MAYDAY issued over Irish Sea
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Bik_116.80
No, I meant what I said.
I know most people tend to think in terms of a T/D couple and a L/W couple, the reason for this is probably that this is the way most basic PofF texts / notes tend to break it down, normally with the intent of discussing the balancing forces required on the tail.
But it is a little off the mark. You don't have to have a pair of opposed forces, applied to different lines of action, in order to produce a rotational effect.
All you need is one force that doesn't act through the C of G.
In our example we are imagining an aircraft in the cruise. The forces are all cancelling one another out, so the aircrafts flight path is steady. The rotational effects of all the forces are also cancelling one another out - we know this because the aircraft attitude is constant.
i.e. in Newtons laws terms there are no unbalanced forces (so no acceleration) and in angular terms there are no unbalanced moments (so no angular acceleration).
i.e. you can forget about them.
All we are saying is: if we add some more thrust, what is the effect of the extra. You might as well be looking at a space ship in deep space (so forget all the other forces).
If the thrust line doesn't pass through the C of G you'll get angular acceleration.
Like I said, its basic mechanics.
Imagine a glider. No thrust so there is no thrust drag couple. Now stick a pole 5 feet up from the canopy with an airbrake on it. Deploy the airbrake. Glider pitchs nose up, because the line of action of the force is above the C of G of the object.
Hope that helps.
Must dash. Mrs Pit Bull is announcing the readiness of dinner.
CPB.
[This message has been edited by Capt Pit Bull (edited 09 November 2000).]
No, I meant what I said.
I know most people tend to think in terms of a T/D couple and a L/W couple, the reason for this is probably that this is the way most basic PofF texts / notes tend to break it down, normally with the intent of discussing the balancing forces required on the tail.
But it is a little off the mark. You don't have to have a pair of opposed forces, applied to different lines of action, in order to produce a rotational effect.
All you need is one force that doesn't act through the C of G.
In our example we are imagining an aircraft in the cruise. The forces are all cancelling one another out, so the aircrafts flight path is steady. The rotational effects of all the forces are also cancelling one another out - we know this because the aircraft attitude is constant.
i.e. in Newtons laws terms there are no unbalanced forces (so no acceleration) and in angular terms there are no unbalanced moments (so no angular acceleration).
i.e. you can forget about them.
All we are saying is: if we add some more thrust, what is the effect of the extra. You might as well be looking at a space ship in deep space (so forget all the other forces).
If the thrust line doesn't pass through the C of G you'll get angular acceleration.
Like I said, its basic mechanics.
Imagine a glider. No thrust so there is no thrust drag couple. Now stick a pole 5 feet up from the canopy with an airbrake on it. Deploy the airbrake. Glider pitchs nose up, because the line of action of the force is above the C of G of the object.
Hope that helps.
Must dash. Mrs Pit Bull is announcing the readiness of dinner.
CPB.
[This message has been edited by Capt Pit Bull (edited 09 November 2000).]
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Bally Heck,
Can't remember the name of the aircraft you saw, but I saw an article in Flight about 7-8 years ago about it. Its a Russian aircraft that they used for serious rough field work, IIRC the idea was the engines above the wing gave a lot more protection from FOD, also the take off roll was reduced as the jet efflux over the wing helped create more lift.
Of course I could be wrong
http://www.airliners.net/open.file?id=119353
Edit to add link to picture
[This message has been edited by Pdub (edited 09 November 2000).]
Can't remember the name of the aircraft you saw, but I saw an article in Flight about 7-8 years ago about it. Its a Russian aircraft that they used for serious rough field work, IIRC the idea was the engines above the wing gave a lot more protection from FOD, also the take off roll was reduced as the jet efflux over the wing helped create more lift.
Of course I could be wrong
http://www.airliners.net/open.file?id=119353 Edit to add link to picture
[This message has been edited by Pdub (edited 09 November 2000).]
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Bally Heck,
Did you mean the Fokker VFW 614??
here's a pic of it: http://www.airliners.net/open.file?id=74490
Did you mean the Fokker VFW 614??
here's a pic of it: http://www.airliners.net/open.file?id=74490
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Mach78--
It's illegal to listen to just about anything in the UK, unless you have a licence. According to the Radiocommunications Agency:
"Although it is not illegal to sell, buy or own a scanning or other receiver in the UK, it must only be used to listen to transmissions meant for GENERAL RECEPTION. The services that you can listen to include Amateur and Citizens' Band transmissions, licensed broadcast radio and weather and navigation broadcasts.
