Forces in a descent

[mathewc]

While presenting Ex. 8 today, I was informed that the aircraft is NOT in a state of equilibrium while in a power off descent. Now, FGU states "Of the four forces, thrust is now absent and a state of equilibrium must be maintained by lift, drag, and weight only." It would seem to me that in a power off descent while maintaining a constant airspeed, the aircraft will be in equilibrium due to the fact there is no acceleration and the forces are balanced (re: fig. 35 FGU p.34).

This website also claims that the forces are in equilibrium: http://www.auf.asn.au/groundschool/umodule1b.html

One of our instructors tried a mirror image of the lift and drag square and this shows how the horizontal component of lift (thrust in this case) is unopposed by drag. This just seemed to confuse me more.

Can someone please explain to me how an aircraft in a constant airspeed power off descent would NOT be in equilibrium?

[iflyforpie]

I'm pretty sure it is in equilibrium.

You aren't speeding up, slowing down, or changing direction. Thrust is replaced by the portion of weight that acts parallel to the aircraft's flight path.

[BTD]

Agreed, Equilibrium. There is no acceleration on the aircraft. The fact that the earth is there and you are closing in towards it is irrelevant.

[Cat Driver]

If you don't know the answer will you just remain stuck in space and never get back to earth when you lose all power?

[Crusty]

+1 with iflyforpie. For some reason many people never get this. It's in the same vein as people telling me that airplanes climb "because lift is greater than weight". That always makes me cringe.

[Shiny Side Up]

If you don't know the answer will you just remain stuck in space and never get back to earth when you lose all power?

Only if you miss the earth when you throw yourself at it.

[loopa]

Whenever the aircraft is in a state of a CONSTANT CLIMB, or DESCENT, the overall system is in equilibrium. Why do I say the overall system?

Looking at the airplane in a whole, the airplane is in equilibrium. However, some forces are smaller than others, and other forces are larger. But together they add up to an equilibrium state.

Example. Our four forces are LIFT WEIGHT THRUST DRAG.

In a constant climb.

Is Lift equal to Weight? Is Thrust = to Drag? The answer to both of those is NO. Why?

The rearward component of weight acts as drag. So in other words, the THRUST has to overcome the rearward weight that is produced by the airplanes current attitude.

So far, we have established that THRUST > DRAG

Now why is LIFT < WEIGHT?

Quite simple, Lift is produced perpendicular to the chord line. When the aircraft is pitched upwards, the LIFT vector no longer cancels the weight. So then in THEORY, the plane should be descending right? But why isn't it?

Well let's think for a second, AHA !! Thrust is acting on an angle, hence it has a FORWARD and a VERTICAL component. The Vertical component of thrust acts as lift and hence you maintain the desired state of equal balance.

Just to summarize if the last bit didn't make sense.

THRUST > DRAG
LIFT < WEIGHT

The above is in EQUILIBRIUM. Why? Because the overall quantity of the 4 vectors, Lift, Weight, Drag, and Thrust add up to 0 Hence why the plane is in a constant climb.

This is my theory on it... see if you can figure out why the airplane is in a state of equilibrium in a constant descent.

Remember, now we don't have THRUST. So how else can we balance the forces of the airplane. We have LIFT, WEIGHT, and DRAG. I'll give you a hint, you use LIFT and DRAG in order to equate to Weight... I pretty much gave it away.

Cheers !

[mathewc]

Wait wait, hold up here. I AGREE that it is in equilibrium. It makes perfect sense that it is in equilibrium.

What I don't understand is how the forces in a descent ARE NOT IN EQUILIBRIUM. My CFI is telling me that yes in a climb, the aircraft is in a state of equilibrium, but the forces NOT in equilibrium in a descent.

It makes sense that if you are in unaccelerated flight, the forces balance out. In a descent the forward component of lift counteracts the force of drag resulting in a constant airspeed. The resultant force of lift and drag opposes weight so the rate of descent is constant.

I do appreciate the input into this problem. Thanks!

[ETOPS]

First of all, lets establish that we're talking about a steady state climb/descent.

My CFI is telling me that yes in a [steady]climb, the aircraft is in a state of equilibrium, but the forces NOT in equilibrium in a [steady]descent.

It then follows that EDITED or something has been miscommunicated.

[loopa]

Wait wait, hold up here. I AGREE that it is in equilibrium. It makes perfect sense that it is in equilibrium.

What I don't understand is how the forces in a descent ARE NOT IN EQUILIBRIUM. My CFI is telling me that yes in a climb, the aircraft is in a state of equilibrium, but the forces NOT in equilibrium in a descent.

It makes sense that if you are in unaccelerated flight, the forces balance out. In a descent the forward component of lift counteracts the force of drag resulting in a constant airspeed. The resultant force of lift and drag opposes weight so the rate of descent is constant.

I do appreciate the input into this problem. Thanks!

You just answered your own question.

The FORCES are not in equilibrium. Name one reason why they wouldn't be in equilibrium... I think a pretty good reason to start with would be that you don't even have THRUST to begin with... But if you revisit what I mentioned earlier by saying that the OVER ALL SYSTEM is in equilibrium, you will see that you don't need all forces to be in equilibrium in order to achieve equilibrium state of an aircraft. Why? Because you can use the other 3 forces at hand to do something called, ENERGY MANAGEMENT. You can use the other forces to achieve or "MANAGE" a TOTAL state of equilibrium.

Look at page 34 in the FGU at FIG35. Lift acts forward of weight. You see the RW line? It's basically the resultant lift and then the weight vector. I want you to take the RW LINE, and with your pencil, draw the same line from L downwards and call it LS. You will see that there is a gap between W and S. The distance from W to S is the "forward momentum" produced by weight hence giving you forward momentum or so called (forward speed). The resultant of Lift and Drag gives you a vector that counter acts weight and thus, maintains a constant vertical speed. With me so far?

Now, how do you achieve this state in an aircraft? How do you put the airplane into such a profile where you manage your lift in order to get the constant descent? You do something called Pitch. If you lose thrust, maintaining the same attitude will result in a loss of airspeed. If not corrected for, eventually an increase in vertical speed. So what do you do in order to maintain the same speed when you lose the THRUST vector? You pitch down, and maintain a balance between LIFT and DRAG. Once you find this balance, you trim for it, and maintain that BALANCE. With this BALANCE you have a certain rate of descent. Now, replace BALANCE with the word "forward speed" and the equation to why the plane is in equilibrium is solved.

[777FO]

wow, i cannot believe a CFI would know such things. the cfi where i work would hear this question, delegate it to the asistant cfi and run out the back door to tim hortons with his tail between his legs. but i guess thats what happens when retired hobbyists play pretend at advanced ftu\'s.

[reunionnais97432]

In a glide without Thrust, the Weight component along the flight path must supply the propulsive force and balance drag. Indeed, in a glide there are only 3 Forces (lift, weight and drag).
The forward component of weight(W sinY(gamma)) is a product of descent angle (Y); the greater the descent angle , the greater the forward component of weight.
The forward component of Weight must balance Drag for the aircraft to be in steady glide. It follows if Drag is reduced and Lift remains constant, the required balance of forces can be achieved at a smaller descent angle.

[Oholio]

Only if you miss the earth when you throw yourself at it.

42 is the answer...

[. ._]

If you are in a steady descent and the ground comes up, pull off the power, pull back on the stick, and wait for the mains to slam on the runway.

That's all I know.

Is equilibrium on any of the TC tests? I forget.

-istp
250 TT
Comm Multi-IFR

[mathewc]

Alright, I think we're all on the same page here. Thanks for the replies!