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Which of the following vectors is refered to as the Ground Speed?

The green one I expect is TAS, the blue one would be Vertical Rate

enter image description here

Note that for simplicity I did not draw them as geodetic curves, but the black one should be shorter than the gray one because of the earth's curvature.

Does the Ground Speed represent the horizontal speed relative to surface at the sea level?

Ex: Helicopter climbing verticaly will have GS=0?

Does the Earth's curvature affect the GS? As the length of the gray vector will be longer the higher the altitude is. Does that mean a plane flying at 1km will have higher GS than a plane flying the same total speed at 10km?

edit:

I'm simulating aircraft movement along a straight trajectory by constant speed. To simplify, the input is start coordinates & altitude and finish coords & alt.

That means the plane can ascend or descent during simulation. I know the total distance between A & B and distance at sea level. So first thing I tought of was:

GS = distanceAtSL / (totalDistance / speed)

In relation to image:

I know the length of Green line and I can calculate how long it will take the plane to reach the end. The time to finish is constant for each line & I know the length of the black line so using formula speed = distance / time I should be able to get the "Ground Speed".

Correct me if I'm wrong.

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  • $\begingroup$ The earth's radius is about 6370 km. This means a difference of about 0.1% in distance over ground between the plane at 1km and the plane at 10km. $\endgroup$
    – fooot
    May 23, 2017 at 21:17
  • $\begingroup$ Well a 2100 km/h at 10km flat is 2096,704227 km/h projected at sea level, according to Geodesy library i'm using. Point is what value would be refered to as the Ground Speed. I forgot its at 50°N 1°E as earth is not sphere $\endgroup$
    – Dave_cz
    May 23, 2017 at 21:29
  • $\begingroup$ @mins ground level = Sea level. That's what I need. I do not account for wind or pressure, the input speed is simply the speed the plane moves between point A & B including the altitude difference, so GS is simply the projection to ground where the alt difference is not accounted for. Like looking from the orbit right above the plane (Top Down View). $\endgroup$
    – Dave_cz
    May 23, 2017 at 22:48
  • $\begingroup$ 2100 km/h? what planes are you considering? $\endgroup$
    – Federico
    May 24, 2017 at 8:17
  • $\begingroup$ When you are considering a) absolute velocity vector relative to sea level surface, versus b) component of above that is actually parallel to sea level surface, versus c) --? -- component of above that is actually parallel to instantaneous slope of earth below??- I'm not sure that there really is a universally accepted definition. $\endgroup$
    – quiet flyer
    Oct 1, 2022 at 23:55

3 Answers 3

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enter image description here
(Source)

Ground speed

Ground speed equipment on-board aircraft measure the speed against the ground. Except in the case shown above where distance information is derived from a ground station (DME station).

Curvature

Ground speed is not derived from true airspeed. When using GPS, or an inertial navigation system to measure the ground speed, it will be the speed across the ground. It is the rate of change of the latitude and longitude, so you don't need to include the curvature.

Once an aircraft (or crew) know the ground speed and TAS, the wind information is then known (how fast the mass of air the aircraft is in is moving).

True airspeed

True airspeed is the speed inside a mass of air. In a zoom climb in an F-15, TAS will be pointing almost the same direction as the [upward] nose of the F-15.

If the mass of air is moving due to head- or tail-wind, then this wind speed is subtracted from (or added to) the horizontal TAS to get the ground speed. (But again, this is not how ground speed is calculated, because there is no direct measurement for the horizontal TAS.)

Vertical speed is how fast the plane is climbing/descending. It's measured by sensing the rate of change of the static air pressure.

RE: edit

I'm simulating aircraft movement along a straight trajectory by constant speed.

This is not how airplanes fly in real life. But for the sake of your model, and if you don't have wind simulation, then the known green distance and the time it takes to move from A to B will do just fine to compute the ground speed.

