Apply force to bottom of object given rotation

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I have a box and I want to apply a force in the direction that the bottom of the box is facing, given any Pitch, Roll, or Yaw rotation. At a resting point where pitch, roll, and yaw are 0, I know the downward force would be: x = 0, y = 0, z = -n, as the box is laying flat.

I also have the box's quaternion (qx, qy, qz, qw) available if that is helpful.

Here is a visual example showing my question:

A box with a pitch of 0

A box with a pitch of 90

How can I get this force using the objects rotation?

trigonometry rotations

share|cite|improve this question

edited Jul 25 at 19:33

asked Jul 25 at 18:33

funcs

11

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw The force will be applied in a loop, so as the rotation changes, the direction of the force needs to change with it to always be applied to the top of the box. At any given time I need to be able to get the direction the force needs to be in using the current rotation
  – funcs
  Jul 25 at 18:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw you are correct, sorry for my misuse of the terms!
  – funcs
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I am looking to get the X, Y, and Z direction of force (in my particular case, Z is less necessary)
  – funcs
  Jul 25 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there no way to calculate the force from the present orientation alone? I don't think that's what I'm trying to find. The goal is the simulated gravity relative to the objects orientation.
  – funcs
  Jul 25 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I believe my problem is much for simple than your interpretation of it. Lets just say for example my box's current orientation is: Pitch=90, Roll=10, Yaw=20. There is 0 force presently on the object (I need to calculate it and apply it myself!). Using that orientation information, can I calculate what the force needs to be to push on the top of the object. I also have the box's quaternion available if that is helpful.
  – funcs
  Jul 25 at 19:31
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

I have a box and I want to apply a force in the direction that the bottom of the box is facing, given any Pitch, Roll, or Yaw rotation. At a resting point where pitch, roll, and yaw are 0, I know the downward force would be: x = 0, y = 0, z = -n, as the box is laying flat.

I also have the box's quaternion (qx, qy, qz, qw) available if that is helpful.

Here is a visual example showing my question:

A box with a pitch of 0

A box with a pitch of 90

How can I get this force using the objects rotation?

trigonometry rotations

share|cite|improve this question

edited Jul 25 at 19:33

asked Jul 25 at 18:33

funcs

11

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw The force will be applied in a loop, so as the rotation changes, the direction of the force needs to change with it to always be applied to the top of the box. At any given time I need to be able to get the direction the force needs to be in using the current rotation
  – funcs
  Jul 25 at 18:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw you are correct, sorry for my misuse of the terms!
  – funcs
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I am looking to get the X, Y, and Z direction of force (in my particular case, Z is less necessary)
  – funcs
  Jul 25 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there no way to calculate the force from the present orientation alone? I don't think that's what I'm trying to find. The goal is the simulated gravity relative to the objects orientation.
  – funcs
  Jul 25 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I believe my problem is much for simple than your interpretation of it. Lets just say for example my box's current orientation is: Pitch=90, Roll=10, Yaw=20. There is 0 force presently on the object (I need to calculate it and apply it myself!). Using that orientation information, can I calculate what the force needs to be to push on the top of the object. I also have the box's quaternion available if that is helpful.
  – funcs
  Jul 25 at 19:31
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I have a box and I want to apply a force in the direction that the bottom of the box is facing, given any Pitch, Roll, or Yaw rotation. At a resting point where pitch, roll, and yaw are 0, I know the downward force would be: x = 0, y = 0, z = -n, as the box is laying flat.

I also have the box's quaternion (qx, qy, qz, qw) available if that is helpful.

Here is a visual example showing my question:

A box with a pitch of 0

A box with a pitch of 90

How can I get this force using the objects rotation?

trigonometry rotations

share|cite|improve this question

edited Jul 25 at 19:33

asked Jul 25 at 18:33

funcs

11

I have a box and I want to apply a force in the direction that the bottom of the box is facing, given any Pitch, Roll, or Yaw rotation. At a resting point where pitch, roll, and yaw are 0, I know the downward force would be: x = 0, y = 0, z = -n, as the box is laying flat.

I also have the box's quaternion (qx, qy, qz, qw) available if that is helpful.