It is an offence to listen to any other radio services unless you are authorised by the Secretary of State to do so. "
The full document's at http://www.radio.gov.uk/document/ra_info/ra169.htm and gives chapter and verse.
Welcome to Edwardian Britain...
R
It's illegal to listen to just about anything in the UK, unless you have a licence. According to the Radiocommunications Agency:
"Although it is not illegal to sell, buy or own a scanning or other receiver in the UK, it must only be used to listen to transmissions meant for GENERAL RECEPTION. The services that you can listen to include Amateur and Citizens' Band transmissions, licensed broadcast radio and weather and navigation broadcasts.
It is an offence to listen to any other radio services unless you are authorised by the Secretary of State to do so. "
The full document's at http://www.radio.gov.uk/document/ra_info/ra169.htm and gives chapter and verse.
Welcome to Edwardian Britain...
R
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Oh Dear
Where do I start. Ok chaps I think it was the Antonov. It was parked up at EDI a couple of days ago….maybe still there.
Capt. Pit Bull. Not convinced with your C of G theory. The C of G is acting vertically down and not in the direction of thrust and drag. ie it is opposing lift. I go for the thrust drag couple myself. If any egghead out there has a convincing thesis. Happy to listen.
Latest. I wouldn’t worry to much about listening in without a licence. Consenting adults in private. If I was the telecoms agency and it bothered me, I could get thousands of punds of fines in a trawl of airport viewing galleries. Surprisingly they seem to have better things to do. Might get arrested listening to the ATIS in my house. Oh no…I have a licence…or is it the radio…Do
Where do I start. Ok chaps I think it was the Antonov. It was parked up at EDI a couple of days ago….maybe still there.
Capt. Pit Bull. Not convinced with your C of G theory. The C of G is acting vertically down and not in the direction of thrust and drag. ie it is opposing lift. I go for the thrust drag couple myself. If any egghead out there has a convincing thesis. Happy to listen.
Latest. I wouldn’t worry to much about listening in without a licence. Consenting adults in private. If I was the telecoms agency and it bothered me, I could get thousands of punds of fines in a trawl of airport viewing galleries. Surprisingly they seem to have better things to do. Might get arrested listening to the ATIS in my house. Oh no…I have a licence…or is it the radio…Do
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Captain Pit Bull, Following this with interest. I agree with your example of the glider and the airbrake, but I have one question. I am flying along in my Tiger Moth, engine working, straight and level, thrust/drag, lift/weight all happily balanced. The engine quits, so I no longer have "thrust" from that source. I am now in a biplane glider
I remember the mantra "convert speed to height". I do that. Naturally, as per instruction's warning voice, I always had a forced landing field selected and now I want to land on it. So now glider, I reconvert height to speed, or potential energy to (I hope I have this right) kinetic energy. I establish a new, albeit now descending, flight stability. Gosh, I am doing this well, I think I am going to make it, and here at my lap top my pulse rate is also steady.
Everything nicely balanced. Constant speed. I still have weight, and I am not dropping like a stone, but at steady speed, so I must have lift, and it must balance the weight. I still have drag, and I am not decelerating, so the drag must be balanced by ...................? Would we not in the normal case call it "thrust" or "thrust equivalent"?
I remember the mantra "convert speed to height". I do that. Naturally, as per instruction's warning voice, I always had a forced landing field selected and now I want to land on it. So now glider, I reconvert height to speed, or potential energy to (I hope I have this right) kinetic energy. I establish a new, albeit now descending, flight stability. Gosh, I am doing this well, I think I am going to make it, and here at my lap top my pulse rate is also steady.
Everything nicely balanced. Constant speed. I still have weight, and I am not dropping like a stone, but at steady speed, so I must have lift, and it must balance the weight. I still have drag, and I am not decelerating, so the drag must be balanced by ...................? Would we not in the normal case call it "thrust" or "thrust equivalent"?
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Davaar,
I have flown Tigers ,I think pulling back on the stick after an engine failure would only cause you to slow down rather than climb
If in your example lift still equalled weight you wouldn't be descending
Speed is constant in a descending glide because the weight vector always points toward the ground therefore giving you a 'forward' component to weight which balances drag.
