Distance Between Two Planes With Different Turn Radiuses

[Badmachine]

Is there a method for calculating the distance between two planes turning at a common point, traveling at the same speed and simultaneously performing turns with two different turn radiuses?

For example:

Plane A begins a turn with a 10,000 foot turn radius at waypoint "X".
Plane B begins a turn with a 20,000 foot turn radius at waypoint "X".

Each plane has a different Rate of Turn (in degrees per second) after 5 seconds.

What is the distance between the two planes after 5 seconds?

[glad rag]

Dunno about your equation but this F16/F4 comparison image is telling.

http://www.tonyrogers.com/news/images/0104viper5.jpg

[Microburst2002]

You need to know turn radius and speed (or derive speed from turn radius and bank angle).

An then, the maths can be complex, because the geometry is complex, as seen in that wonderful picture.

[rudderrudderrat]

Hi Badmachine,

Suppose you are both flying with an airspeed of V ft per sec.
The distance travelled around each arc in t seconds = Vt feet.
Number of Radians travelled around each respective arc = Vt/R.
Therefore heading change = Vt/R * 180/pi.

For a small time of 5 secs assume the heading changes are also small.

So plane A's heading change = Vt/10,000 * 180/pi. ....(AH)
Plane B's heading change = Vt/20,000 * 180/pi. ....(BH)
So the change in heading of aircraft B relative to aircraft A = (BH)-(AH)

i.e. If you are in aircraft A and in 5 secs you change your heading by 15 degrees, then aircraft B will have only changed heading by 7.5 degs

Since you both start on the same heading then the separation after 5 secs = about half of 5*V* sin{(BH)-(AH)}

For small angles you could use 1 in 60 rule, so separation after 5 secs = about half of 5*V*{(BH)-(AH)}/60

Approximately.

[Capt Pit Bull]

Calculation of an exact answer would be doable by a good A level maths or Physics student;

Method would be something like this:

Envisage both aircraft heading north at the waypoint. Set up an X Y coordinate system about the waypoint. Perhaps: X is aircraft distance east of the start point. Y is aircraft distance North of the start point.

You then some equations that tell you where each aircraft is in space at time T. This is not that tricky if you view it as circular motion and invoke basic trigonometry.

Proportion of turn completed at time T = Speed X Time / circumference of turn

(convert TAS in knots into feet per second)
(circumference = 2 x pi x Turn Radius (10 or 20K depending on the aircraft)

The multiply by 360 to find out how many degrees around the turn the aircraft has gone, which since we started on North is actually the heading. Call it HDGa for aircraft A etc

You can then work out X and Y positions in space, bearing in mind the centre of the turn is 10 or 20K to the east of the waypoint: The Y distances may be negative, don't worry about it just apply + - rules rigorously.

Aircraft A:
Xa = 10,000 - (10,000 * Cos HDGa)
Ya = 10,000 * Sin HDGa

Aircraft B:
Xb = 20,000 - (20,000 * Cos HDGb)
Yb = 20,000 * Sin HDGb

Then work out the difference between them, first in each dimension

Xseperation = Xa - Xb
Yseperation = Ya - Yb

Then two dimensionally by using pythagoras

Seperation = square root of ((Xseperation^2) + (Yseperation^2))

For two aircraft at 250 kts, one with a 10 K radius of tuen the other 20K, after 5 seconds the position differnce would be 111.2 feet.
(of course that's a silly level of accuracy).

(I think thats correct, haven't check it rigorously)

pb

[rudderrudderrat]

Hi Pitbull,

At 250 kts (420 ft per sec), in 5 secs I reckon plane A changes heading by 12 degs, plane B by 6 degs; so difference = 6 degs divergence.

My sums give, 1/2 of 5*420 * Sin 6 = 110.3 feet Approximately.
So I agree with your maths.

rrr

[Microburst2002]

Well, maybe you don't need a Nobel prize to do that, but since the airplanes follow differet paths, the distance calculation between both airplanes involves not only distance and time, but also trayectory.

Could it be useful to use polar coordinates, instead of cartesian ones? With the origin in the point where both turns start?

The problem is I don't remember how to translate cartesian into polar nor the opposite.

[Mark 1]

Pretty straightforward really.

After time T, each aircraft has gone through an arc of a circle of angle=(V*T/R) radians (assuming consistent units)
V=velocity
T=time
R=radius of turn

In cartesian coordinates, the X-distance travelled = R*Sin(angle)

and the Y distance = R*(1-Cos(angle))

The distance between the two is sqrt((x1-x2)^2+(y1-y2)^2) corrected for their initial position and separation.

About 5 minutes in Excel should give you the answer

[rudderrudderrat]

Hi MB2002,

Could it be useful to use polar coordinates, instead of cartesian ones? With the origin in the point where both turns start?

Thanks - Good lateral thinking. Relatively speaking we could use either aircraft as a frame of reference. At the point where both turns start, initially the aircraft are accelerating towards their respective centre of arc along a common line. (for a short period of time anyway)
acceleration (a) = V^2 * 1/R.
Plane A's a = V^2 * 1/10,000; Plane B's a = V^2 * 1/20,000.
Acceleration of A relative to B (a')= 2V^2 * 1/20,000 - V^2 * 1/20,000 = V^2 * 1/20,000.

Since both planes start at the same point and speed, their relative initial velocity = 0.
Therefore separation distance = 1/2 a' t^2.
Separation after 5 secs = 1/2 * V^2 * 1/20,000 * 25.

Dist = 111.3 feet Approximately.

V = 421.9 ft per sec (250 kts)

[3 Point]

What degree of accuracy do you need? Could just make a scale drawing on graph paper and measure the distance - no maths no headache!!

KISS!!

Of course I understand that this has limitations but, might suit your needs.

