gdalgorithms-list

Re: [Algorithms] New research in SH rotations.

[Grégory J. <g....@fr...>]

This is very interesting indeed, thanks for pointing it out !

I wish I had some time to implement the paper and compare the
performance/results to my existing implementation !

Grégory Jaegy

-----Message d'origine-----
De : Robin Green [mailto:rob...@gm...] 
Envoyé : lundi 21 septembre 2009 18:59
À : Game Development Algorithms
Objet : [Algorithms] New research in SH rotations.

I received in the mail this morning a print of a paper from a
researcher in "Ambisonics" (the science of multidirectional or
spherical sound field recording and playback), Michael Chapman, who
claims to have new insight into producing rotation matrices for
Spherical Harmonics based on his research into their symmetries:

 
http://ambisonics.iem.at/symposium2009/proceedings/ambisym09-chapman-shsymme
tries.pdf/at_download/file

Along with this was a reference to a recent 2006 paper on efficient SH
rotations from Pinchon et al where they claim to have improved on
Kautz, Sloan and Snyder's methods not just in generation but in
compact storage as well:

   http://www.math.univ-toulouse.fr/Archive-MIP/publis/files/06.30.pdf

Just in case anyone still cares about efficient rotation of SH functions.

- Robin Green.

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[Algorithms] New research in SH rotations.

[Robin G. <rob...@gm...>]

I received in the mail this morning a print of a paper from a
researcher in "Ambisonics" (the science of multidirectional or
spherical sound field recording and playback), Michael Chapman, who
claims to have new insight into producing rotation matrices for
Spherical Harmonics based on his research into their symmetries:

   http://ambisonics.iem.at/symposium2009/proceedings/ambisym09-chapman-shsymmetries.pdf/at_download/file

Along with this was a reference to a recent 2006 paper on efficient SH
rotations from Pinchon et al where they claim to have improved on
Kautz, Sloan and Snyder's methods not just in generation but in
compact storage as well:

   http://www.math.univ-toulouse.fr/Archive-MIP/publis/files/06.30.pdf

Just in case anyone still cares about efficient rotation of SH functions.

- Robin Green.

Re: [Algorithms] solving for multiple matrices

[Andras B. <and...@gm...>]

Hah, you're right. I've never realized you could do that. While the  
operations required are indeed trivial, it doesn't seem like I can express  
it with simple matrix operations. So the code to do it will be somewhat  
non-trivial, but that's fine. Thanks!

Andras

On Sat, 19 Sep 2009 08:41:58 -0600, Staffan Langin
<sta...@ep...> wrote:

> Hi there Andras,
>
> C = Y^-1 * D * Y can be linearized by multiplying by Y (from the left).
>
> <=>
>
> Y*C=D*Y
>
> The equation-system, Y*C=D*Y, is trivial to reformulate to the more  
> common
> form Ax=b.
>
>
> Best regards,
>
> Staffan Langin
>
>
>
> -----Original Message-----
> From: Andras Balogh [mailto:and...@gm...]
> Sent: den 17 september 2009 21:38
> To: gda...@li...
> Subject: [Algorithms] solving for multiple matrices
>
> Hi, I have a chain of transformations with multiple unknown (but fixed!)
> transforms. What I do know is the end result transformation and some of
> the transformations in between, and I know these for multiple frames. So
>   from here, I'd like to compute the unknowns. Here it is in more formal
> version:
>
> I'd like to find 2 unknown matrices X and Y. I have 4 known matrices A1,
> A2, B1 and B2, and also know this:
> A1 = X * B1 * Y
> A2 = X * B2 * Y
>
> I can compute X from the first equation:
> X = A1 * Y^-1 * B1^-1
>
> And substitute it into the second:
> A2 = A1 * Y^-1 * B1^-1 * B2 * y
>
> Assigning:
> C = A1^-1 * A2
> D = B1^-1 * B2
>
> Then it becomes:
> C = Y^-1 * D * Y
>
> Now, how do I solve this for Y? This form lookes strangely familiar, but  
> I
> cannot figure out what to do from here (wish I knew how to Google this  
> ;).
> Hopefully there's an analytic solution to this. Any ideas?
>
> Thanks,
>
>
>
> Andras
>
> ----------------------------------------------------------------------------
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Re: [Algorithms] Unwrapping a cliff mesh.

[Jon W. <jw...@gm...>]

Marius Elvert wrote:
> the mesh is always vertically aligned. (i.e. the world "up" axis can
> always be mapped to V). Also, I don't mind if different, non-connected
> parts of the UV map are overlapping.  Basically, I just want to remove

For the non-cylindrical edges, how about you pick one triangle as 
"reference" and then unfold all the other triangles to that plane? Then 
map UV to the coordinates of vertices in that plane. As long as your 
geometry is heightmap-like (modulo the steep cliff feature) this should 
be possible.
For the cylinder shapes, though, you have to cut them at some point; 
they will not fully tile unless you're willing to sometimes 
significantly scale/distort the mapping. You might be able to get away 
with dual mappings, and ramp from one to the other using blending, if 
it's really important.

Sincerely,

jw


-- 

  Revenge is the most pointless and damaging of human desires.

[Algorithms] Unwrapping a cliff mesh.

[Marius E. <mar...@go...>]

Hi,
I have a terrain mesh generator that generates cliffs in the mesh,
very similar to the terrain used in Warcraft 3.

See the brown parts in this:
http://img178.imageshack.us/img178/8168/terrainwithcliff.jpg

Now I want to generate UV coordinates for using a tiled texture on it.
I figured that it might be a lot easier than generic unwrapping since
the mesh is always vertically aligned. (i.e. the world "up" axis can
always be mapped to V). Also, I don't mind if different, non-connected
parts of the UV map are overlapping.  Basically, I just want to remove
seams as good as possible from the mesh, while applying the tiled
texture.
Can anybody think of a good algorithm to do that? Possibly even one
that would generate no seams at all? (This should be possible since
the cliffs are, at least on one level, at most topologically equivalent to
a cylinder)
My only idea so far is to compute a "perfect" projection for each triangle
and stuff the edge conditions into a linear equations solver, possibly
with a finite-field for the "wrap-around" effect, but that seems like a
huge overkill, especially since I want to generate this at runtime.

