• Create Account

# Rotation of a Bounding Box around a point that is not the objects centre

Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

11 replies to this topic

### #1Collins1987  Members   -  Reputation: 117

Like
0Likes
Like

Posted 22 February 2013 - 08:03 AM

hey,

Ive been trying to work out how to rotate a bounding box about a position that is not the objects centre and perform collisions based on SAT Collision Dectection. I have the collisions working but i cannot figure out how to rotate the bounding box into the right location as all I seem to be able to manage is to rotate the bounding box around the objects centre position.

This is what i have so far:

        float Rot = Entity.getRotation();
D3DXVECTOR2 pos = D3DXVECTOR2(Entity.getPosition().x,Entity.getPosition().y);
Centre = Entity.getCentre() + pos;

D3DXVECTOR2 X(cosf(Rot), sinf(Rot));
D3DXVECTOR2 Y(-sinf(Rot), cosf(Rot));

X *= Entity.getWidth()/2;
Y *= Entity.getHeight()/2;

Corner[0] = (Centre - X) - Y;
Corner[1] = (Centre + X) - Y;
Corner[2] = (Centre + X) + Y;
Corner[3] = (Centre - X) + Y;


This code rotates about the objects centre position but i cant mange to rotate about a position say 15 pixels in the positive X direction from its own centre.

Any Ideas on how I would accomplish this?

### #2Tournicoti  Prime Members   -  Reputation: 689

Like
3Likes
Like

Posted 22 February 2013 - 08:23 AM

It's possible to decompose this rotation around a fixed point into 3 transformations :

- first a translation, so that the fixed point becomes the origin

- then a 'standard' rotation (around the origin)

- then the inverse of the translation , to move back the box

Hope it can help ?

### #3Collins1987  Members   -  Reputation: 117

Like
0Likes
Like

Posted 22 February 2013 - 10:28 AM

Thanks for that but this is just rotating the box then moving it back to the old position.

I am trying to rotate the bounding box so that its is rotated around a point on a fixed circular distance away from the origin and cannot figure out how to rotate the bounding box to the same position around the circle as the object instead of just rotating around the point where the object was before it was rotated around the fixed point. if that makes any sense?

### #4BornToCode  Members   -  Reputation: 771

Like
0Likes
Like

Posted 22 February 2013 - 10:44 AM

If you want to do that then what you can is after you rotate the camera around the point you want to rotate then you need move the camera back based on the camera inverse look at direction in local space multiply by the distance factor you want. Hope that helps. Let me give you an example, let's say the camera look at direction is 0,0,1 and you position in on the point and rotate it by let's say 90 degrees clockwise around the Y axis (0,1,0) then your new look at direction for the camera in local space would be now (1,0,0) taking the inverse of that gives you (-1,0,0). Now you can take that inverse and multiply by the distance factor your want, let say it is 5. So we can (-1,0,0)*5 and we now get (-5,0,0). Take the value and add it to the camera position. As you notice i am using an camera as an example, but the same thing applies to your objects, the camera is no different then any of your objects.

Edited by BornToCode, 22 February 2013 - 10:50 AM.

### #5Scouting Ninja  Members   -  Reputation: 442

Like
0Likes
Like

Posted 04 March 2013 - 05:07 PM

You where on the right track, its like this:

it's Trigonometry.

Edited by Scouting Ninja, 04 March 2013 - 05:08 PM.

### #6Collins1987  Members   -  Reputation: 117

Like
0Likes
Like

Posted 06 March 2013 - 05:51 AM

have tried using this formula there is two things wrong with it, first you cant set a 2D vector with just one value and second it creates a bounding box that is massive compared to the actual entity. Also when drawing these bounding boxes to check to see if it has worked the size of them changes dependent on the rotation of the parent object? Any ideas how to solve this or what I have done wrong?

D3DXVECTOR3 orientation = Entity.getOrientation();
float Rot = Entity.getRotation();
D3DXVECTOR2 pos = D3DXVECTOR2(Entity.getPosition().x,Entity.getPosition().y);
Centre = Entity.getCentre() + pos;

D3DXVECTOR2 X(Entity.getPosition().x + (168 * (sinf(Rot))), 0);
D3DXVECTOR2 Y(0, Entity.getPosition().y + (168 * (cosf(Rot))));

X.x -= Entity.getWidth()/2;
Y.y -= Entity.getHeight()/2;

Corner[0] = (Centre - X) - Y;
Corner[1] = (Centre + X) - Y;
Corner[2] = (Centre + X) + Y;
Corner[3] = (Centre - X) + Y;


Where entity Position = 496, 200.

entity centre = 512, 216

rotation point is = 512, 368

### #7Scouting Ninja  Members   -  Reputation: 442

Like
0Likes
Like

Posted 06 March 2013 - 06:17 AM

Oh!

