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-   -   Definition of ground speed (https://www.pprune.org/tech-log/418855-definition-ground-speed.html)

ATCast 22nd Jun 2010 20:43

PBL, many thanks for that summary of this thread, captures the discussion very effectively. I wish more people provided such overviews, especially on those 20+ pages rumour threads.

But I am even more grateful for the reference you provided. :ok:
It's a big help. And it's the definition that works best for most applications I think. (The context that provoked this question is the GPS reported velocity accuracy.)

Genghis, BOAC, Wizofoz, Pugilistic Animus and others with thoughtful contributions to this thread, thank you as well.

ATCast

ATCast 22nd Jun 2010 20:50


Originally Posted by PBL

Quote:
Originally Posted by LH2
Wouldn't that be "perpendicular to the local vertical" or equivalently, "perpendicular to the local gradient of the gravity field"? Is the parenthesised phrase part of the quote from the book, or is that yours?
Nope, perpendicular to the geodetic latitude. The "local gravity field" is known as the astronomic latitude, apparently, and it is not necessarily the same. Indeed, it is explicitly distinguished in Figure 2.2 of the reference I gave. (The third possibility is "perpendicular to the geocentric latitude", referenced to the "center of the ellipsoid and mass center of Earth", which is different again.)
Full points again for PBL!

Genghis the Engineer 22nd Jun 2010 22:06

Somebody's attempt to regularise the whole thing, with maths.

(Aeronautical Journal. 111 (1120) 381-388, June 2007).

G

Checkboard 22nd Jun 2010 23:45


It's a big help. And it's the definition that works best for most applications I think.
Might just point out, it isn't a "reference definition", but an interpretation of a diagram by an anonymous poster. :hmm:

I don't think it even goes towards answering the question.

Pugilistic Animus 22nd Jun 2010 23:54

PBL could you post your reference?...

No I don't design IRS/INS/GPS systems,...but I think that it is the position solution that gets "jiggy wit it":)

Ghenghis that article was for calibration. techniques..interesting though


at the exact same GS two crafts one above the other should be over the same point at all times assuming constant Vw and TAS

and if one look at the definition for TAS I think the answer is apparent

but,
I'm like quintuple confused now :\

PBL 23rd Jun 2010 06:35

ATCast, I'm glad it helped answer your question!

PA, if you are asking me to scan the Figure in, I would rather suggest looking at the entire Sections 2.2-2.4, pp 23-32, because they also include the equations of motion associated with the quantities in the diagram.

amazon.com (the US branch; local country branches of amazon might be different) has scanned the book, and if you have an account with them, you can search it. The relevant sections in Chapter 2, The Navigation Equations, are 2.2 Geometry of the Earth; 2.3 Coordinate Frames; 2.4 Dead-Reckoning Computations.

Concerning Gratton's paper; yes, it is not nav, but he calculates true airspeed using vector triangles in a way that is only correct if ground velocity is taken to be the the horizontal component of the resultant of the air velocity vector with the wind vector (where by "horizontal" I mean in a plane parallel to the tangential plane of the reference ellipsoid). So that is indirect confirmation, even though not explicit definition.

Checkboard, Kayton and Fried is a canonical reference; I am not really anonymous; and I suggest you may rely on my interpretation.

PBL

pattern_is_full 23rd Jun 2010 07:23

Ground Speed can only be measured relative to - and at - the ground. The surface. Terra Firma.

If it takes 5 hours to fly from directly above point A on the ground to directly above point B on the ground, and A-B is 1000 nm, then the ground speed was 200 kts. Period. It does not matter whether the trip was made at ground level or FL320 or in a low earth orbit of 320 nm.

One's actual speed through the Universe or through the "ether" may have to be higher at FL320 or 320 nm than at the ground, due to flying an arc following the curvature of the Earth at a wider radius (or fighting heavier headwinds, or no significant atmosphere at all) - but actual speed is not Ground Speed.

It is possible an E6-B or GPS or other calculator does not take into account that wider radius (and thus longer distance at altitude) so it may not tell you the correct IAS or CAS to fly to achieve a given ground speed, or it may display an inaccurate calculated (as opposed to actual) Ground Speed. But that is an error in the calculation algorithm - not a "higher" or "lower" ground speed itself needed to cover a given ground track in a given time.

Pugilistic Animus 24th Jun 2010 19:16

Relative Velocity - Ground Reference

I went to these muppets for the definitive answer:E

:}

LH2 26th Jun 2010 14:51


Originally Posted by PBL (Post 5768402)
Nope, perpendicular to the geodetic latitude

Indeed it is according to your reference, which I've now looked up. But my question arises from the fact that, in a gyrocompassing Inertial Navigation System, the "down" channel is aligned with the local vertical[1], and the corresponding velocity equations provide a "basic description of the Earth relative velocity evolution in a local level navigation frame"[2] (my emphasis).

