How does an INS/IRS work?
 

The general principle of all inertia navigation systems is that the system measures the aircraft's inertia movement from an initial position as a great circle track direction and distance to continuously determine it's up-to-date position.
The compnents of an INS are
1 Accelerometers
2 Gyroscopes
3 Position Computer

The aircraft moves in three dimensions, but the navigation equipment is only interested in acceleration in the horizontal plane. Therefore, the key to the whole INS arrangement is the accelerometers.

With this point;
"The aircraft moves in three dimensions, but the navigation equipment is only interested in acceleration in the horizontal plane. Therefore, the key to the whole INS arrangement is the accelerometers."
How does the aircraft measure if accelerating when climbing etc it is only interested in acceleration in the horizontal plane?

Quite simply, it does measure acceleration in all three axes, the navigational output to the pilot is in two.

GF

Basically, like galaxy flyer said. INS uses stabilised platform which should be always parallel to earth's surface - so that the accelerometers only measure horizontal movement. IRS is much better, because it uses mathematical calculation to correct output from the accelerometer by using the attitude information - it's easier to do a little calculation than to have a mechanically stabilised platform completely parelel to the earth's surface for the entire time of flight.

Thanks guys :)

Schuler loop
 

What it is;
"Schuler loop i only common to a stable-platform INS that has been programmed to remain horizontal as the aircraft moves around the surface of the earth. The error exists at the first accelerometer level, i.e., acceleration, which can be passed up through the integration to affect velocity and distance, resulting in a distance error during the schuler loop cycle, but the error returns to zero at the end of the cycle."
Can anyone come up with a simpler explanation of schuler loop, i've tried trusty google but still finding schuler loop hard to understand
Thanks

http://www.pprune.org/tech-log/11907...er-effect.html

Have a read of this if you haven't already, it may help.

The earth wobbles, at a frequency of about 1 wobble per 78 minutes.

An INS doesn't naturally wobble, so therefore it appears to wobble relative to the earth.

This wobble is called the Schuler cycle.

A Schuler loop is a resonant circuit set to the period and amplitude, but in antiphase, to the Schuler cycle.

Adding the output of the Schuler loop to the INS output makes the output correct, relative to the surface of the earth.

G

But note that a strapdown IRS keeps a mathematical equivalent of the stable plaform and reacts to setup errors or gravitational errors in the same way as a stable platform INS - with bounded position errors at the Schuler frequency

It is not true to say an IRS does not display Schuler frequency errors

Dick

Akafrank07

Whatever referece notes you are using, I suggest you get a better set.

That statement is not true

Dick

A lot of rubbish has been said about the schuler oscillation... Much of it found in pilot literature...

It is actually very simple.
If you have a platform that is horizontal when you are in London, it will not be horizontal anymore when you reach Australia, if you keep it in a fixed position (relative to space).
And gyros provide fixed positions relative to space only. So we need some sort of mechanism that will tie the platform relative to the round earth.

In INS systems this is done by measuring acceleration and thereby the speed that we move with.
So if you are moving with XX kts direct south, a tie-mechanism will tilt the platform xx deg/hour so it is kept horizontal with regards to the earth.

So far so good... The problem is then, that this tie mechanism has no clue weather an acceleration is actually caused by the airplane moving, or because the platform is not horizontal.
If the airplane is parked, and the platform for some reason is not horizontal, you will pick up an acceleration due to earth gravity.. This will be interpreted as if the plane was actually moving, and the tie-mechanism will start tilting the platform. The tilting will continue until the platform passes the horizontal, and starts tilting the other way. This is then interpreted as a deceleration, and eventually an acceleration in the opposite direction, causing the tie-mechanism to now tilt the platform back again...And we then have an oscillation....
The frequency of this is quite slow (84 minutes), and unless you dampen this the INS/IRS wold be worthless.
Schuler was a German engineer who claimed that INS/IRS system would be impossible to build, because an accelerometer, is unable to figure out if it is measuring actual acceleration or "false" acceleration due to a tilted platform. He actually published a science paper about this :-D

lasseb

If you "dampen" the response to errors due to incorrect levelling how do you allow the primary function to continue un-damped?

You are not dampening a response, and it is not incorrect leveling. The platform will always oscillate. That's the nature of a feedback circuit. The idea is just to subtract a known (error) oscillation from the incoming signal.

