Calculating Impact Force During Landing of Aircraft

Zero if the pilot has a fine touch.

sir i m doing a project on hard landing of aircraft so i have to calculate the force that will be acting on landing gear of aircraft at the time of impact. i want to know the formulae to calculate that force .

i also want to know what other type of forces acts on landing gear struts during the landing impact ????

You will need to calculate the force required at the tire contact patch to spin up the tires/wheels. This force, acting along the length of the plane must be resisted by the gear, which has a moment based upon it's height. As you can see from the puff of smoke from tires on landing, this spinup process occurs quite quickly, leading to high impulsive loads.

Braking force must also be resisted in a similar way. 1G is a reasonable maximum braking acceleration to use for educational value.

Wouldn't it still be 1g, even with a perfect three point landing?

sir you are absolutely rite but i want to know methedology of calculating this impact force and how much of it will be transferred to aircraft body ????

landing force is greatest at speed when shock of a still wheel touches the moving runway, the down force is not the main shock to the airframe, I have made landings so soft there was no scritch noise on touchdown but those are rare, I had wondered about a lightweight wheel speed motor to match airspeed to runway so that tires would not hit hard, skid and last longer, on a hard landing the shocks do lesson the downward stress but can do nothing about the backward force placed on whole gear and frame, I know of no program or calculation for landing variations and stress.

Sincerely
Mitch retired Peugeot mechanic and built two Rans experimental ultralite two seaters

I read somewhere that some extensive investigation has been done on pre-spinning wheels before touchdown to reduce tyre wear.

From what I remember the results were that pre-spinning the wheels had no discernible effect on tyre life ..... I know, it did not seem to make sense to me either!

NO.

When either Acceleration or decelerations forces are involved there will always be change in the gravity force (G,) acting on the body. 1 G is only possible at total rest or straight line equal velocity movement.

Let's not split hairs.

Dear Mr.lyn,

What I understand - while the aircraft is touching down, the rear landing gars will touch and the nose point should have a slope of 7 Degrees or so so that the front landing gear will touch-down. Hence all the landing gears touching down simultaneously may not be advisable, though technically it is possible.

Further the value of "g" depends upon the velocity.

DHAYANANDHAN.S

My experience is that no two landings are ever exactly alike.

There is no 'one formula' because the variables are infinite.

Crosswinds would put the plane at a yaw angle into the wind putting side forces on the landing gear, remaining fuel or cargo would change the weight of the plane causing different compression rates of the shocks and tires, tire pressure differences caused by temperature differences from high altitude compared to low altitude flight would create different rolling resistance placing different stresses on the suspension, cold air compared to hot air would require different pitch angles on approach changing the angle of contact and stresses on the suspension relative to the runway, different runway characteristics caused by variations in the surface or weather would increase of lessen rolling and braking resistance.

But in reality, if the pilot can pull off a perfect 'three point landing' and roll to a stop without touchng the brakes, the total forces are simply exactly what the plane weighs when it touches down. Any other landing has infinite variables.

Agree with 129 ........ In addition Omair, you need to first specify what all information you have. ...... You cannot have the forum do your project for you. We can guide you, but the work is totally yours.

sir i m doing work on my project that iz all what i have done

sir i have weight of the aircraft , i know sink rate of the aircraft , as far as my knowledge and concerns are i have found out a formula that iz f=ma(decelaration)

as far as i have done my work i m phrasing it down

decelaration time = distance aircraft moved/sink rate as unit are (ft/(ft/s))

then i know vf-vi=at , vf will be zero so rest every thing i know i can find "a" which iz decelaration

then putting this value in f=ma i can fing impact force

so this iz what all i have done sir, guide me that i m doing it rite or not , secondly i m not sure in calculation of distance i user in first equation that how to calculate that as that would be the distance shock absorber has compressed or u can say moved ,

that all what i have sir now i seek ur opinion sir

Deceleration of a typical U.S. Navy F/A-18 Hornet: 140 knots to 0 knots in less than 2 seconds.

Remember, the 140 knot difference is difficult to quantify. The aircraft carrier is (ideally) traveling into the wind, creating an artificially high headwind. After stopping on the carrier deck, the airplane still traveling forward at the velocity of the ship.

Satisfaction not guaranteed, results may vary.

One of the formulas you want is: F = ma. Another one is Hooke's Law for springs.

sir can i know the hook's law procedure that u r relating please ???

Omair,

Lemme try answering ur question in the other thread "Impact Force" here.

You have Mass of the aircraft. Sinkrate, I presume you are refering to the velocity of the aircraft (since the units u mention is ft/s). I would presume sink rate should also give you the drop in altitude, time taken for it and the velocity of the aircraft.

You need to break your question in two parts.

First, find out what is the force that the aircraft will experience when it touches down, assuming the entire aircraft is one rigid frame. To do this you need to know the deceleration (in ft/sq.s). If u do not have the deceleration then you need to know velocities at two different points and the distance between them. From this u cud find out the deceleration.

After u have the force. then you move to the strut. Find the 'K' factor and apply to it, to find the force that will be transferred to the body.

All the Best.

thankx sir it was really a good help

sir can i use (vf - vi = at) or (vf^2 - vi^2 = 2as) as i alredy mention ?

i know vf will bo zero at time of impact as impact is for very small amount amoun of time vi is vertical velocity !

but sir to find force i need to know "s" or "t" that is where i m stuck please guide me through this sir ?

and one more thing sir this "s" whether it would be height of aircraft dropped to the ground or it will be the distance shock absorber get compressed after impact ?

regars

Work your way thru some of these lectures.

http://spiff.rit.edu/classes/phys311.old/lectures/

Hi You Americans are such tarts some times ,haven't you guys realized that not everyone has english as their first language , basically it is not their mother tongue . Don't be so inconsiderate of other culture Omair wants advice on an engineering problem not a english lesson or is engineering only for english speaking people . Regards Proud South African

I thought that a "tart" was a cute British girl?

The language the OP uses is no relationship to anyones "mother tongue". It is an artificial language used for "texting" on the telephone, and it is not appreciated here. Many times we will not even try to decode such gibberish, and ignore the entire thread. This includes ALL of us and not just Americans.

We realize that some people do have problems with English, and we work with that, but decoding texting to help someone is not worth the time.

* A period is needed after "Hi".

* This is an inappropriate usage of the word "tarts".

* Spaces belong AFTER commas, not before them.

* No spaces before periods.

* The second sentence should read "Don't be so inconsiderate of other cultures."

* The work "English" is capitalized, for example: "... the English language ..."

* "a english lesson" is not correct. It should read: "... an English lesson ..."

* There are other more minor errors that will not be discussed here.

* South Africans can go to the USA or the UK for an education.

And "Hooke's Law" is not spelled "Hook's"; especially weird since it needed only to be copied correctly.

doesn't it depend on the pilot?

1. You must never have landed in an airplane. Sometimes the impact is substantial.

2. Please turn off the rude all-caps.

3. It's shock "absorbers," not "observers." Just copy the the OP's word correctly.

4. Shock absorbers reduce the impact forces by spreading them out over time and distance; but they do not eliminate those forces.

5. A key variable here is the vertical component of the descent velocity. A nosedive landing has more impact than a nearly asymptotic landing. A lot more.