Draw an aeroplane with a nose down attitude,draw in a weight vector pointing straight down,draw in a lift vector the same way you would if the a/c is S&L .There are also going to be vectors for vertical component of lift and drag but to keep it simple and if you don't understand vector addition simply think of the two vectors as pieces of string and 'pull' them .Which way does the aeroplane move?
I have flown a jet with the same configuration as the 145 (F28)and increasing /decreasing power just causes acceleration or deceleration,after a while of course the aircraft will climb/descend but more as a function of IAS variance and therefore lift variance than a 'pitch couple'.
MM.
I have flown Tigers ,I think pulling back on the stick after an engine failure would only cause you to slow down rather than climb
If in your example lift still equalled weight you wouldn't be descending
Speed is constant in a descending glide because the weight vector always points toward the ground therefore giving you a 'forward' component to weight which balances drag.
Draw an aeroplane with a nose down attitude,draw in a weight vector pointing straight down,draw in a lift vector the same way you would if the a/c is S&L .There are also going to be vectors for vertical component of lift and drag but to keep it simple and if you don't understand vector addition simply think of the two vectors as pieces of string and 'pull' them .Which way does the aeroplane move?
I have flown a jet with the same configuration as the 145 (F28)and increasing /decreasing power just causes acceleration or deceleration,after a while of course the aircraft will climb/descend but more as a function of IAS variance and therefore lift variance than a 'pitch couple'.
MM.
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Thanks, mountain man. Perhaps example was ill-chosen. I always thought myself that the convert speed to height routine had more to it so piety than practical value, given the speeds and drag of a Tiger. Let me put that aside. Although of course in other types one could well convert speed to height, it is not really material here.
I recognise right away your point about the weight, the lift amd the descent, because at one time I looked at it your way, and then later changed my mind. Maybe I should change it back again. My current reasoning is that there are four forces in balance, two opposing two. If any falls out of balance there is an acceleration in that pair. If it is a thrust/drag imbalance there is a speed acceleration/deceleration until balance is again reached. If it is a weight/lift imbalance there is a vertical accelaration/deceleration until balance is again reached.
The basic proposition is that an aircraft in any stable flight condition is in balance.
I see what you say about the weight vector in a steady descending glide. As you say,this element gives a forward component to weight, and balances drag. I agree. But is that not tantamount to saying that weight provides part of the thrust? The aircraft is not accelerating, and drag is not compensated by any balancing force from any source other than the descent. In a glider the weight element is the only source, in a powered descent it is a partial source.
Perhaps I did not express it well, but that really was what I was addressing to Captain Pit Bull. I think he had said that a glider does not have a thrust/drag pair, but it seems to me that it does, and the source of the "forward" component that creates the balance, commonly called thrust, is exactly the forward weight vector that you describe.
Thank you for your patience.
I recognise right away your point about the weight, the lift amd the descent, because at one time I looked at it your way, and then later changed my mind. Maybe I should change it back again. My current reasoning is that there are four forces in balance, two opposing two. If any falls out of balance there is an acceleration in that pair. If it is a thrust/drag imbalance there is a speed acceleration/deceleration until balance is again reached. If it is a weight/lift imbalance there is a vertical accelaration/deceleration until balance is again reached.
The basic proposition is that an aircraft in any stable flight condition is in balance.
I see what you say about the weight vector in a steady descending glide. As you say,this element gives a forward component to weight, and balances drag. I agree. But is that not tantamount to saying that weight provides part of the thrust? The aircraft is not accelerating, and drag is not compensated by any balancing force from any source other than the descent. In a glider the weight element is the only source, in a powered descent it is a partial source.
Perhaps I did not express it well, but that really was what I was addressing to Captain Pit Bull. I think he had said that a glider does not have a thrust/drag pair, but it seems to me that it does, and the source of the "forward" component that creates the balance, commonly called thrust, is exactly the forward weight vector that you describe.
Thank you for your patience.
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Davaar,
descending is a verticle acceleration ,ie it has direction and magnitude.
Lift is a function of IAS,wing area,wing shape (camber etc),Angle of Attack among other things (Cl=1/2RhoVsquared.S).In a gliding a/c such as your Tiger IAS is less than crz,wing area hasn't changed,neither has air density(in a measureable sense) so it can be seen from the Lift Formula that lift must be less,unless of course you want to increase AOA which is a VERY short term view followed closely by stall/spin.Therefore in a gliding descent Lift<Weight
There is no thrust from the engine but drag still exists but probably less due to less Form Drag from lower IAS.If you are at the optimum glide speed this equates to the maximum coefficient of lift so ideal AOA for min induced drag.
