Why does an aircraft descend quicker when it is lighter?
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Fair point Keith.
I am proberly focusing to much on the max rate because thats what in real life causes you most problems with decent managment. High and need to get the height off can be a major issue where as below profile might be fuel hungry but it cause alot less problems.
O aye and if you use my theory in a tech interview you get the job
I am proberly focusing to much on the max rate because thats what in real life causes you most problems with decent managment. High and need to get the height off can be a major issue where as below profile might be fuel hungry but it cause alot less problems.
O aye and if you use my theory in a tech interview you get the job
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Heavy car, top of hill, release brakes, will be going faster at bottom of hill than light car due to stored (potential) energy!
It's no more complicated than that!
It's no more complicated than that!
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The original question was essentially "Why is the rate of descent less and the distance required to descend more, when an aircraft is heavy, compared to when it is light.
I'm not at all sure how your heavy and light car analogy explain this phenomenon.
I'm not at all sure how your heavy and light car analogy explain this phenomenon.
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Its exactly the same thing except I have used very basic theory where as you have resorted to using the greek alphabet.
At Vmo what is the drag going to be at various weight say 50% 75% and 100% of MTOW. In relation to each other?
Code:
L/D Data for 737-700 @ 0.78/280/250 KIAS UNITS Weight 40.0 50.0 60.0 70.0 (1000 kgf) L/D 15.4 17.8 19.6 20.6 Drag 2.6 2.8 3.1 3.4 (1000 kgf) ERR 2.6 2.8 3.1 3.4 (250kts)*(1000kgf) ERR' 0.65 0.56 0.51 0.49 (25kts) ERR(fpm)1621 1405 1277 1213 (fpm) ERR = Energy Removal Rate ERR' = Energy Removal Rate Per Unit Weight ERR'(fpm) = Energy Removal Rate Per Unit Weight (expressed in feet per minute)
Or put another way when are you going to have maximum energy removal from the system?
But maximum per unit weight energy removal rate is highest for the lightest aircraft. Interestingly enough it has the same units as speed and can be conveniently expressed in fpm.
Your graph is fine and my graph is fine but there are differences.
1. Your graph has speed on its horizontal axis, my graph has angle of attack on horizontal axis
2. Your graph doesn't depict L/D directly, my graph does.
The way to infer L/D curve from your graph is by assuming that lift is constand and speed is varying, that way L/D curve from your graph will be reciprocal of total drag curve.
Now a few arguments why my graph includes the parasite drag as well as induced drag just like your graph. My graph sure doesn't break them down in two components.
At very high speeds where angle of attack approaches zero, lift induced drag must be zero, your graph says that. My graph at zero angle of attacks shows a positibe coefficient of drag, this is parsite drag coefficient.
Please note that my graph doesn't depict lift or drag, instead it depicts the lift and drag coefficients, thats why the shape of the curve is different from yours.
As you can see the max drag is at the highest airspeed which is limit by Vmo. Therefore to get max rate of decent you need to go as fast as possible. Which is where my very simple equation comes into play.
Last edited by jimmygill; 17th Apr 2010 at 21:11.
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It's interesting that this debate is all about drag.
I say interesting because it is not drag that matters in determining the sink rate in a glide. It is the power required.
I say interesting because it is not drag that matters in determining the sink rate in a glide. It is the power required.
Its not the drag that matters.
Its not the power required that matters.
Its the power required per unit weight of the aircraft that matters here, this quantity has a direct relation with the all so familiar Lift to Drag ratio.
Heavy car, top of hill, release brakes, will be going faster at bottom of hill than light car due to stored (potential) energy!
It's no more complicated than that!
It's no more complicated than that!
My instructor always stressed on Keep It Simple and Stupid. But this is analogy way too simplistic.
You have controls in aircraft using which you can set its speed for a descent.
I haven't seen any car with down the hill cruise control while in neutral gear.
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At very high speeds where angle of attack approaches zero, lift induced drag must be zero, your graph says that. My graph at zero angle of attacks shows a positibe coefficient of drag, this is parsite drag coefficient.
Free body diagram of the two cases of weight.
Energy in arrow forward
Energy out Drag arrow to the back.
Energy in the system potential inside the box.
Energy out arrow pointing down (gravity)
Energy in pointing up (lift)
Max airspeed equals max drag which is the same for both as the Vmo doesn't change with the weight. As per the graph I showed if you look at the total drag.
As soon as you go away from Max Drag you are creating more variables that you have to account for and also will affect the result. Which is why I took the situation of Vmo because it zero's out or decreases a multiple of variables to insignificants. Induced drag tends towards insignificant so your variation in L/D ratios can be ignored the AoA of the wing can be ignored. By taking any other speed other than maximum you are into calculus to work out what all the different variables will actually mean and how they relate to each other.