6. I read somewhere that aircraft are designed with a safety factor of 1.67. Unless the airframe yields on impact, the force is thus less than 1.67 times the loaded weight of the plane.

7. As Lynlynch notes, the lower limit of impact force equals the loaded weight. This is rarely if ever attained, but a good landing comes close. AAAH!

Wrong! Shock absorbers don't "disappear" the energy, instead it has to go somewhere. Part will be heat, but part will go into the aircraft structure.

This is like writing:

THERE ARE NO IDIOTS BECAUSE PEOPLE HAVE COMPUTERS

SHOCK OBSERVERS

link

There are two components to the deceleration:

The forward motion only has an effect when the wheel brakes are applied and the nose is forced down on the front wheel. This is alleviated by the back thrust of the engines, and airbrake drag.

The downward motion is the sink rate (initial velocity) at the moment of touchdown. Lousy flare, high sink rate. Downward deceleration is by spring compression, hydraulic resistance in the oleos, and tyre distortion.

The downward force will be that due to the weight of the plane PLUS the force due to deceleration of the plane mass from initial downward velocity to zero downward velocity. Approximately, this is a linear deceleration, so if you know the distance moved by the struts in compression you're in like Flynn (Errol Flynn, Hollywood actor).

The only equations you need, as already indicated, are:

v*v = 2*a*s

F = M*a

F = M*g

how fluid properties can be replaced with spring constant "k" and damping coefficient ?

if we want to replce a oleo pneumatic shock absorber by spring mass damper system

In my opinion if you don't punch the oleos through something - and/or walk away - it's a good one.

How you calculate the heat expended in oleos is well beyond my level of caring.

I wonder why not to accelerate the wheels right before landing, it would be easy with wheel covers with small wings to catch the wind. At least reaching one half of the speed would reduce to 1/4 the kinetic impact, due to the square factor. Jaime Soto Figueroa Chile

The aircraft descends at a given speed, the sink rate. When it lands, it decelerates to a sink rate of zero. The relevant formula is[br[]br] v^2 = u^2 + 2as

where v = final velocity (zero in this case)
u = initial velocity (sink rate)
a = acceleration (the value you want for the next formula)
s = distance (the movement of the piston in the shock absorber)

Having calculated a, you can then apply the formula

P = ma

where P = force (this is the one you want)
m = mass (of the plane)
a = acceleration (as previously calculated)

So far, very simple. The only snag is calculating s, which is a function of the shock absorbers (the greater the sink rate, the larger is s). If the undercarriage were suspended by springs alone, Hooke's law would apply (if you knew the spring rate), but as there is an element of shock absorber as well, there is no way one can determine s without reference to a chart linking suspension travel to initial sink rate for that particular suspension module. If you cannot manage that then I fear this project may be beyond your capabilities.
The good news is that the forward speed of the aircraft is irrelevant to the calculation in the first instance.

I believe your logic is good. Please also consider this simplistic approach:

F= Mv where ''v' is the vertical component of the impact velocity (sink rate)

The horizontal component we will ignore at this time because it will have minimal effect on the structure of the plane and will be dissipated over a longer period of time while the plane rolls on the runway.

The impact force transfered to the structure of the plane will be: 'F' minus 'load capacity' of the shock absorber assembly.

Hello OMAIR.

The formula you are looking for does not exist. And as one poster already pointed out, too many variable are involved.

This is why the 3 major variable are assumed pre-set. and usually are taken out of the airplane manual. The following will always take the "heaviest case scenario"...

1. MLW, The maximum allowable A/C landing weight. This weight takes into account not only the weight applied on the wheels, but also structural load on the airframe, wings etc. This figure will sometimes require fuel dumping due to the possibility of exceeding the MLW, While MTOW (maximum take-off weight) is not exceeded.

2.Landing Gear construction: I.E number of wheels for calculating the weight distribution. (max weight per touch point) and is being calculated for the main landing gear only. Good pilot try not to execute a three point landing if they can only avoid it...But again, if they do, it will be less strain on the main landing gear but much more so on the passenger and the A/C structural integrity.

Touch-down speed: The speed in which a full weight contact is made with the runway. Safety margin is added for possible "Heavy Landing" which are unavoidable in many cases.

environmental conditions: ISA day (International standard day) No wind, no runway slope. and dry runway.

All these will grossly determine the so called "Aircraft foot print" to certify a runway for a maximum permissible A/C weight operations.

If you are into the design of landing gear, than again formulas are being used (and ignored) but the good old "Drop-Test" procedure is what you want... You drop a loaded landing gear to destruction in multiple array of landing conditions.

From your question I assume you are trying to design a runway and not an airplane Landing gear. you are probably a construction engineer rather than an aeronautical one...But maybe not. Good luck anyhow.

Wangito.

sir actually i have to do drop test of aircraft with landing gear assembly ! i m doing it on ansys ls dyna but i could not find right combination for modelling a landing gear assembly in ansys ! if u guide me in that it will be helpfull to me .

omair, try this site from NASA this may help you with what you seek. www.wseas.us/e-library/conferences/2008/malta/ecc/ecc20.pdf

If not do a search of NASA file and I am sure you will find what you seek.

You must look at a particular landing gear assembly. Some have several feet of travel from hanging mode (with bogey wheels tilted) to the static load condition (1G load). The force required to compress the gear is not linear.

Accelerations vary in different parts of the plane's structure, because in a hard landing, the entire plan flexes.

At very least, you will need to know the dynamics of the landing gear.

Springs and shock absorbers do not deduct any force. They spread out the time over which the force acts thus reducing the peak stress. So the answer to part two of the question is 100%

When any cross wind is present the plane will not be completely aligned with its direction of flight. If the flight path is in line the the runway, the undercarriage won't be. In an ideal landing the aircraft is flared just before touchdown so the the tail wheel touches down first. This aligns the main under carriage with the runway and eliminates side forces. The side forces are primarily taken up by the tyre which scuffs. On hard strips this is a very expensive way to land an aircraft. On grass strips the problem is not as acute as the turf takes most of the wear.

Who's talking about taildraggers?

I don't know if you have tried it yourself but landing tailwheel first would be an ideal way to wreck the aircraft if done regularly. The tailwheel is much smaller and less shock-proofed than the main landing gear. The conventional landing of a tail-dragger is with one of two techniques, either main wheels first and ease back onto the tailwheel, or, for the skilled pilot, a three-pointer with all three wheels together.

That's my boy!!!

Like me he doesn't know that airplanes were converted from Tail-dragger's to nose wheels!!!

Look at my avatar and you will understand!

Wangito

Since you have only a limited information you can use the first approach and add at the energy of the mass the initial kinetic energy. What you compulsory need is the estimation of the spring rate. Without it all computations are not valid. The relationship you want to use will not give the right answer since at landing you do NOT have a constant deceleration but a variable one due at least to the tires stiffness. The kinetic energy is to be found by yourself as function of the vertical velocity which is at its turn function of the plane velocity and the slope as indicated in above sketch. Although not precise it is the most logical simplified model to use. As for the help with your program this is not in the frame of the CR4 principles: I am ready to give you hints so that you can find the right path but I will NEVER make your work and your work is also to learn how to use the soft for a simulation. This you have to learn on your OWN.

What is the schematic of your landing gear ?

If you want to use the fluid compressibility you should correct it with the expansion of the cylinder wall which will decrease the overall spring constant.

