JowettTalk

Would any body know what fuel
sender if any fits the Javelin tank ?
A early MGA was someone idea but
is this more of a match with the Jowett gauge.
were as I was trying to get one to line up with the tank
flange hole wise my new gauge is 0 to 180 ohms

Yes, just about any '50s to early '60s Lucas sending unit fits the tank. The length and the position of the float arm needs to be modified. Older plastic floats will not survive ethanol and should be replaced. Moss Motors sells them.

The Moss Motors replacement unit tends to be rather poorly made and not worth the time or money.

The unit should operate in the 0-70 ohms range. Anything else will give inaccurate readings. The gauge can be tuned to operate with any unit, but requires great care not to damage very fine wires inside the gauge.

Add a ground (earth) directly to one of the mounting screws of the unit to assist in proper function.

For anyone interested in a multi-page dissertation in getting your sender/gauge to function well: https://mgaguru.com/mgtech/electric/fg110.htm

Hi,

Just had a look on eBay. There seems to be some modern senders available. Look for eBay reference number 333139607425. These are around £11, complete with gauge, which you probably don't need?

Not sure what the resistance reading of the potentiometer would be, but might be worth a look?

Not original I know, but who would know what you have buried in your fuel tank?

All the best,

David

I would probably go for the ones with a float around a column such as this
300mm-Fuel-Level-Sending-Boat-Marine-Truck-Water-Level-Sender-Sensor-240-33ohms
The length would need to be correct for the tank of course whereas there is some adjustment, I would think, on the arm and float type.
Getting the resistance correct needs some elementary ohms law, but if Scott is correct and the gauge needs to see a resistance of 70 ohms for empty and zero for full, then you would need to supply a different voltage to the gauge to get the desired reading when empty and put a resistance in parallel with the sender to reduce the 30 ohms when full.
The resistance of the meter I measured is 100 ohms.
The extra resistor in series with the gauge is R then the combined resistance when full is 100+30xR/(30+R)

The voltage required is (100+30xR/(30+R)) x .12 with the above sender when on full. By putting R from below gives V= 15.16842934

When on empty this voltage applied should give the same reading with the old sender. However, the combined resistance is now 100+240xR/(240+R)

The equation now becomes .0706 amps=( (100+30xR/(30+R)) x .12 )/ (100+240xR/(240+R))
Moving the divisor across and multiplying by R on both sides.
.0706((100+240xR/(240+R))= (100+30xR/(30+R)) x .12 )
.0706(240xR)-.12x(240+R)(30xR/(30+R))=(1185.6+4.94R)
16.944xR)(30+R)-(240+R)(3.6xR)=(1185.6+4.94R)(30+R)
508.32R+16.944RR-3.6RR - 864R=35568+148.2R+1185.6R+4.94RR
The quadratic equation becomes.
8.404RR-1689.48R-35568=0
RR-201.0328415R-4232.270347=0
R=100.5164208+_SQRT ( 10103.55084 +4232.270347)
R=100.5164208+_SQRT ( 14335.82119)
R=100.5164208+119.7322897
R=220.2487105

So in conclusion the resistance going to earth from the sender side of the meter needs to be 220 ohms and the voltage needs to be raised to 15 volts. This could be done with some inverter electronics.