Principles of Flight
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Principles of Flight
Hi to you all, I am hoping someone out there can put me out of my misery on this question which came up on yesterday's CPL PoF exam. I have spent considerable time trying to fathom it out and even spoken to a PoF ground training instructor and other than the general thought that it is to do with the lift equation Lift = Cl1/2rho(Vsquared)S and the answer must be a reduction I have got nowhere.
The wording may not be exact but the gist of the question is:
An aircraft is traveling at 2xVs and encounters a gust which causes a load factor of 2. If that aircraft had only been traveling at 1.3xVs when it encountered that gust, what would the load factor have been?
answers given were:
(a) 2
(b) 2. (something I don't remember the figure precisely say 2.3)
(c) 2. (something more as above say 2.6)
(d) 1.69
I gave the answer 1.69 on the basis of the logic above but can someone please help with a clear explanation of how to solve it.
Thanks
Nigel
The wording may not be exact but the gist of the question is:
An aircraft is traveling at 2xVs and encounters a gust which causes a load factor of 2. If that aircraft had only been traveling at 1.3xVs when it encountered that gust, what would the load factor have been?
answers given were:
(a) 2
(b) 2. (something I don't remember the figure precisely say 2.3)
(c) 2. (something more as above say 2.6)
(d) 1.69
I gave the answer 1.69 on the basis of the logic above but can someone please help with a clear explanation of how to solve it.
Thanks
Nigel
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At anything less than 1.41Vs you would have stalled before hitting a load factor of 2.
The lift at a given angle of attack is proportional to the square of the speed. Stall occurs at stall AoA. At Vs and 1g the AoA is stalling AoA, and lift = weight.
So if you are flying at 1.41 Vs then at the stalling AoA your lift = weight * 1.41^2 so lift = 2* weight at the stall so you can just maintain aerodynamic flight at 2g.
However if you are flying at 1.3 * Vs then the lift at stalling AoA is only 1.3^2 * weight. 1.3^2 is 1.69, so as the aircraft pulls through 1.69g it stalls, and never exceeds that load factor.
Have you ever seen the limitations in the AFM that above a certain speed you must not use full deflection of the controls? The reason is that above that speed it is possible to exceed the g limitations. The speed is actually your stall speed at the limiting load factor for the aircraft (typically 3.8, 4.4 or 6g for light aircraft depending on category). Below that speed it is not aerodynamically possible to over-stress the aircraft*, as you will stall first. Note that speed, Va or manoeuvre speed, actually reduces with reduced all-up-weight (i.e. if light you have to be more careful) as you will not stall until a lower speed at a given load factor, including the maximum certified load factor or g limit.
* Not quite true. Most will overstress earlier if there is significant aileron input, but high g while actually rolling is rarely encountered unless attempting aerobatics.
The lift at a given angle of attack is proportional to the square of the speed. Stall occurs at stall AoA. At Vs and 1g the AoA is stalling AoA, and lift = weight.
So if you are flying at 1.41 Vs then at the stalling AoA your lift = weight * 1.41^2 so lift = 2* weight at the stall so you can just maintain aerodynamic flight at 2g.
However if you are flying at 1.3 * Vs then the lift at stalling AoA is only 1.3^2 * weight. 1.3^2 is 1.69, so as the aircraft pulls through 1.69g it stalls, and never exceeds that load factor.
Have you ever seen the limitations in the AFM that above a certain speed you must not use full deflection of the controls? The reason is that above that speed it is possible to exceed the g limitations. The speed is actually your stall speed at the limiting load factor for the aircraft (typically 3.8, 4.4 or 6g for light aircraft depending on category). Below that speed it is not aerodynamically possible to over-stress the aircraft*, as you will stall first. Note that speed, Va or manoeuvre speed, actually reduces with reduced all-up-weight (i.e. if light you have to be more careful) as you will not stall until a lower speed at a given load factor, including the maximum certified load factor or g limit.
