aircraft position calculation

[budipro]

Hi all,

I have a question here:

How do you calculate an aircraft current position (latitude, longitude, and altitude) based on:
- the aircraft last position (latitude, longitude, and altitude),
- the elapsed time (now - last position time),
- the aircraft speed (knot), and
- the aircraft direction (stated using angle from north)?

Thanks a lot.

[Divergent Phugoid!]

Talk me through just why you would want to do that.....

:rolleyes: :rolleyes: :rolleyes:

[Feck]

Ah, quite simple really.

The equipment, and hence the aircraft, knows where it is at all times. It knows this because it knows where it isn't. By subtracting where it is from where it isn't (or where it isn't from where it is, depending on which is greater), it obtains a difference or deviation. The inertial navigation system uses deviations to generate corrective commands to fly the aircraft from a position where it is to a position where it isn't. The aircraft arrives at the position where it wasn't; consequently, the position where it was is now the position where it isn't. In the event that the position where it is now is not the same as the position where it originally wasn't, the system will acquire a variation. Variations are caused by external factors, the discussion of which is beyond the scope of your question.

The variation is the difference between where the aircraft is and where the aircraft wasn't. If the variation is considered to be a significant factor, it too may be corrected by the system. The aircraft must now know where it was. The thought process of the equipment is as follows: because a variation has modified some of the navigational information which the aircraft acquired, it is not sure where it is. However, it is sure where it isn't and knows where it was. It now subtracts where it should be from where it wasn't (or vice-versa) and by differentiating this from the algebraic difference between where it shouldn't be and where it was, it is able to obtain the difference between its deviation and its variation, this difference being called error.

[Ouch, must get a scanner]

[ZH875]

****, looks like you have been copying the Nav Inst training notes from the hallowed 2 SofTT Cosford.

(For the Heavies out there, rename Cosford as 1 SofTT)

[FEBA]

Budipro
****'s excellent and comprehensive reply will go far to solving those DR conundrums in the air that all airborne pro's occaisionally suffer from, especially our elite front line pilots that police the skies of the world so that you and I can sleep soundly.
Most, if not all of our modern high tech fighter aircraft use complex high tech navigational systems like Decca. Because **** is a serving member of the RAF he is not at liberty to disclose this name, merely refering to it as 'the system'. I am not a serving member therefore I am and this has been approved by Claire Short.
The complex algorithms are based on the mathematical probabilities of where the aircraft may be as opposed to where it wasn't or where it should be assuming of course that the air plot was based on a position where the aircraft could be. This mathematical formula is know as Sodanos Confluence and makes heavy use of the principal of latitudinal and longtitudanal convergenge and the influence of temperature deviation with latitude on the boiling point of angles, ie a right angle at 90N will boil at 100 Degs C whereas at the equator it's more like 104.32 C.
Another rule that comes in handy is the 1 in 60 rule, this means that for every 60 nms or 60 degs off course you are you only have 1 chance to correct it (especially at low level in mountainous terrain, CFIT etc etc)
Good luck

[monkeytamer]

Further to Fecks Reply. Would it not be simpler to use/buy GPS.
Where LKP is (normally) where you where 1 second ago.

[FJJP]

FEBA, shame on you - you have forgotten the fundimental principle that the correlolis force affects the boiling point. The boiling point is affected by the rotation of the earth; therefore the boiling point at the equator is equal to root T in deg Celcius divided by the square of the hypotinuse of the sine of the latitude. Also, the cube root of the longitude multiplied by the isoginal added to the atmospheric pressure in hectopascals is equal to the cube root of the longitude multiplied by the isoginal added to the atmospheric pressure in millibars.

Then add the pilot's weight divided by the size of his right boot and voila! The temperature in degrees absolute.

QED.. I think

[hifiman]

FJJP

Don't forget the corrections for Precession, Nutation and Granularity! Although all of these are potentially overshadowed by the plotting errors if you apply GOAT (Greater Observed Altitude Towards) wrongly.

[MrBernoulli]

.................. but only on days ending with the letter 'Y'.

[Nozzles]

And remember the definition of one horsepower: the power a standard horse uses on an ISA day at sea level to pull 1kg of water at 100 deg C one meter horizontally in one second.

[FEBA]

Sadly I see that some of you (nozzles in particular) are not taking this seriously.
FJJP yes I confess I had forgotten the effects of correolis on latitude and longtitude, it's such a long time since nav school and its possible that it was covered in the afternoon lecture after we had a few beers for lunch. Anyway at the equator correolis accounts for nowt but temperature does. The effect of temperature distortion on the lat long grid can make for huge navigation errors which is why the Nigerian airforce fly mostly a night to counteract this.
Also no one seems to have picked up on z and the angle of dip and the effect that trim can have on the IRS (no more <> 25% MAC).

