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.