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Thread Subject:
I am trying to solve an impossible problem?

Subject: I am trying to solve an impossible problem?

From: Paulo

Date: 22 Oct, 2010 10:41:03

Message: 1 of 6

Hi everyone.

I have been around a problem that I cannot figure out a solution (if there is one) which is related with sphere orientations and rotations. I already searched in many places, including here in this Newsgroup but without success.

Let me introduce my problem. Given a black sphere (binary image so black sphere and white background) I need to define a orthogonal 3D axis on it. At a first glance seems easy, we can define the 3D axis in the middle of the sphere because it is a well known point. The problem rises when I need to rotate the sphere, because the previous 3D axis must rotate accordingly with the rotation of the sphere. So basically, I'm trying to find the sphere (shape) orientation along rotation movements. In the book that I'm reading, Volumetric Image Analysis, this problem is called orientation ambiguities and it is a section in the middle of Moments chapter, but unfortunately the book goes just till the ellipsoids, resolving problems of 2 ambiguities. The sphere case has 3 ambiguities because the shape of it cannot define any of the axis implicitly.

I thought in the following: with the point in the middle of the sphere, I make a plan that contains that point and with this plan, I can have the normal vector. With this normal vector I can project it on the plane giving another vector orthogonal with the previous one. To find the 3rd vector I just need to calculate the product between these vectors. But when the sphere is rotated I don't know how to define again the plane to give the same vectors.

Could you please help me?

Subject: I am trying to solve an impossible problem?

From: Cris Luengo

Date: 22 Oct, 2010 12:48:03

Message: 2 of 6

"Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9rpnv$l5q$1@fred.mathworks.com>...
> Hi everyone.
>
> I have been around a problem that I cannot figure out a solution (if there is one) which is related with sphere orientations and rotations. I already searched in many places, including here in this Newsgroup but without success.
>
> Let me introduce my problem. Given a black sphere (binary image so black sphere and white background) I need to define a orthogonal 3D axis on it. At a first glance seems easy, we can define the 3D axis in the middle of the sphere because it is a well known point. The problem rises when I need to rotate the sphere, because the previous 3D axis must rotate accordingly with the rotation of the sphere. So basically, I'm trying to find the sphere (shape) orientation along rotation movements. In the book that I'm reading, Volumetric Image Analysis, this problem is called orientation ambiguities and it is a section in the middle of Moments chapter, but unfortunately the book goes just till the ellipsoids, resolving problems of 2 ambiguities. The sphere case has 3 ambiguities because the shape of it cannot define any of the axis implicitly.
>
> I thought in the following: with the point in the middle of the sphere, I make a plan that contains that point and with this plan, I can have the normal vector. With this normal vector I can project it on the plane giving another vector orthogonal with the previous one. To find the 3rd vector I just need to calculate the product between these vectors. But when the sphere is rotated I don't know how to define again the plane to give the same vectors.
>
> Could you please help me?

I don't really understand your problem. A sphere is rotationally symmetric. It doesn't change when you rotate it. You can define a random set of orthogonal axes in the middle of the sphere, it will always be the "correct" axes.

Cris.

Subject: I am trying to solve an impossible problem?