[BEagle]

Not forgetting improbability functions and the angle of dangle.....

From x1, y1 if you proceed at v knots in a direction of theta wrt True North for t hours, you'll.......oh bolleaux, this is boring. Depends on whether we're talking rhumb line, great circle etc. Spherical trig is for navigators and aerosytems course grads. Easy schoolboy trig I can do, wacky maths, err - nope.

Dimensional incongruity and orthogonal transcendantalism may also apply......

Ask Dr Who - and I'll keep Sarah Jane amused..:E

[witchdoctor]

Do any of you guys write questions for the JAA ATPL syllabus by any chance?:} :E

[Nozzles]

Seeing as you're being so harsh and critical, FEBA, answer me this one:

If I do a vertical take-off and do a perfect level attitude hover, how come I don't end up going round the World once every 24 hours?

[zedder]

budipro,
To answer your question at the top of the thread, you can't. This is because of that potentially infinite variable of wind. Based on the parts of the equation that you have, you can calculate where you end up for zero wind. But as many a pilot/nav has discovered to their cost, where you actually end up is affected by how much wind you have.

It should be remembered though that if you have too much wind, you are generally too distracted to care where you are!

Do you have wind?!

[insty66]

Nozzles
The answer to your question is..... You do it's just the world comes with you:\

From you stated occupation I would have thought that you would have taken it for granted:E :ok:

Adapted from the "How many fighter pilots does it take to change a lightbulb? " gag:}

[Pontius Navigator]

I see you need a navigator to answer the question.

If you remember the question said you had a position, speed, time and course. You calculate the position using either track and ground speed based on an assumed wind or you use airplot and the assumed wind vector.

In both cases the output is the dead reckoning position.

****'s answer is how an inertial navigator knows where it is based on where it was and where it knows it isn't.

[sarboy w****r]

A slightly less facetious answer if you are truly after some help...

I'm sure there are plenty more people who are better qualified to answer, but there is some software in use that does something along the lines of what you're after.

The formulae at the heart of the program take a given position (lat/long), IAS/direction and wind and will give the position (lat/long) that you will be at after a given elapsed time.

Calculations are based on rhumb lines (great circle calculations easy enough to do though), and no TAS/IAS conversions are made (not really relevant for helicopters!).

No account is taken of altitude.

Can also be used to provide intercept solutions for multiple moving targets too.

The software is used (but not owned) by the RAF, but I'd rather not publish the source code.

What do you need the answer for?

If you'd rather just know the theory, then I suggest you search the internet for the formulae that will give you a lat/long position that is a given distance and bearing from another lat/long. Try searching for "sperical trigonometry", "bearing distance latitude longitude", "geodetic calculations" or the like, as I don't have the formulae to hand, and I'm not so short of a social life that I need to know it off by heart... There are lots of such sites. The distance you will travel you know based on the elapsed time multiplied by speed. From memory, you will need to convert angles and distances into radians before using the formulae.

E.g. if you are travelling at 100 kts (quiet in the cheap seats, those of you who fly jets), for 2 hours on a track of 360 then you will get from the formulae the lat/long position that is 200 nm north of the start position.

If you wish to consider wind then simply convert the vectors representing the IAS/bearing of the aircraft and the wind from polar notation into cartesian (or rectangular) form.

Again, for the FJ brethren, polar notation means angle/magnitude form (bearing/distance). I do wish you'd worked harder at school...

You can't add polar vectors directly, but you can when they are expressed in cartesian form. Add the IAS and wind vectors together and then convert back to polar notation so that you can substitute the answer (which represents the resultant of the wind and IAS vectors combined) into the formula used to calculate the new position.

E.g. if you have a vector, v, representing your speed/bearing as 360 degrees/10 kts, convert this to cartesian form, where the positive y direction is north, and the positive x direction is east. The vector is now v = 10y.

If the wind (vector w) is 090 degrees / 10 kts, then this translates to w = -10x. The vector is negative, as 090 means that the winds blows from that direction, whereas your aircraft vector represents travel in the direction of 360 degrees.

Add the 2 vectors together to get the resultant, r.

r = -10x + 10y.

This is intuitive, as if we are heading north with a wind from the 3 o'clock, we expect to be blown left (draw the vectors to see what I mean).

Convert back to polar form:

|r| = ((-10)^2 + (10)^2)^1/2

i.e. use Pythagoras to determine the vector length.

