How hard does a 747 land???
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How hard does a 747 land???
Hey there! (my first post on this forum)
I was interested to know what rate (in fpm) does a 747 touch the runway at in real life??
In flightsim, i touch them down at avrg. 140feet per minute. (this represents how hard the wheels touch the ground)and i was wondering (all you 747 pilots out there) what rate (in fpm) would be considered a good landing?
Any help will be MUCH appreciated.
regards, Andrew (sydney, Australia)
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Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
I was interested to know what rate (in fpm) does a 747 touch the runway at in real life??
In flightsim, i touch them down at avrg. 140feet per minute. (this represents how hard the wheels touch the ground)and i was wondering (all you 747 pilots out there) what rate (in fpm) would be considered a good landing?
Any help will be MUCH appreciated.
regards, Andrew (sydney, Australia)
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
Guest
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Boeing build them to withstand a no-flare landing ie upto about 800fpm on a heavy weight landing.
Good landings on the 747 are when you eliminate ALL drift and the final descent rate is such that the spoilers deploy at the end of the touchdown. If you hold off the main gear, when the trucks are no longer tilted (giving an a/c on ground signal to the spoilers) you will end up dropping another 2-3ft spoiling an otherwise great touchdown.
No figures, I'm afraid but you know when it's a good'un.
Good landings on the 747 are when you eliminate ALL drift and the final descent rate is such that the spoilers deploy at the end of the touchdown. If you hold off the main gear, when the trucks are no longer tilted (giving an a/c on ground signal to the spoilers) you will end up dropping another 2-3ft spoiling an otherwise great touchdown.
No figures, I'm afraid but you know when it's a good'un.
Guest
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what do you mean by "the end of the touchdown?" Are you able to give me a little bit of a ballmark figure? :-) Like what rate would slam the wheels straight into the ground and what would drag them along the ground?
You see i'm only 15 and it's gonna be a while before in get in the seat of a 747
Thanks , regards, andrew
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
You see i'm only 15 and it's gonna be a while before in get in the seat of a 747
Thanks , regards, andrew
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
Guest
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Have you not started trigonometry yet ? Get stuck into it.
An aircraft flying down an ILS approach is approaching the ground at a 3 degree (more or less) angle.
So if it is flying in still air at a speed of S knots (1 knot =6080 feet per hour) then its vertical speed V is defined by the equation
V / S = sin(3 degrees)
where sin(3) = .0523
So if the aircraft is doing 150kt, its vertical speed is 795 fpm in still air, which fits with C1's comment.
If he is correct, then the 747 is designed to fly down the glide slope and be planted on the ground without any relief due to flare or ground effect.
(Edited to reverse the definition of V and S)
[This message has been edited by EchoTango (edited 01 December 2000).]
An aircraft flying down an ILS approach is approaching the ground at a 3 degree (more or less) angle.
So if it is flying in still air at a speed of S knots (1 knot =6080 feet per hour) then its vertical speed V is defined by the equation
V / S = sin(3 degrees)
where sin(3) = .0523
So if the aircraft is doing 150kt, its vertical speed is 795 fpm in still air, which fits with C1's comment.
If he is correct, then the 747 is designed to fly down the glide slope and be planted on the ground without any relief due to flare or ground effect.
(Edited to reverse the definition of V and S)
[This message has been edited by EchoTango (edited 01 December 2000).]
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FAA certification criteria requires the Aircraft meet a gross weight rate of descent 6' per second(360') and Max'landing' weight, rate of descent of 10/per second(600') Airbus require a Max'landing'weight,rate of ( 560' per minute)
Normal touchdowns are in the order of 1G,about 2/300' per minute..
Normal touchdowns are in the order of 1G,about 2/300' per minute..
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Thaks guys! This info is really helpful
Concerning Trig. we have started it but only to find out say one angle of a right angle triangle and how long an unknown length is etc..
Is there a way to convert Feet per Minute in a G reading??
Keep the replies coming!
regards, Andrew
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
Concerning Trig. we have started it but only to find out say one angle of a right angle triangle and how long an unknown length is etc..
Is there a way to convert Feet per Minute in a G reading??
Keep the replies coming!
regards, Andrew
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
Guest
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I'm not sure if you are trying to find out how to land a 747 decently, or just mucking around with to sim, but...
