips <at> reflectionsinteractive.com>
To: <gdalgorithms-list <at> lists.sourceforge.net>
Sent: Tuesday, April 01, 2003 5:04 PM
Subject: RE: [Algorithms] game networking security

> 
> 
> 
> > -----Original Message-----
> > From: Research (GameBrains) [mailto:research <at> gamebrains.com]
> > Sent: 01 April 2003 04:28
> > To: gdalgorithms-list <at> lists.sourceforge.net
> > Subject: Re: [Algorithms] game networking security
> > 
> > 
> > For the record, at least one console manufacturer won't let 
> > code be downloaded and run, only data.  I am still checking 
> > on the other two console makers policies, but if one doesn't 
> > allow this, then program downloads are not an option for us anyway.
> > 
> 
> Has said manufacturer made any specific statements about what 
> their definition of code is?   The post you responded to was 
> basically advocating downloading tokenised scripts, which is 
> on the codier end of grey, but given the amount of discussion 
> about data-driven code we've had here,  even replacing level 
> files that contain a few attributed entities can drastically 
> affect the execution path of your engine..
> 
> 
> 
> 
> 
> -----------------------------
> -Virus scanned and cleared ok
> 
> 
> -------------------------------------------------------
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Jamie Fowlston | 1 Apr 12:06 2003

RE: [Algorithms] transforming vectors...

<snip>
I claim, that if we only have uniform scales and rotations
then:
      M = SA = AS

...
</snip>

That's of course true, as a uniform scale is just multiplication by a
scalar, you needn't go into matrix form at all :)

Jamie

-----Original Message-----
From: gdalgorithms-list-admin <at> lists.sourceforge.net
[mailto:gdalgorithms-list-admin <at> lists.sourceforge.net]On Behalf Of
Willem H. de Boer
Sent: 01 April 2003 09:49
To: gdalgorithms-list <at> lists.sourceforge.net
Subject: RE: [Algorithms] transforming vectors...

Yes, you're absolutely right. As my final post to this thread
before I get chased away for bringing math to the peaceful and
quiet community that is GDAlgorithms:

I claim, that if we only have uniform scales and rotations
then:
      M = SA = AS

and thus:

      S^{-1} A^{-1} = A^{-1} S^{-1}

Then:
	M^{-1} = (SA)^{-1}
             = (AS)^{-1}
             = S^{-1} A^{-1}
             = S^{-1} A^t					(1)
And:
	M^-t   = (M^{-1})^t
             = (S^{-1} A^t)^t
             = (A^t S^{-1})^t
             = (S^{-1})^t (A^t)^t
             = S^{-1} A						(2)

Using (1),(2) in the proof in my previous post shows that you
can just use M to transform normal vectors, rather than
its inverse transpose M^-t, if and only if the matrix M contains
rotations and _uniform_ scales. And _only_ if you _always_
normalise your normals afterwards do normals transform
exactly the same under either transformation.

Willem H. de Boer

> -----Original Message-----
> From: Jamie Fowlston [mailto:jamief <at> qubesoft.com]
> Sent: 31 March 2003 18:24
> To: gdalgorithms-list <at> lists.sourceforge.net
> Subject: RE: [Algorithms] transforming vectors...
>
>
> No.
>
> 	(M^{-1})^t = (S^{-1} A^t)^t
> 		     = (S^{-1})^t (A^t)^t
> 		     = S^{-1} A
>
> is not true. t(AB) = t(B)t(A), so
>
> t(inv(M)) = t(inv(S) t(A))
> 	= t(t(A)) t(inv(S))
> 	= A inv(S)
>
> (_not_ inv(S) A)
>
>
> Regardless, my previous post showed that if
>
> 	a = Sx
> 	b = S^{-1}x
>
> then a and b are not in general colinear.
>
> Jamie
>
>
> -----Original Message-----
> From: gdalgorithms-list-admin <at> lists.sourceforge.net
> [mailto:gdalgorithms-list-admin <at> lists.sourceforge.net]On Behalf Of
> Willem H. de Boer
> Sent: 31 March 2003 17:16
> To: gdalgorithms-list <at> lists.sourceforge.net
> Subject: RE: [Algorithms] transforming vectors...
>
>
> >And it wouldn't work for any scalings (apart from the trivial
> >one: all 1's) either. Because a matrix that contains a scaling
> >in whatever direction is not an orthogonal matrix. Before anyone
> >starts objecting: With orthogonal I mean M^T M = I, it's what
> >many people would call orthonormal.
>
> Actually, it _would_ work for scalings, as I've just found out.
>
> [People not interested in maths should click away, NOW]
>
> Suppose M is a transformation (an NxN matrix), composed of two
> other transformations M = SA, where both S and A are NxN matrices.
> S is a scaling matrix, and is thus diagional. And A is an orthogonal
> matrix (or orthonormal, if you will). Now we want to "prove"
> that for a given vector v:
>
> 	Mv . Nv = |Mv| |Nv|				(1)
>
> Or, the vector a=Mv, is colinear but not necessarily of same magnitude
> to the vector b=Nv, where N is the inverse transpose of M. Let's
> calculate the inverse transpose of M, firstly, its inverse:
>
> 	M^{-1} = (S A)^{-1} = S^{-1} A^{-1} = S^{-1} A^t
>
> The last bit comes from the fact that the inverse of an orthogonal
> matrix is just its transpose. Now let's take the transpose of M^{-1}
>
> 	(M^{-1})^t = (S^{-1} A^t)^t
> 		     = (S^{-1})^t (A^t)^t
> 		     = S^{-1} A
>
> The last bit coming from the fact that the transpose of a diagonal
> matrix (S^{-1}, in this case, the inverse of S) is the same
> matrix. And the transpose of the transpose of a matrix A, is
> just A.
>
> So, now back to the original statement (1). We can rewrite the vector
> a as:
> 	a = Mv = (SA)v
> and b:
> 	b = M^-tv = (S^{-1} A)v
>
> Now we now that both a and b transform similarily under A:
> so, let's call Av, x, both equations reduce to:
>
> 	a = Sx
> 	b = S^{-1}x
>
> So, yes, you're assumption was right. a, and b will be colinear
> but of different magnitudes, if and only if M is a transformation
> that consists only of rotations, and scalings (even non-uniform ones).
>
> Willem
>
>
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>
>
>
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Gmane