aircraft position calculation
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aircraft position calculation
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.
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.
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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]
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]
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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
****'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
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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
Then add the pilot's weight divided by the size of his right boot and voila! The temperature in degrees absolute.
QED.. I think
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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.
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.
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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.
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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).
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).
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..
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..
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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?
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?
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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?!
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?!
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 
Adapted from the "How many fighter pilots does it take to change a lightbulb? " gag
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
Adapted from the "How many fighter pilots does it take to change a lightbulb? " gag
I don't own this space under my name. I should have leased it while I still could
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.
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.
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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
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
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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 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
Instead, look at this site 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