Just This Once, I wrote the below on a commute before I read your last post, which is mostly right with the trig and it seems like you’ve been playing coy on us with a better understanding of equilibrium and acceleration that you had let on. I think in the last sentence you meant to say that weight (not lift) is partly working in opposition to drag. Lift and drag are always perpendicular to each other by definition, so no component of either can add or subtract to the other.
Bottom line, you must agree that weight, no matter the descent rate, remains the same and continues to equal the sum of the upward forces (from jet thrust and wings) in a VTOL vertical descent.
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A little while ago I identified the exact confusion that’s happening, and there’s a grain of truth in what you’re saying, but not valid in the way that you think. I hesitate to clarify it, because it’s likely that you’ll miss the subtlety and go “aha, I was right all along and here you finally admit it!”
But anyway,
You connected the discussion about forces in vertical flight to the more typical (and ingrained in us) scenario of reducing thrust to introduce a descent in normal airplane flight. It’s of course true that more thrust is required to climb, and less to descend. As I showed earlier, for a descent, that’s a consequence of drag. In an airplane-centered frame of reference, the more descent, the more weight is aligned with your thrust, the less thrust required to offset the drag. Put another way, the more descent, the more drag resists your vertical motion, the less thrust you need.
Your reasoning is: well since descent=less thrust is true in mostly-forward flight with a descent, why not in vertical flight? Let me remind you here that even though thrust is less, all the forces are still in equilibrium!
(The missing thrust now is balanced by drag that wasn’t there before. This is further complicated by that the total drag for the same IAS is still the same. So how can drag be the same yet more at the same time? The total vertical force is now Lift times the cosine of the descent angle, plus Drag times the sine of the descent angle. Or, the same reason that in a quartering crosswind, the headwind and crosswind components don’t add up to the total wind)
If we take it to an extreme case of a vertical descent (still in a normal airplane, 90 degrees pitch into the ground) all this talk of trig and components simplifies out. Total downward force = thrust + weight. Total upward force = drag. For a constant downward speed, since we must be in equilibrium, Thrust+Weight = Drag. Simple as that. Still in equilibrium, still at 1G (into your straps, not indicated on the G meter).
Now we move into a VTOL in purely vertical descent. Being that it obeys the same laws of physics, as long as it’s descending at a constant rate, the forces must still be in equilibrium, meaning that the total upward force must still equal the total downward force. Now there is only one downward force, weight, which is unchanging. The upward force is of course there, and part of it is thrust, with the remainder being drag (drag now acting upward). (It’s the same 3 forces as the last scenario, but with thrust acting upward instead of downward.) Weight = Thrust+Drag.
What this means is that the faster the descent rate, the higher the drag, the less thrust is needed. But the total upward force (thrust+drag) is still the same! The subtlety is that at descent rates near landing, the descent rate is miniscule compared to the other scenarios. The associated drag is correspondingly miniscule, and essentially all the weight is balanced by thrust. But keep in mind that even if you take an exact accounting, exactly all the weight is balanced by the total upward force.