Here is a visual example showing my question:

A box with a pitch of 0

A box with a pitch of 90

How can I get this force using the objects rotation?

trigonometry rotations

share|cite|improve this question

edited Jul 25 at 19:33

asked Jul 25 at 18:33

funcs

11

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 19:33

edited Jul 25 at 19:33

edited Jul 25 at 19:33

asked Jul 25 at 18:33

funcs

11

asked Jul 25 at 18:33

funcs

11

asked Jul 25 at 18:33

funcs

11

11

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw The force will be applied in a loop, so as the rotation changes, the direction of the force needs to change with it to always be applied to the top of the box. At any given time I need to be able to get the direction the force needs to be in using the current rotation
  – funcs
  Jul 25 at 18:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw you are correct, sorry for my misuse of the terms!
  – funcs
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I am looking to get the X, Y, and Z direction of force (in my particular case, Z is less necessary)
  – funcs
  Jul 25 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there no way to calculate the force from the present orientation alone? I don't think that's what I'm trying to find. The goal is the simulated gravity relative to the objects orientation.
  – funcs
  Jul 25 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I believe my problem is much for simple than your interpretation of it. Lets just say for example my box's current orientation is: Pitch=90, Roll=10, Yaw=20. There is 0 force presently on the object (I need to calculate it and apply it myself!). Using that orientation information, can I calculate what the force needs to be to push on the top of the object. I also have the box's quaternion available if that is helpful.
  – funcs
  Jul 25 at 19:31
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw The force will be applied in a loop, so as the rotation changes, the direction of the force needs to change with it to always be applied to the top of the box. At any given time I need to be able to get the direction the force needs to be in using the current rotation
  – funcs
  Jul 25 at 18:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw you are correct, sorry for my misuse of the terms!
  – funcs
  Jul 25 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I am looking to get the X, Y, and Z direction of force (in my particular case, Z is less necessary)
  – funcs
  Jul 25 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there no way to calculate the force from the present orientation alone? I don't think that's what I'm trying to find. The goal is the simulated gravity relative to the objects orientation.
  – funcs
  Jul 25 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mvw I believe my problem is much for simple than your interpretation of it. Lets just say for example my box's current orientation is: Pitch=90, Roll=10, Yaw=20. There is 0 force presently on the object (I need to calculate it and apply it myself!). Using that orientation information, can I calculate what the force needs to be to push on the top of the object. I also have the box's quaternion available if that is helpful.
  – funcs
  Jul 25 at 19:31
  
  

  
  
  
  

@mvw The force will be applied in a loop, so as the rotation changes, the direction of the force needs to change with it to always be applied to the top of the box. At any given time I need to be able to get the direction the force needs to be in using the current rotation
– funcs
Jul 25 at 18:51

@mvw The force will be applied in a loop, so as the rotation changes, the direction of the force needs to change with it to always be applied to the top of the box. At any given time I need to be able to get the direction the force needs to be in using the current rotation
– funcs
Jul 25 at 18:51

@mvw you are correct, sorry for my misuse of the terms!
– funcs
Jul 25 at 18:54

@mvw you are correct, sorry for my misuse of the terms!
– funcs
Jul 25 at 18:54

@mvw I am looking to get the X, Y, and Z direction of force (in my particular case, Z is less necessary)
– funcs
Jul 25 at 18:57

@mvw I am looking to get the X, Y, and Z direction of force (in my particular case, Z is less necessary)
– funcs
Jul 25 at 18:57

Is there no way to calculate the force from the present orientation alone? I don't think that's what I'm trying to find. The goal is the simulated gravity relative to the objects orientation.
– funcs
Jul 25 at 19:09

Is there no way to calculate the force from the present orientation alone? I don't think that's what I'm trying to find. The goal is the simulated gravity relative to the objects orientation.
– funcs
Jul 25 at 19:09

@mvw I believe my problem is much for simple than your interpretation of it. Lets just say for example my box's current orientation is: Pitch=90, Roll=10, Yaw=20. There is 0 force presently on the object (I need to calculate it and apply it myself!). Using that orientation information, can I calculate what the force needs to be to push on the top of the object. I also have the box's quaternion available if that is helpful.
– funcs
Jul 25 at 19:31

