Hey, Guys, I'm stumped It

Just guessing but:

lets assume your leads have about .15 ohm resistance.

then its a 0.22 R from 1 to 2, a second 0.22R from 1 to 3 (they both read about 0.37) so from 2 to 3 its 0.44R + the lead reistance of 0.15 giving a readfing of 0.59.

MK

If you short 2 and 3 do you get 0.22 ohms between 1 and the short?

Yes, 0.22 is tricky to measure.
it isn't easy to measure these very low resistor values anywhere close to their real value with 2-wire meters. Even if you push hard on the contacts.

"The measurement of very large or very small quantities is always difficult, and resistance measurement is no exception. Values above 1GΩ and values below 1Ω both present measurement problems."

"When measuring larger resistance values this additional lead resistance error can be ignored, but as you can see from the chart below, the error becomes significantly higher as the measured value decreases, and totally inappropriate below 10Ω."

 phoenixcomm , does it have pins at the other side too?

(ah, saw in your post that you mentioned them)

https://www.shutterstock.com/image-photo/cement-power-resistor-three-pins-on-1570576969

phoenixcomm said:

and Please look at the logo, so I can figure out who made the stupid thing..

Looks similar to the old Pana Sonic 

You seem to be spot on with the logo (source):

Thank you dave

I was thinking the same except I would suggest shorting pin 1 to pin 2.  This would put the 0.38 OHM resistance in parallel with the 0.6 OHM resistance which of course is very close to the 0.22 OHM.

Apologies for the bad drawing, I had to do it in MS paint without a stylus/pen.
Assuming the connections are as shown below

R12 = R13 = 0.38 ohms
And R23 = 0.6 ohms.
However as per the specs, the entire resistor should be 0.22 ohms.
As mentioned by dowg, if you short 2 and 3, the resistors will be in parallel and value would be 0.38/2 = 0.19 ohms. 

R23 = 0.6 ohms also makes sense, as both resistors will be in series and so 0.38*2 = 0.76 which is close to 0.6 with not so precision DMM.
Also, to measure such small resistance values, you would need precision ohmeter/DMM otherwise the lead resistances from the probes will not give you a correct reading.

As you've got two meters, perhaps just do your own four-wire measurement.

Set a PSU to constant-current, use one of your meters to measure the current through the resistor, the other to measure the voltage across the resistor, and work out the resistance from Ohm's law.