> when I am in a hurry... measure the cold winding resistances initially, then load the trafo and periodically disconnect and measure again. Plug in the tempco of copper and you have an average temp change.
How do you do it when you are NOT in a hurry???????
> measure the winding R
Yes, omitting the "in operation" part, that's a good first-guess. For 10VA-20VA windings, you want copper resistance no more than about 20% of load resistance (assuming a reference resistor load; we'll correct for rectifiers later). The secondary resistance can be measured directly (DVMs may not like high-Volt windings; use analog). You could measure the primary resistance and do math to reflect it to the secondary; in all practical situations you can assume the (reflected) primary resistance is similar to the secondary resistance, so just double the measured secondary resistance.
> I assume the thickness of the wire used would limit current
Usually not directly. Transformers can pass a LOT of current short-term: we are usually nowhere near the point that the wire melts instantly. The "12V 1.5A" transformer below can probably put 6 Amps into a short-circuit. Of course that's zero power in the "load", and 90 Watts of heat leading to 560 degree C temperature rise.... we can't do that for very long. Bigger transformers can dump relatively more short-circuit current, and at the 1,000VA size can do it for many-many seconds. (A good 10,000VA pole-transformer, rated 240V 40A, can probably dump well over 1,000A or 240,000VA for 5 or 10 seconds, except they are fused to protect the investment.)
> a current hungry tube
A) That's the hard way
B) Tubes suck DC; the transformer is AC
Use power resistors. An assortment of 10W resistors is cheaper than a cheap tube.
Example:
Transformer gives 15VAC no-load. Secondary resistance is 1Ω. Assume the (reflected) primary resistance is the same. Total resistance is 2Ω. This should be no more than 20% of load resistance, or load resistance should be at least four times higher than transformer total resistance. 4*2Ω= 8 ohms load. Output voltage drops just like a resistor voltage divider: 8Ω/(2Ω+8Ω)= 0.8 times the no-load voltage, 0.8*15V= 12V loaded. And now the AC current will be 12VAC/8Ω= 1.5 Amps AC. The heat dissipation inside the transformer is roughly 2Ω*1.5A^2 or 4.5 Watts. Such a transformer is probably 2"x2"x2", surface area 16 square inches. Taking 100 deg C per Watt per square inch, we have 100*4.5/16= 28 degree C temperature rise, not very hot.
That's the rating for AC resistive use, heaters or lamps. With the popular rectifier-capacitor method of making DC, the voltage goes up 1.4 times and the current flows in BIG spikes which cause more heating than you would expect. For DIY, the DC current should be no more than half of the AC current rating. So 1.5A AC, only pull 0.75A DC. (It is possible to derive a number like 0.6 or 0.8 and "prove" that you can run a little more DC. In practice, small transformers will sag too bad to allow pushing much past 0.5. And DIY should always be rounded-up, or else we might as well buy penny-pinched commercial gear.)
