View Full Version : What happens to lift in a climb and why?!?
21st Dec 2003, 04:18
I am a little confused - perhaps someone could explain/ tell me where to find the answer...
When an aircraft is in level unaccelerated flight, lift equals weight. But, when an aircraft is in an unaccelerated vertical climb, lift has decreased to virtually zero - is this so? If this is the case, how is the lift decreasing, if lift acts at 90deg to the relative airflow - which is acting vertically downwards in a vertical climb, and the aircraft is unaccelerated? I understand, from a vector diagram that the lift decreases but how is it doing this??
Why, well, how does an aircraft produce less lift in an unaccelerated climb than it does in level, unaccelerated flight, if the speed in both situations is the same?
I hope my questions have made sense...
PS Happy Christmas y'all!
21st Dec 2003, 04:39
It's all to do with angle of attack of the wing section to the direction of airflow.
In a fully symetrical wing with zero AoA, lift is zero. It therefore needs to be angled slightly relative to the direction of airflow in order to produce lift.
In a more typical wing section, or airfoil, the higher curvature of the upper surface will produce some lift at zero AoA.
So in your scenario where the aircraft is in unaccelerated vertical flight, for the (non symetrical) airfoil to produce no lift, it will need a slightly negative AoA.
22nd Dec 2003, 05:16
I think this answer has something to do with the thrust vector. I remeber this Q from a long time ago. Obviously written by someone with too much spare time on his hands.
Simply put, the thrust in level flight is horizontal. If not, in the climb for example, there is a vertical element. The vertical element generates a small upward reaction, therefore less lift is required in the climb and therefore less of an angle of attack is required.
If you have problems getting your head round this, consider an aircraft with more thrust than mass - i.e an F15. The pilot can maintain a vertical climb where all the weight is being supported by the engine thrust and none by the wings. Therefore the lift generated by the wings is zero and the anle of attack zero as well.
Remember that the rearward component of lift also acts as drag.
Saw a great thing on Discovery wings ... some of the airshow acro pilots in prop planes can just about hang vertically on the prop .... that's cool as anything. :ok:
23rd Dec 2003, 06:44
Since we're talking about unaccelerated flight then there must be no excess off force in any direction so if one force substitutes for another, then the other must reduce by some fraction.
I think you'll find it easier to understand if you think of the forces as:
Weight: Acts vertically towards the centre of the Earth ie downwards
Lift: Acts at 90 deg to the flight path. The lift is produced by having the aircraft maintain an attitude that gives the wings an angle of attack, thus producing the lift. If the wing has appropriately shaped asymmetrical upper & lower surfaces then Lift can be produced at a reduced angle of attack. Reduce the angle of attack enough (even to a 'negative' angle of attack in the case of a 'normal' asymmetrically shaped aerofoil) and lift reduces to zero.
Drag: The cost of moving a body through the air and of producing lift. Combination of all forces acting rearwards along the flight path.
Thust: Acts forwards along the flight path.
In level flight - a horizontal flight path - the usual situation is:
Lift = Weight and they act at 90 deg to Thrust = Drag
So, the aircraft's weight is being opposed wholly by Lift. Any rearwards forces (Drag) is countered by Thrust.
Now think of a vertical flight path (aircraft pointing 'up'):
Weight: Still acts downwards. This happens to be rearwards.
Drag: Still acts rearwards along the flight path
Thrust: Still acts forward along the flight path. This is now 'up'.
Lift: Still by definition 'a force acting at 90 deg to the flight path'...BUT (!) if there was any amount of force acting at 90 deg to the flight path then it would cause the flight path to deviate from vertical. Looking from outside the aircraft there must be NO horizontal forces for the flight path to be vertical. To achieve this the fuselage is held in an attitude that ensures the wings are at an angle of attack in the relative airflow that results in zero lift.