Now determine the angular argument of r:

ang(r) = arctan(10/(-10))

This will give you the angle that r makes with the positive x axis. Convert this to the angle that the vector makes with the positive y axis (north) by subtracting from 90, and normalise such that the angle is positive and less than 361, i.e. 315 degrees.

If you draw the vectors out on a piece of paper, you'll see what I mean (r = 315 degrees, 10*sqrt(2) knots).

This is more reliable than just using the sine/cosine rules, as these won't work (without bodging) if the aircraft and wind vectors are coincident i.e. if they and the resultant don't form a triangle.

If you're still struggling then I think you should ask for some advice, rather than plough on without understanding what the formulae represent in real terms.

Here endeth the lesson on basic vector arithmetic...

Why do I now feel like the teacher in Ferris Bueller's Day Off...? Anyone...? Anyone...?
...

SBW

[Fox_4]

LINS GPS

Or is that too obvious?!

[Bo Nalls]

Ok, I could sit here typing in lots of equations and definitions to explain this problem, but I can't be bothered. The answer is not as easy as it seems, it being dependant upon which world model you are using, how far you are travelling, which method being used (great circle, rhumb line, parallel, traverse or composite) etc.

Instead, look at this site (http://williams.best.vwh.net/) where if you click on 'aviation formulary' you will get lots of formula and if you click on 'on-line calculator' you can use a javascript program to calculate an answer to your question.

Hope this helps :ok: :ok: :ok:

[FEBA]

Nozzles
I'm getting seriously concerned about your lack of theoretical knowledge. It's bloody obvious man, you can't get enough fuel into a Harrier to hover for 24 hrs ......... jeez !!

Sarboy doodahs
Thanks for the Montessori maths lesson but how does that prove the unquestionable logic of Fecks theory that 'if you know where you are not, then you must know where you are' or 'if you don't know where you are then you must know where you are not' Revd Dodgson school of navigation.

[whiz]

Posted by sarboyw****r
I don't have the formulae to hand, and I'm not so short of a social life that I need to know it off by heart...

Judging by the length of your post your social life needs looking at old chap ? ;)

[sarboy w****r]

Ah Whiz,

Such is life on a 24 hour SAR shift such as yesterday that I found myself with nothing better to do at 2200 than answer random questions...

And as for the Montessori maths lesson, well I did pick up a thing or two whilst I was spending your taxes at university! Thanks, they paid for quite a few beers...

Hic!

SBW

[Mr C Hinecap]

Chaps. I hope there is more. I have learned far more from this thread than all the rest added together over the last couple of months! As a blunt and thick as a thick thing, I always wondered how those plane thingys stayed in the air, never mind the moving around thing. This is better than Television for Schools ever was! Thanks and much humbler now. :(

[noprobs]

Nozzles,

Look in your FRCs. That complicated table in there showing engine limitations is the giveaway. The reason for the short time limits operating at lift ratings is just because of earth rotation. If you stay around in the hover for too long, your landing spot will have moved away. It's happened to many a good mate in the past. Mexe pads seem to move faster than most surfaces.

C Hinecap,

To get the full explanation of how the plane thingys stay in the air, you need to read the article first seen in the Wittering View in 1982. The joint authors, Harvey Elling and V T Owen, managed to condense a very technical subject into a form that most could comprehend. Basically, it comes down to conventional aircraft moving fast enough to run into small lighter-than-air creatures called Bernoullis, which tend to stick to the wings. To manoeuvre, flappy bits on the wings (ailerons, spoilers, whatever) are moved, causing the Bernoullis to fall off. In the case of the Harrier, there is another effect when the production of sudden huge volumes of noise under the aircraft causes it to jump up in surprise, though as it gets used to this, it tends to settle back down unless some Bernoullis are collected. Best vertical results are achieved after long periods of conventional flight, when lack of fuel lulls the aircraft into a false sense of security that the noise is about to cease. Then it can really be startled away from the ground. For more details, refer to the eminent aerodynamicists and their original work.

Hope this helps.

[Rattus]

MORNINGTON CRESCENT!

[Nozzles]

insty66,

A very valid point, sir. In that case the laws of circular motion tell me that the further from the centre (and therefore the surface) of the earth one hovers, the faster one is going. At what altitude would one go supersonic in the hover?

eh?
eh?

[StopStart]

noprobs

Your discussion of the Bernoulli, or Bernoullis Kineticus as it is better known at CFS and the Natural History Museum, brings to mind their involvement in the generation of thrust by propellers.

As you are no doubt aware, propeller blades can be quite sharp. Indeed they are sharp enough to damage or even kill Bernoullis. Over the years Bernoullis have come to recognise this fact and as such as have developed an inate fear of them. The sight of a turning propeller will cause any Bernoullis ahead of it to vacate the area. The speed with which they move is enough to briefly generate a vaccuum. As nature abhors a vacuum, the propeller and the associated mass of aircraft is then drawn foward into the space created by the departing creatures. The more Bernoullis you can scare, the faster forward you go.