1st. Because of the height of the cockpit during the flare, there is a tendency not to. At least, not in time that is. Soooo, in addition to all the wonderful eqpt. the 747 has installed, is the GPWS, whose several modes include one that calls out, amongst other things, your radio altitude, in specific incriments. THE must important one is the '50' foot call, followed by 40, 30, 20, 10. The technique we are taught is to begin your flare maneuver right after 50'. The rate of the following callouts probably are more important than the actual callouts, as they tell of how fast you may still be sinking. If in the end, you make a good landing,you get compliments all around. You should really remember those, because one day, things will not go very well..... and you will mutter some foul words, followed by this terrible silence in the flight deck during the taxi in. Then, you know your FPM, whatever they were, were in excess.
Have fun.
1st. Because of the height of the cockpit during the flare, there is a tendency not to. At least, not in time that is. Soooo, in addition to all the wonderful eqpt. the 747 has installed, is the GPWS, whose several modes include one that calls out, amongst other things, your radio altitude, in specific incriments. THE must important one is the '50' foot call, followed by 40, 30, 20, 10. The technique we are taught is to begin your flare maneuver right after 50'. The rate of the following callouts probably are more important than the actual callouts, as they tell of how fast you may still be sinking. If in the end, you make a good landing,you get compliments all around. You should really remember those, because one day, things will not go very well..... and you will mutter some foul words, followed by this terrible silence in the flight deck during the taxi in. Then, you know your FPM, whatever they were, were in excess.
Have fun.
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Hi Andrew
You ask is there a way to convert feet per second into G
Feet per second is a speed
A speed is always how long it takes to cover a distance.
G is not a speed it is an acceleration (or deceleration)
An acceleration (or deceleration) is how long it takes you to change speed.
To go from a speed to an acceleration (or deceleration) you need some more information
Stick with me and you should be able to work out any examples of this type that you need for your simulator.
You must be consistent with your units for speed. Feet per second would be best for your landing problem
Sorry if you know this next stuff!
Speed first:
Examples are metres per hour, metres per second, feet per minute, yards per year, inches per 10 years – all these are speeds because they have a distance covered divided by a time taken.
To change from one set of speed units to another is easy. Take the extreme one of inches per 10 years as an example:- divide the inches by 12 and you have changed the speed to feet per 10 years. Multiple the 10 years number by 10 and the speed becomes feet per year. Multiply that by 365 and the answer becomes feet per day. I’m sure you get the drift.
So lets say the speed was 60 inches per 10 years
that would become 60/12 = 5 feet in 10 years.
To get from feet per 10 years to feet per second you would have to divide that answer by 10 then 365 then 24 and finally 60. Clearly not many feet per second!
Now accelerations.
These are speeds divided by time. G is a rate of change of speed of 32.2 feet per second EVERY second. So if you jump off a cliff (ignoring air drag) thanks to gravity being termed 1G you will fall such that you pick up a speed of 32 feet per second for every second you are falling.
Now to your 747. As has been explained by Echo Tango you can convert your speed down the 3 deg ILS glide path (using trig) to a corresponding rate of descent (by the way useful speed unit conversions are that 45 kt = 76 feet per second while 60 mph = 88 ft per second)
So the big question is how do you convert this vertical speed of X feet per second into an acceleration or deceleration. The answer is YOU have to decide in what distance you are going to change your vertical SPEED from what it is on the approach to what it is after landing ie ZERO
This of course means making an assumption about how much the undercarriage legs will shorten due to the impact. That will vary from aircraft to aircraft, but I am guessing with a jumbo that from first touchdown ( of the rear tyres) to a fully compressed leg case with all tyres on the ground the centre of gravity of the aeroplane could have come down 4 feet (could be 6 feet I really don’t know)
Before we go the final lap, let me say that you must use your common sense with all these calculations about speed/distance/acceleration/time and so on. By that I mean understand whether the answer is going to be smaller or bigger when you make a change. This will tell you whether you have to divide by a factor or multiply by the same number.
Now the last lap. The greatest G when landing from a given rate of descent will be if the undercarriage of your aircraft only shortens a little bit. That means your vertical speed has to change VERY QUICKLY. This will feel like a very heavy landing and will mean lots of G
Let us say you decide your aeroplane has an undercarriage that shrinks 4 feet during the landing.