So: Weight AND Drag both act downwards and at the same time rearwards along the (vertical) flight path. Thrust is the only force that is acting upwards so it must be opposing BOTH Weight & Drag.
In effect Thrust has completely taken over the job that Lift used to do, while still having to do its original anti-Drag function.
At any climb angle less than vertical then some amount of thrust is substituting for a portion of Lift.
Does this help?
24th Dec 2003, 02:10
The lift decreases because the pilot probably wants it to! The pilot must reduce angle of attack so that lift is zero, otherwise the aircraft would perform a loop, and then you'd no longer have vertical flight (as you describe)! Speed is not all that's involved in creating lift you know... ;-)
24th Dec 2003, 19:22
You may like to have a play with my "Four Forces" software - it's a work-in-progress but the intention is that it will help with visualising this sort of situation... all feedback appreciated...!
Four Forces web page (http://www.oview.co.uk/fourforces)
24th Dec 2003, 20:36
Hey Interesting program.
I've only played with it for a few moments but I noticed that the wing section seems to be fully symetrical? It will be interesting to have a go with a section with some +ve camber.
Thought I might try a simpler answer to the original question....
>When an aircraft is in level unaccelerated flight, lift equals
The vertical component of lift equals weight.
>But, when an aircraft is in an unaccelerated vertical climb, lift
> has decreased to virtually zero - is this so?
yes but... if your plane has a wing section with some camber to it then lift is only zero at a slightly negative angle of attack. So although the plane is climbing vertically the plane and wing will not be pointing exactly straight up.
The same applies in a vertical dive. To get zero lift a plane with a camberd wing has start by pitchig nose down and continue pitching slightly beyond the vertical until the angle of attach is slightly negative.
Note that the program posted above seems to model a wing with a _symetrical_ section. In this case the wing produces zero lift at a zero angle of attack and _will_ point vertically up/down in a vertical climb/dive.
How did I work out that its symetrical section? - by putting it in a vertical dive and noting that Cl =0 when AoA = zero.
24th Dec 2003, 22:12
As already suggested, the difference in the forces is that a component of thrust in the climb is now in the vertical direction, thus adding to lift.
(I have a good diagram of this if some one can explain how i post it without a URL)
Without the whole derivation, i hope this adds food for thought
Merry Christmas, :ok:
25th Dec 2003, 18:57
Simplest way I've ever seen it explained is:
An aircraft climbs (at constant speed) due to excess thrust, not increased lift :D
Ok, I think I have grasped what needs to be in the explanation ....
Consider an aircraft in a steady unaccelerated climb or descent ....
1. The vertical component of lift MUST balance weight or the aircraft would be accelerating upwards or downwards.
2. Further, if the aircraft is CLIMBING, the thing that makes it climb is excess THRUST. If it is descending, then you may reduce thrust.
3. If you are climbing, then the flightpath, being angled upwards, yields some rearward component of lift that must be overcome by thrust also.
25th Dec 2003, 23:47
Generally true but your last comment about a rearwards component of Lift isn't quite right.
It depends on what you mean by 'Lift'. If you're referring to the Total Reaction ie the sum of the aerodynamic forces produced by a wing, then yes, you could say it the way you did.
What you have really referred to is the Total Reaction ie total aerodynamic force produced by the wing. This total force is what gets resolved into a component called Induced Drag acting parallel to, and rearwards along, the flight path, and another component at 90 deg to the flight path called Lift. There can be no rearwards component of the part called Lift - by definition it acts at 90 deg to the flight path.
Consider an aircraft in a steady vertical climb BUT ONE THAT HAS HAD ITS WINGS REMOVED.
Thrust has to counter both Weight, and whatever Drag is present (all parasite drag since no lift is being produced to cause induced drag to be present).
Now add some wings BUT keep the aircraft attitude in an orientation that prevents the wings producing Lift. Like the no-wing case, there still won't be induced drag. There will still be all the other parts of Drag that afflict the structure and are lumped together in the term Parasite Drag, so Thrust must still be used to counter both (parasite) Drag & Weight.