Bernoulli scaring can be increased by adding more blades or making them sharper. Pioneering aviators found it difficult to attain much speed as their propellers weren't very sharp and the Bernoullis had yet to develop a fear of them.

Interestingly, even a stationary prop can generate fear in local Bernoullis. As the aircraft is parked and chocked it can't be drawn forward by the departing Bernoullis. This leads to a localised thinning of the air around the aircraft. The effects of this rarefied atmosphere can be seen to be best effect on, say, an early morning at Akrotiri. The crew may appear unsteady on their feet or drowsy and some of them be lying on the pan, under the wing, apparently unconscious. The younger, more inexperienced co-pilots often suffer the worst from this, what is basically a form of altitute sickness, and may often be seen vomiting on the nosewheel.

[FEBA]

SS
Amazing, please explain what this has to do with navigation eh!!

[BEagle]

Transport wander!

[StopStart]

FEBA

Absolutely nothing. It was addressed to noprobs and the references he made in his post.
That said, surely this has to be first?? Somebody actually querying the relevance of a post to a thread title? Surely the idea of this place is to get away from the opening subject in as few moves as possible; or am I missing the point?

Now get back in your ditch.

;)

[Mr C Hinecap]

FEBA - Shhhushhhhhh! I'm learning things here. This is good stuff. Tell me - is it only the Aircrew that know about this stuff, or do the Ginger Beers know it too? Or was it classified? I can imagine that enemy forces would wish to control these Bernoulli thingys. Can they be controlled in any other way? It would make a very powerful anti aircraft weapon - perhaps firing extra Bernoullis in the path of aircraft?

Or am I thinking too much. :8

[insty66]

Nozzles
Having considered your question I have concluded that it is a very good one:ouch:








A very small amount of thought brings me to this conclusion.

Going supersonic means travelling faster than the sound you are making.
Therefore if you are not travelling with respect to the earth, you can't leave your noise behind, so no matter your altitude you will never go supersonic:\ :ok:

Especially in those things:E

[Beeayeate]

Nozzels

If I do a vertical take-off and do a perfect level attitude hover, how come I don't end up going round the World once every 24 hours?

Mmm. . . because you would be, essentially, in a very lo geo-synchronous orbit. :hmm:

Anyway, you're already doing 19 miles per second just standing still!

[BEagle]

Err, nope. Can't be geosynchronous at anything other than in an equatorial orbit at an altitude of 36000 km. At that height gravitational and centripetal forces are in balance when the vehicle velocity is exactly equal to the earth's rotational rate.

[Beeayeate]

Quite! But why spoil it with facts? The grav and centrip forces you mention would obviously be equated by the thrust he would be producing. :rolleyes: :rolleyes:

Anyway, Noz didn't say where he was doing it or how high he was a'hovering. ;)

[Stuff]

Anyway, you're already doing 19 miles per second just standing still!

Wow, think about the extra wing lift this is going to give. 29,000,000 J / 29,003,000 N ? :uhoh:

[Nozzles]

Insty and beeayeate,

You guys are scaring me. I'm getting the horrible feeling that you thought my questions were serious. Here's another one that doesn't need to be answered:

If the space shuttle is doing 19 miles per second in space where sound can't travel, is it going supersonic?????

[Beeayeate]

Nozzels

If you thought my response was serious . . . ! :hmm:

[FEBA]

Nozzles
For your information this thread is a serious one if it wasn't you wouldn't find contributers like Beagle or Stop Start posting on it, would you?
Hyper,super,trans sonic problems are not ones that we would normally associate with SHar pilots, any M number with a decimal point in front of it is more appropriate for you guys, that and a book of nostalgia.

Stop Start
Thank you for the helpful advice regarding my ditch. I read yesterday in the Sunday Times that Lord Hoon of Chilwell has withdrawn all ditches from service as part of his military cut back campaign. Troops will not be put at risk by this cut back as they will be able to make ample use of naturally occuring geological features such as casms rifts etc etc. Getting back to the navigational point of this thread, has anyone tried to get out of Rouen on the road to Dieppe recently?

Beagle
Transport wander!
I'm looking on my keyboard to see how close D is to K

[insty66]

I would just like to echo Beeayeate's sentiments:oh:

[Nozzles]

FEBA,

This EX SHAR pilot was at M1.6 last Thursday
:D

Beagle,

What's causing the centripetal force?

[FEBA]

Nozz
I was on the M25 yesterday ;)