You have said you touchdown at 140 ft/min that is 140/60 = 2.3 ft/ sec
With these two “facts” we put these numbers in the appropriate equation of motion, which for feet per second units is
G = Vsquared/2H
Where G is the G of your touchdown (which we are tying to calculate)
V is YOUR declared touchdown vertical speed (must be in ft/sec)
H is the distance travelled in feet (the amount YOU decide the undercarriage squats)
So: the EXTRA G of touchdown for your numbers would be = 2.3 x 2.3 / 2 x 4 = 0.66
(a tad heavy according to bastud above)
Or if the undercarriage was a 6 ft job it would become 2.3 x 2.3 / 2 x 6 = 0.44
Any help?
JF
You ask is there a way to convert feet per second into G
Feet per second is a speed
A speed is always how long it takes to cover a distance.
G is not a speed it is an acceleration (or deceleration)
An acceleration (or deceleration) is how long it takes you to change speed.
To go from a speed to an acceleration (or deceleration) you need some more information
Stick with me and you should be able to work out any examples of this type that you need for your simulator.
You must be consistent with your units for speed. Feet per second would be best for your landing problem
Sorry if you know this next stuff!
Speed first:
Examples are metres per hour, metres per second, feet per minute, yards per year, inches per 10 years – all these are speeds because they have a distance covered divided by a time taken.
To change from one set of speed units to another is easy. Take the extreme one of inches per 10 years as an example:- divide the inches by 12 and you have changed the speed to feet per 10 years. Multiple the 10 years number by 10 and the speed becomes feet per year. Multiply that by 365 and the answer becomes feet per day. I’m sure you get the drift.
So lets say the speed was 60 inches per 10 years
that would become 60/12 = 5 feet in 10 years.
To get from feet per 10 years to feet per second you would have to divide that answer by 10 then 365 then 24 and finally 60. Clearly not many feet per second!
Now accelerations.
These are speeds divided by time. G is a rate of change of speed of 32.2 feet per second EVERY second. So if you jump off a cliff (ignoring air drag) thanks to gravity being termed 1G you will fall such that you pick up a speed of 32 feet per second for every second you are falling.
Now to your 747. As has been explained by Echo Tango you can convert your speed down the 3 deg ILS glide path (using trig) to a corresponding rate of descent (by the way useful speed unit conversions are that 45 kt = 76 feet per second while 60 mph = 88 ft per second)
So the big question is how do you convert this vertical speed of X feet per second into an acceleration or deceleration. The answer is YOU have to decide in what distance you are going to change your vertical SPEED from what it is on the approach to what it is after landing ie ZERO
This of course means making an assumption about how much the undercarriage legs will shorten due to the impact. That will vary from aircraft to aircraft, but I am guessing with a jumbo that from first touchdown ( of the rear tyres) to a fully compressed leg case with all tyres on the ground the centre of gravity of the aeroplane could have come down 4 feet (could be 6 feet I really don’t know)
Before we go the final lap, let me say that you must use your common sense with all these calculations about speed/distance/acceleration/time and so on. By that I mean understand whether the answer is going to be smaller or bigger when you make a change. This will tell you whether you have to divide by a factor or multiply by the same number.
Now the last lap. The greatest G when landing from a given rate of descent will be if the undercarriage of your aircraft only shortens a little bit. That means your vertical speed has to change VERY QUICKLY. This will feel like a very heavy landing and will mean lots of G
Let us say you decide your aeroplane has an undercarriage that shrinks 4 feet during the landing.
You have said you touchdown at 140 ft/min that is 140/60 = 2.3 ft/ sec
With these two “facts” we put these numbers in the appropriate equation of motion, which for feet per second units is
G = Vsquared/2H
Where G is the G of your touchdown (which we are tying to calculate)
V is YOUR declared touchdown vertical speed (must be in ft/sec)
H is the distance travelled in feet (the amount YOU decide the undercarriage squats)
So: the EXTRA G of touchdown for your numbers would be = 2.3 x 2.3 / 2 x 4 = 0.66
(a tad heavy according to bastud above)
Or if the undercarriage was a 6 ft job it would become 2.3 x 2.3 / 2 x 6 = 0.44
Any help?
JF
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Having worked these calcs for an accident investigation, I can tell you that the strut extension for most of the larger widebodies is about 3', you could safely use 1 meter I think. Those numbers are for the DC-10/MD-11, but I would be surprised if the others are much different.
