Can anyone help me with this BAK question?
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Can anyone help me with this BAK question?
Hi all, I am now doing a home study on the Bob Tait BAK and I encountered a question that I can't figure out by myself.
(quotation from Bob Tait Textbook) "Rate of climb in a wind blowing from right to left condition: Assuming both aircraft climb at the same IAS, Power and Weight, the rate of climb for the aircraft flying headwind (aircraft A) is the same as the aircraft flying tail wind (aircraft B)." (quotation from Bob Tait Textbook)
Baby Tait says both aircraft have the same rate of climb but I don't understand why. My understanding is aircraft A should have a higher rate of climb as it is flying headwind, the wind should have helped aircraft A to climb faster because of the extra lift provided by the headwind.
Question One: Why both aircraft will have the same rate of climb?
Question Two: How can the aircrafts have the same IAS and Power given that one is flying headwind and one is flying tail wind? In my understanding, the one flying headwind should have higher IAS under the same power setting.
I am a bit confused now and would be very appreciated if anyone can help me with this question. This question is staying in my mind for more than 3 months....
p.s. I understand why the angle of climb is steeper for aircraft A.
(quotation from Bob Tait Textbook) "Rate of climb in a wind blowing from right to left condition: Assuming both aircraft climb at the same IAS, Power and Weight, the rate of climb for the aircraft flying headwind (aircraft A) is the same as the aircraft flying tail wind (aircraft B)." (quotation from Bob Tait Textbook)
Baby Tait says both aircraft have the same rate of climb but I don't understand why. My understanding is aircraft A should have a higher rate of climb as it is flying headwind, the wind should have helped aircraft A to climb faster because of the extra lift provided by the headwind.
Question One: Why both aircraft will have the same rate of climb?
Question Two: How can the aircrafts have the same IAS and Power given that one is flying headwind and one is flying tail wind? In my understanding, the one flying headwind should have higher IAS under the same power setting.
I am a bit confused now and would be very appreciated if anyone can help me with this question. This question is staying in my mind for more than 3 months....
p.s. I understand why the angle of climb is steeper for aircraft A.
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You need to consider the very last sentence you wrote.
There are 2 factors.
Angle and Rate.
I can just see myself explaining this so badly.....................at this time of night.
Aerodynamically, the aeroplane does not care what direction the wind is travelling. It is only concerned with its speed through the air.
Lets assume 2 numbers for the story...75 knots = 500fpm climb
So if you are climbing at 75 knots in still air, the aeroplane only see's 75 knots of airflow.
If you are climbing at 75 knots with ten knots of headwind, then the aeroplane see's 75 knots of airflow.
If you are climbing at 75 knots with ten knots of tailwind, the aeroplane still see's 75 knots.
For rate of climb, you do not need to consider the direction of speed of the wind. The aeroplane will travel through that pocket of air and generate the performance you wish.
This is RATE of CLIMB, or descent.
Where the wind comes in, is how it affects your distance over the ground....or ANGLE of CLIMB.
IF you have a headwind, and you fly the aeroplane at 75 knots climb, it will cover less ground distance than nil wind, so the angle is steeper....but you are still climbing at 500fpm....from sea level that is 4 minutes to 2000 foot....and without a calculator...........say 5 NM for nil wind, and 4nm for 15 knots head wind.
If you have a tailwind, it remains at 4 minutes to 2000, but now it is going to take 7nm.
Summary.
Rate= performance through the air
Angle = performance over the ground
There are 2 factors.
Angle and Rate.
I can just see myself explaining this so badly.....................at this time of night.
Aerodynamically, the aeroplane does not care what direction the wind is travelling. It is only concerned with its speed through the air.
Lets assume 2 numbers for the story...75 knots = 500fpm climb
So if you are climbing at 75 knots in still air, the aeroplane only see's 75 knots of airflow.
If you are climbing at 75 knots with ten knots of headwind, then the aeroplane see's 75 knots of airflow.
If you are climbing at 75 knots with ten knots of tailwind, the aeroplane still see's 75 knots.
For rate of climb, you do not need to consider the direction of speed of the wind. The aeroplane will travel through that pocket of air and generate the performance you wish.
This is RATE of CLIMB, or descent.