Your 737 data proves the point, the only variable that is constant is the speed. Drag is different for each weight, the energy removed is different for each weight. You have 4 variables which can't prove a relationship between two of the variables.
Where as my solution the drag is the same for all weights, the energy removed is the same and the speed is the same. We only have two variables the weight and the rate of decent. Which you can prove the relationship.
It is a very simple problem which is given to second year Mech Eng students to teach them how use freebody diagrams for pratical problem solving and manipulations of variables to form simplified systems from complex ones and to also learn how to choose boundary conditions to find a relationship.
Yes you could go into into L/D ratios and all that but they would fail because they wouldn't have proved the relationship.
Now with your 737 data get a set of weights where the drags the same and the energy removal rate is the same and the speed is the same. Then we will be talking about the same thing. I will save you some time it won't happen but if you try Vmo it will be quite close.
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Actually I have worked out how you could do it.
You would have to do some none dimensionless analysis so that you could normalise the rate of decent against the other variables.
A bit like using reynolds number in the wind tunnel.
Its 15 years since I have done any of that in anger so your on your tod.
You would have to do some none dimensionless analysis so that you could normalise the rate of decent against the other variables.
A bit like using reynolds number in the wind tunnel.
Its 15 years since I have done any of that in anger so your on your tod.
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Holy Moly.
Although this one may seem counter-intuitive when you first think about it, most students are happy enough with Rate Of Climb = Excess Power / Weight.
The easiest way to deal with descending is just to think of it as a rate of climb that happens to be negative.
Rate Of Descent = Power Deficit / Weight.
Immediately you can see that increasing the weight makes the descent rate less.
pb
Although this one may seem counter-intuitive when you first think about it, most students are happy enough with Rate Of Climb = Excess Power / Weight.
The easiest way to deal with descending is just to think of it as a rate of climb that happens to be negative.
Rate Of Descent = Power Deficit / Weight.
Immediately you can see that increasing the weight makes the descent rate less.
pb
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Capt. Pit Bull & the others.
Dear all,
I have actually had the same question on my mind last week when I was reading through Gary Bristow's "Ace" book as well. I remember in school, we learned that "the heavier aircraft reaches the ground faster than a lighter aircraft".
It stood out to me when this question came out, and I can now SOMEWHAT understand/appreciate why --- drag?
Please help me (and others) understand this. So, to keep it simple, the heavier the aircraft, the more lift it requires hence the higher AoA thus the more induced drag it generates.
So if the lighter & heavier aircraft had the SAME AoA, the heavier aircraft would descend faster due to momentum/kinetic energy?
This has been bugging me since last week! Please help, anybody! Thanks!
@jfkjohan
I have actually had the same question on my mind last week when I was reading through Gary Bristow's "Ace" book as well. I remember in school, we learned that "the heavier aircraft reaches the ground faster than a lighter aircraft".
It stood out to me when this question came out, and I can now SOMEWHAT understand/appreciate why --- drag?
Please help me (and others) understand this. So, to keep it simple, the heavier the aircraft, the more lift it requires hence the higher AoA thus the more induced drag it generates.
So if the lighter & heavier aircraft had the SAME AoA, the heavier aircraft would descend faster due to momentum/kinetic energy?
This has been bugging me since last week! Please help, anybody! Thanks!
@jfkjohan
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@jfkjohan
What you learnt at the school is correct. The actual context of the learning was this:
Airplane B is heavier than Airplane A. Both are same make. Which one will descend faster for best range glide.
Same make means same L/D characteristics.
In order for these aircrafts to maximise the glide distance each of them should glide at maximum L/D. L/D is purely an aerodynamic variable, function of airplane shape and angle of attack only.
Maximum L/D is achieved at a fixed angle of attack for a fixed shape of the aircraft.
Now the problem is constraining the Angle of Attack. If A and B fly at the same angle of attack, and if B has to produce more lift the only way B can do this is by flying at a higher speed.
Yes the heavier aircraft glides at a higher speed for maximum range glide.
But for a speed restricted descent, the heavier aircraft may have a higher or lower sink rate depending on the L/D ratio at which the heavy aircraft is operating.
I will try to build up on the simple sentence you started.
Thanks to kieth for pointing out the mistake.
@Capt Pitt Bull
Your hypothesis work for low angle of attack regime and fails in high angle of attack regime. Even though it is easy to explain, it is wrong.
"the heavier aircraft reaches the ground faster than a lighter aircraft"
Airplane B is heavier than Airplane A. Both are same make. Which one will descend faster for best range glide.
Same make means same L/D characteristics.
