Peter wrote:
Even an atomic bomb going off is in equilibrium. The radiation pressure is balanced by the acceleration of the various bits of mass departing in all directions. The radiation pressure is “considerable” but that’s not the point, either
And I think it’s a bit unhelpful to think of equilibrium that way (it kind of loses its meaning). In terms of aircraft, everything is in equilibrium so long as the first derivative of velocity (acceleration) in all directions is zero. You are briefly not at equilibrium when you initiate a climb, or while changing the forward speed of the aircraft etc. – at those particular times, one force is greater than its opposing force for at least a brief period of time. The rest of the time they all balance out.
LeSving wrote:
You can decompose forces any way you like.
No, you can’t. You need to break down mg, in mg_x’ and mg_y’, but relative to the stream velocity. I was a bit unclear about that. mg_y’ will be perpendicular to the stream velocity. Lift only needs to counteract mg_y’ (y = the perpendicular to the stream velocity, not the vertical). mg_x’, the mg component that acts in the direction of travel, is counteracted by the increase in thrust. That then results in the only correct answer that lift is smaller than weight.
LeSving wrote:
Then you pitch up again, until the thrust vector is parallel to the x axis. Now the aircraft will gain altitude rapidly, but the speed will be lower than 150 kts.
I think you’re creating a theoretical situation that in practice is never used. Your scenario involves an aircraft that in cruise has a thrust vector pointing down. Or if you want to put it in another way, an aircraft that uses a high speed for climbing, but a slow speed for cruise. That would be very inefficient, and normal operation will not involve a scenario like that. I’m starting to think that in theory you have a point, but from a practical flying perspective it does not make sense, and I can see why that answer is counted wrong.
Archie wrote:
No, you can’t
Of course you can. You can transform it to a coordinate system on Mars if you want (it will be a rather complex transformation, but perfectly doable and “legal”). What you are thinking about is transforming all forces to a local coordinate system following the free stream velocity. Nothing wrong with that, it’s just that it becomes difficult to see the forces when the aoa changes.
Below I have made the same thing in “earth” coordinates and “aircraft coordinates” where aircraft coordinates have x axis along the longitudinal axis. The forces becomes exactly the same. Just put in some numbers and see for yourself. The “cruise climb” example is essentially the same as “aircraft coordinates” since for the special case the coordinate system coincide with earth coordinates.
Archie wrote:
Or if you want to put it in another way, an aircraft that uses a high speed for climbing, but a slow speed for cruise. That would be very inefficient, and normal operation will not involve a scenario like that
This is the normal situation for a high performance aircraft, like an RV for instance (high power to weight ratio). It climbs like a rocket (2000 + ft/min) at best climb speed, the top speed is 4 times the stall speed, it is very efficient and can “cruise climb” at high velocity. Also, many modern microlights are like that. It is just a result of having high power to weight ratio and an efficient aerodynamic airframe. An RV-4 will fly perfectly well with a 100 HP Rotax 912, but it’s much more fun with a 180 HP 360 
Archie wrote:
Yes, but now imagine the F-16 wanting to climb (the question doesn’t address S&L flight, it talks about a steady climb). The F-16 will need to pitch up, or increase power
Only as a transient. Once established in a climb (at a higher airspeed), the pitch could be much lower than 30 degrees up, and the thrust could very well be smaller also since the power it takes to fly at a 30 degrees angle of attack is considerable. I still think this is the greatest weakness of the question – the thrust vector can be angled differently depending on CL (and thus speed), and that overshadows the whole point in the question.
I agree that under normal circumstances, where climbs are performed at a more nose-up attitude than level flight, as both the direction of travel and the angle of attack is turned nose-up, the question and its right and wrong answers are arguably meaningful.
But the point of the F-16 scenario was to give an example where the assumptions for the question do not hold true. Another example could be if you increase the speed from level flight leading to a climb without changing the airplane’s attitude. This is LeSving’s scenario in #30 above. In that case the trust vector could be assumed to be unchanged, while the lift would have to counter both weight and the downward component of the drag, so lift would have to increase.
The question is interesting, it has a point, but the point does not seem to survive variations in the set-up, and not being all-important for a PPL pilot to understand, the question is still not going to make it to the question bank in my country.
LeSving wrote:
Of course you can.
You can, but you need to do it right. Your breakdown of the lift vector is not correct. This is how the principle works:
Lift is less than weight in a steady climb.
huv wrote:
Another example could be if you increase the speed from level flight leading to a climb without changing the airplane’s attitude. This is LeSving’s scenario in #30 above. In that case the trust vector could be assumed to be unchanged, while the lift would have to counter both weight and the downward component of the drag, so lift would have to increase.
Yes, but again a very unusual scenario. Tell me which aircraft has a higher IAS for climb than for cruise in it’s POH. Tell me which pilot accelerates to climb (whilst pushing o the yoke to maintain the same nose attitude) and then slows down again to cruise (whilst releasing forward pressure on the yoke). LeSving tries to tell us that is the norm for RV’s. I don’t believe him. It’s very, very unusual and whilst perhaps theoretically possible, and if this comes up in a PPL student’s mind, he has either been trained wrong, or he is not doing what he is taught to do.
p.s. LeSving, notice now that in your second drawing in post #33 you did draw the vectors relative to the airflow. Like in the above picture, it becomes obvious that Lift only needs to be as large as mg_y, which is less than weight.