What you are doing is deducting a very low amplitude/low frequency signal from the input. This signal is (or rather should) correspond to the oscillation of the platform. The signal is generated internally in the INS/IRS computer, based on a math-model of the system.

Measuring raw input from the platform when an aircraft is parked would tell you that the aircraft is moving forth and back (both N/S and E/W) with an 84 minutes oscillation period. Basically it would tell you that you where driving around in a circle ;-). The INS/IRS computer has a model of this signal internally and deducts it from the input signal. If the modeled signal is a perfect match to the actual oscillation of the platform, you'll have a perfect INS/IRS, that tells you that you are actually parked when you are parked ;-)

This compensating circuit is working both in platform and strap-down units. It's just easier to imagine using a platform. but the accelerometer errors exists in both scenarios.

To add confusion, some models of the INS actually fed the compensating signal to the platform leveling system in stead of deducting it from the input. This will cause the platform to hold still even in parked scenarious. I guess that computers where to slow these days to handle all the calculations, so the more that where done mechanically the better.
Both approaches will work.

EDIT:
The best explanation out there I have found is a NASA document describing low orbiting satellites. They have an initial paragraph on schuler. Very good stuff..

I am apparently to stupid to figure out the link to the PDF, but search for "SCHULER PERIOD IN LEO SATELLITES ", and you will get it.

lasseb

Continuing, for simplicity, with the stable platform INS, if in ALIGN the levelling is carried out without error and there is no gravity element being sensed by the N/S and E/W accelerometers, why would the platform oscillate?

I think you are saying that if there is a leveling error and the platform is oscillating then the response is memorised, carried forward into the NAV regime and removed from all subsequent navigation computations. Do I read you correctly?

Edit. I think I see that you are assuming a rough initial alignment leading to error which is then computed out or, in older systems, fed back to the platform control. Am I getting warmer?

Hi Dick
This is actually harder to describe en words than I thought :-D.

What I'm trying to emphasize here is that the schuler error signal is not read from the platform, or memorized, or anything, it is strictly calculated/generated in the computer (based on a math model) and deducted from the incoming signals from the accelerometers almost at the very start of the calculations.

So the signal going to the integrators is = (SENSOR_DATA - SCHULER_DATA_MODEL.)

Regarding initial oscillation you are kind of correct. In theory if the platform is perfectly aligned no error should ever be present, and the platform should never oscillate. There are 2 problems with this. The first is that there is no such thing as perfect, and even small noise levels in the signal wiring will introduce this error. Also mathematical noise in the computer will do this, since we are not dealing with infinite decimals.

Secondly, because we have introduced the schuler compensations, that will actually make the platform oscillate if starts with no oscillation :-D.

Consider the following:
Lets say that the platform is perfectly aligned, dead-level, perfect wires, perfect computer, no oscillation.
The signal input to the integrators, are now the signal from the accelerometers - (which is zero), but deducted for the current schuler period value. So if the platform is perfect level, the input to the integrators would give us the 84 min oscillation signal (or rather the inverted value). This would then lead to the integrators giving us speed, which leads to the tie-mechanism trying to tilt the platform, and then we have the oscillation...

You could say, that because we have introduced the schuler compensation, the platform must oscillate all the time,
If we where only flying around in a 200NM radius (or so) from home base all the time, we could skip the entire schuler compensation mechanism, and use the same tie-mechanism as an attitude indicator. We would then only need to compensate for heading, and it would be much easier.

Hope this answers the question :-)

Lars

Yes, it's a word thing. I now think I understand what you say, and I am happy to find that I have not been wildly wrong, Thanks a lot

Thank a lot of the explanations - genesis the engineer Dick and Lasseb, i now understand it a bit better now, but obviously a quite complicated subject! Dick and Lasseb you two guys know your ****.:)
N.B. Dick i got this from 'ace the tech pilot interview' i understand there is meant to be a few mistake from this book, but sadly it seems to be what the airlines are using, as a couple of my friends that are now with the airline studied this book religiously though for my own sake and others reading trends i am trying to find out where these mistake are in the book, thanks for your help Dick.

Hey guys,

What is the vertical separation minimum below 30,000 ft?
a. 500 ft.
b. 1,000 ft.
c. 2,000 ft.
d. It depends whether or not RVSM is applied.