Thanks,

Marius

Re: [Algorithms] solving for multiple matrices

[Staffan L. <sta...@ep...>]

Hi there Gino,

As you most probably already know, Cholesky decomposition can only be
applied to a symmetric, PSD matrix. Even if the matrices only contain
rotational transformations, I don't believe Y^T * D * Y generally would
result in a symmetric, PSD matrix. In particular I don't believe the
assertion C = U^T * U and D = V^T * V is generally true for matrices
containing rotational transformations, since that implies that C and D are
symmetric and PSD.


Best regards,

Staffan Langin

-----Original Message-----
From: Gino van den Bergen [mailto:gin...@gm...] 
Sent: den 18 september 2009 09:59
To: Game Development Algorithms
Subject: Re: [Algorithms] solving for multiple matrices

If Y can be considered a rotation then Y is orthogonal and thus Y^-1 = 
Y^T,  in which case this equation can be solved through Cholesky 
decomposition:

For C = Y^T * D * Y, let's decompose

C = U^T * U, and
D = V^T * V

then

U^T U = Y^T * V^T * V * Y
            = (V * Y) ^T * (V * Y)

this gives U = V * Y

and thus Y = V^-1 * U

Basically you are taking the square root of a matrix.

Hope this helps,

Gino 



Andras Balogh wrote:
> Both X and Y matrices represent a simple translation and rotation. For the

> case of the Y matrix, the translation part will likely to be very small,  
> so I could probably pretend it's only rotation.
>
> What I would really like though, is to find a solution, where I could use

> more than two equations, eg:
> A1 = X * B1 * Y
> A2 = X * B2 * Y
> A3 = X * B3 * Y
> ...
> An = X * Bn * Y
>
> And then compute a least squares solution from this over-constrained  
> system.
> BTW, when I said that I'm looking for an analytical solution, I just meant

> something that is not based on an iterative approach. As long as I can get

> to a part where I have to solve a large system of over-constrained linear

> equations, I'm home. Unfortunately, I don't know how to make this linear..
>
>
> Andras
>
>
>
>
> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:
>
>   
>> Andras Balogh wrote:
>>     
>>> Then it becomes:
>>> C = Y^-1 * D * Y
>>>
>>> Now, how do I solve this for Y? This form lookes strangely familiar,  
>>> but I
>>> cannot figure out what to do from here (wish I knew how to Google this  
>>> ;).
>>> Hopefully there's an analytic solution to this. Any ideas?
>>>
>>>
>>>       
>> That's the formula for applying a rotation in the reference frame of
>> another rotation.
>>
>> Do you know anything more about these matrices than that they are
>> matrices? Are they supposed to contain no scale? No translation? If you
>> can formulate them as quaternions, writing out the analytical answer is
>> a lot simpler :-)
>>
>> Sincerely,
>>
>> jw
>>
>>
>>
>>     
>
>
>
>
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Re: [Algorithms] solving for multiple matrices

[Staffan L. <sta...@ep...>]

Hi there Andras,

C = Y^-1 * D * Y can be linearized by multiplying by Y (from the left).

<=>

Y*C=D*Y

The equation-system, Y*C=D*Y, is trivial to reformulate to the more common
form Ax=b.


Best regards,

Staffan Langin



-----Original Message-----
From: Andras Balogh [mailto:and...@gm...] 
Sent: den 17 september 2009 21:38
To: gda...@li...
Subject: [Algorithms] solving for multiple matrices

Hi, I have a chain of transformations with multiple unknown (but fixed!)
transforms. What I do know is the end result transformation and some of
the transformations in between, and I know these for multiple frames. So
  from here, I'd like to compute the unknowns. Here it is in more formal
version:

I'd like to find 2 unknown matrices X and Y. I have 4 known matrices A1,
A2, B1 and B2, and also know this:
A1 = X * B1 * Y
A2 = X * B2 * Y

I can compute X from the first equation:
X = A1 * Y^-1 * B1^-1

And substitute it into the second:
A2 = A1 * Y^-1 * B1^-1 * B2 * y

Assigning:
C = A1^-1 * A2
D = B1^-1 * B2

Then it becomes:
C = Y^-1 * D * Y

Now, how do I solve this for Y? This form lookes strangely familiar, but I
cannot figure out what to do from here (wish I knew how to Google this ;).  
Hopefully there's an analytic solution to this. Any ideas?

Thanks,



Andras

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Re: [Algorithms] solving for multiple matrices

[Andras B. <and...@gm...>]

Well, I'm not sure how a different factorization will help, but I'll look  
into that. At the same time, I'm also starting to consider an iterative  
approach. What do you guys think? It seems to me that this is basically an  
inverse kinematics problem, right? Any pointers/advice, in case I'll  
decide to take that route?

Thanks,

Andras


On Fri, 18 Sep 2009 10:44:12 -0600, Bill Baxter <wb...@gm...> wrote:

> That was my first thought too, but you don't need to a *Cholesky*
> decomp on C and D.  Another kind of factorization will do, like a
> diagonalization --
> http://en.wikipedia.org/wiki/Square_root_of_a_matrix
>
> --bb
>
> On Fri, Sep 18, 2009 at 7:51 AM, Andras Balogh <and...@gm...>  
> wrote:
>> Hmm, I just looked up Cholesky decomposition, and it says you can only  
>> do
>> that if the matrix is symmetric and positive definite. Since the  
>> original
>> definition of C = A1^-1 * A2, where A1 and A2 both contain arbitrary
>> translation and rotation, I don't think C will be symmetric. Or am I
>> missing something?
>>
>> Andras
>>
>>
>> On Fri, 18 Sep 2009 01:58:41 -0600, Gino van den Bergen
>> <gin...@gm...> wrote:
>>
>>> If Y can be considered a rotation then Y is orthogonal and thus Y^-1 =
>>> Y^T,  in which case this equation can be solved through Cholesky
>>> decomposition:
>>>
>>> For C = Y^T * D * Y, let's decompose
>>>
>>> C = U^T * U, and
>>> D = V^T * V
>>>
>>> then
>>>
>>> U^T U = Y^T * V^T * V * Y
>>>             = (V * Y) ^T * (V * Y)
>>>
>>> this gives U = V * Y
>>>
>>> and thus Y = V^-1 * U
>>>
>>> Basically you are taking the square root of a matrix.
>>>
>>> Hope this helps,
>>>
>>> Gino
>>>
>>>
>>>
>>> Andras Balogh wrote:
>>>> Both X and Y matrices represent a simple translation and rotation. For
>>>> the
>>>> case of the Y matrix, the translation part will likely to be very  
>>>> small,
>>>> so I could probably pretend it's only rotation.
>>>>
>>>> What I would really like though, is to find a solution, where I could
>>>> use
>>>> more than two equations, eg:
>>>> A1 = X * B1 * Y
>>>> A2 = X * B2 * Y
>>>> A3 = X * B3 * Y
>>>> ...
>>>> An = X * Bn * Y
>>>>
>>>> And then compute a least squares solution from this over-constrained
>>>> system.
>>>> BTW, when I said that I'm looking for an analytical solution, I just
>>>> meant
>>>> something that is not based on an iterative approach. As long as I can
>>>> get
>>>> to a part where I have to solve a large system of over-constrained
>>>> linear
>>>> equations, I'm home. Unfortunately, I don't know how to make this
>>>> linear..
>>>>
>>>>
>>>> Andras
>>>>
>>>>
>>>>
>>>>
>>>> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...>  
>>>> wrote:
>>>>
>>>>
>>>>> Andras Balogh wrote:
>>>>>
>>>>>> Then it becomes:
>>>>>> C = Y^-1 * D * Y
>>>>>>
>>>>>> Now, how do I solve this for Y? This form lookes strangely familiar,
>>>>>> but I
>>>>>> cannot figure out what to do from here (wish I knew how to Google  
>>>>>> this
>>>>>> ;).
>>>>>> Hopefully there's an analytic solution to this. Any ideas?
>>>>>>
>>>>>>
>>>>>>
>>>>> That's the formula for applying a rotation in the reference frame of
>>>>> another rotation.
>>>>>
>>>>> Do you know anything more about these matrices than that they are
>>>>> matrices? Are they supposed to contain no scale? No translation? If  
>>>>> you
>>>>> can formulate them as quaternions, writing out the analytical answer  
>>>>> is
>>>>> a lot simpler :-)
>>>>>
>>>>> Sincerely,
>>>>>
>>>>> jw
>>>>>
>>>>>
>>>>>
>>>>>
>>>>
>>>>
>>>>
>>>> ------------------------------------------------------------------------------
>>>> Come build with us! The BlackBerry&reg; Developer Conference in SF, CA
>>>> is the only developer event you need to attend this year. Jumpstart  
>>>> your
>>>> developing skills, take BlackBerry mobile applications to market and
>>>> stay
>>>> ahead of the curve. Join us from November 9&#45;12, 2009. Register
>>>> now&#33;
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>>>> Archives:
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>>>>
>>>
>>>
>>> ------------------------------------------------------------------------------
>>> Come build with us! The BlackBerry&reg; Developer Conference in SF, CA
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>>> your
>>> developing skills, take BlackBerry mobile applications to market and  
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>>> ahead of the curve. Join us from November 9&#45;12, 2009. Register
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>>
>>
>>
>> ------------------------------------------------------------------------------
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>> ahead of the curve. Join us from November 9&#45;12, 2009. Register  
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>
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Re: [Algorithms] solving for multiple matrices

[Bill B. <wb...@gm...>]

That was my first thought too, but you don't need to a *Cholesky*
decomp on C and D.  Another kind of factorization will do, like a
diagonalization --
http://en.wikipedia.org/wiki/Square_root_of_a_matrix

--bb

On Fri, Sep 18, 2009 at 7:51 AM, Andras Balogh <and...@gm...> wrote:
> Hmm, I just looked up Cholesky decomposition, and it says you can only do
> that if the matrix is symmetric and positive definite. Since the original
> definition of C = A1^-1 * A2, where A1 and A2 both contain arbitrary
> translation and rotation, I don't think C will be symmetric. Or am I
> missing something?
>
> Andras
>
>
> On Fri, 18 Sep 2009 01:58:41 -0600, Gino van den Bergen
> <gin...@gm...> wrote:
>
>> If Y can be considered a rotation then Y is orthogonal and thus Y^-1 =
>> Y^T,  in which case this equation can be solved through Cholesky
>> decomposition:
>>
>> For C = Y^T * D * Y, let's decompose
>>
>> C = U^T * U, and
>> D = V^T * V
>>
>> then
>>
>> U^T U = Y^T * V^T * V * Y
>>             = (V * Y) ^T * (V * Y)
>>
>> this gives U = V * Y
>>
>> and thus Y = V^-1 * U
>>
>> Basically you are taking the square root of a matrix.
>>
>> Hope this helps,
>>
>> Gino
>>
>>
>>
>> Andras Balogh wrote:
>>> Both X and Y matrices represent a simple translation and rotation. For
>>> the
>>> case of the Y matrix, the translation part will likely to be very small,
>>> so I could probably pretend it's only rotation.
>>>
>>> What I would really like though, is to find a solution, where I could
>>> use
>>> more than two equations, eg:
>>> A1 = X * B1 * Y
>>> A2 = X * B2 * Y
>>> A3 = X * B3 * Y
>>> ...
>>> An = X * Bn * Y
>>>
>>> And then compute a least squares solution from this over-constrained
>>> system.
>>> BTW, when I said that I'm looking for an analytical solution, I just
>>> meant
>>> something that is not based on an iterative approach. As long as I can
>>> get
>>> to a part where I have to solve a large system of over-constrained
>>> linear
>>> equations, I'm home. Unfortunately, I don't know how to make this
>>> linear..
>>>
>>>
>>> Andras
>>>
>>>
>>>
>>>
>>> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:
>>>
>>>
>>>> Andras Balogh wrote:
>>>>
>>>>> Then it becomes:
>>>>> C = Y^-1 * D * Y
>>>>>
>>>>> Now, how do I solve this for Y? This form lookes strangely familiar,
>>>>> but I
>>>>> cannot figure out what to do from here (wish I knew how to Google this
>>>>> ;).
>>>>> Hopefully there's an analytic solution to this. Any ideas?
>>>>>
>>>>>
>>>>>
>>>> That's the formula for applying a rotation in the reference frame of
>>>> another rotation.
>>>>
>>>> Do you know anything more about these matrices than that they are
>>>> matrices? Are they supposed to contain no scale? No translation? If you
>>>> can formulate them as quaternions, writing out the analytical answer is
>>>> a lot simpler :-)
>>>>
>>>> Sincerely,
>>>>
>>>> jw
>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>>
>>> ------------------------------------------------------------------------------
>>> Come build with us! The BlackBerry&reg; Developer Conference in SF, CA
>>> is the only developer event you need to attend this year. Jumpstart your
>>> developing skills, take BlackBerry mobile applications to market and
>>> stay
>>> ahead of the curve. Join us from November 9&#45;12, 2009. Register
>>> now&#33;
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>>>
>>
>>
>> ------------------------------------------------------------------------------
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>> ahead of the curve. Join us from November 9&#45;12, 2009. Register
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>
>
>
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Re: [Algorithms] solving for multiple matrices

