AI4QJ wrote:
Going further, I am
still trying to consider how the extra angle can also be absorbed "into" an
impedance discontinuity.

I have started a phasor diagram of it but it is not
finished yet. Maybe a Smith Chart explanation will
work. All lines are lossless.

On a Smith Chart normalized to 100 ohms, lay out the
10 degrees of 100 ohm line from the infinity point,
i.e. the open-circuit point. The reactance value
is tan(90-10) = 5.67. That means the reactance value
is 5.67*100 = -j567 ohms which has to be the value
at the impedance discontinuity.

Now on a Smith Chart normalized to 600 ohms, lay out
the x degrees of 600 ohm line from the zero point
to the point where -j567/600 is located. Read the
number of degrees required. It is Arctan(567/600)
which is equal to ~43 degrees.

The phase shift at the impedance discontinuity is
therefore 90-10-43 = 37 degrees.
--
73, Cecil http://www.w5dxp.com