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  • 1
    $\begingroup$ So it is the Black vector in my image, as the Earth's curvature might be neglectable, the climb/descent is not. I.e. Rocket ascending at 100 m/s verticaly will have TAS=100 m/s & GS=0 $\endgroup$
    – Dave_cz
    May 23, 2017 at 21:40
  • $\begingroup$ I'm simulating aircraft movement along a straight trajectory by constant speed. To simplify, the input is start coordinates & altitude and finish coords & alt. That means the plane can ascend or descent during simulation. I know the total distance between A & B and distance at sea level. So first thing I tought of was GS = distanceAtSL / (totalDistance / speed) $\endgroup$
    – Dave_cz
    May 23, 2017 at 22:11
  • $\begingroup$ FlightGear is an open source simulation program and includes the code for flight dynamics and earth navigation. $\endgroup$
    – Koyovis
    May 23, 2017 at 22:31
  • $\begingroup$ @mins Straight line is a bit overstatement, It's actualy a geodetic curve. It's straight line from top-down view. $\endgroup$
    – Dave_cz
    May 23, 2017 at 22:56
  • $\begingroup$ @Dave_cz - looking top-down, measure the rate of change of the lat/long. $\endgroup$
    – user14897
    May 23, 2017 at 22:57
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Your case is:

  • An aircraft flies from A to B which 3D coordinates are known
  • The 3D distance flown, AB, is known
  • The flight time is known, or can be calculted from the aircraft acutal speed
  • Wind, temperature, pressure, or any other actual atmosphere element are not taken into account

Your question is:

  • What is the ground speed?

enter image description here

Short answer: 1/ Yes, the ground speed is the speed on the aircraft trajectory projected at sea level. 2/ Sea level is not an ellipsoid of revolution, but a surface close to the geoid of reference. 3/ Using the geoid geometry is however not advisable. It's complex, requires an appropriate GIS library using a geoid (gravitational) model, needs more time for calculation. Sticking to the ellipsoid is usually a good alternative.

When simplified, your problem is not an aviation problem, only a geometry one.

The ground speed is the speed of the orthogonal projection of the aircraft on the brown arc from A' to B'. We can only look for the mean speed because this ground speed is not constant if the green curve flown is not exactly an homothetic transformation of the brown curve, flown at constant speed.

As you said, you need two parameters: Distance and time and you have already one: Time to fly from A to B.

To compute length A'B' of the brown arc, just project A and B orthogonally onto the Earth surface using your 3D library functions and compute the curved distance.

  • By Earth surface, I mean what your 3D library allows, e.g. projection on the ellipsoid, or if required and possible, projection on the geoid.

Having now the length of the brown arc A'B' and the time flown, the ground speed is the ratio of the two.

Note: The circumference of a circle around Earth surface at kilometer above it is only 6.28 km (2π) longer than at Earth level, and each km of height adds only this same quantity. So at 10 km, the difference with sea level is (only) 63 km.

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    $\begingroup$ The library I'm using uses WGS84 elipsoid model. I can ofc. use the lat/long deltas each simulation step to derive the Ground Speed. It should technically give same results as my method as I use the same library to update plane coordinates each step based on speed, bearing & start point. $\endgroup$
    – Dave_cz
    May 24, 2017 at 15:56
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    $\begingroup$ Will mark this as an answer for now. $\endgroup$
    – Dave_cz
    May 24, 2017 at 15:56
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Both the black vector and gray vector. Like land surveying, ground distance is always level. ( "level" curves with the globe, so right angle to any plumb line.) When land surveyors encounter a mountain, angular slope distance is not used, it is effectively stair steps of pure elevation and pure level distance. Keep in mind that the earth is as smooth as a billard ball(57mm), to that scale Mt Everest would be 0.04mm(1.6/1000 inch) and the ovalness of 22km(69,000ft) radius would be 0.1mm(4/1000 inch). Making the error between black and gray at FL350, ~0.084%

More accurately stated, ground speed is angular speed around the earth, approximately converted to an equivalent linear speed at sea level.

The original source of a nautical mile is one minute of a degree of angle around the globe. (So all planets are 21,600 nautical miles in circumference.) And this is why nautical miles are much easier to use for traditional long-distance navigation using trigonometry. Where an exact position is more important than an exact linearized distance, especially due to the influence of currents/wind on the effective travel distance.

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  • $\begingroup$ When navigating, you must decide what is more important; position, ground distance, air/water distance, or course angle.(air/water distance between two fixed points is altered by currents.) IOW Do you care that you traveled 5005miles instead of 4095 miles over the ocean, or do you care more that your current position before descending through the clouds is within one mile of the small atoll with a landing strip? $\endgroup$
    – Max Power
    Oct 1, 2022 at 23:14

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