In order for these aircrafts to maximise the glide distance each of them should glide at maximum L/D. L/D is purely an aerodynamic variable, function of airplane shape and angle of attack only.
Maximum L/D is achieved at a fixed angle of attack for a fixed shape of the aircraft.
Now the problem is constraining the Angle of Attack. If A and B fly at the same angle of attack, and if B has to produce more lift the only way B can do this is by flying at a higher speed.
Yes the heavier aircraft glides at a higher speed for maximum range glide.
But for a speed restricted descent, the heavier aircraft may have a higher or lower sink rate depending on the L/D ratio at which the heavy aircraft is operating.
Please help me (and others) understand this. So, to keep it simple, the heavier the aircraft, the more lift it requires hence the higher AoA thus the more induced drag it generates.
I will try to build up on the simple sentence you started.
- The heavier the aircraft the more lift it requires.
- If the airspeed is constrained only way to increase the lift is by increasing angle of attack.
- An increase in angle of attack will always increase the coefficient* of drag and coefficient of Lift.
- But an increase in AOA may increase the L/D ratio, or it may decrease the L/D ratio.
- The variation in L/D ratio depends on which region of the L/D curve the aircrafts are operating.
- As you can see in the following curve for Angle of Attack between 0-6 L/D increases with increase in AOA. For AOA beyond 6 degrees, L/D reduces with AOA.
- Hence a heavier aircraft is a better glider than a lighter on in low AOA regime. And a lighter aircraft is a better glider in high angle of attack regime.
Thanks to kieth for pointing out the mistake.
@Capt Pitt Bull
Rate Of Descent = Power Deficit / Weight
Last edited by jimmygill; 18th Apr 2010 at 15:06.
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Which isn't the question that has been asked.
It is nothing to do with glide distances or speeds in the glide
It is purely
Its purely asking why the rate of decent is higher when you are lighter
Generally aircraft don't operate on the dirty side of the drag curve unless your on approach. Pit Bulls method works for anything above minimum drag. I know these days of advanced fuel performance the wisdom dictates we don't decend at Vmo but its always at a speed which is faster than minimum drag.
It is nothing to do with glide distances or speeds in the glide
It is purely
Why does an aircraft descend quicker when it is lighter?
Generally aircraft don't operate on the dirty side of the drag curve unless your on approach. Pit Bulls method works for anything above minimum drag. I know these days of advanced fuel performance the wisdom dictates we don't decend at Vmo but its always at a speed which is faster than minimum drag.
Last edited by mad_jock; 18th Apr 2010 at 11:11.
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The question was "why does an aircraft descend quicker when it is lighter?" and the answer is that it doesn't - except in the special case where you required to descend at a fixed airspeed above your minimum drag speed. As you all know if you fly at Vimp or Vimd the heavier aircraft will have a higher rate of descent. If you choose a particular descent angle then again the heavier aircraft, going faster, will have a higher rate of descent. If you can manage a terminal velocity vertical dive the heavier aircraft will go straight down faster.
If you are ordered by ATC to descend at a given airspeed or take it to the limit at Vmo/Vne then to hold the required airspeed the heavier aircraft will have to adopt a shallower descent angle and therefore will have a lower rate of descent.
So, in summary, if you are free to choose your airspeed the lighter aircraft will stay up longer. If you must maintain a fixed airspeed in a cruise descent the heavier aircraft will give you the demanded airspeed in a shallower descent
The reason why the question pops up in interviews and in the CQB is because the aircraft behaviour in a cruise descent - well known in practice - appears paradoxical in theory. Anyway, from Keith and Mad Jock you have all the aerodynamics you could wish for to deploy as your answer
Dick
If you are ordered by ATC to descend at a given airspeed or take it to the limit at Vmo/Vne then to hold the required airspeed the heavier aircraft will have to adopt a shallower descent angle and therefore will have a lower rate of descent.
So, in summary, if you are free to choose your airspeed the lighter aircraft will stay up longer. If you must maintain a fixed airspeed in a cruise descent the heavier aircraft will give you the demanded airspeed in a shallower descent
The reason why the question pops up in interviews and in the CQB is because the aircraft behaviour in a cruise descent - well known in practice - appears paradoxical in theory. Anyway, from Keith and Mad Jock you have all the aerodynamics you could wish for to deploy as your answer
Dick
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I apologise Pitt Bull, your formula is correct
In airplane A and B TAS is same for both,
hence Rate of descent is determined by the Ratio of Drag to the Lift in the case.
Just putting the weight in the denominator is robbing us of accuracy amd lending credence to inappropriate analogies such as a car rolling down the hill.
Code:
Rate Of Descent = Power Deficit / Weight.
= (Drag*TAS) / Weight
= TAS * (Drag/Lift) Lift = Weight (approx)
hence Rate of descent is determined by the Ratio of Drag to the Lift in the case.