A GNSS-derived velocity, on the other hand, will be expressed directly in an Earth-centered Earth-fixed (ECEF) reference frame[3], as long as the standard broadcast or precise ephemerides are used[4] (admittedly I've never heard of an application where a non-standard ephemeris was being used, perhaps referenced to a different system or using non-Keplerian elements, although in theory that would be entirely possible).

Note however, that as long as navigation is taking place in a different reference frame (typically ECEF, for any long range terrestrial navigation) then presumably every relevant output will be transformed to the target system. This, I assume, will be the case in a modern craft with INS-derived velocities before they are presented to the user, which would validate Kayton's definition.

Now, the difference in geoid-referenced to ellipsoid-referenced groundspeed is negligible for all but the most demanding applications. To put this in perspective with a practical example, consider that an aircraft travelling at 400kt G/S will cover 2000nm in 5 hours. Let's assume that said ground speed was referenced to the "wrong" vertical and that the vehicle travelled over relatively rugged terrain resulting in an average vertical deflection in the direction of travel of 20 arc-seconds over the entire distance covered. The corrected groundspeed in this case would be 400*sin(20"), giving an ETA difference in our example of 2000/400 - 2000/[400*sin(20")] = 5*10^-4, or about 1.8 seconds, which is likely well below the measurement precision of most systems.

An interesting discussion, nevertheless.


[1] M. Grewal et al. "Global Positioning Systems, Inertial Navigation, and Integration", 2nd ed., Wiley Interscience, New Jersey, 2007, p31.

[2] R. Rogers, "Applied Mathematics in Integrated Navigation Systems", 3rd ed., AIAA, Blacksburg VA, 2007, p105.

[3] Grewal, p92-93

[4] A. Leick, "GPS Satellite Surveying", 2nd ed., Wiley Interscience, 1995, p487.

p51guy 26th Jun 2010 15:25

An example of " in reference to what" are the geosynchronous orbits of our Directv satellites that have a ground speed of zero but are moving over 20,000 mph through space.

italia458 26th Jun 2010 21:36


Well, it's not this one, as absolute speed does not exist. Speed is the scalar part of velocity which is a relative value. You cannot express speed without specifying speed relative to what.

People will try and say "Relative to a fixed point in space"- this ALSO does not exist.
Yes speed is relative to a point. For this, let's say speed is relative to a point on the surface of the earth. Makes things easy.

Now about the definition of groundspeed. It's just that!... the speed over the ground (surface of the earth). This is going to be true for navigation at any altitude.

Now looking at different altitudes, an aircraft traveling around the earth at 35,000' AGL will travel a further distance relative to the "same" point on the earth as an aircraft flying at 1000' AGL. Therefore, to travel over the surface of the earth at the same speed, the 35000' aircraft will have to travel faster! It will have a faster TAS. The TAS is how fast the aircraft is traveling through the air. Now to understand this concept we should state that we are disregarding density of the air, like TAS does.

In this little diagram I whipped up, the "groundspeed" is the same, but the TAS is not! The TAS would be the definition (2) that the OP wrote. The 1.0 and 1.42 are just rough estimates and nothing here was calculated, it was just made to clearly illustrate a point.

http://lh3.ggpht.com/_R8Od6auTHiw/TC...cUuql4A/11.png

PBL 28th Jun 2010 03:46


Originally Posted by LH2
[PBL:
perpendicular to the geodetic latitude]
Indeed it is according to your reference, which I've now looked up.

Just to clarify in case of misunderstanding: my answer addressed the question what is perpendicular to the reference ellipsoid, and I think the answer is standard, isn't it?

You point out that the accuracy of inertial navigation is dependent amongst other things upon the integrated effects of local gravity over the route of flight, not over some purely mathematical (engineering) construct such as the geodetic latitude. Indeed so, but I haven't worked in navigational avionics so have no idea of the magnitude of the difference between construct and reality, so in particular I can't comment on your numerical example.

Thanks for the references!

PBL

PBL 28th Jun 2010 04:28

Well, italia458, it's nice to have your opinion on what "groundspeed" means, but unfortunately it's wrong! You might like to look up some of the references which have been cited here.