[Nguyen B. <ng...@gm...>]

You gotta be more precise about your definition of "transformation matrix" here.
Normally, we refer to them as "rotation matrix" which is a member of
SO(3) group. This type of group has been extensively studied and there
are many mathematical tools for you to use.

If you refer to them as homogeneous transformation which also contains
translation then they belong to the set SO(3)xR^3. I don't think you
can do a lot here.

Note that when talking about symmetry, people usually mean "rotation
matrix" as homogeneous matrix is not symmetry by definition.

--------------------------------------------------
Binh Nguyen
Computer Science Department
Rensselaer Polytechnic Institute
Troy, NY, 12180
--------------------------------------------------



On Fri, Sep 18, 2009 at 10:51 AM, Andras Balogh <and...@gm...> wrote:
> Hmm, I just looked up Cholesky decomposition, and it says you can only do
> that if the matrix is symmetric and positive definite. Since the original
> definition of C = A1^-1 * A2, where A1 and A2 both contain arbitrary
> translation and rotation, I don't think C will be symmetric. Or am I
> missing something?
>
> Andras
>
>
> On Fri, 18 Sep 2009 01:58:41 -0600, Gino van den Bergen
> <gin...@gm...> wrote:
>
>> If Y can be considered a rotation then Y is orthogonal and thus Y^-1 =
>> Y^T,  in which case this equation can be solved through Cholesky
>> decomposition:
>>
>> For C = Y^T * D * Y, let's decompose
>>
>> C = U^T * U, and
>> D = V^T * V
>>
>> then
>>
>> U^T U = Y^T * V^T * V * Y
>>             = (V * Y) ^T * (V * Y)
>>
>> this gives U = V * Y
>>
>> and thus Y = V^-1 * U
>>
>> Basically you are taking the square root of a matrix.
>>
>> Hope this helps,
>>
>> Gino
>>
>>
>>
>> Andras Balogh wrote:
>>> Both X and Y matrices represent a simple translation and rotation. For
>>> the
>>> case of the Y matrix, the translation part will likely to be very small,
>>> so I could probably pretend it's only rotation.
>>>
>>> What I would really like though, is to find a solution, where I could
>>> use
>>> more than two equations, eg:
>>> A1 = X * B1 * Y
>>> A2 = X * B2 * Y
>>> A3 = X * B3 * Y
>>> ...
>>> An = X * Bn * Y
>>>
>>> And then compute a least squares solution from this over-constrained
>>> system.
>>> BTW, when I said that I'm looking for an analytical solution, I just
>>> meant
>>> something that is not based on an iterative approach. As long as I can
>>> get
>>> to a part where I have to solve a large system of over-constrained
>>> linear
>>> equations, I'm home. Unfortunately, I don't know how to make this
>>> linear..
>>>
>>>
>>> Andras
>>>
>>>
>>>
>>>
>>> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:
>>>
>>>
>>>> Andras Balogh wrote:
>>>>
>>>>> Then it becomes:
>>>>> C = Y^-1 * D * Y
>>>>>
>>>>> Now, how do I solve this for Y? This form lookes strangely familiar,
>>>>> but I
>>>>> cannot figure out what to do from here (wish I knew how to Google this
>>>>> ;).
>>>>> Hopefully there's an analytic solution to this. Any ideas?
>>>>>
>>>>>
>>>>>
>>>> That's the formula for applying a rotation in the reference frame of
>>>> another rotation.
>>>>
>>>> Do you know anything more about these matrices than that they are
>>>> matrices? Are they supposed to contain no scale? No translation? If you
>>>> can formulate them as quaternions, writing out the analytical answer is
>>>> a lot simpler :-)
>>>>
>>>> Sincerely,
>>>>
>>>> jw
>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>>
>>> ------------------------------------------------------------------------------
>>> Come build with us! The BlackBerry&reg; Developer Conference in SF, CA
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>>
>>
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>
>
>
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Re: [Algorithms] solving for multiple matrices

[Andras B. <and...@gm...>]

Hmm, I just looked up Cholesky decomposition, and it says you can only do  
that if the matrix is symmetric and positive definite. Since the original  
definition of C = A1^-1 * A2, where A1 and A2 both contain arbitrary  
translation and rotation, I don't think C will be symmetric. Or am I  
missing something?