* Not quite true. Most will overstress earlier if there is significant aileron input, but high g while actually rolling is rarely encountered unless attempting aerobatics.
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Tim
Thanks for your reply.
I am well aware of the principles involved here and partly based my answer on the limits you mention in the AFM of the Arrow i fly. However what i am looking for is a clear and unambiguous explanation of the method and principles required to answer that examination question.
Once again thank you for taking the trouble to post a reply but if your explanation does fully answer my question then I am afraid I don't fully follow it and would you be good enough to rephrase your explanation so I can have another crack at understanding it.
Meanwhile anyone else care to have a go?
Am I just being thick?
Nigel
Thanks for your reply.
I am well aware of the principles involved here and partly based my answer on the limits you mention in the AFM of the Arrow i fly. However what i am looking for is a clear and unambiguous explanation of the method and principles required to answer that examination question.
Once again thank you for taking the trouble to post a reply but if your explanation does fully answer my question then I am afraid I don't fully follow it and would you be good enough to rephrase your explanation so I can have another crack at understanding it.
Meanwhile anyone else care to have a go?
Am I just being thick?
Nigel
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ATPL Theory Summary And A Good Book Too:
Cl1/2rho(Vsquared)S... nigelisom, here is the "ATPL Theory Summary."
They show and illustrate better than I can, but "n" load factor= 1/cos times the bankangle. The squrae root of the load factor= Vstall increasing factor.
http://www.freelancepilot.nl/ATPL%20summary.pdf
Essentially it is just setting up the rigth equation the right function and plugging in the values.
Mario Asselin has a great book from 1997 titled:
An introduction to aircraft performance
On page 270 on Google books too, you may find a more detailed treatment of load factors with different velocity, wind gusts:
ngust= Lgust/W= 1/2 rho (V+-u)^2 SCl/W considers an horizontal gust.
For a full treatment, purchase this inexpensive book, or go on to Google Books and read page 270, (start on page 269, which begins the topic of flying in adverse weather) among other pages... happy trails!
They show and illustrate better than I can, but "n" load factor= 1/cos times the bankangle. The squrae root of the load factor= Vstall increasing factor.
http://www.freelancepilot.nl/ATPL%20summary.pdf
Essentially it is just setting up the rigth equation the right function and plugging in the values.
Mario Asselin has a great book from 1997 titled:
An introduction to aircraft performance
On page 270 on Google books too, you may find a more detailed treatment of load factors with different velocity, wind gusts:
ngust= Lgust/W= 1/2 rho (V+-u)^2 SCl/W considers an horizontal gust.
For a full treatment, purchase this inexpensive book, or go on to Google Books and read page 270, (start on page 269, which begins the topic of flying in adverse weather) among other pages... happy trails!
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Very thick!
Read and review your materials. This is just simple algebra. Almost forgot: Pilot's Handbook of Aerounautical Knowledge from the FAA is available on Google for free, and is the 2008 edition: Pilot's Handbook of Aeronautical Knowledge
Last edited by jcbmack; 15th Sep 2009 at 20:29.
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Consider two conditions.
Condition 1 is straight and level flight in still air.
Let CL = CL1 and V = V1
Lift 1= CL1 1/2ρ V1squared S
In straight and level flight Lift = weight so we can say that
Weight = CL1 1/2ρ V1squared S
Rearranging gives
S = Weight / CL1 1/2ρ V1squared……………………Equation 1
Condition 2 is in a sudden vertical gust.
Let CL CL2 and V = V2
Lift 2 = CL2 1/2ρ V2squared S
Rearranging this gives
S = Lift 2/ CL2 1/2ρ V2squared …………………….Equation 2
S is the wing area which, for the same aircraft is the same in both cases. So we can combine equations 1 and 2 to give.