PBL

xuejiesandi 28th Jun 2010 05:48

Yerrraaa Eishhhh!! :eek:
All this time I thought Ground Speed is the Speed indicated by GPS :}

BOAC 28th Jun 2010 10:33

Whatever 'high-fallutin' references you come up with, GROUND speed is ALWAYS speed relative to the ground. Nothing else, pure and simple. The OP sought a way of correcting for the circumference of an ellipsoid shell, which is relevant when dealing with travel around an object rather than travel along an object, and the cause of many conundrums and fixes in science and aviation, including such concepts as 'profile rate' which needs to be applied to an inertial attitude reference platform when moving around a body.

To derive either circumferential distance/speed travelled around the shell or convert that to distance relative to the surface of the shell is indeed a problem for the OP (and will also depend on whether the GPS has an altitude function and on which datum that is based), but what he/she refers to is not necessarily GROUNDspeed but some other speed - hence the confusion. As I said earlier, what is sought is to derive 'groundspeed' from 'Tangspeed' or some derivative of Orbital Velocity. (Never did get those postcards....). That can easily be derived by schoolboy mathematics. The moment you begin to 'orbit' just 1mm from the surface of a body the 'speed' link with that surface is broken.

We should think ourselves indeed lucky that our earth is not cuboid - then we'd have some fun..

Genghis the Engineer 28th Jun 2010 10:38


Originally Posted by BOAC (Post 5778365)

We should think ourselves indeed lucky that our earth is not cuboid - then we'd have some fun..

As opposed to the current situation of it being sat on the back of a giant turtle held up by four elephants?

G

PBL 28th Jun 2010 11:04


Originally Posted by BOAC
Whatever 'high-fallutin' references you come up with, GROUND speed is ALWAYS speed relative to the ground.

Yes, well, whatever that means precisely.

If high-falutin' is too rarified, let's take the low-falutin' route.

Consider. The earth is a perfect sphere, with circumference about 21,600nm (if we take a nm to be a second of arc, which is what it was supposed to be). You fly at your preferred altitude of FL370, let's say in a standard atmosphere. I fly at 1,000 ft because I like to bird-watch. Both of us without wind (we should be so lucky!). We are both pretty fast - say we go around the earth in a day. Just under 1,000 kts per (let's not worry about the logistics of refuelling, or about the number of windows shattered in my wake).

But you are about 6 nm higher than I (1 nm is about 6,000 ft) so you actually travel 21,637.7 nm, whereas I travel 21,601 nm. You do nearly 37 more air miles than I! So your TAS is about 1.5 kts faster than mine.

Those NASA "muppets" (as PA dubbed them) says your ground speed is 1.5 kts faster than mine. Myron Kayton says that also. What do you say?

PBL

BOAC 28th Jun 2010 11:06

Well, THAT is easy, requiring a simple application of the Loxodontal Tranformation.

italia458 28th Jun 2010 15:13

PBL you're really confused about this! No, I'm not wrong about ground speed is.

In your little example you just explained to me what True AIRspeed is! NOT, what ground speed is.

LH2 28th Jun 2010 16:03


Originally Posted by PBL (Post 5777918)
Just to clarify in case of misunderstanding: my answer addressed the question what is perpendicular to the reference ellipsoid

Ok, I think I might have misunderstood it to mean that the definition of ground speed as given by Kayton was relative to the ellipsoid (which is also correct, even if that's not what you wished to emphasise!).


You point out that the accuracy of inertial navigation is dependent amongst other things upon the integrated effects of local gravity over the route of flight, not over some purely mathematical (engineering) construct such as the geodetic latitude. Indeed so,
That's correct. Inertial navigation systems incorporate a gravity model in order to account for the differences in the mass distribution of the Earth (as well as centrifugal acceleration due to rotation), and barometric (or GNSS) height aiding to control vertical channel instability. And I have the feeling you already know this perfectly well :)

Note that the "integrated effects of local gravity" are modelled to within the precision of the provided gravity model, and unmodelled residual errors are accounted for and controlled during Kalman filtering. So, as you say, the source of residual errors is neither purely mechanical nor mathematical, but a combination of the design limitations in both.

BOAC,

GROUND speed is ALWAYS speed relative to the ground. Nothing else, pure and simple.
Not so "pure and simple" I'm afraid. You will find that what we call ground speed is in effect the first integral of acceleration in a given reference frame which may or may not be physically referenced[*]. What I think you might be thinking of would be the first derivative of change of position over time, which could be a perfectly valid definition (not without its own set of complications, mind you) but unfortunately is not what is used.

[*] Consider the following practical example: If you were measuring the ground speed of the Eiffel tower you will find that your answer will be either about zero, or ~3cm/yr, depending on whether you were using the ETRS89 or ITRF2008 reference systems (due to different velocity fields). In other words, even defining what "the ground" is presents a major problem, past a certain threshold of accuracy.


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