The correct answer is; b. 1,000 ft, and the reference states:

The vertical separation minimum (VSM) is:
• Within designated airspace (subject to RAN agreement (RVSM)), a nominal 300m (1,000 ft) below FL 410 or a higher level where so prescribed for use under specified conditions, and a nominal 600m (2,000 ft) at or above this level and; and
• Within all other airspace: a nominal 300m (1,000 ft) below FL 290 and a nominal 600m (2,000 ft) at or above this level.

This however doesn’t appear to coincide with the flight levels stated in the ICAO tables (the semi circular rule), in which case you can have an east bound VFR flight on FL 35, a west bound VFR flight on FL 45 and a west bound IFR flight on FL 40, giving a vertical separation of 500 ft.

Am I missing something here?

Anders,

what you are missing is that VFR flights are only seperated in airspace classes A through C (cf. ICAO Annex 10, Appendix 4). Apart from this, you will seldomly find IFR traffic at FL 40 in uncontrolled airspace. Firstly, FL 40 would only exist in a hypothetical country having a transition altitude which is low enough, and secondly such a low flight level would mostly be in uncontrolled airspace.

Edit: Sorry, the part after 'secondly' is rubbish. Forget it, please.

Ok, so basically it’s not considered “separation” outside of controlled airspace?

Those flight levels were just examples, the principle would still stand, I think, if flight levels 135, 145 and 140 were used instead.

Thanks for prompt response.

Hey guys, I have a question for you.

Look at this TAF for Zurich airport
TAF LSZH 211322 22018G35KT 9999 SCT012 BKN030 BECMG 1315 25025G45KT TEMPO 1720 4000 +SHRA BKN025TCU BECMG 2022 25015KT T1815Z T1618Z=

Which of these statements best describes the weather likely to be experienced at 1500 UTC?

B. Meteorological visibility 10 km or more, main cloud base 3000 ft, wind 250°, temperature 18°C

C. Meteorological visibility 10 km or more, main cloud base 1200 ft, gusts up to 45 knots.

The correct answer is supposedly B, whereas C was my selection.

There is the matter of the entries “T1815Z T1618Z” at the end of the TAF which I’m not entirely sure what they mean, but the book states that “TAFs do not contain information on temperature”, and I also ran through the CBT on the subject and couldn’t find anything about it there either, so am I missing something?

Hi Anders,

I edited my previous entry about ,,main cloud base'' being the same as cloud ceiling:

Ceiling is by definition more than SCT, but it turns out that main cloud base is the higher layer if there are multiple layers, and the higher one is thicker.

I pm you a thread from another forum about this very question.

Anders,

my interpretation of the respective group is the forecast temperature for 1500H UTC is 18 °C and the forecast temperature for 1800H UTC is 16 °C.

ICAO Annex 3 states forecast temperatures, though not an element of a standard TAF, may be included in accordance with regional air navigation agreements. If they are given in a TAF it is recommended to indicate the maximum and minimum temperatures for the forecast period.

The format of the temperature group in your example, however, is not in line with ICAO Annex 3 nor the WMO Manual on Codes in their current versions. I take it the example is from a question bank and might be outdated.

If the temperatures in your example were indeed maximum and minimum temperatures at 1500H and 1800H this would have to be indicated today as "TX18/1500Z TN16/1800Z", with "TX" standing for maximum and "TN" standing for minimum temperature.

Ok, that makes sense. Thanks for the responses.

Gen Nav questions
 

Hi

there are a few questions that are confusing me right now with regards to gen nav.

these are departure and convergency. I never seem know which formula to use and when. I understand the 2 formulas. Ch long x sine mean lat and ch long cosine lat. can someone give an example of when I should use both of these.

Right now I'm looking at q' 5566 in the atpl online system.
Given: Position 'A' N60 W020, Position 'B' N60 W021, Position 'C' N59 W020. What are, respectively, the distances from A to B and from A to C?

Why do I use the departure one for this question and not the convergency?

This is probably a pretty stupid question but its totally confusing me.

Thanks in advance

Hi,

Use the departure formula for your example, because it is for calculating distance between two point on the same latitude.

You use the convergency formula for calculating change of direction of a great circle track. Not for calculating distance, unless perhaps the average latitude and the convergency is given, in which case you could calculate the change of longitude.

To put it simple, departure and convergency has little to do with each other. One is for distance, the other for degrees.

In simple terms if the question is mentioning the words distance or position it is highly likely to be a DEPARTURE related question, going east/west along a parallel of latitude.

