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Looking for formula to convert I,j,k vectors


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I need to convert I,j,k vectors into decimal degree angular values for a project I’m working on. Any help would be appreciated.

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Inside Calypso? -> You can chage display

All angles or one combined angle?

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If you have the XYZ nominals I have a spreadsheet that will calculate the IJK vectors for them. Will that help?

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Thanks Clarke

But what I'm looking for is the formula to take a i,j,k vector and convert it to an angle value. I think I have to convert it to a Euler Angle then maybe I'll have that angle, but my tests are not working. 

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Not specific to Calypso. I'm working on another project.

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So you want one combined angle? I think ijk are from 0 to 1 and it should be calculated with sin,cos and tan mainly - there should be formula on google.
So what? you want to convert each ijk to angle or that combinated one?

 

To find the angles associated with a vector given in IJK notation (I, J, K), you need to calculate the direction cosines and then use the inverse cosine function (arccos). The direction cosines are found by dividing the components of the vector by its magnitude. 
 
Here's a breakdown:
1. Calculate the Magnitude of the Vector:
  • The magnitude of a vector (I, J, K) is calculated as: sqrt(I^2 + J^2 + K^2). 
     
2. Calculate the Direction Cosines:
  • The direction cosine for the x-axis (angle with the x-axis) is: I / magnitude.
  • The direction cosine for the y-axis (angle with the y-axis) is: J / magnitude.
  • The direction cosine for the z-axis (angle with the z-axis) is: K / magnitude. 
     
3. Calculate the Angles:
  • To find the angle associated with each axis, use the arccos function:
    • Angle with x-axis = arccos(I / magnitude).
    • Angle with y-axis = arccos(J / magnitude).
    • Angle with z-axis = arccos(K / magnitude). 
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