If the aircraft attitude is changed so that the wings now produce Lift (ie a force at 90 deg to the flight path) then there will also be some additional amount of Drag: Induced Drag. This would also have to be overcome by Thrust in order to maintain speed. Bear in mind that any force at 90 deg to the vertical flight path will cause the flight path to deviate from vertical.
The Lift & Drag being produced are components of the total aerodynamic force the wing produces. That total force is what has a component acting back along the flight path - Induced Drag - plus the desirable Lift.
You could also consider it like this:
We're concerned with unaccelerated flight so that means the sum of all forces in any direction must be balanced by some combination of forces acting to produced an equal size force acting in the opposite direction. Straight & Level flight is the easiest example of this: Lift = Weight, Thrust = Drag. You could could draw a line at any orientation through the aircraft and all the forces on one side of the line will be exactly opposed by the combination of all the forces on the other side of the line.
This applies to unaccelerated motion in any direction: Forward, backward, sideways, vertical or whatever.
In an unaccelerated climb it still applies. All forces on one side of a line must be balanced by all forces on the other side.
Think of a horizontal dividing line: All 'upwards/forwards' forces must be balances by all 'downwards/backwards' forces. Thrust & Lift are the only upwards or forwards forces, and Drag & Weight are the only downwards or backwards ones.
Each group of forces can be combined into a single resultant using standard vector addition. The resultant has the same effect as combination of each of the forces from which it was made.
The Drag/Weight resultant MUST be balanced by a resultant caused by Lift & Thrust. The only way to make a Lift/Thrust resultant that balances Drag/Weight resultant is for Thrust to be larger than Drag & for Lift to be less than Weight. This is because the angles between the four forces varies, therefore the size of each force must vary to keep the resultants the same.
A diagram makes it soooo much easier to explain...
Tinny, if the flightpath is angled up at say 5 degrees in a climb, then some of the lift from the wing acts backwards. In the US, we call that the rearward component of lift, and it adds a little to the total drag from induced and parasite drag.
In anything except level flight, lift can be resolved into a vertical component (=weight in unaccelerated flight) and a component acting forwards (descent) or backwards (climb).
I hear ya about diagrams ...
26th Dec 2003, 05:45
Can I just point out the the original post said _Vertical climb_.
In which case any horizontal component of lift has to balance any horizontal component of drag or thrust (or the plane will move horizontally as well). If we are talking about a well designed plane where all the drag and thrust lines act through the CoG then the horizontal component of lift really will be zero when climbing verticaly as he said.
The answer is simply that for 'perfect' aircraft to make a vertical climb the wing has to be flying at an AOA at which lift is zero and the four forces program shows this nicely. If you put the plane into a vertical climb or dive all the horizontal vectors disappear and thrust and drag point straight up and down.
However in most real world aircraft the zero lift AoA is not zero degrees so to make a vertical climb the wing has to fly at a slightly negative AoA.
26th Dec 2003, 23:06
No, Kabz, it doesn't.
'Lift' is the resolved portion of the total force and acts at 90 deg to the flight path. It's not possible to resolve a parallel component from a vector that acts at 90 degrees to the plane in question.
The Total Force is what is inclined. The rearwards component gets resolved into a force labelled as 'Drag' ie induced drag that acts parallel to the flight path.
Yes, Lift is inclined 'backwards' in relation to the (earth referenced) horizontal plane however the components of forces as we label them are defined around the (aircraft referenced) flight path.
27th Dec 2003, 00:31
I agree with you Tinstaafl, but just out of interest...
Is there a common name for the resolved portion of the total force that acts vertically wrt to the ground (eg the component that opposes gravity).
In S&L flight this component is the same as the lift component but not in a climb. Do the books have a name for it?
27th Dec 2003, 06:43
Erm.....the 'vertical component of Lift'? :confused: :O