Archie wrote:
You can, but you need to do it right
You never give up! It is 100% correct. It doesn’t matter how you do it. The ONLY reason lift (of the wing) can be smaller than mg in a steady climb is that something else takes part in the lifting process. How you decompose the forces has nothing to do with it. That is pure geometry, transformation from one coordinate system to another. We do that to make things easier to calculate and understand, but you cannot fool good old Newton with it
Now, that something else is usually thrust from the engine. It could also be pull on the towing line for a glider. If the thrust does not point up, then the wings have to take care of it. The main point is that thrust does not have to point up for the aircraft to climb (in a steady climb). This is the normal situation for a glider being towed, and in particular for a glider being winched. It is also the situation for any aircraft doing a “cruise climb”.
I agree the normal climb for an aircraft is with a high nose, thus thrust pointing upward unloading the wings. This will give best climb rate and also best climb angle, because this is the most efficient. But, it is not the only way, and that is the main point, and why the original question is poor. It lacks definition.
Archie wrote:
Tell me which pilot accelerates to climb (whilst pushing o the yoke to maintain the same nose attitude) and then slows down again to cruise (whilst releasing forward pressure on the yoke). LeSving tries to tell us that is the norm for RV’s. I don’t believe him.
I never said such a thing. I said the speed will be slightly slower in “cruise climb” compared with the 150 kt cruise. It s just a matter of how much excess thrust you have.
LeSving wrote:
It is also the situation for any aircraft doing a “cruise climb”.
LeSving wrote:
I never said such a thing. I said the speed will be slightly slower in “cruise climb” compared with the 150 kt cruise. It s just a matter of how much excess thrust you have.
A “cruise climb” is usually not defined as “climbing with the nose pitch attitude zero”. It’s a faster climb, but not as fast as a cruise, somewhere in between. In a “cruise climb” all aircraft will have their nose pitched up a little or more, in other words, thrust takes up part of the weight, and the wings are unloaded slightly, and lift is less than weight.
If the speed of an RV is slightly slower in the climb as you say, it means the nose has to be pitched up, (in fact the pilot will have to have trimmed slightly nose up for the speed to decrease), meaning thrust again will take up part of the weight, and the wings are unloaded, and lift is less than weight.
I now agree with you on the exceptional situations (i.e. a glider winch up, or a very high speed flight), but I think it’s pretty obvious the question doesn’t ask about a glider winch-start, not? In every normal climb, the nose will be slightly or more pitched up depending on the type of climb, and lift will be less than weight.
Think about it again, if the engine is pointing horizontal during climb as you say, it will be pointing down during any type of cruise (when you reduce power to stop climb). This in itself is very inefficient. You can correct for this by trimming nose up, and reducing power slightly, until the engine is pointing straight forward again, but now you are cruising at a lower speed than the speed that you were climbing, and I have never seen an aircraft operated like this, or a pilot heard talking about this. So whilst your theory might sound true, in practice it provides no valid answer to the question.
We may just have to agree to disagree on this…
Archie wrote:
This in itself is very inefficient
I don’t see efficiency being part of the question. Besides, engines are typically pointed down 1-2 degrees anyway and also pointing to the right the same amount. An RV is set up pointing to the right and zero degree inclination from the horizontal axis. This will soon end up being 1-2 degrees down because the rubber mounts will sag slightly. This also have to be seen in relation to the inclination of the wing. RV 6 and 9 have 1 degree inclination while the 3, 4 and 8 have 0.5 degree. An RV-6 will therefore cruise with more down-thrust than the -4 even though they have the same wings and the same tail. The reason is Vans thought the -6 and -9 would be used more for cross country and high altitude cruise and be more heavily loaded than the tandems, which would be used more for acro and shorter trips with lighter load. With 1 degree, the -6 and -9 will cruise more efficiently when heavily loaded and at high altitude while the others will cruise more efficiently when loaded light and going faster at lower alts. Talking about efficiency with this 1/2 degree is of course rather academic for a light aircraft, but you see the point here; Going faster than optimal cruise speed will make the thrust vector pointing down and this optimal is different for the -4 and the -6 and varies with load. How much down is purely academic, the point is that an aircraft can climb steady, and the lift produced by the wings is larger than the weight.
Or you can simply set out a notch or two with flaps.
The physics of the question is confused. It is not climb than unloads lift from the wings, it is thrust. The question should be reformulated or be more precise with more information.
Hey, we’re talking PPL student level here 
Can I recommend this website to anyone who wants to learn more about the physics and forces acting on an aeroplane in flight:
p.s. that’s interesting info with regard to how RV’s are built. A slightly different topic perhaps, but when you say the inclination of the wing is 1 degree, and the engine is down 1-2 degrees, in relation to what are you talking here? I’m asking the question, because the normal cruise AoA is ~4° for pretty much every aircraft. Where are the engine and the wings pointing in that case? Mind you the RV is a pretty unique aircraft in many respects.
