I think I figured it out.
TomGeorge yes, I'm referring to that post.
Its a voltage divider
The voltage drop of the resistor is result of a voltage divider between the resistor and the linear regulator.
If the input voltage is 17v and the Linear voltage regulator is a 5v regulator, then that leaves 12 to spread between the resistor and the LVO.
The LVO can said to have the resistance of its load.
Thus if the R1 is 330R and LVO load is 10k ohms, then the Voltage drop across R1 can be found by using the voltage divider equation.
R1(vdrop) = 12v x (330/10330)
When I physically tested it, my results were 0.5v off but the input voltage was not exactly 17v so that might account for it.
How it influences current
As I raised the value of R1 the current barely changed at all, as if R1 was not even there.
But if I made R1 VERY large, then it started to affect the current.
Maybe R1 only starts to seriously affect the current when it drops so much voltage that there is not enough left for the LVO to do its job (drop out voltage I think (2v) )
I may very well be wrong.
I still don't understand what math I need in order to figure out how a resistor will affect the current.
I'll keep experimenting or make the final circuit and change the resistor if the thing gets destroyed.
(Electronics are cheap)