But as I mentioned before the tire is the most effective stiffness since it is less than the damper stiffness.

You should make an explanatory sketch since without no further comment is possible.

Using the fluid bulk modulus and the cylinder deformation under pressure you may compute the equivalent spring stiffness starting with the original definition of a spring stiffness: C= change of force versus stroke. Make a trial on your own and bring equations and results indicating as well the input data. CR4 participants will have a look and correct you but do not expect us to make your work.

A fascinating excursion into mythical mechanics. Do I have to point out that the force pulling the aircraft down of 1G is largely balanced by-wing generated lift, so that the sink rate is a few m/sec rather than the 50-odd m/sec resulting from free fall in unopposed gravity. The actual force on the suspension at touch-down is far less than 1G.
For those with less mechanical insight I offer some real-world experiences for comparison. The UK Highway Code gives a table for vehicle braking distances which assumes that emergency braking produces a deceleration of 0.7G. Even that is felt as an unpleasant sensation. A pilot who allowed that deceleration force up his passengers' spines would be relieved of his job forthwith

I gave a simplified model which of course can be made as complex as one wishes. For the scope of this student I think it has not be made too complex and that it is enough as it is. But if you want to consider the portance then you should introduce its value at the contact and as every body knows that the lower the speed the lower the portance and how it decreases along the braking distance. It can be done but it is worthwhile for this case ? Now you mention an "emergency braking deceleration" of 0.7g. It is the HORIZONTAL deceleration since "braking" is active only when the plane already rolls on the tarmac. Look please in the papers you have which is the VERTICAL maximal allowed deceleration which also should figure somewhere. May I also remind you that the fall speed is equal to the product v=g*t so that the odd 50 m/s you mention will occur after 5 s of FREE fall which was NEVER considered in the model!!! When on ground due to braking portance goes to zero the weight of the plane becomes more and more supported by the landing gear.

Sorry, nick name, but you are missing the point completely. Your mythical calculation of a (minimum) 2G upward force on the aeroplane necessarily implies that at some point in the landing the aircraft is going to accelerate upwards at (at least) 9.8m/sec^2. This simply does not happen. My reference to the 0.7G experienced during heavy braking of a road vehicle is not part of any calculation of the descent of an aircraft. It is a reference to the real world experience of a human. If your net upward force of 1G is applied to a plane, it is also applied to the passengers. If any passenger experiences that force up his spine he will never want to fly again.

dear phph001, Thank you for the comment. It is of course possible that my limited knowledge of English (it is neither my mother nor my father language) is the reason for not getting the point however my knowledge of physics and mathematics being independent of the used language is sure. 1- which is the myth which can be related to my calculation? 2- have you understood the equations? or do you need further explanations ? 3- on a scale you can verify the rightness of the "mythical" calculation, better with a simple spring one. If you use an electronic one it can happen that the sampling frequency and the conversion time could damp the peak. Electronic scales have load cells and since those are stiffer the own frequency is higher. In such a test there is NO upwards acceleration. Your assumption is not a good sign for understanding of physics. 4- I agree with the 0.7g limit in HORIZONTAL deceleration, this is taken care of at landing since the passengers are obliged to have their safety belts on. But as you recognize this deceleration is NOT on the vertical. 5- You assume -wrongly- that a plane is a stiff structure. It is NOT, when on board a plane I would suggest you take a window seat and look at the wings as well in flight as at landing you will be surprised to notice how much they bend. Do not forget that the OP asked for the LANDING GEAR loading. Between landing gear and passengers there are A LOT of compliances which will strongly damp (dynamically) the amplitudes. Every element in the chain has a low-pass function for frequencies higher than its own. 6- The model was thought for a student so that it has to be SIMPLIFIED with only the essential elements. I have as well the feeling that in your activity you were not very often -if ever- confronted with simulations. It could of interest to have a look at this modern approach which is more and more used so far that to day we can speak of "digital prototyping" which is only possible due to simulation. Please make the scale test and after you see the result peak weight indicated > end weight convince your self that you should change you point of view.

Sorry, nick name, it is no good hiding behind a different mother language. If you look at my post #26 you will see that I have already given an accurate account of the physics of the model in a simple form, with an explanation of where it needed extending and, in particular, emphasising that the forward movement and deceleration was a separate matter. Whichever language is your native one, you need to read what has already been said.
Second, I have the advantage, which you obviously have not, of having landed an aircraft. The planes concerned were gliders, which do not have wheeled undercarriage but a skid under the fuselage with rudimentary shock absorbers. A net 1G upward force directly on the fuselage would certainly bounce the aircraft and the pilot inside. It would also result in the pilot being bounced from the airfield by the chief flying instructor.

Thanks again although you did not get the point you bring with your new comment a lot of water to my mill!

1- have you made the "scale test" ? most probably not since you consider that I am so far away from reality that it is not worthwhile to consider it.

2- You mentioned that what you landed was a glider with a very rudimentary shock absorber. This means your aircraft did NOT have any spring and this makes the difference you are not aware off.

I will not go into details but as you will notice on above scheme without a spring the behaviour is TOTALLY different. I do not have the time to make a simulation and show you the differences but if one considers a constant "d" the deceleration is NOT constant as you assumed. It is an "e^(-t/τ) function. Where "τ = M/d" with "M" the object mass and "d" the damping coefficient. This "τ" is the parameter which determine the overall behaviour. And if you think it is correct since the kynetic energy is proportional to the mass. The initial velocity enters as well in the final relation as "g". Now let us have a look at "d". Imagine that the damper has a high "d", at the contact velocity "wo" a resistant force F= d*wo will be generated. this is a braking force which will impose a deceleration to the falling object. But this force will also reduce the speed since a higher part of the incoming energy is transformed in heat. S the stroke will be shorter since the speed decay will be more important. The mass will move in the gravitational field and make a work but this one will be small in comparison with the energy destroyed by damper. Now the fluid in the damper is not stiff and the cylider neither so that this cushion will play the role of a stiff spring but due to the leak through the orifices pressure cannot build up as much as in a only spring sytem. Result is a lower peak force. And considering a glider with its struture it is better for the passenger that it is so.

If you look above you will notice that the "over weight" is due to the movement and to the fact that the spring is loaded, in your case energy is destroyed from the start so that the system has different dynamics.

No, I did not land an aircraft but this will not stop me thinking.

No comment on my other remarks ?

Trying to think big picture, we have an aircraft (MASS) with a sink rate(velocity) which is reduced to zero vertical velocity in a period of time giving an acceleration (deceleration) .... f=ma then gives the vertical loading vector

Horizontal loading - wheels have a rotational inertia of 0 at touchdown they are accelerated up to a rotational speed and thus inertia equivalent to the aircraft speed.

force is applied at the circumference (point of contact with the runway) to accelerate the wheels to this rotational inertia over a period of time again f=ma will give you the horizontal loading vector.

vector add the two .......

How all this is absorbed and distributed is another and far more complex question as several people have discussed.

