Why heavier aircrafts take longer to slow down in the air?
No need to discuss momentum or energy.
Thanks for that - I was beginning to worry that when I did my aero engineering degree I must have missed a lecture where they explained that momentum was a force.
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Dave,
We're debating with a person who claims to be an engineer designing aero engines yet writes such nonsense as (see his first post #15 on this thread):
Then states his objective as:
Regards,
HN
We're debating with a person who claims to be an engineer designing aero engines yet writes such nonsense as (see his first post #15 on this thread):
... we can't use the F=MA formula as it doesn't allow us to take account of the velocity of the aircraft at the point of forward thrust being replaced with momentum / inertia.
... if the objects have different mass as the drag of each object equals the mass of the object acceleration will become zero
... if the objects have different mass as the drag of each object equals the mass of the object acceleration will become zero
water sufficiently muddy i hope.
HN
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Using Newton (forces, inertia, acceleration, momentum) is complicated here, because airplanes move a lot of air. It is not just ann ideal body moving in outer space.
For F=m.a we need to know the force, which is D, and it is a complex dorce to work with. It doesn't behave linearly.
For KE= 1/2 m v2, we don't know how much energy is put in the air, so it is difficult to calculate decceleration rate based on that.
The best approach, I deem, is the empirical formulae of L and D, which are based on the empirically obtained values of CL and CD. And then Newton.
Mass increases, Weight increases proportionally, AoA increases proportionally, CL increases proportionally, CD increases but less than proportionally to CL (at typicall AoAs), so D increases less than mass does. Newton's second and Galileo's equations then take us to the conclusion: The heavier a/c will use more time and distance to slow down.
For F=m.a we need to know the force, which is D, and it is a complex dorce to work with. It doesn't behave linearly.
For KE= 1/2 m v2, we don't know how much energy is put in the air, so it is difficult to calculate decceleration rate based on that.
The best approach, I deem, is the empirical formulae of L and D, which are based on the empirically obtained values of CL and CD. And then Newton.
Mass increases, Weight increases proportionally, AoA increases proportionally, CL increases proportionally, CD increases but less than proportionally to CL (at typicall AoAs), so D increases less than mass does. Newton's second and Galileo's equations then take us to the conclusion: The heavier a/c will use more time and distance to slow down.
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hen I did my aero engineering degree I must have missed a lecture where they explained that momentum was a force.
But hey you can't let the facts get in the way of an attack now can you.
what i said and repeated was that a force is derived from the momentum to make the glider go up against gravity.
In this circumstance there are 4 forces acting on the glider not 3.
The momentum is not a force it creates a force on the mass to push it in a direction momentum has a quantity and a vector.
i assume you can prove this claim that i am not an engineer designing aero engines
But you certainly made this one, when addressing me:
So to say you clearly have a background in the subject when all you rely on is google is a bit rich and highly insulting to those of that rely do have a background in aeronautical engineering.
Pot ? Kettle ? Black ?
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Bye,
I'm just quoting the facts as you wrote them in your posts. It is up to the reader to form his/her opinion about your sincerity in this debate.
If you're not playing the game of trolling, then take the sound advice offered to you in good faith in post #75.
Regards,
HN
I'm just quoting the facts as you wrote them in your posts. It is up to the reader to form his/her opinion about your sincerity in this debate.
If you're not playing the game of trolling, then take the sound advice offered to you in good faith in post #75.
Regards,
HN
Last edited by HazelNuts39; 24th Nov 2012 at 12:16.
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Bye,
You have complained that nobody will explain the basic aerodynamics to you but you yourself gave a reference to a document that does just that in post #27. Is it unreasonable to expect that you have read this?
You are a design engineer and own a company that designs aero. engines so you ought to be able to put some numbers in the equations given in your reference, or employ someone who could do it for you, so if I give you some realistic values you can check what others are saying for yourself.
For an aircraft of the B737/A320 class:
Cdo (profile drag coefficient) 0.018
K (induced drag factor) 0.0425
Wing area 1300 sq.ft
Weights between 180,000 and 120,000 lb
Over to you
You have complained that nobody will explain the basic aerodynamics to you but you yourself gave a reference to a document that does just that in post #27. Is it unreasonable to expect that you have read this?
You are a design engineer and own a company that designs aero. engines so you ought to be able to put some numbers in the equations given in your reference, or employ someone who could do it for you, so if I give you some realistic values you can check what others are saying for yourself.
For an aircraft of the B737/A320 class:
Cdo (profile drag coefficient) 0.018
K (induced drag factor) 0.0425
Wing area 1300 sq.ft
Weights between 180,000 and 120,000 lb
Over to you
Last edited by Owain Glyndwr; 24th Nov 2012 at 12:36.
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Adding any energy to the system (thrust) was mooted when it was eliminated in the original question. The question had everything to do with reducing velocity, "slowing down". This involves dissipating energy, not accreting it.
I think calling momentum 'thrust' is not appropriate. Momentum is the result of thrust, not its own precursor. The identified heavy aircraft is unpowered by definition.
all respect.
I think calling momentum 'thrust' is not appropriate. Momentum is the result of thrust, not its own precursor. The identified heavy aircraft is unpowered by definition.
all respect.