Happy landings.

3 Point

[Microburst2002]

Thanks Rudderrudderrat

Lateral thinking sounds better that sciolism :)

But it's 20 years since the last time I saw such coordinates. I remember they were used in problems with graphs similar to those of the F4 and F16. That's why it occurred to me.

Anyway...

What is the purpose of finding this distance? ATC separation?

[rudderrudderrat]

Hi MB2002,

Badmachine has asked at least 2 similar questions - but seems to have disappeared.

It shows that you can think in terms of TAS and air mass, or ground speed and earth, or simply relative headings or relative speeds and get similar answers.

[Badmachine]

Thanks Capt. Pit Bull. :ok:

Your helpful equations seem to make sense. Was surprised at how little distance is generated by two high speed aircraft (traveling at 799f/s) and turning with a angle of bank (AoB) difference of 5 degrees between the two (assuming my calculations are correct).

By the way ...

Is there a way to calculate the separation distance, if each aircraft begins the type of turns discussed, but at different times? (for example, a turning delay of 5 seconds between each aircraft)

Traveling nearly a mile, the separation distance between two high speed aircraft is only 41 feet.

T = speed*time/circumference of turn.

Aircraft A: 20 degrees AoB: (r: 54,795); [(799)5/344,287]*360 = 4.17

Aircraft B: 25 degrees AoB: (r: 42,771); [(799)5/268,738]*360 = 5.35

Aircraft A:

Xa (20 degrees AoB)= 54,795 - (54,795 * Cos (4.17)) = 145
Ya (20 degrees AoB) = 54,795 * Sin 4.17 = 3985

Aircraft B:

Xb (25 degrees AoB)= 42,771 - (42,771 * Cos (5.35)) = 186
Yb (25 degrees AoB) = 42,771 * Sin 5.35 = 3988

Xseperation = Xa - Xb
Yseperation = Ya - Yb

Xseperation = 145 - 186 = -41
Yseperation = 3,985 - 3,988 = -3

Seperation = square root of ((Xseperation^2) + (Yseperation^2))

(-41)^2 + (-3)^2 = 1690^1/2 = 41

[rudderrudderrat]

Hi Badmachine,

1) The heading difference between the aircraft after 5 secs is only 1.18 degs (5.35 – 4.17).
For small travel around arc: Separation ~ distance travelled * sin average heading difference.

Separation ~ 5 * 799 * sin (1.18/2) = 41.1 feet approximately.

[Badmachine]

Obtained some unexpected results. Logically, as turn circumferences shrink, the number of degrees traveled around the circumferences at a given speed (799f/s) increases. The distances between each turn also increase with bank angle increases yet the distances are suprisingly small.

Turn Separation Between 15 and 20 Degrees of Bank

Aircraft Turn A: 15 degrees: (r: 211,601); [(799)8/1,329,528]*360 = 1.73

Aircraft Turn B: 20 degrees: (r: 155,783); [(799)8/978,813]*360 = 2.35

Aircraft Turn A:

Xa (15 deg.) = 211,601 - [211,601* Cos (1.73)] = 96.44
Ya (15 deg.) = 211,601* Sin 1.73 = 6,388.15

Aircraft Turn B:

Xb (20 deg.) = 155,783 - [155,783* Cos (2.35)] = 131
Yb (20 deg.) = 155,783* Sin 2.35 = 6,387.68

X-separation = Xa - Xb
Y-separation = Ya - Yb

X-separation = 96.44 - 131 = -34.56
Y-separation = 6,388.15 - 6,387.68 = .47

Separation = square root of [(X-separation^2) + (Y-separation^2)]
(-34.56)^2 + (.47)^2 = 1,194.61^1/2 = 34.56 feet

Turn Separation Between 20 and 25 Degrees of Bank

Aircraft Turn A: 20 degrees: (r: 155,783); [(799)8/978,813]*360 = 2.35

Aircraft Turn B: 25 degrees: (r: 121,600); [(799)8/764,035]*360 = 3.01

Aircraft Turn A:

Xa (20 deg.) = 155,783 - [155,783* Cos (2.35)] = 131
Ya (20 deg.) = 155,783* Sin 2.35 = 6,387.68

Aircraft Turn B:

Xb (25 deg.) = 121,600 - [121,600* Cos (3.01)] = 167.76
Yb (25 deg.) = 121,600* Sin 3.01 = 6,385.24

X-separation = Xa - Xb
Y-separation = Ya - Yb

X-separation = 131 – 167.76 = -36.76
Y-separation = 6,387.68 - 6,385.24 = 2.44

Separation = square root of [(X-separation^2) + (Y-separation^2)]
(-36.76)^2 + (2.44)^2 = 1,357.25^1/2 = 36.84 feet

Turn Separation Between 25 and 30 Degrees of Bank

Aircraft Turn A: 25 degrees: (r: 121,600); [(799)8/764,035]*360 = 3.01
Aircraft Turn B: 30 degrees: (r: 98,215); [(799)8/617,103]*360 = 3.72

Aircraft Turn A:

Xa (25 deg.) = 121,600 - [121,600* Cos (3.01)] = 167.76
Ya (25 deg.) = 121,600* Sin 3.01 = 6,385.24

Aircraft Turn B:

Xb (30 deg.) = 98,215 - [98,215* Cos (3.72)] = 206.93
Yb (30 deg.) = 98,215* Sin 3.72 = 6,372.25

X-separation = Xa - Xb
Y-separation = Ya - Yb

X-separation = 167.76 – 206.93 = -39.17
Y-separation = 6,385.24 - 6,372.25 = 12.99

Separation = square root of [(X-separation^2) + (Y-separation^2)]
(-39.17)^2 + (12.99)^2 = 1,357.25^1/2 = 41.26 feet