Overall in a glide Lift<Weight and the 'forward' component of weight = Drag.
By the way congrats to the 145 crew ,well done !
MM.
PS While I've not done a lot of gliding,gliders climb because the air they are descending in is going up at a greater rate.
descending is a verticle acceleration ,ie it has direction and magnitude.
Lift is a function of IAS,wing area,wing shape (camber etc),Angle of Attack among other things (Cl=1/2RhoVsquared.S).In a gliding a/c such as your Tiger IAS is less than crz,wing area hasn't changed,neither has air density(in a measureable sense) so it can be seen from the Lift Formula that lift must be less,unless of course you want to increase AOA which is a VERY short term view followed closely by stall/spin.Therefore in a gliding descent Lift<Weight
There is no thrust from the engine but drag still exists but probably less due to less Form Drag from lower IAS.If you are at the optimum glide speed this equates to the maximum coefficient of lift so ideal AOA for min induced drag.
Overall in a glide Lift<Weight and the 'forward' component of weight = Drag.
By the way congrats to the 145 crew ,well done !
MM.
PS While I've not done a lot of gliding,gliders climb because the air they are descending in is going up at a greater rate.
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While you are all discussing Theory of Flight (It must be true "Get 3 Pilots together and you'll get 4 opinions") no one has commented on TripleIRS ' question
"Whilst I don't doubt that the crew did a good job, I was wondering whether their decision to continue all the way to MAN was a sound one. With a control difficulty such as that, I would have thought that Land ASAP would have been the better decision. They seemed to have regained sufficient control whilst still over N. Ireland. Comments??"
And as for Growlers statement that they had identified the problem etc - well the Alaska Air lads had the problem identified but they still ended up dead. As far as I remember they pulled (and reset) the appropriate CB. With some prodding, unfortunately, from their Maintenance.
No point in being a Dead superhero!
------------------
The Stamp is mightier than the Toolbox!!
"Whilst I don't doubt that the crew did a good job, I was wondering whether their decision to continue all the way to MAN was a sound one. With a control difficulty such as that, I would have thought that Land ASAP would have been the better decision. They seemed to have regained sufficient control whilst still over N. Ireland. Comments??"
And as for Growlers statement that they had identified the problem etc - well the Alaska Air lads had the problem identified but they still ended up dead. As far as I remember they pulled (and reset) the appropriate CB. With some prodding, unfortunately, from their Maintenance.
No point in being a Dead superhero!
------------------
The Stamp is mightier than the Toolbox!!
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34DD’s little book of odds:
Davaar 10 000 to 1 ON
MM 1000 to 1 AGAINST
But I will still fly with MM ‘cos life proves everyday that (we) pilots don’t have to know how an aeroplane works in order to use one
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Forward CGs are better than aft ones
Davaar 10 000 to 1 ON
MM 1000 to 1 AGAINST
But I will still fly with MM ‘cos life proves everyday that (we) pilots don’t have to know how an aeroplane works in order to use one
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Forward CGs are better than aft ones
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WideBodiedEng:
Yes, I was thinking what a pity this had turned into an ego-massaging exercise by some!!
Still, I guess that's much more important than discussing what I thought was an important flight safety issue.
It seems that, understandably, the crew were seeking advice from the Isle of Man over their Company Frequency. Still in an Emergency situation, with evident control difficulties, it seems very strange that they should have continued to MAN.
Yes, I was thinking what a pity this had turned into an ego-massaging exercise by some!!
Still, I guess that's much more important than discussing what I thought was an important flight safety issue.
It seems that, understandably, the crew were seeking advice from the Isle of Man over their Company Frequency. Still in an Emergency situation, with evident control difficulties, it seems very strange that they should have continued to MAN.
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Everyone:
34DD is spot on when he says that we don't have to understand every aspect of an aircraft inorder to use it.
In science you basically have a model of how a system behaves. By necessity, that model is usually a gross oversimplification of what is going on. But as long as the model holds good, for the operating regime that we are interested in, who cares?
Well, inadequacies in your model only become a problem when you try and extrapolate the behaviour of the system outside of the regime the model was designed to cover. At which point, one of two things happens:
1) Either the system still behaves as predicted by the model (you slap yourself on the back!).