Of course that would only apply if the struts actually bottom out, and if they do bottom out any residual acceleration will affect the structure in a much different manner.
One other important consideration: The amount of lift being developed by the wings just prior to touchdown must be factored in. If you are at 1.2 g decelerrating just prior it's far different than being at .8g...
[This message has been edited by Prof2MDA (edited 02 December 2000).]
Of course that would only apply if the struts actually bottom out, and if they do bottom out any residual acceleration will affect the structure in a much different manner.
One other important consideration: The amount of lift being developed by the wings just prior to touchdown must be factored in. If you are at 1.2 g decelerrating just prior it's far different than being at .8g...
[This message has been edited by Prof2MDA (edited 02 December 2000).]
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Andrew,
John Farley gave you the good oil.
Prof2MDA is a bit off in his last para. He is mixing up the deceleration experienced by the passenger with the deceleration experienced by the aircraft.
Think about sitting on a piece of aluminium which is allowed to free-fall to the ground. The aluminium is called airplane. Result. Sore bum. Because the aluminium is incompressible, the deceleration experienced by it, and your bum, are the same.
Put a piece of sponge between your bum and the aluminium. Your deceleration is slower. It takes longer to reduce your vertical velocity to zero. So your bum has an easier landing. The aluminium has an easier landing because your bum takes longer to add its full weight.
Put a piece of sponge between the aluminium and the ground. The aluminiun will now have a softer landing. And if you still have the sponge under your bum, it will be softer still. (so far as you are concerned)
The point I am trying to make is that landing deceleration forces depend on what you are in the system which is decelerating.
Mr. Boeing could design aircraft with rigid seats, so that you (the passenger) became effectively part of the airplane. Then the deceleration you experience on landing will depend on the rate at which the vertical velocity will be absorbed by the oleos and tyres.
But Mr. Boeing is stuck with a system where (I'm guessing) 90% of the landing system is rigid (ie airplane and freight), with only the oleos and tyres to arrest its vertical velocity on landing. The other 10% (you and me on our padded seats) have the added deceleration distance of the depth of our cushioned seats. So the rigid airplane will experience one deceleration, and the floating passengers will experience another (lesser) deceleration.
See Tech Log thread "G and its real meaning" November 2000
I REPEAT. If you want to fly airplanes, get stuck into trigonometry. IT IS VERY IMPORTANT TO YOU, AND NOT DIFFICULT.
John Farley gave you the good oil.
Prof2MDA is a bit off in his last para. He is mixing up the deceleration experienced by the passenger with the deceleration experienced by the aircraft.
Think about sitting on a piece of aluminium which is allowed to free-fall to the ground. The aluminium is called airplane. Result. Sore bum. Because the aluminium is incompressible, the deceleration experienced by it, and your bum, are the same.
Put a piece of sponge between your bum and the aluminium. Your deceleration is slower. It takes longer to reduce your vertical velocity to zero. So your bum has an easier landing. The aluminium has an easier landing because your bum takes longer to add its full weight.
Put a piece of sponge between the aluminium and the ground. The aluminiun will now have a softer landing. And if you still have the sponge under your bum, it will be softer still. (so far as you are concerned)
The point I am trying to make is that landing deceleration forces depend on what you are in the system which is decelerating.
Mr. Boeing could design aircraft with rigid seats, so that you (the passenger) became effectively part of the airplane. Then the deceleration you experience on landing will depend on the rate at which the vertical velocity will be absorbed by the oleos and tyres.
But Mr. Boeing is stuck with a system where (I'm guessing) 90% of the landing system is rigid (ie airplane and freight), with only the oleos and tyres to arrest its vertical velocity on landing. The other 10% (you and me on our padded seats) have the added deceleration distance of the depth of our cushioned seats. So the rigid airplane will experience one deceleration, and the floating passengers will experience another (lesser) deceleration.
See Tech Log thread "G and its real meaning" November 2000
I REPEAT. If you want to fly airplanes, get stuck into trigonometry. IT IS VERY IMPORTANT TO YOU, AND NOT DIFFICULT.
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EchoTango,
No, I am afraid it is you that is incorrect here. My data is backed up by Boeing, NTSB engineers and 3 separate physicists who assisted me in analysis.