Andras


On Fri, 18 Sep 2009 01:58:41 -0600, Gino van den Bergen  
<gin...@gm...> wrote:

> If Y can be considered a rotation then Y is orthogonal and thus Y^-1 =
> Y^T,  in which case this equation can be solved through Cholesky
> decomposition:
>
> For C = Y^T * D * Y, let's decompose
>
> C = U^T * U, and
> D = V^T * V
>
> then
>
> U^T U = Y^T * V^T * V * Y
>             = (V * Y) ^T * (V * Y)
>
> this gives U = V * Y
>
> and thus Y = V^-1 * U
>
> Basically you are taking the square root of a matrix.
>
> Hope this helps,
>
> Gino
>
>
>
> Andras Balogh wrote:
>> Both X and Y matrices represent a simple translation and rotation. For  
>> the
>> case of the Y matrix, the translation part will likely to be very small,
>> so I could probably pretend it's only rotation.
>>
>> What I would really like though, is to find a solution, where I could  
>> use
>> more than two equations, eg:
>> A1 = X * B1 * Y
>> A2 = X * B2 * Y
>> A3 = X * B3 * Y
>> ...
>> An = X * Bn * Y
>>
>> And then compute a least squares solution from this over-constrained
>> system.
>> BTW, when I said that I'm looking for an analytical solution, I just  
>> meant
>> something that is not based on an iterative approach. As long as I can  
>> get
>> to a part where I have to solve a large system of over-constrained  
>> linear
>> equations, I'm home. Unfortunately, I don't know how to make this  
>> linear..
>>
>>
>> Andras
>>
>>
>>
>>
>> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:
>>
>>
>>> Andras Balogh wrote:
>>>
>>>> Then it becomes:
>>>> C = Y^-1 * D * Y
>>>>
>>>> Now, how do I solve this for Y? This form lookes strangely familiar,
>>>> but I
>>>> cannot figure out what to do from here (wish I knew how to Google this
>>>> ;).
>>>> Hopefully there's an analytic solution to this. Any ideas?
>>>>
>>>>
>>>>
>>> That's the formula for applying a rotation in the reference frame of
>>> another rotation.
>>>
>>> Do you know anything more about these matrices than that they are
>>> matrices? Are they supposed to contain no scale? No translation? If you
>>> can formulate them as quaternions, writing out the analytical answer is
>>> a lot simpler :-)
>>>
>>> Sincerely,
>>>
>>> jw
>>>
>>>
>>>
>>>
>>
>>
>>
>> ------------------------------------------------------------------------------
>> Come build with us! The BlackBerry&reg; Developer Conference in SF, CA
>> is the only developer event you need to attend this year. Jumpstart your
>> developing skills, take BlackBerry mobile applications to market and  
>> stay
>> ahead of the curve. Join us from November 9&#45;12, 2009. Register  
>> now&#33;
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>> _______________________________________________
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>> Archives:
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>>
>
>
> ------------------------------------------------------------------------------
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> is the only developer event you need to attend this year. Jumpstart your
> developing skills, take BlackBerry mobile applications to market and stay
> ahead of the curve. Join us from November 9&#45;12, 2009. Register  
> now&#33;
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Re: [Algorithms] solving for multiple matrices

[Gino v. d. B. <gin...@gm...>]

If Y can be considered a rotation then Y is orthogonal and thus Y^-1 = 
Y^T,  in which case this equation can be solved through Cholesky 
decomposition:

For C = Y^T * D * Y, let's decompose

C = U^T * U, and
D = V^T * V

then

U^T U = Y^T * V^T * V * Y
            = (V * Y) ^T * (V * Y)

this gives U = V * Y

and thus Y = V^-1 * U

Basically you are taking the square root of a matrix.

Hope this helps,

Gino 



Andras Balogh wrote:
> Both X and Y matrices represent a simple translation and rotation. For the  
> case of the Y matrix, the translation part will likely to be very small,  
> so I could probably pretend it's only rotation.
>
> What I would really like though, is to find a solution, where I could use  
> more than two equations, eg:
> A1 = X * B1 * Y
> A2 = X * B2 * Y
> A3 = X * B3 * Y
> ...
> An = X * Bn * Y
>
> And then compute a least squares solution from this over-constrained  
> system.
> BTW, when I said that I'm looking for an analytical solution, I just meant  
> something that is not based on an iterative approach. As long as I can get  
> to a part where I have to solve a large system of over-constrained linear  
> equations, I'm home. Unfortunately, I don't know how to make this linear..
>
>
> Andras
>
>
>
>
> On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:
>
>   
>> Andras Balogh wrote:
>>     
>>> Then it becomes:
>>> C = Y^-1 * D * Y
>>>
>>> Now, how do I solve this for Y? This form lookes strangely familiar,  
>>> but I
>>> cannot figure out what to do from here (wish I knew how to Google this  
>>> ;).
>>> Hopefully there's an analytic solution to this. Any ideas?
>>>
>>>
>>>       
>> That's the formula for applying a rotation in the reference frame of
>> another rotation.
>>
>> Do you know anything more about these matrices than that they are
>> matrices? Are they supposed to contain no scale? No translation? If you
>> can formulate them as quaternions, writing out the analytical answer is
>> a lot simpler :-)
>>
>> Sincerely,
>>
>> jw
>>
>>
>>
>>     
>
>
>
> ------------------------------------------------------------------------------
> Come build with us! The BlackBerry&reg; Developer Conference in SF, CA
> is the only developer event you need to attend this year. Jumpstart your
> developing skills, take BlackBerry mobile applications to market and stay 
> ahead of the curve. Join us from November 9&#45;12, 2009. Register now&#33;
> http://p.sf.net/sfu/devconf
> _______________________________________________
> GDAlgorithms-list mailing list
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> https://lists.sourceforge.net/lists/listinfo/gdalgorithms-list
> Archives:
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>   

Re: [Algorithms] solving for multiple matrices

[Andras B. <and...@gm...>]

Both X and Y matrices represent a simple translation and rotation. For the  
case of the Y matrix, the translation part will likely to be very small,  
so I could probably pretend it's only rotation.

What I would really like though, is to find a solution, where I could use  
more than two equations, eg:
A1 = X * B1 * Y
A2 = X * B2 * Y
A3 = X * B3 * Y
...
An = X * Bn * Y

And then compute a least squares solution from this over-constrained  
system.
BTW, when I said that I'm looking for an analytical solution, I just meant  
something that is not based on an iterative approach. As long as I can get  
to a part where I have to solve a large system of over-constrained linear  
equations, I'm home. Unfortunately, I don't know how to make this linear..