2) The system doesn't behave as expected, the model has been shown to be imperfect.
When 2 occurs, you try a more complex model. Your simpler model either gets binned completely , or, since it is simple, you remember it as a way of introducing the system to students that are trying to get to grips with it.
The problem most pilots have with principles of flight is that they do not have a solid scientific education. Therefore most PofF books, by necessity, provide very simple explanations for what is going on.
view of a starting point for studying PofF.
This doesn't really represent a problem. The models are good enough to give everyone a rough idea of whats going on so that they can go and start poling the aeroplanes.
But don't be to hopeful in expecting the models to acurately predict more complex situations. You may even find that the pure behaviour of the system is the exact opposite of what you have been told, because secondary effects may be predominant.
The case in point:
You can say with absolute certainty (I'm not going to prove it here, because its lengthy) that when the line of action of thrust is above the line of action of drag a nose down tendancy will result (and vice versa), but you can only say this for sure when in the cruise with T = D).
It does not hold true when T <> D.
Lets extend my glider example. Having opened the airbrake on the top of the pole, you push forward on the control column to maintain the attitude. I.E. you have provided a pitching moment to counteract the pitching moment caused by the airbrake.
Now attach a JATO pod halfway up the pole. Its below the drag line (because the airbrake is providing the majority of the drag). Its still above the C of G. Fire the JATO pod. Aircraft will pitch nose down because the thrust is applied above the C of G. The fact that the drag line is above it is irrelevant.
CPB
34DD is spot on when he says that we don't have to understand every aspect of an aircraft inorder to use it.
In science you basically have a model of how a system behaves. By necessity, that model is usually a gross oversimplification of what is going on. But as long as the model holds good, for the operating regime that we are interested in, who cares?
Well, inadequacies in your model only become a problem when you try and extrapolate the behaviour of the system outside of the regime the model was designed to cover. At which point, one of two things happens:
1) Either the system still behaves as predicted by the model (you slap yourself on the back!).
2) The system doesn't behave as expected, the model has been shown to be imperfect.
When 2 occurs, you try a more complex model. Your simpler model either gets binned completely , or, since it is simple, you remember it as a way of introducing the system to students that are trying to get to grips with it.
The problem most pilots have with principles of flight is that they do not have a solid scientific education. Therefore most PofF books, by necessity, provide very simple explanations for what is going on.
view of a starting point for studying PofF.
This doesn't really represent a problem. The models are good enough to give everyone a rough idea of whats going on so that they can go and start poling the aeroplanes.
But don't be to hopeful in expecting the models to acurately predict more complex situations. You may even find that the pure behaviour of the system is the exact opposite of what you have been told, because secondary effects may be predominant.
The case in point:
You can say with absolute certainty (I'm not going to prove it here, because its lengthy) that when the line of action of thrust is above the line of action of drag a nose down tendancy will result (and vice versa), but you can only say this for sure when in the cruise with T = D).
It does not hold true when T <> D.
Lets extend my glider example. Having opened the airbrake on the top of the pole, you push forward on the control column to maintain the attitude. I.E. you have provided a pitching moment to counteract the pitching moment caused by the airbrake.
Now attach a JATO pod halfway up the pole. Its below the drag line (because the airbrake is providing the majority of the drag). Its still above the C of G. Fire the JATO pod. Aircraft will pitch nose down because the thrust is applied above the C of G. The fact that the drag line is above it is irrelevant.
CPB
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Davaar,
Your arguement is flawed. I suspect this is because your basic definitions of Lift is wrong.
Its often worth remembering that the behaviour of aerodynamics will provide us with an overall aerodynamic force on the aircraft.
We call this the Total Reaction. Most text book introduce it on page 3 and then never mention it again.
Lift and Drag are one way of splitting the total reaction into separate components. We do this in order to simplify our models.
But there are times when it is convenient to merge lift and drag back together again and just think about the total reaction.
Gliding is the case in point. With Thrust taken out of the equation, and lift + drag merged into one force, that leave only 2 forces to look at:
Weight (acting down)
Total Reaction.
Your premise that there are no unbalanced forces is spot on.
Your premise that Weight is balanced (solely)by Lift is false.
What is happening is that Weight must be equal to Total reaction.
If you want to see what is happening with lift and drag your definition of them must be sound.