Sorry, nick name, this just gets weirder. You are now trying to suggest that the reason a glider can be landed softly by comparison with a wheeled aircraft is because it lacks springing and damping. If that were the case, then why does any aircraft, indeed any wheeled vehicle, bother with suspension elements at all?
The time has come to take this in hand. A glider pilot is compelled, by the nature of his aircraft, to land accurately, with a touchdown sink rate of a few ft/sec, because otherwise he will cause himself discomfort and possibly the aircraft some damage. The pilot of a wheeled aircraft can allow himself considerably more latitude. An Airbus pilot tells me that the desirable sink rate on touchdown is in the range 50-200 ft/sec, which would wreck a glider, yet his passengers experience no discomfort. No, that is not due to the wings flexing on touchdown; it is because of the suspension.
First, let us do the experiment with a spring balance you imagine I am avoiding. I take a 1kg bag of sugar and drop it from a height onto the pan. The needle goes way past the 1kg mark and may even hit the endstop. That is no surprise, with the kinetic energy of the drop added to the original G force. However for the second part of the experiment, I pick up the sugar and lower it gently into the pan. The needle goes slowly to the 1kg mark and stops there. The compressed spring does not generate a 2G upward force at any stage. That is what is happening with the aircraft undercarriage. The shock absorber is actually absorbing energy from the moment of touchdown until the spring reaches the point of maximum compression. The equation in your post #41

E1 = E2

is, to put it simply, false. Furthermore, there is no point in trying to impress either myself or the OP with exponential equations, because they are probably wrong too. I gave him a simple account of the physics because, to put it simply, I thought he would struggle with a more complicated one. Unless you are privy to the engineering of the suspension element you have no justification for any deeper analysis either. Why do you suggest that I am wrong to leave it with an averaged deceleration? Because you have not considered the possibility that the suspension element contains variable-rate springs and variable shock absorber rates which make your calculations meaningless?
I suggest that the next time you post you consider the real-world effects of your suggestions (and in the meantime please don't design any aeroplanes).

"Sorry, nick name, this just gets weirder. You are now trying to suggest that the reason a glider can be landed softly by comparison with a wheeled aircraft is because it lacks springing and damping (1). If that were the case, then why does any aircraft, indeed any wheeled vehicle, bother with suspension elements at all? The time has come to take this in hand. A glider pilot is compelled, by the nature of his aircraft, to land accurately, with a touchdown sink rate of a few ft/sec, because otherwise he will cause himself discomfort and possibly the aircraft some damage (2). The pilot of a wheeled aircraft can allow himself considerably more latitude. An Airbus pilot tells me that the desirable sink rate on touchdown is in the range 50-200 ft/sec, which would wreck a glider, yet his passengers experience no discomfort. No, that is not due to the wings flexing on touchdown(3); it is because of the suspension. First, let us do the experiment with a spring balance you imagine I am avoiding. I take a 1kg bag of sugar and drop it from a height onto the pan. The needle goes way past the 1kg mark and may even hit the end stop. That is no surprise, with the kinetic energy of the drop added to the original G force. However for the second part of the experiment, I pick up the sugar and lower it gently into the pan. The needle goes slowly to the 1kg mark and stops there. The compressed spring does not generate a 2G upward force at any stage (4). That is what is happening with the aircraft undercarriage. The shock absorber is actually absorbing energy from the moment of touchdown until the spring reaches the point of maximum compression. The equation in your post #41 E1 = E2 is, to put it simply, false. (5) Furthermore, there is no point in trying to impress either myself or the OP with exponential equations, because they are probably wrong too. (6) I gave him a simple account of the physics because, to put it simply, I thought he would struggle with a more complicated one. Unless you are privy to the engineering of the suspension element you have no justification for any deeper analysis either(7). Why do you suggest that I am wrong to leave it with an averaged deceleration? (8)Because you have not considered the possibility that the suspension element contains variable-rate springs and variable shock absorber rates which make your calculations meaningless?(9)" I suggest that the next time you post you consider the real-world effects of your suggestions (and in the meantime please don't design any airplanes). Dear ph², I accept your comment and will answer point by point, the reference numbers are in the text in brackets. What you write shows again that you do NOT understand neither the model nor the equations. I apologize for the long answer but there are so many aspects which have to be clarified that I took the time to do it. It is my last comment to your reaction you may write what you want after, say again that all I wrote is wrong or false I shall not react since even a contradictory discussion can be pursued with somebody who understands the arguments and it is unfortunately not your case. 1- You mentioned it has no spring. I wrote that the fact that it has only a damper does change the behaviour and this is compatible with what I wrote. I may give you an explanation: no spring no reaction force proportional to displacement( thus no swinging system, a damper gives a reaction proportional to velocity thus a 1st order differential equation which gives an exponential result). 2- Of course because the resistant force generated by a damper is proportional to the speed, so that a too high initial speed at contact will generate higher forces and a stronger vertical "braking" effect. 3- In your reaction you are so blinded that you did not understand that I mentioned the wings ONLY to explain that a plane is NOT rigid but very compliant. You should read AND try to understand what it means. 4- You are AGAIN either unable to get the point or do not want to accept an idea contrary to yours. If you look with ATTENTION at what I wrote you see that I never said that the mass has to be dropped from a height neither have I said it has to be put "gently" on the scale. The assumption for this model was that the mass was at contact with the scale but that the weight was not yet on the scale and let to fall free from this relative height =0. The assumption for the model and for the equations related to the model was that the mass was in contact with the scale (spring) but has an initial velocity =0. No initial speed was considered in order to simplify the problem. Putting the mass "gently" changes the test basics and you compare apple with pears or vice versa. 5- In your test as you made it or only said you did E1 =0 ! The energy the mass would have gained by the vertical movement in the gravitational field was in fact retained by your hand during the "gentle" approach. It is a further proof that you did not accept to make the effort to understand what was the content. The equation was not "false" but your test unfortunately was not according to the model! What is the "false"? 6- I do not try to impress you with equations since for me those are usual working tools, you felt uneasy because you do not feel home in this field. Any way please DEMONSTRATE that the equations are wrong, show where are the flaws. I could accept you to say that models are incomplete or wrong but you cannot appreciate equations without relating them to the model. You confirm my assumption that you are NOT accustomed to model thinking and related equations. 7- According to an on line dictionary the definition of "privy" does not correspond to the use you make off: " Made a participant in knowledge of something private or secret: was privy to classified information". And I have all rights to make any analysis I want, as many "analysts" I have even the right to write stupid things! ( this does not mean that I consider what you say as stupid it is only the reaction of somebody feeling "out"). 8- I criticized the principle of averaged values because they do not give any indication at all about peak values. 9- Again to the model, to simplify it I considered a linearized behaviour which is not the real one of course but to give an order of magnitude such approximations are often used in design in the first phase. I am aware of the non-linearities since I designed hydraulic dampers (with included springs for the piston return!). As you may be informed such dampers are designed so that the total area for fluid flow decreases over the stroke to adapt the damping capability to the speed reduction. An orifice builds up pressure drop proportional to flow^2 but without a simulation with a dedicated soft the evolution of the speed versus time is difficult to compute(it can be done even on an excel sheet if you know how to do it but with quite important errors) so that when starting a project the behaviour is assumed to be linear and that the damping coefficient is constant at least for a part of stroke. Although the approximated results are not exactly what is obtained at the end in real life they are near enough for a good estimation of other parts and start the project. Such models are NOT "meaningless" but ONLY a first step. Again it shows that in your work you did not have enough contacts with such advanced practices. 10- You are one of those who part the knowledge in 2 : one part is their knowledge, thus the "truth" and the rest which of course is "false". Try to come out of this "black and white" conception and try to accept that you may be wrong at least partly. Nobody detains all truth or all knowledge, neither you nor me. 11- Because I consider the pleasant "chat" as closed I want to explain how a simplified model is generated. The base is the real thing which is analysed qualitatively so that as many influencing factors as possible are listed. The second step is to determine important and less important influences and define those which have to figure in the "model" The third step is to write the equations corresponding to the physics of the phenomenon Finally, the 4t step is to try to solve the equations. Depending on the tools at disposal and the degree of uncertainty accepted at the considered development stage the equations are more or less simplified and solved either on paper or via the use of a computer soft. For dynamical analysis the first approach is based on linear equations since they can be solved with the Laplace transform as normal algebraic equations. Final FULL STOP.

