P
Pete Verdon
Hi,
I'm planning to fit a fuel tank with a contents gauge. The sender that
the tank manufacturer will put in is essentially a variable resistor,
which reads 180 ohms when the tank is full, and 0 ohms when it is empty
with, presumably, a linear resistance in between. There is a gauge that
connects to this - presumably an ammeter and a means of passing current
through the sender?
My problem is that the tank is not square, and so the height of the
sender float is not directly proportional to the amount of fuel actually
remaining. More unfortunately, it's actually "the wrong way round", so
that when the gauge reads half-full there will actually be significantly
less than half a tank of fuel left - perhaps a quarter or less.
I've wired up a couple of PICs and so on in my time, but have very
little experience with basic analogue electronics. Is it possible to
build a passive circuit (perhaps a network of resistors?) that will
"convert" the resistance of the sender so that it's proportional to the
amount of fuel remaining rather than the height of the sender float?
Some sketches of the tank here: http://verdonet.org.uk/stuff/tank.png
Note that the lame attempt at isometric projection actually makes the
tank seem squarer than it is - the slope is in fact very pronounced
indeed, it's basically wedge-shaped in all three dimensions, plus a
sneaky little change of gradient in one side.
This shape means the perfect correction is probably quite complex - but
all I really want to do is pin 3/4, 1/2, 1/4 in their right places. I
can pour water through the tank before installing it, to measure the
"observed" resistance at those points.
Note the "orientation" of the sender is indeed 0 ohms when empty - I
believe the other way round is normal in the US but I haven't made a
mistake! I could special-order the American sender if required, but it
would then make my gauge read backwards unless flipped again as part of
the circuit.
Thanks for any help or suggestions,
Pete
I'm planning to fit a fuel tank with a contents gauge. The sender that
the tank manufacturer will put in is essentially a variable resistor,
which reads 180 ohms when the tank is full, and 0 ohms when it is empty
with, presumably, a linear resistance in between. There is a gauge that
connects to this - presumably an ammeter and a means of passing current
through the sender?
My problem is that the tank is not square, and so the height of the
sender float is not directly proportional to the amount of fuel actually
remaining. More unfortunately, it's actually "the wrong way round", so
that when the gauge reads half-full there will actually be significantly
less than half a tank of fuel left - perhaps a quarter or less.
I've wired up a couple of PICs and so on in my time, but have very
little experience with basic analogue electronics. Is it possible to
build a passive circuit (perhaps a network of resistors?) that will
"convert" the resistance of the sender so that it's proportional to the
amount of fuel remaining rather than the height of the sender float?
Some sketches of the tank here: http://verdonet.org.uk/stuff/tank.png
Note that the lame attempt at isometric projection actually makes the
tank seem squarer than it is - the slope is in fact very pronounced
indeed, it's basically wedge-shaped in all three dimensions, plus a
sneaky little change of gradient in one side.
This shape means the perfect correction is probably quite complex - but
all I really want to do is pin 3/4, 1/2, 1/4 in their right places. I
can pour water through the tank before installing it, to measure the
"observed" resistance at those points.
Note the "orientation" of the sender is indeed 0 ohms when empty - I
believe the other way round is normal in the US but I haven't made a
mistake! I could special-order the American sender if required, but it
would then make my gauge read backwards unless flipped again as part of
the circuit.
Thanks for any help or suggestions,
Pete