Just putting the weight in the denominator is robbing us of accuracy amd lending credence to inappropriate analogies such as a car rolling down the hill.
Last edited by jimmygill; 18th Apr 2010 at 12:45.
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The equation Capt Pit Bull stated is exactly the same as mine but uses different terms. I used a constant of energy lose per min so you could get a rate of decent per min.
Putting the weight on the bottom does nothing to affect the solution of the relationship. It gives us an inverse relationship. Accuracy is an issue with expicit solutions not implicit.
Why do you insist on bring L/D none dimensional ratio into it and TAS its not required and adds nothing to the relationship of rate of decent to weight.
As I said very very common mistake for engineers early on in their academic education and for some it never does click. Thankfully they make very good accountants
Putting the weight on the bottom does nothing to affect the solution of the relationship. It gives us an inverse relationship. Accuracy is an issue with expicit solutions not implicit.
Why do you insist on bring L/D none dimensional ratio into it and TAS its not required and adds nothing to the relationship of rate of decent to weight.
People have focused on the mechanics of the the conversion instead of boxing off the system and applying basic physics laws
Last edited by mad_jock; 18th Apr 2010 at 13:28.
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LMAO Farrell
Or one from my past which I always forget the real reason for the correct answer.
"Which wing section is more prone to Icing a thick section or a thin section?"
MJ's Answer with fingers crossed "a thin section is more prone to icing"
"why is that?"
MJ's answer "I don't have a clue to be honest I just know DC3's don't have any de-icing boots on the wing and they have big thick sections and they have flown all over the world without dropping out the sky due to icing"
"well thats the most orginal answer we have had to that question, and can't fault your logic. Tell me what you know about type IV de-ice fluid?"
Or one from my past which I always forget the real reason for the correct answer.
"Which wing section is more prone to Icing a thick section or a thin section?"
MJ's Answer with fingers crossed "a thin section is more prone to icing"
"why is that?"
MJ's answer "I don't have a clue to be honest I just know DC3's don't have any de-icing boots on the wing and they have big thick sections and they have flown all over the world without dropping out the sky due to icing"
"well thats the most orginal answer we have had to that question, and can't fault your logic. Tell me what you know about type IV de-ice fluid?"
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An increase in angle of attack will always increase the total drag.
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Actually I know its a bit of a thread diversion.
Why is a thin section more prone to icing than a thick one?
Just in case the interviewer doesn't have a sense of humour like the last one.
Why is a thin section more prone to icing than a thick one?
Just in case the interviewer doesn't have a sense of humour like the last one.
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Why do you insist on bring L/D none dimensional ratio into it and TAS its not required and adds nothing to the relationship of rate of decent to weight.
And rate of descent comes out to be
ROD = TAS / (D/L)
How can one ignore this important non dimensional characteristics?
@keith
I have made the edit in the original post. Thanks for pinting out.
Last edited by jimmygill; 18th Apr 2010 at 17:57.
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Because its irrelvant to forming the solution. It looks sexy and try's to make the solution look aero dynamic, but that is the sum total of its usefulness.
All you do by bring TAS into it is introducing a heap of enviromental variables. Which then cancel each other out because your comparing aircraft in exactly the same situation bar the fact that there weight is different. Again its irrelevent to the solution apart from it makes it look aero dynamic.
You are trying to give an expicit answer to something that doesn't require one.
The point I am making is that if you start bring L/D ratio's and other such factors into play to answer the question. You will just come across as using techno waffle to answer a simple question.
So my answer would be
"The heavier aircraft has more potential energy than the lighter one and at a constant airspeed they both remove the same amount of energy via drag in the decent. Therefore for over a period of time the lighter aircraft would have lost more altitude than the heavy one ie the lighter one has a higher rate of decent"
Thats the answer they are looking for which for wannabies is the answer they should give if they get to a oral tech interview.
All you do by bring TAS into it is introducing a heap of enviromental variables. Which then cancel each other out because your comparing aircraft in exactly the same situation bar the fact that there weight is different. Again its irrelevent to the solution apart from it makes it look aero dynamic.
You are trying to give an expicit answer to something that doesn't require one.
The point I am making is that if you start bring L/D ratio's and other such factors into play to answer the question. You will just come across as using techno waffle to answer a simple question.
So my answer would be
"The heavier aircraft has more potential energy than the lighter one and at a constant airspeed they both remove the same amount of energy via drag in the decent. Therefore for over a period of time the lighter aircraft would have lost more altitude than the heavy one ie the lighter one has a higher rate of decent"
Thats the answer they are looking for which for wannabies is the answer they should give if they get to a oral tech interview.