An Airbus pilot tells me that the desirable sink rate on touchdown is in the range 50-200 ft/sec, which would wreck a glider, yet his passengers experience no discomfort.

10 feet per second (600 fpm) is the maximum landing sink rate that airliners are designed for. In a plane doing about 175 miles per hour on approach, on a standard 3 degree approach path, the vertical speed is 807 fpm. Thus, a pilot could simply slam the plane into the ground without any flare at all, and barely exceed the limit. (The 1.5 safety factor should mean no damage to the plane.)

If we take 125 ft/sec as the average between 50 and 200, this is 7500 fpm vertical speed: about 85 mph straight down. Unlikely that the plane or passengers would survive this. I suspect the Airbus pilot meant 50 - 200 fpm. (1-3 f/s)

Yes, but . . . I am not an aircraft designer, but consider: For design of the landing gear, aircraft chassis/body, etc, etc; the designer will have to discover the MAXIMUM forces that the gear will encounter, including screw-ups, poor landings, and anything else that may happen. Smooth landings are easy, it's these other things that will determine the design.

As I understand it, Commercial Pilots these days are taught to put the main undercarriage firmly onto the runway - sometimes they do it too firmly - but the reason behind that training is that at touchdown the cockpit is still very high with the nose up attitude and it is very difficult for the pilot to know when the main undercarriage is on the ground unless he can feel it.

Experience has taught that a very difficult to identify wheels on the ground in a very smooth touchdown, and it is possible to engage thrust reversers with the aircraft still 10 feet in the air (thinking they are on the ground) with disastrous consequences ......! Firm landings are a much better alternative!

You have said a couple of pertinent things. I maybe should cover first.

The tyre spinning thing was shown to rob the aircraft/pilot, of that deceleration - which serves to pull the nose down, so changes attack angle, so kills lift. Too fast a spin has the opposite effect - exact right spin and 'bounce' is the outcome.

Tyres are less costly than crashes.

For a while certain airlines went for 'soft touch' landings and found 'bounce' instead. This lead to a series of landing stalls and 'oleo punched through stuff' and/or one or several 'running out of runway'.

You don't need the 'screech' to know it's down. Nor is the 'piano key' or 'flare picture' any different to a 'little one' - if you are 'at one with the scale'. Like, if you ride a mower and drive a truck - yet "know" where the edges of each are.

Moving on to other comments; Oleo are primarily "extension" dampers, or "anti bounce".

I.e. "the gear is down and locked" is more than a "three green" protocol. You need it 'down and extended' - meaning the oleo are fully extended by the springs or hydraulic extenders. These tend to be configured for 'not bottomed out' at wet take off weight.

Obviously in large aircraft the 'springs" are not only adjustable but read 'weight and balance'.

Still the MTO setting/limit is well above the maximum landing weight permitted - because much more than 1g equivalent of MTO may punch things through things.

Or basically what will hold an aircraft up on takeoff - can't take the forces of landing.

Which makes this aircraft/piloting even more remarkable

So as I remarked earlier "if you don't punch them through something - it's a good one".

And from the OP pov - its the spring (or extension hydraulics) rates, that apply here.

Meaning see 7

Though this is good too - particularly if #7 is a little mysterious

Bear in mind the vast majority of this is nothing to do with you WAWAUS - but do regard yourself as the excellent catalyst

Thank you 34point5 for your clarification, I was writing from memory of things I had read years ago .... memory fading a little ....

I also thought people may be interested in this 10 sec video clip which shows a heavy landing test where aircraft structural limits were exceeded:

<http://www.youtube.com/watch?v=yp7XkW72Ils>

Well the oleo's seemeed to cope Therefore a 'good one' - I guess - less you're in the dunny

You are correct. I can't see why this is marked as off-topic. In a smooth landing, the vertical force on the gear begins at zero (at the instant of touch-down) and then smoothly increases to a 1G load, (at which point the suspension has collapsed to its static ride height). The further compression beyond the static ride height depends upon how smooth the landing is. There is some degree of roughness that would produce a 2G landing (and consequent large compression of the gear) but that is more apt to happen in carrier landings.

After the wheels have spun up, and before the nosewheel has come down, the wings are still generating considerable lift, so it is not until the plane has gone 100 meters down the runway (or more) with the wheels spinning that the load on the gear reaches the plane's full weight. This can be seen in this landing video.

In an ideal landing, which the video represents very closely if not fully (!!), the upward force exerted by the landing gear will start at a low value and increase asymptotically to the plane's weight, ultimately settling at the plane's weight. Thus the landing gear and fuselage forces must at least equal the weight of the plane.

In any "overshooting" of the asymptotic curve, it will take a larger force than the airplane's weight to restore the landing path to horizontal. How much larger? Up to 1.67 x airplane weight*, nothing should break. But one video in this thread shows a whole tail breaking off, so that could have been 2x or 3x or even more of the plane's weight.

Of the modeling offered on this thread, nick name's appears the best. It would take some specific angles, spring rates, and hydraulic leakage rates to quantify matters much better.

Landings are not generally perfect; unless the aircraft breaks, the force would generally be between 1.0 and 1.67 x the airplane weight.

*I read the 1.67 safety factor somewhere I have long forgotten; if the number is different, these quickie calcs would need to be altered accordingly.

I think 1.67G is low, as a limit. The maximum design sink rate is 10 fps. At one time I knew the FARs pretty thoroughly, but no longer do. In this thread 2.6G is mentioned as a maximum, which sounds about right.

Design max flight loads in commercial planes are only about 2.5 positive and 1.1 negative (if I remember... although I might not). If you consider that a 60 degree banked level turn creates 2G acceleration, you'd have little margin when performing that maneuver. A 60 degree banked turn (spiralling down) is used as an emergency maneuver in the event of a depressurization. (Of course there is a 1.5 safety factor above the max design load.) If your plane ended up inverted, you'd need to be very careful with the elevator control.

The little aerobatic planes I flew were good for 6G positive and 3G negative: less likely to fall apart when handled roughly. (Many purpose-built aerobatic planes are built to higher limits than required.)

Of the modeling offered on this thread, nick name's appears the best.

However, it leads to incorrect conclusions, such as his assertion "This happens when there is NO initial vertical down oriented velocity of the mass, this shows that the force at landing is AT LEAST 2 x plane's weight." If the plane touches down with no vertical velocity, the load on the gear gradually increases to 1 x plane's weight.

He also writes that "The vertical landing velocity is kept at a minimum with a VERY small incidence angle since a too high vertical speed would be not accepted by the contact element which is the TIRE. If you look at the sketch above you see that the landing gear is limited by its tires!"

