md` Suppose we want to convert an Euler intrinsic y-x'-y" rotation into a Tait-Bryan rotation x-z'-y". A typical approach might be to: 1. Compute the rotation matrix for the known Euler intrinsic y-x'-y" rotation angles 2. Symbolically compute the rotation matrix for the x-z'-y" rotation 3. Equate these and solve for the Tait-Bryan x-z'-y" rotation angles e.g. [Conversion Methods](https://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Conversion_formulae_between_formalisms) But is there another approach? For a bit of fun, consider [Napier's Rules for Right Spherical Triangles](https://en.wikipedia.org/wiki/Spherical_trigonometry#Napier's_rules_for_right_spherical_triangles). ![](${await FileAttachment("Napier Phase Angle Better 300.png").url()}) Imagine the +y axis projecting through the point on a sphere we'll denote "A". A +y rotation will sweep the yz plane through the angle illustrated at point A. A subsequent +x rotation would result in the +y axis carving out a line on the surface of the sphere we'll denote "c". We'll denote the point where the +y axis terminates as point "B" (with the xz plan perpendicular to c). Is there a different sequence of rotations we could do to align the +y axis to point B? Let's try an x then z sequence. Returning to our starting point A, a +x rotation would result in the +y axis carving out a line on the surface of the sphere we'll denote "b" terminating at a point we'll denote "C". A subsequent -z rotation would result in the +y axis carving out a line in the surface of the sphere we'll denote "a". Notice, the lines b and a at the point "C" create a right angle on the surface of the sphere due to the x & z axis being orthogonal to each other. The resulting spherical triangle is a right spherical triangle and Napier's Rules can be applied. In a given y-x' rotation sequence - y rotation -> vertex angle A - x' rotation -> side c An x-z' rotation sequence can be computed as - x rotation -> side b - -z rotation -> side a Now, though the proposed Euler intrinsic and the Tait-Bryan target rotations finish on B, the final xy planes are out of phase. The Euler intrinsic y-x' transformation results in an xy plane perpendicular to c. The Tait-Bryan x-z' transformation results in an xy plane parallel to a. These two planes are out of phase by a +y rotation of (90°-B) To make the transformations fully equivalent we can adjust the final y"ᵀᴮ as follows: y"ᵀᴮ = y"+(90°-B) Applying Napier's rules we have:

Variable	Value	Rule
A	y
c	x'
a	sin⁻¹(sin(A)*sin(c))	(R2)
b	cos(a) ? cos⁻¹(cos(c)/cos(a)) : PI/2	(R1)
B	cos⁻¹(sin(A)*cos(b))	(R9)

To yield the Tait-Bryan rotation angles: # From y-x'-y"

x	z'	y"ᵀᴮ
b	-a	y"+(90°-B)

And a more complete set from right hand rule symmetry... ![](${await FileAttachment("Napier Phase Angle Better Symmetry 300.png").url()}) | EI | y-x'-y" | x-z'-x" | z-y'-z" | |----------------|---------|---------|---------| | TB Complement | x-z'-y" | z-y'-x" | y-x'-z" | | A | y | x | z | | c | x' | z' | y' | | b | x | z | y | | a | -z' | -y' | -x' | | Final rotation | y"+90°-B | x"+90°-B | z"+90°-B | | EI | y-x'-y" | x-z'-x" | z-y'-z" | |----------------|---------|---------|---------| | TB Complement | z-x'-y" | y-z'-x" | y-x'-z" | | A | -y | -x | -z | | c | x' | z' | y' | | b | -z | -y | -x | | a | x' | z' | y' | | Final rotation | y"+B-90° | x"+B-90° | z"+B-90° |`