Rotate a cube
Debarun Mukherjee
Greenhorn
Posts: 8
posted 1 year ago
 1
Using Trigonometry, specifically the Pythagorean Theorem h^2 = x^2 + y^2, so the square root of x^2 + y^2 gives you the length from your set vertices.
X = cos(A), Y = sin(A) where A is your rotational angle.
X = cos(A), Y = sin(A) where A is your rotational angle.
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Piet Souris
Rancher
Posts: 1694
42
posted 1 year ago
Rotating a cube around one of its edges is a rotation in 3D,
so that is not an easy thing to do.
I don't know much about JavaFX, but a quick look at the API
gives that you can set a rotation axis to a Node, and a Cube
is a Box, which is a subsubclass of Node, if I understand correctly.
Have you looked at the API's of JavaFx?
so that is not an easy thing to do.
I don't know much about JavaFX, but a quick look at the API
gives that you can set a rotation axis to a Node, and a Cube
is a Box, which is a subsubclass of Node, if I understand correctly.
Have you looked at the API's of JavaFx?
Jan Daermann
Greenhorn
Posts: 3
Debarun Mukherjee
Greenhorn
Posts: 8
posted 1 year ago
Yeah I checked them. I need to rotate the cube by setting the pivot of rotation at one of the corners of the cube. But I cant determine the coordinates of the cube. How to get that? I actually need to rotate 4 cubes placed side by side about the centre of the system. That's why I need to get the coordinates of the common corners of the 4 cubes and set each of their pivot of rotation to that coordinate. How can I figure that centre ?
Debarun Mukherjee
Greenhorn
Posts: 8
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