Where the wind comes in, is how it affects your distance over the ground....or ANGLE of CLIMB.
IF you have a headwind, and you fly the aeroplane at 75 knots climb, it will cover less ground distance than nil wind, so the angle is steeper....but you are still climbing at 500fpm....from sea level that is 4 minutes to 2000 foot....and without a calculator...........say 5 NM for nil wind, and 4nm for 15 knots head wind.
If you have a tailwind, it remains at 4 minutes to 2000, but now it is going to take 7nm.
Summary.
Rate= performance through the air
Angle = performance over the ground
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Keep in mind that the majority of training questions are simplistic and involve assumptions .. whether stated or implied ... so that they can be simplistic.
The inferences with this question is that there exists neither turbulence nor gradient associated with the wind field .. either of which will change the outcome.
However, the main thrust of the question is to get the student to understand, as a starting point, that wind is relevant to navigation (ie G/S) rather than aircraft speed as the pilot sees it in the cockpit (IAS).
The inferences with this question is that there exists neither turbulence nor gradient associated with the wind field .. either of which will change the outcome.
However, the main thrust of the question is to get the student to understand, as a starting point, that wind is relevant to navigation (ie G/S) rather than aircraft speed as the pilot sees it in the cockpit (IAS).
I wonder if they still teach the Heisenberg uncertainty principle? From my astrophysics days essentially you can know the speed of something fairly accurately but the position is quiet uncertain. The opposite is also true. Yeah ok...too much red wine! But back the question, yes you even understand yourself by knowing the angles are different. So headwind, higher angle (or height) with less horizontal distance. Conversely, with tailwind lower angle (height) with greater horizontal distance. Thus h/d is the same.
With regards to speed, I suggest you look at the different types of speeds. There are a multitude of types IAS, TAS, GS, CAS......etc etc etc
With regards to speed, I suggest you look at the different types of speeds. There are a multitude of types IAS, TAS, GS, CAS......etc etc etc
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Thank you very much for your reply, Bandit, John and Jas.
I read your reply carefully and can I have this conclusion?
For the angle of climb, we have to consider the tail wind and head wind as they will affect the GS. Tail wind increase GS while headwind decrease GS, so that the horizontal distance covered by aircraft A & B is different, making their angle of climb different, given that they are having the same rate of climb.
For the rate of climb, we have to consider IAS only. Tail wind and head wind is meaningless when talking about rate of climb. I know that IAS is a good indication of lift available. Regardless the wind direction, as long as both aircraft have the same IAS, they will have the same performance (e.g., same rate of climb).
Can I have this conclusion? Thank you very much!!!
re Bandit: No I can't see anything looks like Heisenberg uncertainty principle in the BAK textbook.
re Jas: I think I know more after reading your explanation!
re John: Your last sentence helped me to connect things together.
I read your reply carefully and can I have this conclusion?
For the angle of climb, we have to consider the tail wind and head wind as they will affect the GS. Tail wind increase GS while headwind decrease GS, so that the horizontal distance covered by aircraft A & B is different, making their angle of climb different, given that they are having the same rate of climb.
For the rate of climb, we have to consider IAS only. Tail wind and head wind is meaningless when talking about rate of climb. I know that IAS is a good indication of lift available. Regardless the wind direction, as long as both aircraft have the same IAS, they will have the same performance (e.g., same rate of climb).
Can I have this conclusion? Thank you very much!!!
re Bandit: No I can't see anything looks like Heisenberg uncertainty principle in the BAK textbook.
re Jas: I think I know more after reading your explanation!
re John: Your last sentence helped me to connect things together.
What Jas said.
Couldn't have put it better myself.
Other than when climbing or descending through a wind gradient, the wind speed will have no effect on the aircraft performance.
Couldn't have put it better myself.
Other than when climbing or descending through a wind gradient, the wind speed will have no effect on the aircraft performance.
The only factor that affects *time* to climb is whatever rate of the climb the plane happens to be doing. Rate of climb is nothing more than speed measured vertically, instead of along the flight path as with IAS. Aircraft have an airspeed that will provide the maximum rate of climb for a given set of conditions & configuration. Change those conditions or configuration, and the speed that provides the best rate of climb will also vary. Fly at a speed faster or slower than the appropriate speed for best rate of climb under those conditions, and rate of climb will be reduced.