Andras




On Thu, 17 Sep 2009 16:32:25 -0600, Jon Watte <jw...@gm...> wrote:

> Andras Balogh wrote:
>>
>> Then it becomes:
>> C = Y^-1 * D * Y
>>
>> Now, how do I solve this for Y? This form lookes strangely familiar,  
>> but I
>> cannot figure out what to do from here (wish I knew how to Google this  
>> ;).
>> Hopefully there's an analytic solution to this. Any ideas?
>>
>>
>
> That's the formula for applying a rotation in the reference frame of
> another rotation.
>
> Do you know anything more about these matrices than that they are
> matrices? Are they supposed to contain no scale? No translation? If you
> can formulate them as quaternions, writing out the analytical answer is
> a lot simpler :-)
>
> Sincerely,
>
> jw
>
>
>

Re: [Algorithms] solving for multiple matrices

[Bill B. <wb...@gm...>]

> -----Original Message-----
> From: Andras Balogh [mailto:and...@gm...]
> Sent: Thursday, September 17, 2009 12:38 PM
> To: gda...@li...
> Subject: [Algorithms] solving for multiple matrices
>
> Hi, I have a chain of transformations with multiple unknown (but fixed!) transforms. What I do know is the end result transformation and some of the transformations in between, and I know these for multiple frames. So
>  from here, I'd like to compute the unknowns. Here it is in more formal
> version:
>
> I'd like to find 2 unknown matrices X and Y. I have 4 known matrices A1, A2, B1 and B2, and also know this:
> A1 = X * B1 * Y
> A2 = X * B2 * Y
>
> I can compute X from the first equation:
> X = A1 * Y^-1 * B1^-1
>
> And substitute it into the second:
> A2 = A1 * Y^-1 * B1^-1 * B2 * y
>
> Assigning:
> C = A1^-1 * A2
> D = B1^-1 * B2
>
> Then it becomes:
> C = Y^-1 * D * Y
>
> Now, how do I solve this for Y? This form lookes strangely familiar, but I cannot figure out what to do from here (wish I knew how to Google this ;).
> Hopefully there's an analytic solution to this. Any ideas?
>
> Thanks,
>
>
>
> Andras

It has the form of a similarity transformation, if that helps.
http://mathworld.wolfram.com/SimilarityTransformation.html

Similarity transforms preserve eigenvalues and consequently the trace
and determinant of the matrix applied to, so if C and D don't have the
same trace and determinant, then your Y doesn't exist.

--bb

Re: [Algorithms] solving for multiple matrices

[Jon W. <jw...@gm...>]

Andras Balogh wrote:
>
> Then it becomes:
> C = Y^-1 * D * Y
>
> Now, how do I solve this for Y? This form lookes strangely familiar, but I
> cannot figure out what to do from here (wish I knew how to Google this ;).  
> Hopefully there's an analytic solution to this. Any ideas?
>
>   

That's the formula for applying a rotation in the reference frame of 
another rotation.

Do you know anything more about these matrices than that they are 
matrices? Are they supposed to contain no scale? No translation? If you 
can formulate them as quaternions, writing out the analytical answer is 
a lot simpler :-)

Sincerely,

jw



-- 

  Revenge is the most pointless and damaging of human desires.

Re: [Algorithms] solving for multiple matrices

[David B. <dbe...@na...>]

This is just a shot in the dark, but the wikipedia entry on "QZ decomposition" looks a lot like your original statement of the problem --
http://en.wikipedia.org/wiki/Matrix_decomposition#QZ_decomposition

I don't know if X & Y are "unitary" matrices though.  I guess that depends on which kinds of transforms you are using.

Hope that helps,

David Bennett
NAMCO BANDAI Games America Inc.


-----Original Message-----
From: Andras Balogh [mailto:and...@gm...]
Sent: Thursday, September 17, 2009 12:38 PM
To: gda...@li...
Subject: [Algorithms] solving for multiple matrices

Hi, I have a chain of transformations with multiple unknown (but fixed!) transforms. What I do know is the end result transformation and some of the transformations in between, and I know these for multiple frames. So
  from here, I'd like to compute the unknowns. Here it is in more formal
version:

I'd like to find 2 unknown matrices X and Y. I have 4 known matrices A1, A2, B1 and B2, and also know this:
A1 = X * B1 * Y
A2 = X * B2 * Y

I can compute X from the first equation:
X = A1 * Y^-1 * B1^-1

And substitute it into the second:
A2 = A1 * Y^-1 * B1^-1 * B2 * y

Assigning:
C = A1^-1 * A2
D = B1^-1 * B2

Then it becomes:
C = Y^-1 * D * Y

Now, how do I solve this for Y? This form lookes strangely familiar, but I cannot figure out what to do from here (wish I knew how to Google this ;).
Hopefully there's an analytic solution to this. Any ideas?

Thanks,



Andras

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[Algorithms] solving for multiple matrices

[Andras B. <and...@gm...>]

Hi, I have a chain of transformations with multiple unknown (but fixed!)
transforms. What I do know is the end result transformation and some of
the transformations in between, and I know these for multiple frames. So
  from here, I'd like to compute the unknowns. Here it is in more formal
version:

I'd like to find 2 unknown matrices X and Y. I have 4 known matrices A1,
A2, B1 and B2, and also know this:
A1 = X * B1 * Y
A2 = X * B2 * Y

I can compute X from the first equation:
X = A1 * Y^-1 * B1^-1

And substitute it into the second:
A2 = A1 * Y^-1 * B1^-1 * B2 * y

Assigning:
C = A1^-1 * A2
D = B1^-1 * B2

Then it becomes:
C = Y^-1 * D * Y

Now, how do I solve this for Y? This form lookes strangely familiar, but I
cannot figure out what to do from here (wish I knew how to Google this ;).  
Hopefully there's an analytic solution to this. Any ideas?

Thanks,



Andras

Re: [Algorithms] mouse smoothing

[James R. <ja...@fu...>]

Ah yes, sorry.  I meant DirectInput et al.


