How do you calculate the vertical angle from a 3d rotation matrix or a quaternion?

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I've got the rotation matrix and quaternion of a mobile device. I'm trying to calculate the vertical angle from it. What I'm trying to get is for example 0° if the device is held in portrait mode and 90° in landscape mode.

Any help would be appreciated. I'm sorry if this was asked before, I don't have any experience with this and didn't really know what to search for.

rotations quaternions

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asked Jul 19 at 21:00

Janko

32

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try starting here.
  – amd
  Jul 19 at 22:11
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I've got the rotation matrix and quaternion of a mobile device. I'm trying to calculate the vertical angle from it. What I'm trying to get is for example 0° if the device is held in portrait mode and 90° in landscape mode.

Any help would be appreciated. I'm sorry if this was asked before, I don't have any experience with this and didn't really know what to search for.

rotations quaternions

share|cite|improve this question

asked Jul 19 at 21:00

Janko

32

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try starting here.
  – amd
  Jul 19 at 22:11
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I've got the rotation matrix and quaternion of a mobile device. I'm trying to calculate the vertical angle from it. What I'm trying to get is for example 0° if the device is held in portrait mode and 90° in landscape mode.

Any help would be appreciated. I'm sorry if this was asked before, I don't have any experience with this and didn't really know what to search for.

rotations quaternions

share|cite|improve this question

asked Jul 19 at 21:00

Janko

32

I've got the rotation matrix and quaternion of a mobile device. I'm trying to calculate the vertical angle from it. What I'm trying to get is for example 0° if the device is held in portrait mode and 90° in landscape mode.

Any help would be appreciated. I'm sorry if this was asked before, I don't have any experience with this and didn't really know what to search for.

rotations quaternions

share|cite|improve this question

asked Jul 19 at 21:00

Janko

32

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 21:00

Janko

32

asked Jul 19 at 21:00

Janko

32

asked Jul 19 at 21:00

Janko

32

32

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try starting here.
  – amd
  Jul 19 at 22:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try starting here.
  – amd
  Jul 19 at 22:11
  

  
  
  
  

Try starting here.
– amd
Jul 19 at 22:11

Try starting here.
– amd
Jul 19 at 22:11

add a comment | 

1 Answer
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There are many sign conventions associated with rotation matrices, quaternions, and Euler angles. It can get confusing.

But hopefully you just want the angle $theta$ between the space-fixed $z$-axis and the body-fixed $z'$-axis, and presumably your $3times3$ rotation matrix $mathbfA$ is defined such that $mathbfr'=mathbfAcdotmathbfr$ (or the same equation, but with the transpose of $mathbfA$). In that case $costheta=A_33$ because it is just the scalar product between the original, and rotated, $(0,0,1)$ vectors. So take the arccos of the $3,3$ element of the rotation matrix.

I advise you to double check that this makes sense for your definitions and coordinate systems!

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edited Jul 20 at 6:21

answered Jul 19 at 22:55

LonelyProf

1415

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

There are many sign conventions associated with rotation matrices, quaternions, and Euler angles. It can get confusing.

But hopefully you just want the angle $theta$ between the space-fixed $z$-axis and the body-fixed $z'$-axis, and presumably your $3times3$ rotation matrix $mathbfA$ is defined such that $mathbfr'=mathbfAcdotmathbfr$ (or the same equation, but with the transpose of $mathbfA$). In that case $costheta=A_33$ because it is just the scalar product between the original, and rotated, $(0,0,1)$ vectors. So take the arccos of the $3,3$ element of the rotation matrix.

I advise you to double check that this makes sense for your definitions and coordinate systems!

share|cite|improve this answer

edited Jul 20 at 6:21

answered Jul 19 at 22:55

LonelyProf

1415

add a comment | 

up vote
0
down vote



accepted

There are many sign conventions associated with rotation matrices, quaternions, and Euler angles. It can get confusing.

But hopefully you just want the angle $theta$ between the space-fixed $z$-axis and the body-fixed $z'$-axis, and presumably your $3times3$ rotation matrix $mathbfA$ is defined such that $mathbfr'=mathbfAcdotmathbfr$ (or the same equation, but with the transpose of $mathbfA$). In that case $costheta=A_33$ because it is just the scalar product between the original, and rotated, $(0,0,1)$ vectors. So take the arccos of the $3,3$ element of the rotation matrix.

I advise you to double check that this makes sense for your definitions and coordinate systems!

share|cite|improve this answer

edited Jul 20 at 6:21

answered Jul 19 at 22:55

LonelyProf

1415

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

There are many sign conventions associated with rotation matrices, quaternions, and Euler angles. It can get confusing.

But hopefully you just want the angle $theta$ between the space-fixed $z$-axis and the body-fixed $z'$-axis, and presumably your $3times3$ rotation matrix $mathbfA$ is defined such that $mathbfr'=mathbfAcdotmathbfr$ (or the same equation, but with the transpose of $mathbfA$). In that case $costheta=A_33$ because it is just the scalar product between the original, and rotated, $(0,0,1)$ vectors. So take the arccos of the $3,3$ element of the rotation matrix.

I advise you to double check that this makes sense for your definitions and coordinate systems!

share|cite|improve this answer

edited Jul 20 at 6:21

answered Jul 19 at 22:55

LonelyProf

1415

There are many sign conventions associated with rotation matrices, quaternions, and Euler angles. It can get confusing.

But hopefully you just want the angle $theta$ between the space-fixed $z$-axis and the body-fixed $z'$-axis, and presumably your $3times3$ rotation matrix $mathbfA$ is defined such that $mathbfr'=mathbfAcdotmathbfr$ (or the same equation, but with the transpose of $mathbfA$). In that case $costheta=A_33$ because it is just the scalar product between the original, and rotated, $(0,0,1)$ vectors. So take the arccos of the $3,3$ element of the rotation matrix.

I advise you to double check that this makes sense for your definitions and coordinate systems!

share|cite|improve this answer

edited Jul 20 at 6:21

answered Jul 19 at 22:55

LonelyProf

1415

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 6:21

edited Jul 20 at 6:21

edited Jul 20 at 6:21

answered Jul 19 at 22:55

LonelyProf

1415

answered Jul 19 at 22:55

LonelyProf

1415

answered Jul 19 at 22:55

LonelyProf

1415

1415

add a comment | 

add a comment | 

 

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