This is not true. Planes suffer structural damage as a result of hard landings even though the tires survive. The model offers no way of concluding that the "landing gear is limited by its tires." The gear can be relatively strong and the tires relatively weak, and vice versa. Unless you know the tires static load limit and the gear's vertical load limit, you cannot make such a statement.

I have to make a remark since you make same error as ph². This is ONLY a SIMPLIFIED model which tries to estimate the loading under a LIMITED assumption. The basics are correct since if at zero initial speed the load can go as 2*G with an initial velocity the kinetic energy will be even higher and the travel of the spring necessary to absorb this energy will be also more important thus the final load bigger. I gave this to the student as a way to estimate the loads. As you see the other picture considers as well the tire elasticity as the damper and its spring in a as well simplified principle sketch. This is the model which although also simplified if elements are considered as linear or constant in their behaviour comes nearer to the real landing gear schematics. Of course different parameters have to be considered as well as the geometry and the deformations under load of the landing gear structure, the own inertia of components and many other. In this simplified model there is no damper so whole kinetic energy + work done are considered to deform the spring till speed is zero. I dare say that if the tires are weak they burst before anything else. This I meant by the "landing gear is limited by the tires". Of course I could have written "landing gear is limited by weakest element in the load transmitting chain". If you prefer then accept this definition in stead of the other. If you consider that my schematic model is not good please present a better one (but also as principle) I shall be delighted to have it for further discussions.

This is ONLY a SIMPLIFIED model which tries to estimate the loading under a LIMITED assumption.

Your model is actually over-complicated yet incorrect.

It seems that I am not a stupid exception, all other examples in the literature are too as stupid as I am. Please look at following pictures and explain why those researchers use same approach?

The only reproach you can make is that I did not consider the tire damping, but the rest is as the other stupid guys work with. As you see even the initial speed is present. To give an explanation for it, I thought tires are as for the ground vehicles whose tires have for many reasons less "damping". I recognize it is an ERROR. Is it soo big to consider the model as totally wrong ?

So that apparently if all other are not totally idiots I am neither ?

If, at the instant of wheel touchdown, a plane's wings fell off, then the plane would drop onto its landing gear and the peak acceleration would approach 2G .

The plane is in the gravitic field so that the gravity is the only acceleration it can suffer from and this one is as far as I know only 1g. So that your plane cannot drop with such an acceleration. If you would write that due to the fact that portance disappears the load on the damper will grow instantly and an the plane will be "braked" with a high force leading to a high "deceleration" I would have accepted but so I have to write that your assumption is unfortunately WRONG.

However, that is not what happens. The wing lift decays slowly as the airplane goes down the runway, with the nose up and with substantial wing angle of attack.

Yes, but as you very well describe the phenomenon AFTER the landing!

You write: The basics are correct since if at zero initial speed the load can go as 2*G

In the assumptions listed at start. I shall explain you further down why I suggested this over simplified model. First I have to discuss your arguments.

That's incorrect. The vertical load on at the top of the landing gear strut (at its attachment pin to the wing spar structure) is, just before touchdown, equal to the landing gear weight. If the normal (on-ground) load is considered positive, then this load is negative.

Correct, the weight of landing gears represents according to following tables and graphs only a very small part of the total plane weigh:

The structural weights are listed in the table for the planes I could find values:

The planes weights are for instance as follows:

Boeing 707 3.28E5 lbs 100*(11216)/328000=3.42%

727 1.61E5 100*(6133)/161000=3.81%

737 1.04E5 100*(4382)/104000= 4.21%

747 9.78E5 100*(31108)/978000=3.18%

I let you verify the orders of magnitude. The "pulling load" is so small in comparison with the compressive one that it can be neglected even in the fatigue computations not only in the strength validations. For the landing gears a pulsating cycle with R=0 is considered.

(We'll imagine that there is a single landing gear wheel and single landing gear strut.)

When the plane first touches down (in the near zero initial speed condition) the load at the pin goes to roughly zero, because the ground now supports the landing gear weight.

Again an error the plane cannot touch down at near zero vertical speed since this speed can only be changed by the ground reaction forces so that the plane touches with full initial descend velocity.

A half-second later, the load goes to 1/4 of the aircraft weight in the positive direction(7). At one second, the load goes to half the aircraft weight, and the spring is compressed to a point roughly half way between fully extended and the static load compression. at 1.5 seconds 3/4 of the aircraft weight is being supported by the gear. At 2 seconds , the full aircraft weight is supported by the gear, and the spring has compressed to its static load length. If the plane touches down at near zero vertical speed, most of the vertical deceleration (from the glide path rate f 500 - 800 fpm) has been accomplished in the air, as the pilot flairs.

I have not found graphs of force versus time but several force versus stroke. All display a maximal force bigger than the static load as pictures show. At contact startspeed is important (relatively) so that the damper will be more effective if a special correction has not been implemented so that the reaction force will grow faste, later the energy has been destroyed by the damper and accumulated in the spring so that velocity decreases and the balance is more direction spring than shock absorbers. This explain the curves below:

The figures are from 3 different sources and it is not thinkable that same source was used for the 3 of them. So that your assumption of a steady reaction force growth is contradicted by the other.

Many small airplanes have only spring struts for landing gear. In an ordinary landing there is no rebound of these spring struts, even though they have no damping: the load gradually increases to the full aircraft load.

Apparently there is a rebound at least according to the report I found and display as follows. You may control as you have the source indicated:

From eng.tips.com 24.02.2005

"aerodog (Aeronautics)" to

Drkwing

In 1964, I designed spring main gears for the Alon Aircoupe. They deflected aft towards the tail. The design replaced the original struts which used Bellville springs. The Bellville's acted as both springs and shock absorbers.

For the first flight, the test pilot was able to log one take off and eight landings. Boing...Boing...Boing, well you get the idea, a pogo stick with wings. There was no damping (shock absorber) in the system.

After we sorted thru the problem, we slimmed down the strut near the axle, so that on landing, the strut would twist causing the tire to scrub and dampen the rebound.

The most interesting aspect of the project was the industrial espionage it took to acquire Cessna's spec on their spring gear. It was a special steel alloy purchased in mill runs. The gears were formed, heat treated to 220,000 psi then shot peened if my memory serves.

If you have ONLY a spring you come to the contested model and it has to be a rebound as following equations show:

Incoming energies are

1- Kinetic energy Ek=M*Wo²/2

2- Work done in the gravitic field Eg=(M*g-P)*z where P= portance force

The body will stop i.e. w=0 when the sum of those energies is equal to the spring deformation energy Esp=C*z²/2

We obtain the equation: C/2*z²- (M*g-P)*z-M*Wo²/2=0

The solution is z* stroke at which the stop occurs. The static stroke is zst=M*g/C.

Putting the condition to have z*/zst=1 leads to the relationship

P/G=Wo/g*ω where ω=(C/M)^0.5

for all other values the "stop stroke" will be different from the static stroke. Since the plane mass and the Wo are never exactly same at all landings there is a high probability that at least for some landings the strokes will be different. As you very well know a system consisting of a mass and a spring will oscillate as long as energy is not taken off this is the role of the dampers or in this case of the shock absorbers.