Emil Persson wrote:
> Actually, Raw Input is what Microsoft recommends for games (except for
> controlling a cursor). I don't know if you confuse Raw Input with some
> legacy API, but this is the relatively new API (since Windows XP) for
> obtaining high resolution input. It's supposed to replace semi-deprecated
> APIs such as DirectInput that are now implemented through Raw Input anyway.
>
> -Emil
>
>
> -----Original Message-----
> From: James Robertson [mailto:ja...@fu...] 
> Sent: 14 September 2009 09:03
> To: Game Development Algorithms
> Subject: Re: [Algorithms] mouse smoothing
>
> Err, you really shouldn't be using raw input any more.  You should be 
> obtaining mouse data through the Windows messages sent to your 
> application/window.  More importantly you shouldn't attempt to by pass 
> any acceleration that Windows provides per the users settings.  (There's 
> nothing worse than trying to play a game with a mouse cursor that crawls 
> across the screen like a snail with a broken leg.)
>
>
> Tibor Klajnscek wrote:
>   
>> I haven't seen anyone mention this, but if you're using raw input 
>> (which is most likely the case), you're bypassing all windows' 
>> sensitivity and acceleration functions already.
>>
>> Cheers,
>> Tibor
>>
>> On 13.9.2009 8:31, Jeff Russell wrote:
>>     
>>> This thread has been totally worth it, just for adding "delta mickey" 
>>> to my vocabulary. Bypassing windows mouse acceleration curves might 
>>> be of interest to developers, actually, I had forgotten about that.
>>> ------------------------------------------------------------------------
>>>
>>>
>>>       
> ----------------------------------------------------------------------------
> --
>   
>>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>>>       
> 30-Day 
>   
>>> trial. Simplify your report design, integration and deployment - and
>>>       
> focus on 
>   
>>> what you do best, core application coding. Discover what's new with 
>>> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>>> ------------------------------------------------------------------------
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>>> Archives:
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>>>       
>> ------------------------------------------------------------------------
>>
>>
>>     
> ----------------------------------------------------------------------------
> --
>   
>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>>     
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>   
>> trial. Simplify your report design, integration and deployment - and focus
>>     
> on 
>   
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>
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>   

Re: [Algorithms] mouse smoothing

[Emil P. <hu...@co...>]

Actually, Raw Input is what Microsoft recommends for games (except for
controlling a cursor). I don't know if you confuse Raw Input with some
legacy API, but this is the relatively new API (since Windows XP) for
obtaining high resolution input. It's supposed to replace semi-deprecated
APIs such as DirectInput that are now implemented through Raw Input anyway.

-Emil


-----Original Message-----
From: James Robertson [mailto:ja...@fu...] 
Sent: 14 September 2009 09:03
To: Game Development Algorithms
Subject: Re: [Algorithms] mouse smoothing

Err, you really shouldn't be using raw input any more.  You should be 
obtaining mouse data through the Windows messages sent to your 
application/window.  More importantly you shouldn't attempt to by pass 
any acceleration that Windows provides per the users settings.  (There's 
nothing worse than trying to play a game with a mouse cursor that crawls 
across the screen like a snail with a broken leg.)


Tibor Klajnscek wrote:
> I haven't seen anyone mention this, but if you're using raw input 
> (which is most likely the case), you're bypassing all windows' 
> sensitivity and acceleration functions already.
>
> Cheers,
> Tibor
>
> On 13.9.2009 8:31, Jeff Russell wrote:
>> This thread has been totally worth it, just for adding "delta mickey" 
>> to my vocabulary. Bypassing windows mouse acceleration curves might 
>> be of interest to developers, actually, I had forgotten about that.
>> ------------------------------------------------------------------------
>>
>>
----------------------------------------------------------------------------
--
>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
30-Day 
>> trial. Simplify your report design, integration and deployment - and
focus on 
>> what you do best, core application coding. Discover what's new with 
>> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> ------------------------------------------------------------------------
>>
>> _______________________________________________
>> GDAlgorithms-list mailing list
>> GDA...@li...
>> https://lists.sourceforge.net/lists/listinfo/gdalgorithms-list
>> Archives:
>> http://sourceforge.net/mailarchive/forum.php?forum_name=gdalgorithms-list
> ------------------------------------------------------------------------
>
>
----------------------------------------------------------------------------
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30-Day 
> trial. Simplify your report design, integration and deployment - and focus
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Re: [Algorithms] mouse smoothing

[Alen L. <ale...@cr...>]

Monday, September 14, 2009, 8:04:49 PM, Tibor wrote:
> I see no reason why we shouldn't use it as there's nothing worse
> than UI cursor acceleration and sensitivity being applied to player
> view/rotation input...

De gustibus non est disputandum. IME, it is best left as an option for
the user.

Alen

Re: [Algorithms] Kinematic Collision

[metanet s. <met...@ya...>]

> (Cue: someone will now insert a reference to a fighter that
> does just that, and I just wasn't aware

Probably you're aware of these -- not perfect, but steps in that general direction:

http://www.fun-motion.com/physics-games/toribash/
http://www.fun-motion.com/physics-games/sumotori-dreams/

AFAIK the recent Fight Night demo I tried was doing some driven-ragdoll stuff, in that when your punch is blocked/deflected your arm seems to move aside in a more realistic way. But I don't think it feeds into the character's position though, it's just one-way/graphical :(

raigan




--- On Mon, 9/14/09, Jon Watte <jw...@gm...> wrote:

> From: Jon Watte <jw...@gm...>
> Subject: Re: [Algorithms] Kinematic Collision
> To: "Game Development Algorithms" <gda...@li...>
> Received: Monday, September 14, 2009, 11:16 AM
> Pierre Terdiman wrote:
> >
> > I think people just make sure the bounding volume is
> big enough to 
> > enclose the whole punch motion.
> 
> That means that the bounding volume expands as part of the
> animation. 
> What do you do when the bounding volume pushes into the
> mountain? 
> Sliding back the entire rest of the character is not ideal
> -- especially 
> if something equally immovable is behind.
> Or the bounding volume of the longest animation determines
> where the 
> character can walk, which means you can't pass through
> doors narrower 
> than 5 meters...
> 
> >
> > Same here. I wonder what's the most advanced thing
> done in games in 
> > that respect, or if all of this is really needed (I
> had the impression 
> > it wasn't, so far)
> 
> If we have someone from Tekken or DoA here, that'd be nice!
> Personally, 
> I find that the foot-backwards-sliding that traditionally
> happens when 
> you punch into blocks, for example, is quite disconcerting.
> I believe 
> that if someone could built a responsive fighter with
> integrated ragdoll 
> physics and full collision detection, it could have the
> opportunity to 
> lift the genre to the next level.
> 
> (Cue: someone will now insert a reference to a fighter that
> does just 
> that, and I just wasn't aware, as I come at this from 
> maker-of-characters and player-of-fighters but not as
> maker-of-fighters)
> 
> Sincerely,
> 
> jw
> 
> 
> 
> -- 
> 
>   Revenge is the most pointless and damaging of human
> desires.
> 
> 
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Re: [Algorithms] mouse smoothing

[Tibor K. <tib...@gm...>]

We're actually using the windows cursor as our cursor (or only it's 
position and rendering our own - forgot exact details), but for the game 
input (delta mickeys, :) ) we're using raw input via WM_INPUT and it 
works great for us. I see no reason why we shouldn't use it as there's 
nothing worse than UI cursor acceleration and sensitivity being applied 
to player view/rotation input...