People get so used to the picture on page 1 of their text book showing lift up, weight down, thrust to the left and drag to the right that they get it fixed in their subconscious that:
Lift is defined as 'Up' relative to the ground, and drag is defined as being horizontal.
This is cobblers and confuses no end of people.
Lift and drag are more correctly defined with reference to the aircrafts flight path:
Try this as a model:
Drag is the component of the total reaction that acts backwards along the flight path.
Lift is the component of the total reaction that acts perpendicular to the flight path.
Ergo they are mutually at right angles.
(a picture is worth a thousand words!)
So our glider has its weight opposed by total reaction. To see whats happening with lift and drag, split the total reaction up based on the flight path.
Lets say the glider is pretty poor, and has a glide angle of 10 degrees. Sketch this. The flight path is inclined at 10 degrees to the horizontal. Our drag line is in the same orientation, tilted up 10 degrees relative to the horizontal. The lift line in at right angles (i.e. 10 degrees from the vertical). If you vector add the lift and drag together, you should end up with a right angle triangle with lift, drag, and total reaction on the sides. Lift<Weight, the shortfall is made up by a vertical component of drag.
Try it again with something that has a very poor L/D ratio. Say a House Brick, falling at terminal velocity. In this case the total reaction is made entirely of drag, so its just (weight of house brick) down, opposed by (drag = weight of house brick) up.
With reference to your earlier post, weight (or any component there of) will never cause a pitching moment. Why? Because weight acts through the CofG. Therefore the distance between its line of action and the CofG is always zero. Therefore it never really causes rotation about any axis, although some explanations for some kinds of stability related design features would suggest that it does - remember my earlier post - they are just simple models to give you a rough idea.
Hope that helps.
Maybe we should move this to Tech Log.
CPB
Your arguement is flawed. I suspect this is because your basic definitions of Lift is wrong.
Its often worth remembering that the behaviour of aerodynamics will provide us with an overall aerodynamic force on the aircraft.
We call this the Total Reaction. Most text book introduce it on page 3 and then never mention it again.
Lift and Drag are one way of splitting the total reaction into separate components. We do this in order to simplify our models.
But there are times when it is convenient to merge lift and drag back together again and just think about the total reaction.
Gliding is the case in point. With Thrust taken out of the equation, and lift + drag merged into one force, that leave only 2 forces to look at:
Weight (acting down)
Total Reaction.
Your premise that there are no unbalanced forces is spot on.
Your premise that Weight is balanced (solely)by Lift is false.
What is happening is that Weight must be equal to Total reaction.
If you want to see what is happening with lift and drag your definition of them must be sound.
People get so used to the picture on page 1 of their text book showing lift up, weight down, thrust to the left and drag to the right that they get it fixed in their subconscious that:
Lift is defined as 'Up' relative to the ground, and drag is defined as being horizontal.
This is cobblers and confuses no end of people.
Lift and drag are more correctly defined with reference to the aircrafts flight path:
Try this as a model:
Drag is the component of the total reaction that acts backwards along the flight path.
Lift is the component of the total reaction that acts perpendicular to the flight path.
Ergo they are mutually at right angles.
(a picture is worth a thousand words!)
So our glider has its weight opposed by total reaction. To see whats happening with lift and drag, split the total reaction up based on the flight path.
Lets say the glider is pretty poor, and has a glide angle of 10 degrees. Sketch this. The flight path is inclined at 10 degrees to the horizontal. Our drag line is in the same orientation, tilted up 10 degrees relative to the horizontal. The lift line in at right angles (i.e. 10 degrees from the vertical). If you vector add the lift and drag together, you should end up with a right angle triangle with lift, drag, and total reaction on the sides. Lift<Weight, the shortfall is made up by a vertical component of drag.
Try it again with something that has a very poor L/D ratio. Say a House Brick, falling at terminal velocity. In this case the total reaction is made entirely of drag, so its just (weight of house brick) down, opposed by (drag = weight of house brick) up.
With reference to your earlier post, weight (or any component there of) will never cause a pitching moment. Why? Because weight acts through the CofG. Therefore the distance between its line of action and the CofG is always zero. Therefore it never really causes rotation about any axis, although some explanations for some kinds of stability related design features would suggest that it does - remember my earlier post - they are just simple models to give you a rough idea.
Hope that helps.
Maybe we should move this to Tech Log.
CPB