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.

Out on HF and heard nobody, but didn't call CQ? Nobody heard you either. 73 de N7GH

Piet Souris

Rancher

Posts: 1844

61

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 sub-subclass 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 sub-subclass 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

It is sorta covered in the JavaRanch Style Guide. |