Einar's and Alec's post got me thinking about this (again). What came out of it is not an end-all, do-all, but might add some insight...
The attached sim shows a way of thinking about the problem. Older cars and airplanes use a fuel gauge system like shown in the "
original circuit" below. There is a voltage regulator that creates a fixed voltage lower than the car's system voltage (12 to 14.5V). I'm not sure exactly what the voltage is, but I used 8V for the purpose of this discussion. The gauge is just an ammeter, which has an internal resistance R2 Again, I don't know exactly what the resistance of the meter is, but am using 50Ω. For the sender, I am using the OP's 0Ω to 360Ω.
For simulation, I step the resistance of the original sender from near 0 to 360Ω in steps of 1Ω, and I plot the current through the meter I(R2). The independent variable in the simulation is the resistance of R1, and the plots show the resistance of R1 along the X-axis.
Note the current through the gauge (meter) I(R2) is somewhat non-linear, but no matter, it is what it is... The meter face on the gauge is marked according to a calibration such that E, 1/8, 1/4, 3/8, ... F make sense with the geometry of the sender wiper position and the shape of the tank. If we replace the sender with one that has half the resistance of the original at every wiper position, the goal is to have the same current flow through the meter as it would have had with the original sender.
So here (thanks to Alec) is a way of doing that. In "
the prototype" circuit, I use a current mirror to duplicate the current flow through the meter, and add that current through the new half-resistance sender. Intuitively, that doubles the current through the half-resistance sender, meaning that V(N) is identical to V(S). The current mirror is an LTSpice voltage-controlled current-source with a transconductance of 1/50 mhos, where the 50 comes from the meter resistance (R3,R4). Note that the plot of I(R4) [the meter in the Prototype] falls perfectly on top of I(R2), showing that they are identical.
For interest sake, I plot the expression V(n)/I(R3) just to show that R3 is indeed (R1)/2. Also note that I plot V(s)-V(n) to prove they are identical.
The "
Implementation" circuit shows how to implement the current mirror using an op-amp, a resistor R7 equal to the meter resistance R6, and an NPN used as a current booster. Again, the goal is to force V(M) to be the same as V(S), thereby making I(R6) identical to I(R2) and I(R4). This circuit needs to be checked for dynamic stability before someone tries to build it, but the DC transfer curve does the right thing.
Note that to use this approach, you would have to know the internal resistance of the meter movement, and you assume that the geometry (arm length, total arm angle of travel, resistance linearity) of the new half-resistance sender is similar to the original. You would also have to bring a wire out of the instrument regulator inside the instrument cluster. If the total resistance change of the new sender is not exactly half, with the appropriate scaling, this approach could still work.