From: Paulo

Date: 22 Oct, 2010 13:39:03

Message: 3 of 6

"Cris Luengo" <cris.luengo@google.for.my.name.to.contact.me> wrote in message <i9s163$dlc$1@fred.mathworks.com>...
> "Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9rpnv$l5q$1@fred.mathworks.com>...
> > Hi everyone.
> >
> > I have been around a problem that I cannot figure out a solution (if there is one) which is related with sphere orientations and rotations. I already searched in many places, including here in this Newsgroup but without success.
> >
> > Let me introduce my problem. Given a black sphere (binary image so black sphere and white background) I need to define a orthogonal 3D axis on it. At a first glance seems easy, we can define the 3D axis in the middle of the sphere because it is a well known point. The problem rises when I need to rotate the sphere, because the previous 3D axis must rotate accordingly with the rotation of the sphere. So basically, I'm trying to find the sphere (shape) orientation along rotation movements. In the book that I'm reading, Volumetric Image Analysis, this problem is called orientation ambiguities and it is a section in the middle of Moments chapter, but unfortunately the book goes just till the ellipsoids, resolving problems of 2 ambiguities. The sphere case has 3 ambiguities because the shape of it cannot define any of the axis implicitly.
> >
> > I thought in the following: with the point in the middle of the sphere, I make a plan that contains that point and with this plan, I can have the normal vector. With this normal vector I can project it on the plane giving another vector orthogonal with the previous one. To find the 3rd vector I just need to calculate the product between these vectors. But when the sphere is rotated I don't know how to define again the plane to give the same vectors.
> >
> > Could you please help me?
>
> I don't really understand your problem. A sphere is rotationally symmetric. It doesn't change when you rotate it. You can define a random set of orthogonal axes in the middle of the sphere, it will always be the "correct" axes.
>
> Cris.

Hi Cris. The problem is, imagine that you have a south pole and you need to identify it. At the first time, you set this pole and an axis pointing it. If you rotate this image you lose the orientation of the sphere and consequently where is the south pole. This is an example.

I hope that now it is more clear for everyone.

Thank you.

Subject: I am trying to solve an impossible problem?

From: Matt J

Date: 22 Oct, 2010 14:11:04

Message: 4 of 6

"Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9s45n$qdi$1@fred.mathworks.com>...

> Hi Cris. The problem is, imagine that you have a south pole and you need to identify it. At the first time, you set this pole and an axis pointing it. If you rotate this image you lose the orientation of the sphere and consequently where is the south pole. ========

The solution is to not throw away the rotation parameters. For example if you express the rotation as a 3x3 rotation matrix R, the columns of R will always be your rotated axes. You can then use these axes to find the south pole.

Subject: I am trying to solve an impossible problem?

From: Cris Luengo

Date: 22 Oct, 2010 14:19:03

Message: 5 of 6

"Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9s45n$qdi$1@fred.mathworks.com>...
> "Cris Luengo" <cris.luengo@google.for.my.name.to.contact.me> wrote in message <i9s163$dlc$1@fred.mathworks.com>...
> > "Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9rpnv$l5q$1@fred.mathworks.com>...
> > > Hi everyone.
> > >
> > > I have been around a problem that I cannot figure out a solution (if there is one) which is related with sphere orientations and rotations. I already searched in many places, including here in this Newsgroup but without success.
> > >
> > > Let me introduce my problem. Given a black sphere (binary image so black sphere and white background) I need to define a orthogonal 3D axis on it. At a first glance seems easy, we can define the 3D axis in the middle of the sphere because it is a well known point. The problem rises when I need to rotate the sphere, because the previous 3D axis must rotate accordingly with the rotation of the sphere. So basically, I'm trying to find the sphere (shape) orientation along rotation movements. In the book that I'm reading, Volumetric Image Analysis, this problem is called orientation ambiguities and it is a section in the middle of Moments chapter, but unfortunately the book goes just till the ellipsoids, resolving problems of 2 ambiguities. The sphere case has 3 ambiguities because the shape of it cannot define any of the axis implicitly.
> > >
> > > I thought in the following: with the point in the middle of the sphere, I make a plan that contains that point and with this plan, I can have the normal vector. With this normal vector I can project it on the plane giving another vector orthogonal with the previous one. To find the 3rd vector I just need to calculate the product between these vectors. But when the sphere is rotated I don't know how to define again the plane to give the same vectors.
> > >
> > > Could you please help me?
> >
> > I don't really understand your problem. A sphere is rotationally symmetric. It doesn't change when you rotate it. You can define a random set of orthogonal axes in the middle of the sphere, it will always be the "correct" axes.
> >
> > Cris.
>
> Hi Cris. The problem is, imagine that you have a south pole and you need to identify it. At the first time, you set this pole and an axis pointing it. If you rotate this image you lose the orientation of the sphere and consequently where is the south pole. This is an example.
>
> I hope that now it is more clear for everyone.
>
> Thank you.

If there is a South pole, it's no longer just a sphere: it's a sphere with a pole!