So in your example,which funnily enough I worked through with a student yesterday. A to B is a departure question. So Dep = Ch Long (mins) x cos lat.
1 * 60 = 60 * cos 60 (.5) = 30 nm. However A to C is not departure as the positions are on the same meridian. It is testing do you know at 1 degree of Latitude is 60 nm so A to C = 60 nm.

Now if the question mentions the words like great circle, rhumb line, initial or final track highly likely to be a CONVERGENCY problem. I always teach my students tips similar to the ones mentioned.

Pressure error question
 

TOTAL PRESSURE ERROR

"As an aircraft moves through the air, a static pressure disturbance is generated in the air, producing a static pressure field around the aircraft. At subsonic speeds, the flow perturbations due to the aircraft static pressure field are nearly isentropic and do not affect the total pressure. As long as the total pressure source is not located behind a propeller, in the wing wake, in a boundary layer, or in a region of localized supersonic flow, the pressure errors due to the position of the total pressure source are usually negligible. Normally, the total pressure source can be located to avoid total pressure error."

"As an aircraft moves through the air, a static pressure disturbance is generated in the air, producing a static pressure field around the aircraft."

What is this static pressure field around the aircraft i would of thought this was just the free air around the aircraft, the air which is not impacted by the aircraft which would be then dynamic pressure right?

It then goes on to say: "at subsonic speeds, the flow perturbations due to the aircraft static pressure field are nearly isentropic and do not affect the total pressure."

I understand that isentropic basically means the same all round though i can't grasp what this paragraph mean either can anyone simplify it?

Basically there is a thin layer around the aircraft in which it feels like static pressure - a good way to demonstrate is put your hand out of the window in a moving car and put it flat against the door, you will feel no dynamic pressure, only static. Hence, static ports are usually small holes on the side of the fuselage.

To measure the PITOT or toal pressure, the Pitot tube must stick out far enough so it it outside this static layer. Remember that PITOT (total) pressure is dynamic pressure PLUS static. For altitude read out you need static pressure, for airspeed you need dynamic, dynamic pressure is created by the movement through the air.

Some thoughts..
As you probably are familiar with the bernoulli effect (Pt=Ps+Pd), the distribution of dynamic and static pressure over a moving surface depends on many variables (aoa, airflow angle, camber, material, etc.) f.ex. at the stagnation point of a wing the dynamic pressure will be 0 and the static equal to Total! So the a/c form can interfere with the ''normal'' airflow and that will cause pressure disturbances, mostly increased static pressure because of the deceleration of the air molecules that come in contact with the airframe . At subsonic speeds the air is thought to be isentropic, meaning that it can adiabatically change state (Kinetic to Potential energy without heating loss and vice versa). At supersonic speeds and above (I think even transonic), the air can not longer be seen as an isentropic, incompressible fluid and that means it will produce an error that will affect Total pressure readings!

Hope I didn't mess it up! ;)

Thanks Richard and Da-20 monkey that explains things better. as you can see by my user name I am an ex PTC student and I am now doing atpl's on my own for the time being. It is hard trying to figure out everything on your own :)

thanks again

Total pressure in a compressible medium is not dynamic + static pressures.

When air (M ≤ 1) is brought to rest isentropically at the head of the pitot probe, the total pressure will be given by a function such as (6) in the graphic on this Glenn Research Center page, Isentropic Flow Equations

The subject of (6) in the above is Ps/Pt (s=static; t=total) and will always take a value equal to or less than unity. Hence, Pt ≥ Ps or Pt - Ps ≥ 0. This latter quantity, the difference between total and static pressure, is often said to be dynamic pressure. It is in fact something else called impact pressure. If you look at (5) in the above reference, which is the definition of dynamic pressure (defined in this form by Euler and not Bernoulli), you can see the difference in the behaviour of these two functions.

Airspeed indicators map impact pressure (not dynamic pressure) to calibrated airspeed. The function which provides that mapping assumes an isentropic process, meaning as given by Jetpipe that the process leading to a total pressure is an adiabatic and reversible one, i.e. heat is neither added nor removed and no dissipative effects occur. Obviously if these conditions are not satisfied the airspeed value will be erroneous. For these reasons it is quite important that both the pitot probes and static ports are designed, and located, to ensure minimum dissipation and heat addition.

