Separation Of 2 Planes W/ Different Turn Radiuses

[Aidan Monaghan]

2 planes flying side by side at 800fps turn left simultaneously.

Plane #1's turn radius is 35,000 feet.

Plane #2's turn radius is 55,000 feet.

What is the distance between the 2 planes after 2.5 seconds?

[MechEngr]

One solution is they collide almost immediately, with the tighter radius plane to the starboard of the larger radius plane.

Neglecting that the tighter radius will descend slightly faster if no trim or power changes are made ...

2.5 seconds * 800 fps = 2,000 ft.

Plane #1 will turn (2,000 ft/35,000 ft) radians
Relative to the start position it will go:
along the original direction 35,000 ft *sin(2/35) = 1998.9 feet
sideways from the original direction (35,000 ft - 35,000 ft * cos(2/35)) => 35,000 (1-cos(2/35)) ~= 57 feet

Plane #2 will turn (2000 ft/55,000 ft) radians
along the original direction 55,000 ft *sin(2/55) = 1999.56 ft
sideways from the original direction (55,000 ft - 55,000 ft * cos(2/55)) => 55,000 (1-cos(2/55)) ~= 36.36 ft

subtracting the second delta from the first:

along the original direction = 1998.9 ft - 1999.56 ft = -.648 ft, so plane #1 lags plane #2
sideways to the original direction = 57 ft - 36.36 ft = 20.8 ft so plane #1 moves sideways more than plane #2.

Neglecting the original offset, the change in the distances between the CGs = 20.78 ft and their paths will be diverging, instantaneously, by (2/35-2/55) radians or 1.19 degrees.

[Aidan Monaghan]

Excellent.

Thank you.

[ScepticalOptomist]

Did I venture into the homework thread? 🤣

[340drvr]

African, or European planes?

[FOUR REDS]

I thought the plural of radius is Radii?
🤓

[nonsense]

Originally Posted by ScepticalOptomist

Did I venture into the homework thread? 🤣

Calculate circumferences of circles of radii of 35,000 units and 55,000 units.
Calculate angles in radians represented by arcs of 2.5min x 800 units per minute.
Calculate change in position for each aircraft in Cartesian coordinates (easier trig and easier subtraction/addition).
Subtract vectors for change in relative positions.
Add result to original separation for new separation.

Or just plot the mess in something like AutoCAD.

[ShyTorque]

Originally Posted by nonsense

Calculate circumferences of circles of radii of 35,000 units and 55,000 units.
Calculate angles in radians represented by arcs of 2.5min x 800 units per minute.
Calculate change in position for each aircraft in Cartesian coordinates (easier trig and easier subtraction/addition).
Subtract vectors for change in relative positions.
Add result to original separation for new separation.

Or just plot the mess in something like AutoCAD.

I don’t recall ever having to make such a calculation in flight in 53 years of flying. Or on the ground, for that matter….

[400guy2]

Originally Posted by ShyTorque

I don’t recall ever having to make such a calculation in flight in 53 years of flying. Or on the ground, for that matter….

I assumed this was math homework, based on the units used: fps for speed and feet for turn radius. In a flying context I would expect perhaps knots and miles or rate-of-turn or bank angle.

For a tale of an oddball (for a pilot!) calculation, enjoy "Optimism Takes Work and is Rewarded" https://www.code7700.com/rule_15.htm

[nonsense]

Originally Posted by ShyTorque

I don’t recall ever having to make such a calculation in flight in 53 years of flying. Or on the ground, for that matter….

If it was something you might have to calculate in flight, there would either be a simple rule of thumb which got you a close enough answer, or an instrument...

It's slightly reminiscent of designing four bar linkages, except that the distances between adjacent nodes remain fixed and the rates of movement change.

(Yes, I'm an engineer, not a pilot)

[swh]

Originally Posted by MechEngr

One solution is they collide almost immediately, with the tighter radius plane to the starboard of the larger radius plane.

Neglecting that the tighter radius will descend slightly faster if no trim or power changes are made ...

2.5 seconds * 800 fps = 2,000 ft.

Plane #1 will turn (2,000 ft/35,000 ft) radians
Relative to the start position it will go:
along the original direction 35,000 ft *sin(2/35) = 1998.9 feet
sideways from the original direction (35,000 ft - 35,000 ft * cos(2/35)) => 35,000 (1-cos(2/35)) ~= 57 feet

Plane #2 will turn (2000 ft/55,000 ft) radians
along the original direction 55,000 ft *sin(2/55) = 1999.56 ft
sideways from the original direction (55,000 ft - 55,000 ft * cos(2/55)) => 55,000 (1-cos(2/55)) ~= 36.36 ft

subtracting the second delta from the first:

along the original direction = 1998.9 ft - 1999.56 ft = -.648 ft, so plane #1 lags plane #2
sideways to the original direction = 57 ft - 36.36 ft = 20.8 ft so plane #1 moves sideways more than plane #2.

Neglecting the original offset, the change in the distances between the CGs = 20.78 ft and their paths will be diverging, instantaneously, by (2/35-2/55) radians or 1.19 degrees.

I thought of this differently, if the two aircraft started of 10 ft apart the two graphs would be in Cartesian form
x^2+y^2=z^2
(a+19900)^2+b^2=c^2

where

z=35000
c=55000
t=2.5
s=800

In polar form

x1=z*cos[(s*t)/(2*pi*z)]
y1=z*sin[(s*t)/(2*pi*z)]

x2=c*cos[(s*t)/(2*pi*c)]-19900 {this is to offset the centre of turn 19900 ft left of the starting position, so the aircraft start 10 ft apart}
y2=c*sin[(s*t)/(2*pi*c)]

x1=34998
y1=318

x2=35099
y2=318

Distance between them
Sqrt((x2-x1)^2+(y2-y1)^2)=100 ft.

The curvature of each graph is 1/radius.

Maybe the question is trying to use the Rate of turn formula = speed*[360/(2*pi*r)]

In 2.5 seconds, the inner air fat would have changed direction 3.27 degrees, the outer aircraft 2.08 degrees.

[DaveReidUK]

These kind of calculations are much easier with a slide rule.

[MechEngr]

swh,

At 10 feet apart that would be (a+19990)^2+b^2=c^2

You started them 100 feet apart.

[MechEngr]

Originally Posted by DaveReidUK

These kind of calculations are much easier with a circular slide rule.

Much better than the linear ones.

[EXDAC]

I would suggest that in 2.5 seconds they will still be "side by side".

[DaveReidUK]

Originally Posted by MechEngr

Much better than the linear ones.

Excuse me for a moment, while I unstrap my Dalton Computor from my knee ...

[Mogwi]

I now know why I decided not to do ETPS!!

Mog

[swh]

Originally Posted by MechEngr

swh,

At 10 feet apart that would be (a+19990)^2+b^2=c^2

You started them 100 feet apart.

Apologies for being practical.

a+19990 would have them with overlapping fuselages, which would be normally classified as a collision.

Aircraft normally have these things sticking out from the sides, forget the technical term for them.

[swh]

Originally Posted by EXDAC

I would suggest that in 2.5 seconds they will still be "side by side".

Pretty well as much, just accelerating at slightly different rates.

[ShyTorque]

This is why lookout is so important……