The amount of lift the wing is generating has a direct effect on the force. Consider the difference between contact with the ground under 1g, the wings supporting the exact weight of the aircraft. The force the gear has to absorb is based on the KE=1/2MV^2. Now, consider an object that is not being supported but is actually accelerating, such as a dropped object. The rate of change is only part of the equation, as the amount of force the gear will "feel" normally comes in two parts. Part 1 stops the vertical motion, part 2 the lift is removed and the gear feels the weight of the aircraft due to the normal acceleration of gravity (as you feel on your rear end right now as you sit reading this).
However, the wings are never going to be lifting exactly the weight of the aircraft (and I'm rolling TDL into the weight) during the landing phase due to transients, but if the trend is a normal flare the lift is actually slightly greater than the weight so the aircraft is accelerating upwards (slowing it's vertical speed) and the g will read a bit above 1. If the reverse is occuring the aircraft is accelerating downwards at some rate, which means the gear has to absorb that additional component.
Of course, we must also factor in the pitch transients, but to make that work in your favor with decreasing pitch prior to touchdown it requires precise timing before the above factors outweight it, and of course if you don't time it precisely the recovery is not possible due again to the pitch transient in the reverse direction.
No, I am afraid it is you that is incorrect here. My data is backed up by Boeing, NTSB engineers and 3 separate physicists who assisted me in analysis.
The amount of lift the wing is generating has a direct effect on the force. Consider the difference between contact with the ground under 1g, the wings supporting the exact weight of the aircraft. The force the gear has to absorb is based on the KE=1/2MV^2. Now, consider an object that is not being supported but is actually accelerating, such as a dropped object. The rate of change is only part of the equation, as the amount of force the gear will "feel" normally comes in two parts. Part 1 stops the vertical motion, part 2 the lift is removed and the gear feels the weight of the aircraft due to the normal acceleration of gravity (as you feel on your rear end right now as you sit reading this).
However, the wings are never going to be lifting exactly the weight of the aircraft (and I'm rolling TDL into the weight) during the landing phase due to transients, but if the trend is a normal flare the lift is actually slightly greater than the weight so the aircraft is accelerating upwards (slowing it's vertical speed) and the g will read a bit above 1. If the reverse is occuring the aircraft is accelerating downwards at some rate, which means the gear has to absorb that additional component.
Of course, we must also factor in the pitch transients, but to make that work in your favor with decreasing pitch prior to touchdown it requires precise timing before the above factors outweight it, and of course if you don't time it precisely the recovery is not possible due again to the pitch transient in the reverse direction.
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Prof2MDA
OK. I misunderstood you.
I had made it easy for myself by ignoring flare and ground effect.
Stabilised, on the glide slope, A/c and passenger experience 1g.
At flare, a/c pitches up, wing loading increases, g force on a/c and passenger a little over 1. Little change in speed, but the angle of approach to the ground drops below 3 degrees. If the pilot can halve that angle of approach in the last few seconds to touchdown, he will reduce landing forces by a factor of 4, because acceleration is a function of v squared. But he wants to get the main gear firmly planted on the ground in order to trigger the auto speed brakes, and so get rid of lift quickly. So he wants to fly it into the ground, but at something less than an angle of 3 degrees. (m&v quotes an average touchdown 2-300fpm, which implies an approach angle of about 1 degree)
(Andrew, If you get your 737 of 747 on Flight Sim on a good ILS approach at the recommended speed, but with speed brakes disabled, them do your 140fpm touchdown, but at the moment of touchdown haul the yoke back, it will float or even lift. You will at best use a lot of runway, and at worst fall back on your tail. So a moderately firm landing, close to the threshold, is important. Speed brakes cut in quickly, a/c quickly plants itself firmly on the runway, wheel braking becomes possible quickly and you can vacate the runway earlier.
Compare that technique with that of landing a little Cessna. You can approach at a steep angle because the a/c has lots of drag, then arrange to be at the threshold close to stall speed. Pull power and hold it just off the runway until it will no longer fly)
OK Prof2MDA. You have made me think about what does happen to the undercarriage in a conventional landing, with flare, ground effect etc.
It seems to me that with a fly-it -on approach,
On final, a/c experiencing 1 g vertical +/- a bit as a/c pitches up and down.