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Bye
I don't know what that means exactly, but if I knew that references in one of my posts were wrong I would feel duty bound to issue a correction. As it stands the reference you quoted DOES contain the explanation you are looking for and you can access it and put in the numbers I gave, so you can, if you wish, make an independent verification of what others have been saying here.
Once again, over to you.
Owain, sorry but someone posted that the link i posted contained references that were wrong so i expected them to be corrected before i could use them
Once again, over to you.
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HN39 said the material in the link had errors in it.
He didn't say what they were. and i couldn't see any errors but then i know nothing.
so i was hoping he would point out where the material was in error.
i have worked through with your numbers and thanks for them, and i am wiser for it thanks to you.
oh and forgot to state that FORCE = MOMENTUM / TIME
GB
Edit, having gone thru again, i still can't get why the 2 aircraft at different weights can't have the same L/D ratio. the graph shows they do at 226 kts, and my sums show they do as well with the data you gave.
I can get them to have the same L/d ratio at a certain V every time, so i would be grateful if you would explain again why its not possible and where the graphs are wrong.
GB
He didn't say what they were. and i couldn't see any errors but then i know nothing.
so i was hoping he would point out where the material was in error.
i have worked through with your numbers and thanks for them, and i am wiser for it thanks to you.
oh and forgot to state that FORCE = MOMENTUM / TIME
GB
Edit, having gone thru again, i still can't get why the 2 aircraft at different weights can't have the same L/D ratio. the graph shows they do at 226 kts, and my sums show they do as well with the data you gave.
I can get them to have the same L/d ratio at a certain V every time, so i would be grateful if you would explain again why its not possible and where the graphs are wrong.
GB
Last edited by Bye; 24th Nov 2012 at 16:40.
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Dear mr Bye,
I'm not aware of anything in my writing that calls for the apology you ask for. I've pointed to an incongruence that I perceive between your claim of being an engineer and the understanding of elementary physics that you display in your posts. That leaves two possibilities:
Either you are an engineer and deliberately pretend to misunderstand the physical realities you write about, or -
You really don't understand physics and refuse to learn more about it, which in my mind is not compatible with having had an engineering education.
I do apologize for a mistaken remark that I made regarding the paper referenced in your post #27. That paper discusses speed stability at two arbitrily chosen speeds V1 and V2. When writing my post just before going to bed I mistakenly associated those with the V1 and V2 speeds that are used in the operation of civil transport airplanes. Thinking about it later I realized my error and deleted that remark first thing the next morning.
I'm not aware of anything in my writing that calls for the apology you ask for. I've pointed to an incongruence that I perceive between your claim of being an engineer and the understanding of elementary physics that you display in your posts. That leaves two possibilities:
Either you are an engineer and deliberately pretend to misunderstand the physical realities you write about, or -
You really don't understand physics and refuse to learn more about it, which in my mind is not compatible with having had an engineering education.
I do apologize for a mistaken remark that I made regarding the paper referenced in your post #27. That paper discusses speed stability at two arbitrily chosen speeds V1 and V2. When writing my post just before going to bed I mistakenly associated those with the V1 and V2 speeds that are used in the operation of civil transport airplanes. Thinking about it later I realized my error and deleted that remark first thing the next morning.
the rest of us can't hope to match your grasp of aeronautics
and i missed the bit where i claimed that it [momentum] was [a force]
Your post #43:
in a glider the only thrust you have is in fact the momentum you have
yes its called thrust because the momentum is PUSHING the glider upwards against gravity
Yet you claim i said that momentum was a force, sorry i don't see it, i see the word thrust, but not the word force
So presumably the thrust that's produced by the engines you design isn't capable of making the aircraft move.
Words fail me.
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hi bye,
I see no libel. I also see an ephemeral chance for truce.
IMO. Thrust is a force that is not contained in the object to which it is applied.
So, to apply the word Thrust to the system under discussion, we look for a separate object, divorced from the airframe, such that it is subject to, and gains from, the force contained in the system.
The Airmass is acted upon, and reacts to, the entry of the aircraft into its separate region. The aircraft imparts energy to the airmass, the airmass reacts, the airplane slows. Accelerating the airmass produces momentum therein, the aircraft accelerates negatively, etc.
As I say, ephemeral. But not wrong, and in the proper spirit, etc.
bye, bye
I see no libel. I also see an ephemeral chance for truce.
IMO. Thrust is a force that is not contained in the object to which it is applied.
So, to apply the word Thrust to the system under discussion, we look for a separate object, divorced from the airframe, such that it is subject to, and gains from, the force contained in the system.
The Airmass is acted upon, and reacts to, the entry of the aircraft into its separate region. The aircraft imparts energy to the airmass, the airmass reacts, the airplane slows. Accelerating the airmass produces momentum therein, the aircraft accelerates negatively, etc.
As I say, ephemeral. But not wrong, and in the proper spirit, etc.
bye, bye
Last edited by Lyman; 24th Nov 2012 at 17:53.