Lift 2/ CL2 1/2ρ V2squared = Weight / CL1 1/2ρ V1squared
Multiplying both sides by 1/2ρ gives
Lift 2/ CL2 V2squared = Weight / CL1 V1squared
Rearranging gives
Lift 2/ Weight = CL2 V2squared / CL1 V1squared
But Lift2 / Weight = new load factor so
New load factor = CL2 V2squared / CL1 V1squared
If we assume that V1 is VS then we have
New load factor = CL2 V2squared / CLmax Vs squared
If we also assume that the aircraft stalls in the gust then we have
New load factor = CLmax V2squared / CLmax Vs squared
Cancelling out Clmax from the top and bottom of the right side gives
New load factor = V2squared / Vs squared
Taking the figures Vs = 1 and V2 = 1.3 Vs = 1.3 gives
New load factor = 1.3 squared / 1 squared = 1.69
So the load factor in the gust is 1.69
BUT WE HAVE ASSUMED THAT BOTH AIRCRAFT STALLED.
If one or both of them did not stall then we cannot assume that both were at CLmax and so we cannot give a definite value for load factor. We can however say that the gust cannot (theoretically) produce a load factor greater than 1.69 when flying at 1.3vs.
Condition 1 is straight and level flight in still air.
Let CL = CL1 and V = V1
Lift 1= CL1 1/2ρ V1squared S
In straight and level flight Lift = weight so we can say that
Weight = CL1 1/2ρ V1squared S
Rearranging gives
S = Weight / CL1 1/2ρ V1squared……………………Equation 1
Condition 2 is in a sudden vertical gust.
Let CL CL2 and V = V2
Lift 2 = CL2 1/2ρ V2squared S
Rearranging this gives
S = Lift 2/ CL2 1/2ρ V2squared …………………….Equation 2
S is the wing area which, for the same aircraft is the same in both cases. So we can combine equations 1 and 2 to give.
Lift 2/ CL2 1/2ρ V2squared = Weight / CL1 1/2ρ V1squared
Multiplying both sides by 1/2ρ gives
Lift 2/ CL2 V2squared = Weight / CL1 V1squared
Rearranging gives
Lift 2/ Weight = CL2 V2squared / CL1 V1squared
But Lift2 / Weight = new load factor so
New load factor = CL2 V2squared / CL1 V1squared
If we assume that V1 is VS then we have
New load factor = CL2 V2squared / CLmax Vs squared
If we also assume that the aircraft stalls in the gust then we have
New load factor = CLmax V2squared / CLmax Vs squared
Cancelling out Clmax from the top and bottom of the right side gives
New load factor = V2squared / Vs squared
Taking the figures Vs = 1 and V2 = 1.3 Vs = 1.3 gives
New load factor = 1.3 squared / 1 squared = 1.69
So the load factor in the gust is 1.69
BUT WE HAVE ASSUMED THAT BOTH AIRCRAFT STALLED.
If one or both of them did not stall then we cannot assume that both were at CLmax and so we cannot give a definite value for load factor. We can however say that the gust cannot (theoretically) produce a load factor greater than 1.69 when flying at 1.3vs.
Last edited by Keith.Williams.; 16th Sep 2009 at 15:19.
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Keith
Thanks that is much appreciated. I will give all that further thought this evening.
I still can't help thinking that this is all a bit convoluted for a 1 point question in a 45 minute exam. There must be a shortcut to work this out, unless all they were after was that the load factor reduced and rely on there only being 1 answere that fits.
Who can know the mindset of the CAA!
Nigel
Thanks that is much appreciated. I will give all that further thought this evening.
I still can't help thinking that this is all a bit convoluted for a 1 point question in a 45 minute exam. There must be a shortcut to work this out, unless all they were after was that the load factor reduced and rely on there only being 1 answere that fits.
Who can know the mindset of the CAA!
Nigel
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Simple Plug In, Equation Rearrangement:
Every point matters when you are in the sky! Just use the mauals and see the equations for vertical and horizontal gust.