@mvw I believe my problem is much for simple than your interpretation of it. Lets just say for example my box's current orientation is: Pitch=90, Roll=10, Yaw=20. There is 0 force presently on the object (I need to calculate it and apply it myself!). Using that orientation information, can I calculate what the force needs to be to push on the top of the object. I also have the box's quaternion available if that is helpful.
– funcs
Jul 25 at 19:31

 | 
show 1 more comment

1 Answer
1

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up vote
0
down vote

I expanded my search terms to finding a force direction from a quaternion and was able to find an excellent answer by a user named Dobbs:

forward vector:

x = 2 * (xz + wy)

y = 2 * (yz - wx)

z = 1 - 2 * (xx + yy)

up vector:

x = 2 * (xy - wz)

y = 1 - 2 * (xx + zz)

z = 2 * (yz + wx)

left vector:

x = 1 - 2 * (yy + zz)

y = 2 * (xy + wz)

z = 2 * (xz - wy)

Using the "forward vector" with a negative magnitude I was able to apply a constant force onto the top of my object, no matter the orientation.

share|cite|improve this answer

answered Jul 27 at 17:26

funcs

11

add a comment | 

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

I expanded my search terms to finding a force direction from a quaternion and was able to find an excellent answer by a user named Dobbs:

forward vector:

x = 2 * (xz + wy)

y = 2 * (yz - wx)

z = 1 - 2 * (xx + yy)

up vector:

x = 2 * (xy - wz)

y = 1 - 2 * (xx + zz)

z = 2 * (yz + wx)

left vector:

x = 1 - 2 * (yy + zz)

y = 2 * (xy + wz)

z = 2 * (xz - wy)

Using the "forward vector" with a negative magnitude I was able to apply a constant force onto the top of my object, no matter the orientation.

share|cite|improve this answer

answered Jul 27 at 17:26

funcs

11

add a comment | 

up vote
0
down vote

I expanded my search terms to finding a force direction from a quaternion and was able to find an excellent answer by a user named Dobbs:

forward vector:

x = 2 * (xz + wy)

y = 2 * (yz - wx)

z = 1 - 2 * (xx + yy)

up vector:

x = 2 * (xy - wz)

y = 1 - 2 * (xx + zz)

z = 2 * (yz + wx)

left vector:

x = 1 - 2 * (yy + zz)

y = 2 * (xy + wz)

z = 2 * (xz - wy)

Using the "forward vector" with a negative magnitude I was able to apply a constant force onto the top of my object, no matter the orientation.

share|cite|improve this answer

answered Jul 27 at 17:26

funcs

11

add a comment | 

up vote
0
down vote

up vote
0
down vote

I expanded my search terms to finding a force direction from a quaternion and was able to find an excellent answer by a user named Dobbs:

forward vector:

x = 2 * (xz + wy)

y = 2 * (yz - wx)

z = 1 - 2 * (xx + yy)

up vector:

x = 2 * (xy - wz)

y = 1 - 2 * (xx + zz)

z = 2 * (yz + wx)

left vector:

x = 1 - 2 * (yy + zz)

y = 2 * (xy + wz)

z = 2 * (xz - wy)

Using the "forward vector" with a negative magnitude I was able to apply a constant force onto the top of my object, no matter the orientation.

share|cite|improve this answer

answered Jul 27 at 17:26

funcs

11

I expanded my search terms to finding a force direction from a quaternion and was able to find an excellent answer by a user named Dobbs:

forward vector:

x = 2 * (xz + wy)

y = 2 * (yz - wx)

z = 1 - 2 * (xx + yy)

up vector:

x = 2 * (xy - wz)

y = 1 - 2 * (xx + zz)

z = 2 * (yz + wx)

left vector:

x = 1 - 2 * (yy + zz)

y = 2 * (xy + wz)

z = 2 * (xz - wy)

Using the "forward vector" with a negative magnitude I was able to apply a constant force onto the top of my object, no matter the orientation.

share|cite|improve this answer

answered Jul 27 at 17:26

funcs

11

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 17:26

funcs

11

answered Jul 27 at 17:26

funcs

11

answered Jul 27 at 17:26

funcs

11

11

add a comment | 

add a comment | 

 

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