In that situation, not only would you be descending, but your rate of descent would be increasing, because the net upward lift is less than the mass of the airplane. Therefore gravity takes over and the airplane accelerates downward.

Yep. I think the issue is -steady state=, not =level, climbing or descending=.

Leave out the steep turn for a moment. In terms of the VSI, if the VSI is =constant=, we're at 1 G, whether level, climbing or descending. If the VSI is =changing=, an upward change is an increase in load, a downward change a decrease in load.

The key to understanding this is to understand that there is a difference between a descent in which the =rate= of descent is the =same= (say 500 fpm) and a descent in which the rate of descent is =changing= (say 50 fpm the 1st second, 100 fpm the second second, 150 fpm the third second, etc)

That's easy to demonstrate in an airplane in flight. Push down aggressively on the yoke and you'll feel the negative G (decreased load); let the airplane settle in a 500 fpm descent and, once established, the G will go back to 1.

The steep turn adds load. But it adds load to the load that otherwise exists.

If the load factor is 1 - whether level or in a stable 500 fpm descent, you are increasing the load (and you =are= pulling back on the stick in order to maintain that 500 fpm descent)

If, OTOH, the load factor is, say -1, because you are in a descent that is constantly increasing, then the load factor is less than 1 and the steep turn will tend to neutralize it.

There's actually a maneuver that was in the private PTS for a while a few years ago that demonstrates this: the emergency descent. The first step in the procedure (after reducing power) was to aggressively push forward on the stick to start the descent and add a 45° bank to neutralize the negative Gs. Then, once you stabilized the descent rate, you needed to wither decrease the bank or increase the back pressure.

I'm sure tgrayson will step in with a better, more technical answer.