In: Physics
Complete a crash force analysis for the problem below:
A regional jet aircraft crashes on a rolling hill declining at 5o relative to the line of flight. It was determined the flight path angle was 17 degrees the time of the crash. Analysis of the airspeed indicator showed the airspeed was 140 knots. Initial impact occurred with the aircraft nose 3 degrees above the horizon. The impact scar was measured to be 1 foot at its deepest and 273 feet long. The fuselage is crushed 12 inches vertically and 60 inches horizontally.
Method of Solution:
1. Sketch known quantities
2. Convert known quantities to standard dimensional units
a) distances in feet
b) velocities in feet/second
3. Determine magnitude of velocity component
a) parallel to impact surface
b) perpendicular to impact surface
4. Determine stopping distance
a) parallel to impact surface
b) perpendicular to impact surface
5. Determine acceleration components
a) select the crash pulse
b) parallel to impact surface
c) perpendicular to impact surface
6. Determine resultant acceleration magnitude and direction, with respect to impact surface
7. Determine direction of resultant acceleration with respect to aircraft axes
8. Estimate survivability potential
for the givcen crash force analysis
decline, alpha = 5 deg
flight path angle, gamma = 17 deg
v = 140 knots
pitch angle = beta = 3 deg
now
a. the floowing is the sketch of the described scene
b. a. horizontal length of scar, l = 273 ft
vertical, d = 1 ft
fuselage crushed vertically, y = 12 in = 1 ft
horizontally, x = 60 in = 5 ft
b. v = 140 knots = 236.293 ft/s
c. a. parallel to impact surface, vp = v*cos(17 + 5) = 219.08705 ft/s
b. perpendicular velocity of impact = v| = v*sin(23) = 92.9969 ft/s
d. a. parallel to mipact surface, stoppinf distance = l + x = 278 ft
b. perpendicular to impact surface, stopping distance = d + y = 2 ft
e. acceleration parallel to the impact surface = vp^2/2*(l + x) = 86.32938 ft/s^2
perpendicular to crash surface = v|^2/2(d + y) = 2161.78502 ft/s^2