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Nigelisom
Yes there is an easier way for the exam, but you you did say
For the exam just remember the final bit of the process.
New load factor = V2squared / Vs squared
But as I said, the question is defective anyway.
Yes there is an easier way for the exam, but you you did say
but can someone please help with a clear explanation of how to solve it.
New load factor = V2squared / Vs squared
But as I said, the question is defective anyway.
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Keith
I have just had time to sit and work through your explanation and yes that fully answers my question and I now understand how to address the problem.
Once again I thank you for taking the time and trouble to set that all out and I walk away a (slightly) wiser man. I am also relieved that my intuitive feel of the situation was correct (at least I got that one correct). I am much happier now that I can put the numbers in place for the problem as well, as it always worries me not knowing how to go about solving a problem.
Thanks again also to everyone else who took the trouble to try to help.
Now back to the long wait to find out how many of the rest of the paper I got right!
Nigel
I have just had time to sit and work through your explanation and yes that fully answers my question and I now understand how to address the problem.
Once again I thank you for taking the time and trouble to set that all out and I walk away a (slightly) wiser man. I am also relieved that my intuitive feel of the situation was correct (at least I got that one correct). I am much happier now that I can put the numbers in place for the problem as well, as it always worries me not knowing how to go about solving a problem.
Thanks again also to everyone else who took the trouble to try to help.
Now back to the long wait to find out how many of the rest of the paper I got right!
Nigel
Read your advisory circulars and the Pilot's Handbook of Aeronautical Knowledge.
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It is of course possible that the examiner is being really devious….
All gust load calculations questions have previously been concerned with vertical gusts. This is why my earlier explanation considered only the vertical gust scenario.
But it is possible that this new question is concerned with horizontal gusts.
If so the solution takes the following form.
Horizontal gusts will not change angle of attack or CL, but they will change the dynamic pressure and lift. Because the angle of attack remains constant the wing will not stall.
At 2Vs the gust caused load factor to double.
This means that V squared increased by a factor of 1.41 from 2Vs to 2.82Vs.
So the gust speed was 0.82 Vs
At 1.3 Vs adding 0.82 gives a new speed of 2.12 Vs
New load factor = New V squared / Old V squared
So new load factor = 2.12 squared / 1.3 squared = 2.66
So the new load factor is 2.66
If this was the examiner’s intention then it really was a devious question because most candidates, being familiar with previous questions would have assumed that it was a vertical gust.
Hmmmmmmm.......The examiner -Instructor arms race moves on
All gust load calculations questions have previously been concerned with vertical gusts. This is why my earlier explanation considered only the vertical gust scenario.
But it is possible that this new question is concerned with horizontal gusts.
If so the solution takes the following form.
Horizontal gusts will not change angle of attack or CL, but they will change the dynamic pressure and lift. Because the angle of attack remains constant the wing will not stall.
At 2Vs the gust caused load factor to double.
This means that V squared increased by a factor of 1.41 from 2Vs to 2.82Vs.
So the gust speed was 0.82 Vs
At 1.3 Vs adding 0.82 gives a new speed of 2.12 Vs
New load factor = New V squared / Old V squared
So new load factor = 2.12 squared / 1.3 squared = 2.66
So the new load factor is 2.66
If this was the examiner’s intention then it really was a devious question because most candidates, being familiar with previous questions would have assumed that it was a vertical gust.
Hmmmmmmm.......The examiner -Instructor arms race moves on
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Gee thanks Keith, just when I was feeling happy that I had cracked it!
The trouble with these exams is that you will never know which ones you got wrong so lose the learning opportunity that would give. It would be really nice to get a resume of the errors for future reference.
Looking again at the problem I bet your second thought is correct as for the vertical gust you don't need to know the load factor of the 2xVs situation. That doesn't stop them including it as a red herring of course!
