Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Euler angles
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Proper Euler angles=== Assuming a frame with [[unit vector]]s (''X'', ''Y'', ''Z'') given by their coordinates as in the main diagram, it can be seen that: :<math>\cos (\beta) = Z_3.</math> And, since :<math>\sin^2 x = 1 - \cos^2 x,</math> for <math> 0<x<\pi </math> we have :<math>\sin (\beta) = \sqrt {1 - Z_3^2}.</math> As <math>Z_2</math> is the double projection of a unitary vector, :<math>\cos(\alpha) \cdot \sin(\beta) = -Z_2,</math> :<math>\cos(\alpha) = -Z_2 / \sqrt{1 - Z_3^2}.</math> There is a similar construction for <math>Y_3</math>, projecting it first over the plane defined by the axis ''z'' and the line of nodes. As the angle between the planes is <math>\pi/2 - \beta</math> and <math>\cos(\pi/2 - \beta) = \sin(\beta)</math>, this leads to: :<math>\sin(\beta) \cdot \cos(\gamma) = Y_3,</math> :<math>\cos(\gamma) = Y_3 / \sqrt{1 - Z_3^2},</math> and finally, using the [[inverse trigonometric functions|inverse cosine]] function, :<math>\alpha = \arccos\left(-Z_2 / \sqrt{1 - Z_3^2}\right),</math> :<math>\beta = \arccos\left(Z_3\right),</math> :<math>\gamma = \arccos\left(Y_3 / \sqrt{1 - Z_3^2}\right).</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)