Because rate of climb is only concerned with vertical movement, anything that moves the plane horizontally, such as wind, has no effect on the rate of climb.
As a thought experiment, think of a helium balloon at floor level in a sealed container on a train that isn't moving.
* Release the balloon at floor level and it will climb at a particular rate.
* Start the train moving, release the balloon again and it still will climb at the same rate.
* Now imagine a wind that exactly matches the train's speed & direction. What happens in the container if you release the balloon again? Nothing different to before.
* Now imagine there's a window in the container's side. If you put your hand out the window you wouldn't feel any wind *because the train & the wind are all moving in the same direction at the same speed*. The balloon's rate of climb is unaffected.
* No matter how much of the container's structure is removed - a bit at a time until it's all gone, for example - there is still no effect on the balloon's rate of climb. No matter how fast the train goes, or the balloon within the moving airmass, the horizontal motion doesn't affect the vertical.
If you were standing at the side of the rail line watching the balloon's *angle* of climb through all of this then you would see a dramatic difference in its angle of climb. 90 deg when still, and becoming shallower & shallower as the horizontal speed increased. Whether the balloon's horizontal speed was caused by the train, or because it was within a moving airmass is irrelevent. Angle of climb is a measure of 'rise' over 'run'. The greater the 'run' for a fixed 'rise', the shallower the angle.
Because rate of climb is only concerned with vertical movement, anything that moves the plane horizontally, such as wind, has no effect on the rate of climb.
As a thought experiment, think of a helium balloon at floor level in a sealed container on a train that isn't moving.
* Release the balloon at floor level and it will climb at a particular rate.
* Start the train moving, release the balloon again and it still will climb at the same rate.
* Now imagine a wind that exactly matches the train's speed & direction. What happens in the container if you release the balloon again? Nothing different to before.
* Now imagine there's a window in the container's side. If you put your hand out the window you wouldn't feel any wind *because the train & the wind are all moving in the same direction at the same speed*. The balloon's rate of climb is unaffected.
* No matter how much of the container's structure is removed - a bit at a time until it's all gone, for example - there is still no effect on the balloon's rate of climb. No matter how fast the train goes, or the balloon within the moving airmass, the horizontal motion doesn't affect the vertical.
If you were standing at the side of the rail line watching the balloon's *angle* of climb through all of this then you would see a dramatic difference in its angle of climb. 90 deg when still, and becoming shallower & shallower as the horizontal speed increased. Whether the balloon's horizontal speed was caused by the train, or because it was within a moving airmass is irrelevent. Angle of climb is a measure of 'rise' over 'run'. The greater the 'run' for a fixed 'rise', the shallower the angle.
Trick Q ??
Que ? why then in the initial statement is the wind stated as "blowing from right to left" ??
And then talk about 'head wind' and 'tail wind' when you have a cross wind ???
Was there a diagram with this question?
And then talk about 'head wind' and 'tail wind' when you have a cross wind ???
Was there a diagram with this question?
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A simply way to explain this is an analogy I give my students about two lovers Jonny and Mary who live across the river from each other. During the summer the river does not flow and the river is 10 m wide. Jonny swims at 2m per second so it takes him 5 secs to swim across in a straight line.
During winter rains the river flows and when Jonny swims across he does not swim in a straight line due the river flowing but he still swims at 2m per second . His rate is the same but only the angle is affected . The river flowing only affects his angle he swims across the river.
That's why angle is affected by wind not the rate of climb/ descent .
And so the story goes...
During winter rains the river flows and when Jonny swims across he does not swim in a straight line due the river flowing but he still swims at 2m per second . His rate is the same but only the angle is affected . The river flowing only affects his angle he swims across the river.
That's why angle is affected by wind not the rate of climb/ descent .
And so the story goes...
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Remember Lift is less than weight in a climb.
And can't you convert a power off descent to a momentary power off climb? No thrust component to add to the vector so it is all lift which is greater than weight in this case.
Just asking.
So - momentarily lift increases, probably above weight, and you accelerate in the vertical direction (initial ROC=0, new ROC=positive, means acceleration). When you're established on the new constant ROC, lift is less than weight (due to vertical component of thrust).