On 14.9.2009 9:03, James Robertson wrote:
> Err, you really shouldn't be using raw input any more.  You should be
> obtaining mouse data through the Windows messages sent to your
> application/window.  More importantly you shouldn't attempt to by pass
> any acceleration that Windows provides per the users settings.  (There's
> nothing worse than trying to play a game with a mouse cursor that crawls
> across the screen like a snail with a broken leg.)
>
>
> Tibor Klajnscek wrote:
>    
>> I haven't seen anyone mention this, but if you're using raw input
>> (which is most likely the case), you're bypassing all windows'
>> sensitivity and acceleration functions already.
>>
>> Cheers,
>> Tibor
>>
>> On 13.9.2009 8:31, Jeff Russell wrote:
>>      
>>> This thread has been totally worth it, just for adding "delta mickey"
>>> to my vocabulary. Bypassing windows mouse acceleration curves might
>>> be of interest to developers, actually, I had forgotten about that.
>>> ------------------------------------------------------------------------
>>>
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Re: [Algorithms] Kinematic Collision

[Jon W. <jw...@gm...>]

pontus birgersson wrote:
> > Physics. Because you wanted to include a physics engine. If you don't
> > use a physics engine, you don't have that option.
>
> Your reasoning seems to be from the perspective of reaction to an 
> attack e.g. one character punches another character transfers a force 
> upon that character which moves him and when doing that the 
> appropriate frames of some animation will render. This makes perfect 
> sense and is definetly an interesting solution to try out.
>

No, my reasoning is from the attacker's point of view: Attack -> intent 
-> physics -> movement -> animation.

> However I'm more concerned with movement of the character that is 
> throwing the punch, because the punch may be extra ordinary and in 
> order to look good the character needs to be translated in a precise 
> and non-linear fashion. When this is the case, the force based move is 
> really hard to get exact enough because of it's smooth nature.
>

Agreed -- a hard case. You could key the animation on movement of some 
bone that you specify (the fist, say) -- or just do what you said: drive 
a ragdoll using animated forces, and make sure that the integrator 
and/or kinematic controller makes it look good.


Sincerely,

jw


-- 

  Revenge is the most pointless and damaging of human desires.

Re: [Algorithms] Kinematic Collision

[pontus b. <her...@ho...>]

> > > However, in general, you'll want the animation to be driven by the
> > > movement, and not the other way around. Each frame, you determine how
> > > far you've traveled since the last frame, and forward the animation an
> > > appropriate amount to avoid foot sliding. (The animation is marked up,
> > > or you measure this based on certain named bones).
> 
> > Not sure exactly how this would benefit a fighting move though. Where 
> > would the initial movement come from if not from the animation?
> >
> 
> Physics. Because you wanted to include a physics engine. If you don't 
> use a physics engine, you don't have that option.

Your reasoning seems to be from the perspective of reaction to an attack e.g. one character  punches another character transfers a force upon that character which moves him and when doing that the appropriate frames of some animation will render. This makes perfect sense and is definetly an interesting solution to try out. 

However I'm more concerned with movement of the character that is throwing the punch, because the punch may be extra ordinary and in order to look good the character needs to be translated in a precise and non-linear fashion. When this is the case, the force based move is really hard to get exact enough because of it's smooth nature. 

The only solution I can see to this is to either not allow this kind of movement or doing it by predefined translation, in which case I seem to be loosing physics support for the character. But maybe the implicit integration method could be good for this...as you said it would be good to find out so I guess I have my next home project all layed out before me:)

> Usually, it's the case that quality has to cost, because there is much 
> less competition the more specialization and higher quality you want. 
> The cheap generic crap is just that, cheap. That doesn't mean that 
> everything that's expensive is good, of course, but it does mean that 
> it's generally hard to be consistently good without being expensive.
> 
> Middleware seems to start at the $250 range (old Torque, C4 engine, 
> etc), take a leap to the $5k-$10k+ range (Granny, FMOD, etc), take 
> another leap to the $50k range (physics packages etc), and then have a 
> top strata at $150k+ (full engines, scene graphs, integrated editors, 
> etc). Within that range, I wouldn't think endorphin is entirely out of 
> reason.
Oh don't get me wrong it's pricey but rightfully so, I haven't worked with it but it seems to be a remarkable product.

Pontus

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Re: [Algorithms] Kinematic Collision

[Jon W. <jw...@gm...>]

Pierre Terdiman wrote:
>
> I think people just make sure the bounding volume is big enough to 
> enclose the whole punch motion.

That means that the bounding volume expands as part of the animation. 
What do you do when the bounding volume pushes into the mountain? 
Sliding back the entire rest of the character is not ideal -- especially 
if something equally immovable is behind.
Or the bounding volume of the longest animation determines where the 
character can walk, which means you can't pass through doors narrower 
than 5 meters...

>
> Same here. I wonder what's the most advanced thing done in games in 
> that respect, or if all of this is really needed (I had the impression 
> it wasn't, so far)

If we have someone from Tekken or DoA here, that'd be nice! Personally, 
I find that the foot-backwards-sliding that traditionally happens when 
you punch into blocks, for example, is quite disconcerting. I believe 
that if someone could built a responsive fighter with integrated ragdoll 
physics and full collision detection, it could have the opportunity to 
lift the genre to the next level.

(Cue: someone will now insert a reference to a fighter that does just 
that, and I just wasn't aware, as I come at this from 
maker-of-characters and player-of-fighters but not as maker-of-fighters)

Sincerely,

jw



-- 

  Revenge is the most pointless and damaging of human desires.