Nigel
The trouble with these exams is that you will never know which ones you got wrong so lose the learning opportunity that would give. It would be really nice to get a resume of the errors for future reference.
Looking again at the problem I bet your second thought is correct as for the vertical gust you don't need to know the load factor of the 2xVs situation. That doesn't stop them including it as a red herring of course!
Nigel
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At the risk of putting the cat amongst the pigeons I believe the question that was issued by the JAA was:
Q. An aeroplane maintains straight and level flight at a speed of 2 * VS. If a vertical gust causes a load factor of 2, the load factor n caused by the same gust at a speed of 1.3 VS would be:
(A) n = 1.65.
(B) n = 1.69.
(C) n = 4.
(D) n = 1.3.
With (A) being marked as the correct answer. I have seen an explanation of why the answer 'is' 1.65 rather than 1.69 but it was quite lengthy and certainly too much for a one mark question. What may have happened here is that the CAA received complaints about the original question then one of their resident 'experts' decided to tart it up and modified the answers. Which only goes to show, you can't polish a t*rd.
Q. An aeroplane maintains straight and level flight at a speed of 2 * VS. If a vertical gust causes a load factor of 2, the load factor n caused by the same gust at a speed of 1.3 VS would be:
(A) n = 1.65.
(B) n = 1.69.
(C) n = 4.
(D) n = 1.3.
With (A) being marked as the correct answer. I have seen an explanation of why the answer 'is' 1.65 rather than 1.69 but it was quite lengthy and certainly too much for a one mark question. What may have happened here is that the CAA received complaints about the original question then one of their resident 'experts' decided to tart it up and modified the answers. Which only goes to show, you can't polish a t*rd.
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Alex
I will defer to you on the precise wording of the question but the answers were definately one at l/f 2 two of them at greater than l/f 2 and the last one at l/f 1.69
I hope your version of the wording of the question (including vertical gust) is correct!
nigel
I will defer to you on the precise wording of the question but the answers were definately one at l/f 2 two of them at greater than l/f 2 and the last one at l/f 1.69
I hope your version of the wording of the question (including vertical gust) is correct!
nigel
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The question
Was appealled by Oxford in June 2007 and put on hold.
In the ensuing debate several instructors put in various arguments, but all agreed that the question was unsafe. The CAA eventually decided to allow the appeal and put the question on hold to allow the SET to look at it again.
As Alex has said, the new question is probably an "improved" version of the old one.
But the numbers appear to have changed and one of the new options is correct for a horizontal gust. This raises the question of whether their "improvement action" has actually radically changed the situation.
Q. An aeroplane maintains straight and level flight at a speed of 2 * VS. If a vertical gust causes a load factor of 2, the load factor n caused by the same gust at a speed of 1.3 VS would be:
(A) n = 1.65.
(B) n = 1.69.
(C) n = 4.
(D) n = 1.3.
(A) n = 1.65.
(B) n = 1.69.
(C) n = 4.
(D) n = 1.3.
In the ensuing debate several instructors put in various arguments, but all agreed that the question was unsafe. The CAA eventually decided to allow the appeal and put the question on hold to allow the SET to look at it again.
As Alex has said, the new question is probably an "improved" version of the old one.
But the numbers appear to have changed and one of the new options is correct for a horizontal gust. This raises the question of whether their "improvement action" has actually radically changed the situation.
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This dates from, I think, a March 09 issue of the CQB but the original question dates back to 2003. Because the question is still in the CQB in 2009, apparently unchanged, I would suspect that the CAA said it was 'on hold' in 2007, put it on local hold, didn't process the query back through the SET and then modified the question locally in some wierd way. Dreadful.
What is even worse is that the CAA will no longer allow FTOs to appeal questions in the way that Keith describes. They will only accept appeals from candidates.
What is even worse is that the CAA will no longer allow FTOs to appeal questions in the way that Keith describes. They will only accept appeals from candidates.