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We're debating with a person who claims to be an engineer designing aero engines yet writes such nonsense
Much like someone saying to me that they can't believe I claim to be pilot if I did a landing like that - They're not claiming I'm lying about being a pilot, they're just claiming that as a pilot they can't believe I actually did a landing like that!
And off the record I nail every landing - on the record though...
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Already have one engineering degree, and working toward computer science master's degree.
all this thread is .....mm....weird! I initially (silly me) thought that there is only momentum envolved. Then, ok, if you need to account for different attitude (=different AoA + different profile, obviously) for different weights you can do it too. But after you calculate initial momentum which doesn't change (I don't consider burning fuel here, since that wasn't the initial question and there is no point in making task over-complicated) it shouldn't be too hard to account for different AoA + profile drag too. induced drag will be almost the same, I assume.
I think the thread needs to be closed since there is no point, I am afraid, to continued!
One side tries to stick to physics, others try to develop their own...ehmm...understanding which is a bit strange to me!
All sloppiness that are presented here can't be a sign of good knowledge in physics in any way.
Everything that was said here about legal claims whatever it might be, simply beyond anything logical...don't want to comment on that.
all this thread is .....mm....weird! I initially (silly me) thought that there is only momentum envolved. Then, ok, if you need to account for different attitude (=different AoA + different profile, obviously) for different weights you can do it too. But after you calculate initial momentum which doesn't change (I don't consider burning fuel here, since that wasn't the initial question and there is no point in making task over-complicated) it shouldn't be too hard to account for different AoA + profile drag too. induced drag will be almost the same, I assume.
I think the thread needs to be closed since there is no point, I am afraid, to continued!
One side tries to stick to physics, others try to develop their own...ehmm...understanding which is a bit strange to me!
All sloppiness that are presented here can't be a sign of good knowledge in physics in any way.
Everything that was said here about legal claims whatever it might be, simply beyond anything logical...don't want to comment on that.
Last edited by Sunamer; 25th Nov 2012 at 03:19.
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I don't have an engineering degree, but have been interested in the points raised.
The OP's request related to two similar aircraft, one heavier than the other, and why the heavier aircraft took longer to slow down than the lighter one. Assuming that the initial speeds are the same, the mass of each aircraft along with the L/D ratio (associated with AoA) will become the variable. Discount the flight idle fuel usage and thrust which are equal for both aircraft, then we are essentially dealing with kinetic energy.
EK = ½mv²
The L/D ratio along with AOA will change in a similar (though different) fashion for both aircraft as they slow, so it can be assumed that the Mass and V² are the only real factors we are dealing with. SLOWING is the operate word, and it follows without too much argument that the aircraft with the higher mv² product will always take longer to slow than the lighter one.
My 3 year old grandson proved it to me when he set up two parallel lines of Thomas wooden railway and placed one engine and a carriage on one line, then one engine and a wagon loaded with sand on the other. He pushed them off together, and you know the answer - he observed it also.
The OP's request related to two similar aircraft, one heavier than the other, and why the heavier aircraft took longer to slow down than the lighter one. Assuming that the initial speeds are the same, the mass of each aircraft along with the L/D ratio (associated with AoA) will become the variable. Discount the flight idle fuel usage and thrust which are equal for both aircraft, then we are essentially dealing with kinetic energy.
EK = ½mv²
The L/D ratio along with AOA will change in a similar (though different) fashion for both aircraft as they slow, so it can be assumed that the Mass and V² are the only real factors we are dealing with. SLOWING is the operate word, and it follows without too much argument that the aircraft with the higher mv² product will always take longer to slow than the lighter one.
My 3 year old grandson proved it to me when he set up two parallel lines of Thomas wooden railway and placed one engine and a carriage on one line, then one engine and a wagon loaded with sand on the other. He pushed them off together, and you know the answer - he observed it also.
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yes. all that is true, but
the thing that needs to be accounted for also is:
unlike in the case with two different-weighted railroad engines, that have different mass but identical railways, identical speed and identical friction (drag is the same here)
two identical planes with different masses will have different AoA due to the fact that once needs more lift to keep AC at the same altitude if AC has more mass, than derivation of AoA = aerodynamic profile will be (probably) less efficient and this will create more profile drag (not induced, though). Since it creates more drag, it probably in some cases will be enough to counteract more momentum of heavier AC and thus the result will be the same = two aircraft will slowdown the same way.
everything depends upon particular coefficients and values. There is no general solution here to say that yes, it will slow down slower (or faster), I am assuming.
the thing that needs to be accounted for also is:
unlike in the case with two different-weighted railroad engines, that have different mass but identical railways, identical speed and identical friction (drag is the same here)
two identical planes with different masses will have different AoA due to the fact that once needs more lift to keep AC at the same altitude if AC has more mass, than derivation of AoA = aerodynamic profile will be (probably) less efficient and this will create more profile drag (not induced, though). Since it creates more drag, it probably in some cases will be enough to counteract more momentum of heavier AC and thus the result will be the same = two aircraft will slowdown the same way.
everything depends upon particular coefficients and values. There is no general solution here to say that yes, it will slow down slower (or faster), I am assuming.
What my solicitor has said is that the intent is clear that his intention was to discredit my standing as a professional engineer on a public forum.