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You guys are trying to make this way too complicated. Tait is saying that rate of climb is independent of the direction the airmass you are in is moving. It is feet/minute which has nothing to do with miles travelled.
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Sometimes a good way to think about these things in the future can be to take them to the extreme. Consider your climb speed is about 70kts and you are climbing at 1000fpm. If you had a 70kt headwind then you would be climbing vertically so more or less hovering over one spot on the ground. Now after 1 minute you are at 1000ft and havent moved an inch so your climb angle is absolutely vertical!
Lets change that now to a 70kt tailwind. You are still climbing at 1000fpm so after 1 minute you are still at 1000ft therefore your RATE of CLIMB is exactly the same. Your angle however is far flatter as you have been travelling forwards at 140kts for that minute which has taken you well away from where you began.
P.S. The wind moving left to right and vice versa is not relevant in a question like this until later on in your studies when you will learn about effective TAS.
Lets change that now to a 70kt tailwind. You are still climbing at 1000fpm so after 1 minute you are still at 1000ft therefore your RATE of CLIMB is exactly the same. Your angle however is far flatter as you have been travelling forwards at 140kts for that minute which has taken you well away from where you began.
P.S. The wind moving left to right and vice versa is not relevant in a question like this until later on in your studies when you will learn about effective TAS.
patagonian -
I can't answer that 
But in a constant speed climb, Max rate of climb is purely rated to excess thrust.
Lift is less than weight in a climb.
Like trim set says - Take it to an extreme-
You are in a jet with buckets of thrust.
You are climbing vertically at a constant 120kts on the dial. The Lift component is 90 deg to thrust, and weight is directly opposite thrust.
Maximum ROC = Maximum excess thrust.
So what did I miss in my studies?
But in a constant speed climb, Max rate of climb is purely rated to excess thrust.
Lift is less than weight in a climb.
Like trim set says - Take it to an extreme-
You are in a jet with buckets of thrust.
You are climbing vertically at a constant 120kts on the dial. The Lift component is 90 deg to thrust, and weight is directly opposite thrust.
Maximum ROC = Maximum excess thrust.
Jas,
Your explanation was 'luvly'....
Even poor ole moi followed that like it was written and very well 'mind pictured' illustrated.
Thanks for doing an excellent job for Mr K. - a 'rare' event here on PPRuNe these days.....
Cheeerssss
An Old not so bold, pilot.....
Your explanation was 'luvly'....
Even poor ole moi followed that like it was written and very well 'mind pictured' illustrated.
Thanks for doing an excellent job for Mr K. - a 'rare' event here on PPRuNe these days.....
Cheeerssss
An Old not so bold, pilot.....
Question
Katsunlo,
You are obviously early in your studies and good on you for showing some initiative and coming on here for some advice. BUT, if you are currently undertaking flying training, your first port of call should always be your instructor. It could be your very new and haven't yet established that rapport, but your instructor will be very keen to help you, provided you're putting in the work your end. That's what, he or she is there for. If you feel that your instructor isn't prepared to assist, then it really is imperative that you discuss this with the CFI. If you have been on the relevant flying lesson that topic should have been covered in the brief, and your understanding assessed. Trust me, your CFI will want to know about these gaps in learning so he can improve the schools systems.
Another bit of advice. Don't sit on questions for too long. Get them addressed. As time goes on and you get more experienced it gets harder to go back and get clarification on the more basic stuff.
All the best with your studies
You are obviously early in your studies and good on you for showing some initiative and coming on here for some advice. BUT, if you are currently undertaking flying training, your first port of call should always be your instructor. It could be your very new and haven't yet established that rapport, but your instructor will be very keen to help you, provided you're putting in the work your end. That's what, he or she is there for. If you feel that your instructor isn't prepared to assist, then it really is imperative that you discuss this with the CFI. If you have been on the relevant flying lesson that topic should have been covered in the brief, and your understanding assessed. Trust me, your CFI will want to know about these gaps in learning so he can improve the schools systems.
Another bit of advice. Don't sit on questions for too long. Get them addressed. As time goes on and you get more experienced it gets harder to go back and get clarification on the more basic stuff.
All the best with your studies