Re: Calculating new coordinates in 3D after moving an object
- From: matt271829-news@xxxxxxxxxxx
- Date: Sat, 20 Oct 2007 16:46:58 -0700
On Oct 20, 5:05 pm, andreas <anpa...@xxxxxxxxx> wrote:
Good evening.
I have a question on how to calculate an objects coordinated after altering its position.
Example:
A triangle with coordinates;
P1 : -4, -4, 4 (x, y, z)
P2 : -3, -4, 4
P3 : -3, -4, 5
Let's say I want to rotate the line P2-P3 about 14 degrees around the axis P1-P2.
How can I find out the new coordinates of P3?
It would be a simple task using trigonometry if it it was only a 2D space, but 3D is puzzling me a little.
Someone hinted me that "Quaternion" would be the solution, but I tried to understand but didn't. Embarrassingly that area is a little over my understanding of mathematics.
Another question is if the object would be more complex, such as a Dodecahedron.
I assume I first must find the a central point around which the rotation should be done, but since all points are moving...
How can I find out the new coordinates of each and every one?
Any pointers to online resources or direct help are much appreciated.
Thank you.
You don't need to find any "central point": 3D rotations are about an
axis (a straight line), and the axis is all you need to know. Assuming
a points-and-straight-lines object, the complexity of the object makes
no difference to the complexity of the problem, except in the very
obvious sense that you have to calculate the rotation for every point.
The usual method would be to translate (shift) the figure so that the
axis of rotation passes through the origin, perform the rotation, and
then perform the reverse translation. The translations are trivial, so
what you really need to know is how to rotate a point (x,y,z) through
angle theta about a vector (i,j,k) (i.e. about a line between (0,0,0)
and (i,j,k)). To make the things marginally easier, assume also that
the vector has been normalised to have length 1 (i.e. i^2 + j^2 + k^2
= 1).
The formulas below, which are just a matrix multiplication done
explicitly, achieve this. The rotated point is returned in (x',y',z').
You also need to consider the clockwise/anticlockwise issue. Let's
assume the handedness of the coordinate system is such that when
looking towards origin from a point in (+,+,+) octant, the positive x,
y, z axes are anticlockwise in that order. In these formulas positive
theta then corresponds to an anticlockwise rotation when viewed
looking along the vector (i.e. looking along the line from the origin
to (i,j,k)).
x' = x*(i^2 + (j^2 + k^2)*Cos(theta))
+ y*(i*j * (1 - Cos(theta)) + k*Sin(theta))
+ z*(k*i * (1 - Cos(theta)) - j*Sin(theta))
y' = x*(i*j*(1 - Cos(theta)) - k*Sin(theta))
+ y*(j^2 + (i^2 + k^2)*Cos(theta))
+ z*(j*k*(1 - Cos(theta)) + i*Sin(theta))
z' = x*(k*i*(1 - Cos(theta)) + j*Sin(theta))
+ y*(j*k*(1 - Cos(theta)) - i*Sin(theta))
+ z*(k^2 + (i^2 + j^2)*Cos(theta))
.
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