The Parachute Problem Ethan Retherford College of the Redwoods May - PowerPoint PPT Presentation

The Parachute Problem Ethan Retherford College of the Redwoods May 15, 2015

Introduction The “parachute problem” is commonly seen in the beginning of an introductory differential equation course and involves the use of Second or first-order ODEs (ordinary differential equations). The traditional problem to be modeled and analyzed involves a parachutist jumping from an aircraft at a specific height, x 0 , and falling towards Earth under the influence of gravity.

The Basic Model We start with analyzing the parachutist’s free-fall from, x 0 .This is done by applying Newton’s Second Law of Motion, Σ F = ma .

Free-fall and parachute deployed stages of the jump Parachute Deployed Free − fall F d = − k 1 v F d = − k 2 v x 0 x 0 F g = − mg F g = − mg Assume the parachute deploys instantaneously at t d

Deriving the differential equations � 0 ≤ t < t d k 1 , k = k 2 , t ≥ t d Σ F = ma F g + F d = ma − mg − kv = ma ma + kv = − mg

Deriving the differential equations First-order system for position and velocity x ′ = v , x (0) = x 0 v ′ = − g − k v (0) = 0 . mv ,

Free-fall During free-fall k = k 1 for t < t d v ′ + k 1 mv = − g , v (0) = 0 Solving for v(t), we get the velocity while in free-fall v ( t ) = mg � � e − k 1 t / m − 1 k 1

Free-fall Now, to obtain the position x ( t ) we can sub v ( t ) into the position equation x ′ = v x ′ = mg � � e − k 1 t / m − 1 x (0) = x 0 , k 1 Solving t − m 2 g x ( t ) = x 0 − mg � � e − k 1 t / m − 1 k 2 k 1 1

Parachute is deployed v ′ + k 2 d ) = v ( t + m v = − g , v ( t − d ) Obtaining the general solution and using the initial condition mg = − mg � � e − k 1 t d / m − 1 + Ce − k 2 t d / m k 1 k 2 C = mg e k 2 t d / m + mg e k 2 t d / m � � e − k 1 t d / m − 1 k 2 k 1 Substituting C back into v ( t ), t ≥ t d v ( t ) = mg + mg e − k 2 ( t − t d ) / m � � � � e − k 1 t d / m − 1 e − k 2 ( t − t d ) / m − 1 k 1 k d

Parachute is deployed Use Same method for Position �� m 2 g + m 2 g x ( t ) = x 0 − mg �� � � e − k 1 t d / m − 1 e − k 2 ( t − t d ) / m − 1 t d − k 2 k 1 k 1 k 2 1 − m 2 g + mgt d − mgt � � e − k 2 ( t − t d ) / m − 1 k 2 k 2 k 2 2

Position and Velocity for Entire Jump k 1 t − m 2 g � �  x 0 − mg e − k 1 t / m − 1 , t < t d k 2   1      x ( t ) = �� �� � � m 2 g 1 + m 2 g e − k 1 t d / m − 1 e − k 2 ( t − t d ) / m − 1 x 0 − mg k 1 t d − k 2 k 1 k 2     � � − m 2 g  e − k 2 ( t − t d ) / m − 1 + mgt d − mgt k 2 ,  t ≥ t d  k 2 k 2 2  � � e − k 1 t / m − 1 mg , t < t d  k 1   v ( t ) = k 1 e − k 2 ( t − t d ) / m � � � �  e − k 1 t d / m − 1 e − k 2 ( t − t d ) / m − 1 mg + mg  , t ≥ t d  k 2

Acceleration and Terminal Velocity Acceleration a = v ′ = − g − k mv Terminal velocity, when a = 0 v t = − mg k

Analysis of Basic Model k 1 = 14 kg / s , k 2 = 160 kg / s 10 Position ( x / 200 ) Velocity ( v / 10 ) Acceleration ( a / g ) 5 time ( s ) 10 20 30 40 - 5

Real-World Considerations Goals to meet The parachute deployment is not instantaneous and a certain amount of force is experienced during the process. The snatch force is the acceleration at the first instant when the assembly reaches full extension; the opening shock , or jerk , is the shock produced while the parachute deploys. The snatch force felt when the lines are fully elongated is a heavy, but smooth, tug that is not particularly uncomfortable. While this force depends on the weight of the jumper, it should not exceed 500 lbs ( ≈ 3 Gs for a 165 lb (75 kg ) person ). The landing velocity should be no worse than a free-fall from a 5 ′ (1 . 52 m ) wall–between 4 . 6 and 5 . 2 m / s .

Deployment stage Added Free − fall F d = − k 1 v x 0 F g = − mg Deployment Parachute Deployed F d = − k d v F d = − k 2 v x 0 x 0 F g = − mg F g = − mg

Connecting the constant states in a continuous manner  0 ≤ t < t d k 1 ,   k ( t ) = t d ≤ t < t d + τ d k d ,  t ≥ t d + τ d . k 2 , 

Linear Interpolation k ( t ) ( t d + τ d , k 2 ) 160 ( t d , k 1 ) 14 t 20 23.5 k d ( t ) = k 1 + k 2 − k 1 ( t − t d ) τ

Plots with Linear Interpolation 2 time ( s ) 20 40 60 80 100 - 2 Position ( x / 1000 ) - 4 Velocity ( v / 10 ) Acceleration ( a / g ) - 6

Jerk j ( t ) := da dt = − k ′ ( t ) m v ( t ) − k ( t ) m a ( t )

Linear Interpolation with Jerk 4 2 time ( s ) 20 22 24 26 - 2 Position ( x / 1000 ) Velocity ( v / 10 ) - 4 Acceleration ( a / g ) Jerk ( j / g ) - 6

Cubic Interpolation k ( t ) ( t d + τ d , k 2 ) 160 ( t d , k 1 ) 14 t 20 23.5 � 3 � 2 � t − t d � t − t d k d ( t ) = ( − 2 k 2 + 2 k 1 ) + (3 k 2 − 3 k 1 ) + k 1 . τ d τ d

Plots with Cubic Interpolation 4 2 time ( s ) 20 22 24 26 - 2 Position ( x / 1000 ) Velocity ( v / 10 ) - 4 Acceleration ( a / g ) Jerk ( j / g ) - 6