Sorry my mistake, my C is a bit rusty.(Using python now)

you would use these values in only one vector: D3DXVECTOR2 MID(Entity.getPosition().x + (168 * (sinf(Rot))) ,Entity.getPosition().y + (168 * (cosf(Rot))));

This will be the mid point so then you define the bounding box by adding or subtracting from the mid point:

D3DXVECTOR2 X( D3DXVECTOR2 MID.x + HalfSiseOfBox,D3DXVECTOR2 MID.x - HalfSiseOfBox)

D3DXVECTOR2 Y( D3DXVECTOR2 MID.y + HalfSiseOfBox,D3DXVECTOR2 MID.y - HalfSiseOfBox)

See normally if your bounding box's middle is 0 the first X vector would be say +2 and second -2 and the same for Y

+2_______-2

|               |

|      0       |

|               |

-2_______+2 Is this correct or are you using a deferent bounding box ?

Sorry I mean this

__+2__

|     |     |

-2  0   +2

|     |     |

|__-2__|

Edited by Scouting Ninja, 06 March 2013 - 06:31 AM.

### #8Collins1987  Members   -  Reputation: 117

Like
0Likes
Like

Posted 06 March 2013 - 06:43 AM

Yes that is the type of bounding box I am using just trying to get it placed in the right location for the collision detection to work. I will give it a try and see what happens

Thanks

### #9Collins1987  Members   -  Reputation: 117

Like
0Likes
Like

Posted 06 March 2013 - 07:15 AM

Have tried implementing it this way and its not working still just rotating the bounding boxes into some weird places, would using a matrix transform during drawing of the sprite effect this as directx draws the sprite in the right location but the world only knows the positions without the transform matrix applied to it?

should state that im working with Orientated Bounding Boxes opposed to Axis Aligned Bounding Boxes

Edited by Collins1987, 06 March 2013 - 08:10 AM.

### #10Scouting Ninja  Members   -  Reputation: 442

Like
1Likes
Like

Posted 06 March 2013 - 09:38 AM

Personally I cant see why it isn't working, the equation is meant to rotate around a point, then it would only mean reassemble the bounding box at that point.

Make sure that your using radians and not degrees for rotation,degrees causes 90 degree jumps.(could be the weird jumping you see)

You could do it for each point of the bounding box, but that could get complicated.

You could always test it first by attempting to rotate a sprite around a point.

You could always search: trig rotating around point.

These are the last ideas I can think of with out you sending me the file so I can look at it, if you do get it right please drop a comment here showing how you did it,I am curios now.

### #11Collins1987  Members   -  Reputation: 117

Like
1Likes
Like

Posted 07 March 2013 - 09:23 AM

Ive figured it out really was quite simple to be honest.

involved adding the radius of the circle to be rotated around to the objects position, then when the object had been rotated working out the objects current orientation

and multiplying that by the radius offset and taking it from the position before adding it the rotated corners.

Here's the code that is being used:

D3DXVECTOR2 X(cosf(Rot), sinf(Rot));
D3DXVECTOR2 Y(-sinf(Rot), cosf(Rot));

X *= halfWidth/2;
Y *= halfHeight/2;

Corner[0] = Centre - X - Y;
Corner[1] = Centre + X - Y;
Corner[2] = Centre + X + Y;
Corner[3] = Centre - X + Y;

float offsetX = orientation.x * Offset;
float offsetY = orientation.y * Offset;

D3DXVECTOR2 offset(offsetX, offsetY);

pos -= offset;

Corner[0] += pos;
Corner[1] += pos;
Corner[2] += pos;
Corner[3] += pos;

Centre += pos;


Thanks for all the help! Wouldn't have made it without the pointers in the write direction!

### #12Scouting Ninja  Members   -  Reputation: 442

Like
0Likes
Like

Posted 07 March 2013 - 11:52 AM

Good work!

I see,you multiply the offset after calculating the rotation point and then subtract it from the position.

This will definitely be a useful piece of code thanks!

Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

PARTNERS