Oleo struts (and most shock "absorbers") do not have a lot of compression damping. The majority of damping is on rebound. The less compression damping, the more gradual the transfer of force to the plane's structure. In the interests of simplification, damping can be ignored until the student has his model close.

Interesting ! You suggest in fact my model which you at same time contest ! Not very constant in your opinions aren't you ?

So... a simplified model would not have a glide slope and trig. Every plane has a vertical speed indicator. During the impact, the vertical speed will go from X fpm to 0 fpm, and then the plane will rebound, with the rate of rebound reduced by damping .

May I mention that almost all systems for shock absorbing are effective ONLY in one direction? I write "almost" because my knowledge is limited and may be some are but all solutions I met during the search are effective as dampers ONLY in ONE direction.

The maximum force on the attachments will be reached when the vertical speed first goes to 0.

It seems to be ONLY your opinion, at least from the available literature on the net (see again above pictures) which of course can be obsolete.

How quickly that point is reached depends on the spring rate, the spring preload, and the available travel. If the spring is too soft, then peak force will be higher as the strut comes up against the compression stop.

Even if it is rebound the back stroke will NEVER be as big as the forward stroke since available energy is less than at start and the spring will never be able to accelerate the plane mass full stroke back.

But in above text you neglect totally the shock absorber efficiency and consider that ONLY the spring monitors the landing, if it is so why do people bother to put on so complex devices as shock absorbers are ?

I am sure you know that there is a broad research going on to make aircraft shock absorbers intelligent using piezo actuated valves instead of orifices in order to adapt damper efficiency to all possible situations or some rheological fluids able to be controlled in their viscosity by an electric field. It is according to what you write a stupid action.

Considering the acceleration to be steady state might not get him close enough to satisfy his teacher (and would certainly not get him close enough to design landing gear.) A spread sheet with .1 second increments would work, as would calculus.

To your information there are dynamic simulation packs as Vissim, Simulik or Scilab which allow to make a full simulation of the landing under the wished and specified initial conditions with all constrains one wants to introduce. The equations being (if not simplified, but this you do not like) are too complex to be solved on paper with usual means. If you mean an numerical integration using the Rung-Kutta method then it can be solved on a spread sheet but it is a lot of work and the time step being small if low errors are expected the 0.1 s you suggest is too rough. Any way when somebody writes equations and tries to solve them he follows a MODEL which you say would not be required.

His teacher may want to use a rising rate spring (to better match a gas-filled strut).

Above mentioned software have "blocks" which can compute the power of inputs so that they can fully simulate the polytropic or adiabatic gas compression.

But in any case, it is just a case of determining a peak acceleration given speeds, distances and spring forces. No real model is required.

Now I dare give you an explanation since I consider you as a very bright person having only the difficult situation to be in an environment which gives you the feeling to be the only one to be clever.

There is a German proverb which says : " Other Mother have also clever children!"(You will allow me being European to use proverbs from the Old Continent).

You have one of the highest GA rates so that you match very well the average profile of participants and understand their way of thinking reacting accordingly.

I am sure that you very well understood the model problem but you did not like when Tornado wrote that the model I suggested was the best. The virulent and aggressive reaction gives this feeling.

In fact I am VERY thankful that you reacted this way: it gave me the opportunity to ask myself if I am or not right and to check every thing again. I made a thorough research on the web and came to the result that with the above mentioned error with respect to tires I was not out of the general practice.

I am far from being convinced that I detain the truth and put myself under question every time I have the feeling that something is not OK.

Now to the explanation:

Let us consider the 2 mass system first with the 2 spring rates and the 2 damping constants. Let us assume further that springs are linear .We have to make 2 other definitions, one is that the mass M1(aircraft) has at start a velocity "Wo" and that the portance "P" is constant during the process. This is "almost" true since the portance is function of the plane velocity and this will not change very much during the few seconds the process develops. In fact the horizontal component will stay almost constant and only the vertical will be affected. However this has as result a change since with respect to the wing the attack angle will change :

the angle between horizontal and plane path goes progressively to zero and the polar diagram indicates a modification of the portance and drag coefficients. As following graph shows:

We obtain 2 differential equations as:

M1*z1"= (G-P)-d1*(z1'-z2')-C1*(z1-z2)

M2*z2"= d1*(z1'-z2')-C1*(z1-z2)-d2*z2'-C2*z2

If we apply the Laplace transform, which we can since the equations are linear with constant coefficients, we come to :

L(z1)=(G-P)/{s*[M1*s²+d1*s+C1-(d1*s+C1)²/(M2*s²+(d1+d2)*s+(C1+C2))]}

The equation is complex and it is - as I mentioned in a previous comment - necessary to determine which parameters are more or less important for the domain we will investigate.

- The ratio C2/C1 is high since the cylinder stroke is much more important than the tires deformation let us write C2=C1*q with q>> 1

- We noticed before that the M2 is a few percent of M1

The dynamic behaviour of the M1-C1 system part is determined by the own frequency ω1=(C1/M1)^0.5 and the M2-C2 by ω2=(C2/M2)^0.5. The ratio of the 2 is ω2/ω1=(C2*M1/(C1*M2))^0.5, M1/M2≈25 and we can assume that C2/C1≈10 this means that ω2/ω1≈250)^0.5=16.

Since we investigate the M1-C1 system the M2-C2 system will be loaded in frequencies for which its response is almost static. We can thus assume M2=0.

Since C2>>C1 and d2<<d1 (in any case the shock absorber will "destroy" more energy than the tire) we can assume with no big error that d1+d2≈d1 and that C1+C2≈C2.

The equation becomes with those simplifications:

L(z1)=(G-P)/{s*[M1*s²+d1*s+C1-(d1*s+C1)²/( d1*s+C2)]} here "s" is the Laplace operator.

From this equation it is possible with some algebraic work to define a Laplace transform for the force acting upon the aircraft body

L(F)= L(z1)*C1+( L(z1')*d1+Wo*d1/s

and to pass over to the time domain.

The final function, you are of course not interested to know, is a sum of trigonometric functions (sin and cos) with amplitudes and arguments depending on the system's parameters. It is a long work and not at all interesting for a student project.

It becomes interesting when a first detailed approach has to be done for defining the range of parameters before a simulation with one of above mentioned soft will be done.

So that I tried to define a "worse case":

- No shock absorber whole energy goes in the spring

- The initial velocity is small (this was my error to show it on the model but not use it)

- No portance or the portance becomes small enough to be neglected in a first approximation

Under those assumptions which is the maximal force one could expect ?

This lead to the over simplified model which, I am sure, you very well understood.

But it was a pleasure for you to destroy it even, if not all arguments were pertinent.

So as conclusion:

The contested model was not proposed because I was not aware of the complexity as ,you wanted to present the situation, but because I thought it to be enough for the student to have an order of magnitude for the loads to be expected.

In fact, even the "nom de guerre" you use is a support for my theory: no stupid person would choose such a nickname!

This is only to show that willingly you "misunderstood" my reasons to show that I know nothing about landing of planes.

You flew a plane but not many of the plane designers did it and never the less their work flies.

I consider that this subject does not have the merit to be further discussed.

You may react as you wish no other comment from me can be expected.

Hi NickName,

You seem to have construed my criticism of your model as an attack on you. I certainly don't think of you as "stupid", at least not dramatically so. I'm kidding... I don't think of you as stupid in any way at all -- you are obviously a bright guy.

