How to estimate average current draw?

I'd like to get some insight into estimating the current draw of a welder over a period of say 30 minutes of normal welding. It's a little unusual of a concern, I found out that my power company has a charge for going over a specified current demand over any 30 minute period during the month. It can be just one time, and they add a charge that more than doubles my bill. I'm trying to estimate possible current demands of any of the individual machines in my shop during a realistic 30 minute period, and see if it could add up to what they say I've been using. So far I doubt it, but the welder is still magic to me and is the biggest unknown.

Machine and typical settings:

210EXT
240v
DCEN
65 amps
using foot pedal for pulse-and-dab style

My logic is this so far:

The specs in the manual for this machine show the TIG DC output voltage max at 18.4v. With 65amps, that would be about 1.2Kw.
If my pedal motion was close to a sine wave, and going all the way to zero, then I'd expect average current draw to be 1/2 that. I know it doesn't go to zero usually, but also don't usually go to 100% either. Conservative guess might be that average draw, with pedal motion, might be 3/4 of the possible max which is 0.9kw.
For the actual physical duty cycle over 30 minutes, I can't imagine that being much higher than 25%, considering time between individual welds, moving the next part into place... which gets down to 225 watts average.

Does this seem like correct logic?
Does anybody have better estimates of how you might expect the pedal to effect this, or actual duty cycles?
Has anybody actually measured current draw while working over some period of time?

Voltage depends on arc length. It can go beyond that. Wattage is not true because of loss of efficiency as well. Plus you have inrush loading etc.

As for the rest of it, it seems kind of convoluted circular stuff. I can't follow it. Duty cycle is based on Thermal. It's based off of a defined 10 minute window...nothing more, nothing less. It's not exactly linear.