Hi Jetpipe



"As you probably are familiar with the bernoulli effect (Pt=Ps+Pd), the distribution of dynamic and static pressure over a moving surface depends on many variables (aoa, airflow angle, camber, material, etc.) f.ex. at the stagnation point of a wing the dynamic pressure will be 0 and the static equal to Total! So the a/c form can interfere with the ''normal'' airflow and that will cause pressure disturbances, mostly increased static pressure because of the deceleration of the air molecules that come in contact with the airframe . At subsonic speeds the air is thought to be isentropic, meaning that it can adiabatically change state (Kinetic to Potential energy without heating loss and vice versa). At supersonic speeds and above (I think even transonic), the air can not longer be seen as an isentropic, incompressible fluid and that means it will produce an error that will affect Total pressure readings!"

Why can the air no longer be seen as an incompressible fluid at supersonic speeds and how come it starts to lose heat at supersonic speed?

The air IS a compressible fluid but at low speeds the friction heating is almost negligible so the isentropic equations for incompressible fluids are quite so precise... Check this wiki-page about Total air temperature !

If you need a more academic answer, I think Selfin is the man to ask! :ok:

Selfin
 

PITOT (total) pressure is dynamic pressure PLUS static. [F]or airspeed you need dynamic [] pressure [] created by the movement through the air. Total pressure in a compressible medium is not dynamic + static pressures.

When air (M ≤ 1) is brought to rest isentropically at the head of the pitot probe, the total pressure will be given by a function such as (6) in the graphic on this Glenn Research Center page, Isentropic Flow Equations

The subject of (6) in the above is Ps/Pt (s=static; t=total) and will always take a value equal to or less than unity. Hence, Pt ≥ Ps or Pt - Ps ≥ 0. This latter quantity, the difference between total and static pressure, is often said to be dynamic pressure. It is in fact something else called impact pressure. If you look at (5) in the above reference, which is the definition of dynamic pressure (defined in this form by Euler and not Bernoulli), you can see the difference in the behaviour of these two functions.

Airspeed indicators map impact pressure (not dynamic pressure) to calibrated airspeed. The function which provides that mapping assumes an isentropic process, meaning as given by Jetpipe that the process leading to a total pressure is an adiabatic and reversible one, i.e. heat is neither added nor removed and no dissipative effects occur. Obviously if these conditions are not satisfied the airspeed value will be erroneous. For these reasons it is quite important that both the pitot probes and static ports are designed, and located, to ensure minimum dissipation and heat addition.

When air (M ≤ 1) is brought to rest isentropically at the head of the pitot probeWhat does isotropically mean?


For these reasons it is quite important that both the pitot probes and static ports are designed, and located, to ensure minimum dissipation and heat additionDoes the location affect heat addition because the pitot probe is located far out from the aircraft skin and the static probe is built into the aircraft?


Cheers

Hey guys, need some help again.

Question 1:
On a CVOR the phase difference between the AM and FM signals is 30°. The VOR radial is?
a. 210
b. 030
c. 330
d. 150

Question 2:
For a conventional VOR a phase difference of 090° would be achieved by flying …… from the beacon.
a. West
b. North
c. East
d. South

I put down b. on Q1, which was incorrect, and c. on Q2 which was correct. According to the answers the correct answer for Q1 was c. 330. The only thing I can come up with is that it’s in the “wording” of the question, because they are asking for the difference between the AM and FM signals, and not the difference between the FM and AM signals. Could it be that simple, or am I missing something else?

Did question one include the words TO or FROM?

The 30 degree phase difference will occur on the 030 radial FROM the VOR, which is also the 210 radial TO the VOR.

Nope, the question is basically copy/pasted.

In my copy of CQB15 the correct answer is marked at 030. Is it perhaps just a case of your reference material being wrong?

Quite possibly, it wouldn't be the first time.

gen nav question HELP!
 

Hi can someone please explain this question to me. I don't really understand how it isn't a track change of zero. Its following the same line of latitude but the longitude is changing. wouldn't this mean that its just following a straight line towards the east?

The following points are entered into an inertial navigation system (INS). WPT 1: 60°N 30°W WPT 2: 60°N 20°W WPT 3: 60°N 10°W The inertial navigation system is connected to the automatic pilot on route (1-2-3). The track change when passing WPT 2 will be approximately:

4° decrease
9° increase
zero
9° decrease

Some help here would be great thanks






Am I right in thinking that I need to find the conversion angle?
I.E. find the convergency and half it? that way I know that for every 5degrees it will decrease by that number?