Also, my criticism of your model is not a criticism of the entire model. We both make assumptions regarding the student's intent, experience, and degree of laziness (or otherwise). The FAR's (in the US, which serve as the basis for similar regulations throughout the world) spell out the loads which are to be considered in testing landing gear, so the student's question would seem to indicate that he is quite new to the subject matter, or lazy. I've assumed that he is new to the subject matter.

Based upon that assumption, then it makes sense, I think, to deal only with the nearly vertical loads imposed on the gears. (In actual testing, this is not what happens, because, among other things, the plane must be in the landing attitude, rather than parallel to the ground, etc etc.) I think you made a similar assumption. I also agree with your assumption that we can initially not address the damping effect of the tires, and I am willing to further simplify (for the sake of instructional value) to even eliminate the spring effect of the tires.

I'd like the student to be able to say: "I've modeled this 100,000 lb plane's main gear as vertical struts with constant rate springs (of 6000lb/in) with 2 feet of available compression and with a preload of 25,000 lb. In a 12 fps landing (the test requirement) the peak acceleration goes to 6G, but I know that a "hard landing" is considered to be 2G's. Am I doing something wrong?"

After he has grasped the basics, then we (but preferably his teacher) would add the fore-and-aft and side-to-side structural requirements (to resist spin-up loads, braking loads, side loads from yaw and cross wind landings, etc.) But first he must understand the basics. And your model confuses rather than elucidates in the single most key aspect:

A 1G landing, in which the gear loads gradually increase to the aircraft's weight, is called "greasing it on" etc. A 2G landing, in which the gear experiences a peak load of twice the aircraft's weight, is a hard landing, and is cause for a hard landing inspection, an expensive and time-consuming process. So... orders of magnitude (if we take that to mean powers of 10) are not anywhere near close enough for even this simple evaluation.

If your model or logic states that "This happens when there is NO initial vertical down oriented velocity of the mass, this shows that the force at landing is AT LEAST 2 x plane's weight." then the model is not useful. A landing which commences with a wheel spin-up at essentially 0 fps (obviously the plane cannot land if the vertical speed remains at 0 fps, with the wheels held just an inch off the runway) is considered a greaser, not the hard landing that 2G's suggests. The range (of suspension loads in G's) between "good" and "so bad that there could be damage that must be investigated" is between 1G and 2G.

The relevant part of the FARs concerning landing gear test.

The relevant part of the FARs concerning landing gear design limits.

How these translate into practice with Boeing aircraft.

Now I dare give you an explanation since I consider you as a very bright person having only the difficult situation to be in an environment which gives you the feeling to be the only one to be clever.

Not at all. I considered phph001's first response (#26) to be just fine, and completely adequate to the needs of the student, until such time as the student indicates that he understands that... and can move on to the complexities. In his later response to you he correctly states that "The actual force on the suspension at touch-down is far less than 1G."

As can be seen from simple observation of landing gear during a landing, the plane is a very long way down the runway before the load reaches 1G, at which point the static load extension of the gear has been achieved. The gear strut projection can be used as a scale to indicate gear loading, and in a smooth landing (see my link in previous post) it can be seen that the gear does not rebound later in the landing. In a rough landing, it does, of course, rebound from the length that represents a load as high as 2G, to that representing 1G.

In the landing for which I provided video, one can see the overshoot and rebound in the nose gear. The pilot could have lowered the nose more smoothly (earlier), had he wanted to do so by controlling the elevator. In this case it appears that the elevator was held full back, and that the elevator has stalled, causing a dramatic and divergent change in lift. (This is standard practice, however: there are overriding benefits to holding the wheel back.)

Much of your model is OK, it's really only the numbers that are wrong.

BTW the term acceleration, in aircraft, is used to mean both positive and negative acceleration in the direction aligned with the wings' lift. "Deceleration" is more apt to be used in reference to braking. I don't, as far as I know, use the terms consistently.

You remain one of the people whose responses I think of as being almost always correct, and aplogize if my posts seem blunt.

I flew, BTW, a small plane for several years that had a G meter because it was certified for aerobatics. These have recording needles, and I never landed with more than slight deviations from 1 G (although there may have been several bad landings that were not recorded, because I often did not reset the meter before landing). (Its real use in flight is to see how close to the structural limits you have come in flying a particular maneuver... and you look at it after the fact, because you don't have time to look at it while flying the maneuver).

Guys, Great ...... I liked the exchange you all have had on the subject matter. ...... This in my opinion is what makes a site like this worhty of being a member of.......... I must compliment the originator of this site and all the regular and not so regular contributor .......... One code of conduct that i suggest we all follow (religiously) is to pass a comment only when we are sure of the subject and not just to make our presence felt. ........ My heartfelt admirations to all.

Greetings Sir.
I have been reading up on the topic and came across this thread as I am currently busy designing a retractable landing skid for an unmanned aerial vehicle. However the only design requirements that we have been given is the descent velocity and the approach angle. My problem is however that when I attempt to evaluate the UAV as a second order system of masses, dampers and springs that I have too many variables to reach a sensible conclusion.
I have come across an academic paper published on the subject and in order to calculate the forces acting in on the landing struts in order to design them the author simply used the change in momentum over time in order to calculate the reaction force so F = (Pf-Pi)/t where they assumed the time which it takes for the shock absorbers to compress to be 0.5.
I feel that this isn't a very accurate approach to the problem and I am uncertain if I perhaps misinterpreted the approach to solving the problem. Any advice would be greatly appreciated.

You may be able to measure/estimate the time from youtube videos of small aircraft, drones and models landing.

Thanks I'll keep that in mind. Would it also be acceptable to calculate the forces using a varying time, thus planning for "worst case" scenarios instead of just using one? Of course over design might be an issue..

That sounds reasonable, also on you tube you are likely to find videos of 'heavy' landings -both those that the aircraft survives and those that break things.

You can also search google for 'landing loads' 'landing stresses' 'landing forces'.... 'carrier landing ....' ...etc ... enjoy

"Hard Landings" got me thinking about ELTs or Emergency Locator Transmitters. Of course the real purpose of these transmitters is to activate due to a crash, BUT approximately 95% of "ELT events" are due to hard landings. If you could find out from the manufacturers the deceleration (meters / sec² ) at which their equipment activates, you would have a close approximation of what the numbers should be.

Bill

This might be of help.

try cirrus airplanes, they design parachutes for lost/crashing airplanes , they are looking at small business jets chutes R&D, they must know something about the loads on the bodies of the airplanes. This is a landing gear specialist knowledge. I had a friend who was a 747 landing gear specialist, he can probably help you or refer you to the proper people. But first start with cirrus.

From my basic studies, most of the forces are absorbed by the ground effect just before landing , i think the landing gear mostly stabilizes the planes from air flow. I would not be surprised that the loads are greater when the plane is stoped. This is what i experience landing on my legs from a surfkite lift.

F=ma is the formula we learned.

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At the moment of touchdown, the load on the undercarriage is zero. It quickly increases to the weight of the aircraft as load is transferred off the wings onto the landing gear.

It can be more than the weight of the aircraft if it bounces.

Controlling those forces without detriment to the integrity of the aircraft structure is the job of fiendish contraptions that use springs, dampers and other wonderful things.

It's not easily calculated.

It can be measured using accelerometers inside the aircraft.