According to table 400.5(A)1 10 gauge SOOW is good for 25A (Column A). You can't use Column B because there are more than two current carrying conductors.

SOOW is cord by the way, not cable.

Thanks for the quick response...very helpful! Based on the table it seems the 10/4 cords sold by generator companies are in fact 25A cords rather than 30A (even though they have 30A connectors), is that a safe assumption? If so I would rather put together the longer cord, if all else will be equal, minus a small increase in voltage drop.

Thanks again for the help!

The NEC delegates the responsibility for pre-made extension cords to UL.
I'm not really up on what UL's rules are for conductor sizes and connectors.

They may alaso be playing games with the "7000 W."

Just like fantasy horsepower.

At 100% efficiency ~746 W is 1 HP.

That means you would need at least a ~9.4 HP engine.

When include inefficiency (alternators are pretty far from 100%) the horsepower can increase significantly.

I would be very surprised if you can run 7,000 W of resistive load, let alone inductive.

Thanks again for the input! Yes Brick I sort of took that 7000W number with a grain of salt. While the engine is a 13/14 HP Honda, I'm sure they stretch those ratings for the absolute best scenario etc. That being said I'm starting to think the generator isn't going to be cranking out 30A continuous anyways and the 25A cord is likely sufficient.

Thanks again.

Scott

It probably puts out 7,000 W just as the engine grinds to a halt.

I have a 3 HP Baldor motor on my table saw, that by its locked rotor KVA code is 'Sears' rated over 60 HP.

The motor shaft would fail at that kind of load (it is only 3/4 inch).