During flare, a/c experiencing a little more than 1 g. From 50 feet, that will occur over a period of about 3 to 4 seconds. If we say the the vertical speed drops linearly from 13 to 5 ft/sec in the last 40 feet, we have a decelleration of 1.8 ft/sec/sec, or an additional .06g.
a/c flys into the deck at something less than 3 degrees, until (weight of a/c - lift) = oleo force upwards.
Then as lift decays, more a/c weight settles on undercarriage and further deflection occurs.
Wind gusts, ripples in the runway etc are going to add small +/- g forces during rollout.
It seems likely that with a really heavy landing, the oleos will over-compress, then extend again before lift decays and they compress again. But you would need to know the load / deflection characteristics of the oleos, which I do not. I would suspect that they would be progressive shock absorbers, becoming stiffer as they are compressed more.
Have I got it right ?
John Farley,
Think you missed a bit in calculating the g force at 140 ft/sec. Your answer was .66 feet/sec/sec - not .66g.
So .66/32.2 is a very low g loading.
Regards to all
OK. I misunderstood you.
I had made it easy for myself by ignoring flare and ground effect.
Stabilised, on the glide slope, A/c and passenger experience 1g.
At flare, a/c pitches up, wing loading increases, g force on a/c and passenger a little over 1. Little change in speed, but the angle of approach to the ground drops below 3 degrees. If the pilot can halve that angle of approach in the last few seconds to touchdown, he will reduce landing forces by a factor of 4, because acceleration is a function of v squared. But he wants to get the main gear firmly planted on the ground in order to trigger the auto speed brakes, and so get rid of lift quickly. So he wants to fly it into the ground, but at something less than an angle of 3 degrees. (m&v quotes an average touchdown 2-300fpm, which implies an approach angle of about 1 degree)
(Andrew, If you get your 737 of 747 on Flight Sim on a good ILS approach at the recommended speed, but with speed brakes disabled, them do your 140fpm touchdown, but at the moment of touchdown haul the yoke back, it will float or even lift. You will at best use a lot of runway, and at worst fall back on your tail. So a moderately firm landing, close to the threshold, is important. Speed brakes cut in quickly, a/c quickly plants itself firmly on the runway, wheel braking becomes possible quickly and you can vacate the runway earlier.
Compare that technique with that of landing a little Cessna. You can approach at a steep angle because the a/c has lots of drag, then arrange to be at the threshold close to stall speed. Pull power and hold it just off the runway until it will no longer fly)
OK Prof2MDA. You have made me think about what does happen to the undercarriage in a conventional landing, with flare, ground effect etc.
It seems to me that with a fly-it -on approach,
On final, a/c experiencing 1 g vertical +/- a bit as a/c pitches up and down.
During flare, a/c experiencing a little more than 1 g. From 50 feet, that will occur over a period of about 3 to 4 seconds. If we say the the vertical speed drops linearly from 13 to 5 ft/sec in the last 40 feet, we have a decelleration of 1.8 ft/sec/sec, or an additional .06g.
a/c flys into the deck at something less than 3 degrees, until (weight of a/c - lift) = oleo force upwards.
Then as lift decays, more a/c weight settles on undercarriage and further deflection occurs.
Wind gusts, ripples in the runway etc are going to add small +/- g forces during rollout.
It seems likely that with a really heavy landing, the oleos will over-compress, then extend again before lift decays and they compress again. But you would need to know the load / deflection characteristics of the oleos, which I do not. I would suspect that they would be progressive shock absorbers, becoming stiffer as they are compressed more.
Have I got it right ?
John Farley,
Think you missed a bit in calculating the g force at 140 ft/sec. Your answer was .66 feet/sec/sec - not .66g.
So .66/32.2 is a very low g loading.
Regards to all
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Well, what can i say. I'm overwhelmed with the amount of replies i've got to this topic and it's very interesting to read all the factors involved when a plane lands.
If you have anything further to add then feel free to post some more!
Thanks for the interest
regards, Andrew (sydney, Australia)
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."
If you have anything further to add then feel free to post some more!
Thanks for the interest
regards, Andrew (sydney, Australia)
------------------
Great Quotes:
"Flying is an unnatural act, probably punishable by God."
"London Heathrow has been described as the only building site to have its own airport."
"A good landing is one you can walk away from. A great landing is one you can still use the plane after."