wirewound adjustable resistor. Something I missed?
What you got was exactly as expected.  Think about it.  Drawing out the schematic helps:



R1 (your adjustable resistor end to end) and Rm (the resistance of the voltmeter) form a voltage divider.  The voltage accross Rm is the input voltage times Rm/(R1 + Rm).  I'll let you work that out.

I assumed 10 M&Omega; for the voltmeter resistance, which is probably about right if you have a modern electronic meter.  If this is a old analog meter then it could be less, maybe as low a 1 M&Omega;.  Either way though, it won't change the above answer much.