As said above the square root of an arbitrary matrix is ambiguous and it's hard to say what the scipy implementation actually does.

But:

When you have placement with rotation R and translation t and apply it twice, i.e. get the square you have:

y = R * (R * x + t) + t

y = R * (R * x) + R * t + t

y= (R * R) * x + (R + I) * t

this means you have the rotation Q = R * R and the translation s = (R + I) * t.

Now if you consider a placement with rotation Q and translation s you can easily get the square root of Q (by doing only the half rotation) but for the translation part you have to compute t = (R + I)^-1 * s and the question is if this matrix is always invertible.

Example:

Code: Select all

```
import math
# arbitrarily rotate the object
p=App.ActiveDocument.ActiveObject.Placement
p.Base # Vector (-4.642491320876232, 2.744713842009025, 9.827212698728683)
p.Rotation.Q # (0.14145656219842426, -0.6950597415393837, -0.4084936490538904, 0.5744692640906436)
q=App.Rotation(p.Rotation.Axis, math.degrees(p.Rotation.Angle/2))
# compare the axes and angles
p.Rotation.Axis
q.Axis
p.Rotation.Angle
q.Angle # = p.Rotation.Angle/2
# get q as matrix
r=q.toMatrix()
r.determinant() # = 1.0 ok
m = r + App.Matrix() # (R + I) from above
m_inv = m.inverse()
s = p.Base
t = m_inv.multVec(s)
p_root = App.Placement(t, q)
# p_root * p_root gives the same as p
p_root.multiply(p_root) # Placement [Pos=(-4.64249,2.74471,9.82721), Yaw-Pitch-Roll=(-114.246,-43.0795,90.4892)]
p # Placement [Pos=(-4.64249,2.74471,9.82721), Yaw-Pitch-Roll=(-114.246,-43.0795,90.4892)]
# the same for their matrix representations
p.toMatrix()
p_root.toMatrix().multiply(p_root.toMatrix())
```