Cecil, W5DXP wrote:
"What reflects the other half of the energy?"

The previous example I gave was a 25-ohm resistor-load on a 50-ohm line.
Change the load to 100 ohms. Now the load cannot accept all the current
carried by the incident wave. Lenz`s law says the falling current
generates a rising voltage in an attempt to maintain the current. The
load-generated voltage is in the same phase as the incident voltage so
their sum is greater. Increased voltage across the load reverses phase
and direction of the line-current at the too-high load resistance. Thus,
direction of the reflection is opposite that of the incident wave.

If the load is too small or too large for Zo, some of the incident
energy is reflected by the load. The two processes are analogous. When
the load value is too small, there is a reversal in the phase of the
voltage without change in the phase of the current (1955 Terman page
92). When the load value is too large, there is a reversal in the phase
of the current without change in the phase of the voltage (1955 Terman
page 89).

Those are the necessary and sufficient conditions to reverse the
direction of some of the energy in an incident wave on a transmission
line. For a complete reversal, a short or an open is required.

Best regards, Richard Harrison, KB5WZI