No, serious, if there is something drawn on the surface of the sphere, you can use this drawing to determine the orientation of the sphere. If there's a magnetic field, you can use that to determine the orientation of the planet. If it's just a sphere, a plain sphere, there's no point in arbitrarily assigning one point on it as the South pole, and expect to be able to find it back afterward. You can point at any point on the sphere and claim that it's the same point, nobody will know the difference. If there is a difference, use that to identify your pole.

On the other hand, if this is just a bookkeeping exercise, that is, you have an image with a sphere and a vector pointing to the South pole, and you want to rotate the image, you can rotate the vector at the same time, thereby "remembering" where that point was.

Cheers,
Cris.

Subject: I am trying to solve an impossible problem?

From: Paulo

Date: 22 Oct, 2010 15:05:05

Message: 6 of 6

"Cris Luengo" <cris.luengo@google.for.my.name.to.contact.me> wrote in message <i9s6gn$eh$1@fred.mathworks.com>...
> "Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9s45n$qdi$1@fred.mathworks.com>...
> > "Cris Luengo" <cris.luengo@google.for.my.name.to.contact.me> wrote in message <i9s163$dlc$1@fred.mathworks.com>...
> > > "Paulo " <paulofreitas7@portugalmail.pt> wrote in message <i9rpnv$l5q$1@fred.mathworks.com>...
> > > > Hi everyone.
> > > >
> > > > I have been around a problem that I cannot figure out a solution (if there is one) which is related with sphere orientations and rotations. I already searched in many places, including here in this Newsgroup but without success.
> > > >
> > > > Let me introduce my problem. Given a black sphere (binary image so black sphere and white background) I need to define a orthogonal 3D axis on it. At a first glance seems easy, we can define the 3D axis in the middle of the sphere because it is a well known point. The problem rises when I need to rotate the sphere, because the previous 3D axis must rotate accordingly with the rotation of the sphere. So basically, I'm trying to find the sphere (shape) orientation along rotation movements. In the book that I'm reading, Volumetric Image Analysis, this problem is called orientation ambiguities and it is a section in the middle of Moments chapter, but unfortunately the book goes just till the ellipsoids, resolving problems of 2 ambiguities. The sphere case has 3 ambiguities because the shape of it cannot define any of the axis implicitly.
> > > >
> > > > I thought in the following: with the point in the middle of the sphere, I make a plan that contains that point and with this plan, I can have the normal vector. With this normal vector I can project it on the plane giving another vector orthogonal with the previous one. To find the 3rd vector I just need to calculate the product between these vectors. But when the sphere is rotated I don't know how to define again the plane to give the same vectors.
> > > >
> > > > Could you please help me?
> > >
> > > I don't really understand your problem. A sphere is rotationally symmetric. It doesn't change when you rotate it. You can define a random set of orthogonal axes in the middle of the sphere, it will always be the "correct" axes.
> > >
> > > Cris.
> >
> > Hi Cris. The problem is, imagine that you have a south pole and you need to identify it. At the first time, you set this pole and an axis pointing it. If you rotate this image you lose the orientation of the sphere and consequently where is the south pole. This is an example.
> >
> > I hope that now it is more clear for everyone.
> >
> > Thank you.
>
> If there is a South pole, it's no longer just a sphere: it's a sphere with a pole!
>
> No, serious, if there is something drawn on the surface of the sphere, you can use this drawing to determine the orientation of the sphere. If there's a magnetic field, you can use that to determine the orientation of the planet. If it's just a sphere, a plain sphere, there's no point in arbitrarily assigning one point on it as the South pole, and expect to be able to find it back afterward. You can point at any point on the sphere and claim that it's the same point, nobody will know the difference. If there is a difference, use that to identify your pole.
>
> On the other hand, if this is just a bookkeeping exercise, that is, you have an image with a sphere and a vector pointing to the South pole, and you want to rotate the image, you can rotate the vector at the same time, thereby "remembering" where that point was.
>
> Cheers,
> Cris.

So the title of this thread has some fundamental, I was trying to solve an unrealistic problem.

Thank you all.

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