The Parachute Problem Ronald Phoebus and Cole Reilly College of the - PowerPoint PPT Presentation

1/23 The Parachute Problem Ronald Phoebus and Cole Reilly College of the Redwoods Differential Equations � Spring 2004, Final Project � � � � � �

Objectives • To present a basic model for the Parachute Problem as presented in 2/23 many Differential Equation textbooks. • Solve and explain problems with this basic model. • To introduce an improved model that more accurately depicts a real life sky-diving jump. � � � � � � �

Basic Model Simple application of Newtonian Mechanics. 3/23 F = ma F = F g + F d = ma Where, Fg = − mg � � F d = − kv � � � � �

If deployment occurs at t 0 then, � k 1 , 0 ≤ t < t 0 4/23 k = k 2 , t ≥ t 0 The problem can be expressed as either a second-order differential equation (ODE) for position or as a first-order system of ODE’s for the velocity and position. During the first interval the velocity satisfies the initial value problem mdv dt = − mg − k 1 v, v (0) = 0 . (1) � Solutions for velocity and position are well known. � v ( t ) = mg ( e − k 1 t/m − 1) , � 0 ≤ t < t 0 k 1 � t − m 2 g y ( t ) = y 0 − mg ( e − k 1 t/m − 1) , � 0 ≤ t < t 0 k 2 k 1 1 � �

During the second interval the velocity satisfies the I.V.P. mdv dt = − mg − k 2 v, 5/23 with the initial condition v ( t 0 ) = mg ( e − k 1 t 0 /m − 1) . k 1 Therefore, the equation for velocity is v ( t ) = mg ( e − k 2 ( t − t 0 ) /m − 1)+ mg ( e − k 2 ( t − t 0 ) /m )( e − k 1 t 0 /m − 1) , t ≥ t 0 . k 2 k 1 � Using a numerical solver we can plot these equations over time. � � � � � �

Results Using MatLab we obtain the following graph, 6/23 Velocity and acceleration for first 30 seconds of the jump. 20 0 −20 −40 −60 � −80 velocity acceleration � −100 0 5 10 15 20 25 30 t−axis � � � � �

Skydiving Physics • Based on principles of Fluid Mechanics 7/23 • Relationship between viscous and inertial forces is described by the Reynolds Number. • Re = ρdv/µ • Ranges from 0 for a dust particle or larger objects in less viscous fluids to more than 10 8 for submarine in water. • When Re < 1 viscous forces dominate and the drag is linear in � velocity. � • When Re > 10 3 inertial forces dominate and the drag is quadratic � in velocity. � • Therefore, Reynolds Numbers are essential to consider in the devel- � opment of the model. � �

Calculation of Reynolds Number • Assuming ρ an µ are constant at altitudes appropriate for skydiving. 8/23 Where ρ = 1 kg/m 3 and µ =1.5x 10 − 5 Kg/m/s. • Terminal velocity is a reasonable choice characteristic velocity when determining Reynolds Number. • During free-fall v ≈ 45 m/s ≈ 100 mph. • With chute deployed v ≈ 5 . 3 m/s • Thus, at each stage of the jump � Re > 10 6 � � • Therefore, drag is proportional to the square of velocity for our � model. � � �

Coefficient of Drag • C d is determined by the shape of the body and is found experimen- 9/23 tally or through complex computational analysis. F d = 1 2( C d Aρv 2 ) • Both the skydiver and their equipment generate separate drag forces, therefore, d = 1 d A b + C e � F d = F b d + F e 2 ρ ( C b d A e ) v 2 � � � � � �

Shape Reynolds Number C d Re > 10 3 Hemispherical Shell 1.33 Re > 10 3 10/23 Flat Strip 1.95 Re > 5 x 10 5 Cylinder ≈ 0 . 35 • During free-fall the skydiver is in a horizontal position and can be represented by a flat strip with A ≈ 0 . 5 m 2 . • After parachute deployment the skydiver is in a vertical position and can be represented by a long cylinder with A ≈ 0 . 1 m 2 . � • The canopy can be represented by a hemispherical shell where A ≈ � 43 . 8 m 2 . � � � � �

Time-Line of Jump 11/23 0 t 0 t 1 Free Fall Lines Extend � � Jump Begins Ripcord Pulled Snatch � � � � �

12/23 a 1 A e a 1 A e 1 , 2 2 , 3 t 1 t 2 t 3 Parachute Inflates Parachute Overinflates Final Desent Snatch Parachute Fully Inflates Parachute Inflation � Reaches Steady-State � � � � � �

Area of Skydiver’s Body 13/23 a 1 b 0 b 1 h l 43 . 8 m 2 0 . 5 m 2 0 . 1 m 2 1 . 78 m 8 . 96 m m t 0 t 1 t 2 t 3 97 . 2 kg 10 s 10 . 5 s 11 . 5 s 13 . 2  b 0 , t ≤ t 0     b 0 , t 0 < t ≤ t 1     � A b ( t ) = b 1 , t 1 < t ≤ t 2 �  b 1 , t 2 < t ≤ t 3    �   b 1 , t ≥ t 3   � � � �

Coefficient of Drag for Skydiver 14/23 a 1 b 0 b 1 h l 43 . 8 m 2 0 . 5 m 2 0 . 1 m 2 1 . 78 m 8 . 96 m m t 0 t 1 t 2 t 3 97 . 2 kg 10 s 10 . 5 s 11 . 5 s 13 . 2  1 . 95 , t ≤ t 0     1 . 95 , t 0 < t ≤ t 1     � b ( t ) = C d 0 . 35 h, t 1 < t ≤ t 2 �  0 . 35 h, t 2 < t ≤ t 3    �   0 . 35 h, t ≥ t 3   � � � �

Area of the Equipment 15/23 a 1 b 0 b 1 h l 43 . 8 m 2 0 . 5 m 2 0 . 1 m 2 1 . 78 m 8 . 96 m m t 0 t 1 t 2 t 3 97 . 2 kg 10 s 10 . 5 s 11 . 5 s 13 . 2  0 . 0 , t ≤ t 0     b 1 , t 0 < t ≤ t 1     � A e ( t ) = A e 1 , 2 ( t ) , t 1 < t ≤ t 2 �  A e 2 , 3 ( t ) , t 2 < t ≤ t 3    �   a 1 , t ≥ t 3   � � � �

Coefficient of Drag for Equipment 16/23 a 1 b 0 b 1 h l 43 . 8 m 2 0 . 5 m 2 0 . 1 m 2 1 . 78 m 8 . 96 m m t 0 t 1 t 2 t 3 97 . 2 kg 10 s 10 . 5 s 11 . 5 s 13 . 2  0 . 0 , t ≤ t 0    0 . 35 l t − t 0  t 1 − t 0 , t 0 < t ≤ t 1     � e ( t ) = C d 1 . 33 , t 1 < t ≤ t 2 �  1 . 33 , t 2 < t ≤ t 3    �    1 . 33 , t ≥ t 3  � � � �

Improved Model 17/23 mdv dt = − mg + kv 2 , v (0) = 0 where d A b + C e k = 1 / 2 ρ ( C b d A e ) . Thus,  1 . 95 b 0 , t ≤ t 0   �  1 . 95 b 0 + 0 . 35 b 1 l t − t 0  t 1 − t 0 , t 0 < t ≤ t 1    k = 1 �  0 . 35 b 1 h + 1 . 33 A e 1 , 2 ( t ) , 2 ρ t 1 < t ≤ t 2 �  0 . 35 b 1 h + 1 . 33 A e 2 , 3 ( t ) , t 2 < t ≤ t 3   �    0 . 35 b 1 h + 1 . 33 a 1 , t ≥ t 3  �  � �

Continuity at End Points 18/23 • At t 0 , � t − t 0 � 1 . 95 b 0 = 1 . 95 b 0 + 0 . 35 b 1 l t 1 − t 0 Substituting t 0 for t ; � t 0 − t 0 � 1 . 95 b 0 = 1 . 95 b 0 + 0 . 35 b 1 l � t 1 − t 0 � 1 . 95 b 0 = 1 . 95 b 0 � � � � �

• At t 1 , 1 . 95 b 0 + 0 . 35 b 1 l = 0 . 35 b 1 h + 1 . 33 A e 1 , 2 ( t 1 ) 19/23 1 . 95 b 0 + 0 . 35 b 1 l − 0 . 35 b 1 h = 1 . 33 A e 1 , 2 ( t 1 ) 1 . 95 b 0 + 0 . 35 b 1 ( l − h ) = 1 . 33 A e 1 , 2 ( t 1 ) 1 , 2 ( t 1 ) = 1 . 95 b 0 + 0 . 35 b 1 ( l − h ) A e 1 . 33 � � � � � � �

• At t 2 , 0 . 35 b 1 h + 1 . 33 A e 1 , 2 ( t 2 ) = 0 . 35 b 1 h + 1 . 33 A e 2 , 3 ( t 2 ) 20/23 A e 1 , 2 ( t 2 ) = A e 2 , 3 ( t 2 ) • At t 3 , 0 . 35 b 1 h + 1 . 33 A e 2 , 3 ( t 3 ) = 0 . 35 b 1 h + 1 . 33 a 1 A e 2 , 3 ( t 3 ) = a 1 � � � � � � �

Equation for A e 1 , 2 ( t ) 21/23 A e 1 , 2 ( t ) = α 0 e β 0 ( t − t 1 ) / ( t 2 − t 1 ) where , α 0 = 1 . 95 b 0 + 0 . 35 b 1 ( l − h ) � a 1 � and , β 0 = ln . 1 . 33 α 0 Note: A e 1 , 2 ( t 1 ) = α 0 e β 0 ( t 1 − t 1 ) / ( t 2 − t 1 ) � A e 1 , 2 ( t 1 ) = α 0 . � And, � A e 1 , 2 ( t 2 ) = α 0 e β 0 ( t 2 − t 1 ) / ( t 2 − t 1 ) � A e 1 , 2 ( t 2 ) = α 0 e ln ( a 1 /α 0 ) � A e 1 , 2 ( t 2 ) = a 1 . � �

Equation for A e 2 , 3 ( t ) 22/23 � � π t − t 2 �� A e 2 , 3 ( t ) = a 1 1 + β 1 sin t 3 − t 2 where , β 1 = 0 . 15 . Note: � � �� πt 2 − t 2 A e 2 , 3 ( t 2 ) = a 1 1 + β 1 sin t 3 − t 2 A e 2 , 3 ( t 2 ) = a 1 (1) � A e 2 , 3 ( t 2 ) = a 1 . � And, � � � �� πt 3 − t 2 A e 2 , 3 ( t 3 ) = a 1 1 + β 1 sin � t 3 − t 2 � A e 2 , 3 ( t 3 ) = a 1 . � �

Results 50 23/23 0 velocity acceleration � −50 0 5 10 15 20 25 30 t−axis � � � � � �