Section 3.5 Recovery Systems: Parachutes 101

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1 Section 3.5 Recovery Systems: Parachutes 101 Material taken from: Parachutes for Planetary Entry Systems Juan R. Cruz Exploration Systems Engineering Branch NASA Langley Research Center Also, Images from: Knacke, T. W.: Parachute Recovery Systems Design Manual, Para Publishing, Santa Barbara, CA, and Ewing, E. G., Bixby, H.W., and Knacke, T.W.: Recovery System Design Guide, AFFDL-TR ,

2 Basic Terminology 2

3 Basic Terminology (2) For our purposed conical and elliptical parachutes are same thing 3

4 Basic Terminology (3) F drag F drag = q C D S 0 q = 1 2 ρ V 2 "incompressible dynamic pressure" C D = "drag coefficient" C D S 0 = "drag area" V 4

5 Basic Terminology (4) In general, under parachute, 2-DOF equations of motion are. (ignore centrifugal & Coriolis forces) V V 2 r + V ν 2 ( F ) drag parachute + ( F ) drag vehicle V r V ν = ( F ) drag parachute + F drag m ( ) vehicle ( F ) drag parachute + ( F ) drag vehicle m sinγ g cosγ Vehicle decelerates very rapidly in horizontal direction 5

6 Basic Terminology (4) Terminal Velocity.. Equilibrium velocity where parachute + vehicle are no longer accelerating V r V ν 0 0 γ 90 ( F ) drag parachute + F drag D 0 parachute = cm ( ) vehicle = m g 1 2 ρ V 2 terminal C S D 2 V terminal = ( ) parachute 0 + ( CD S) ref 2M vehicle g ρ ( C D S) parachute 0 + ( CD S) vehicle ref vehicle = M vehicle g 6

7 Parachute Types We ll be using solid parachutes 7

8 Parachute Shapes Hemispherical parachute: - Deployed canopy takes on the shape of a hemisphere. - Three dimensional hemispherical shape divided into a number of 2-D panels, called gores Angle subtended on the left hand side of the pattern is 60 degrees When all six gores are joined they complete the 360 degree circle. 8

9 Parachute Shapes (2) Conical Parachute - 2-D Canopy shape in form of a triangle 9

10 Parachute Shapes (3) Conical Parachute Gore Shape - 2-D Canopy shape in form of a triangle Higher drag coefficient than hemispherical parachutes, but also less stability 10

Parachute Shapes (4) Elliptical parachute: - Parachute where vertical axis is smaller than horizontal axis - A parachute with an elliptical canopy

11 Parachute Shapes (4) Elliptical parachute: - Parachute where vertical axis is smaller than horizontal axis - A parachute with an elliptical canopy has essentially the same CD as a hemispherical parachute, but with less surface material h r Canopy profile for different height / radius ratios 11

12 Parachute Shapes (5) Comparison of gore shapes for different height : radius ratios 12

13 Parachute Types (2) 13

14 Parachute Types (3) 14

15 Example Calculation: Drogue Chute Terminal Velocity h apogee = h agl + h launch = site ( ) meters 1850 meters ρ apogee = kg m g = µ km 3 r = sec 2 = ( ) 2 m km 2 sec 2 Maximum mass at apogee : m apogee = m launch m fuel = ( ) = kg m apogee g = = Nt 15

16 Example Calculation: Drogue Chute Terminal Velocity (2) Descent rate under drogue, ft/sec Go with minimum value ~ m/sec (50 ft/sec) Vehicle Drag Area.. Rocket is broken into two pieces ( C D S 0 ) vehicle 2 C D ( ) rocket A ref ( ) rocket Double up nominal rocket drag area = ( ) m 2 16

17 Example Calculation: Drogue Chute Terminal Velocity (3) Parachute Drag Coefficient Elliptical Parachute.. Take median value ( C ) 0.76 ± D chute 17

18 Example Calculation: Drogue Chute Terminal Velocity (4) Calculate required chute area: 2 V terminal = V terminal ( S ) 0 parachute = 2 m 2M g vehicle g ρ C D S 0 C ρ D S ( ) parachute + ( C D S ) ( ) parachute 0 vehicle 0 + ( CD S) vehicle ref m g 1 2 ρ V 2 terminal ( C D S ) 0 vehicle = ( C ) D parachute m 2 D 0 = 4 ( S 0 ) parachute π π 12 = = 4.28 ft 18

19 Example Calculation: Drogue Chute Terminal Velocity (5) Drag Chute Areas Versus Terminal Velocity 19

20 Example Calculation: Drogue Chute Terminal Velocity (5) Drag Chute Diameter Versus Terminal Velocity 20

21 Parachute Opening Loads Largest Tensile Load on Vehicle often the Ultimate Design Load Driver Design Tool Verification Tool (Direct Simulation) 21

22 Parachute Opening Loads (2) 22

23 Parachute Opening Loads (3) t inf = n D 0 V 1 k n = canopy fill constant k = decceleration exponent 23

24 Parachute Opening Loads (4) 24

25 Infinite-Mass Inflation Parachute Opening Loads (5) 25

26 Finite-Mass Inflation Parachute Opening Loads (6) 26

27 Pflanz' (1942): Pflanz s Method -introduced analytical functions for the drag area (finite mass inflation approximation) Simple, frst-order, design book type method -Requires least knowledge of the system compared to other methods fight -Assumes no gravity acceleration limits application to shallow path angles at parachute deployment - Neglects entry vehicle drag - Yields only peak opening load - Allows for finite mass approximation Doherr (2003) extended method to account for gravity and arbitrary fight path angles 27

28 F peak = q 1 ( C D S) 0 C x X 1 X r = f ( A,η) Pflanz s Method (2) (finite mass inflation approximation) q 1 Dynamic Deployment C x Shock Load Factor X 1 Opening Force Reduction Factor ( C D S) 0 Nominal Drag Full Inflation η Inflation Curve Exponent A Ballistic Parameter 2 M A = V ( C D S) 0 ρ 1 V 1 τ infl η M V ρ 1 τ infl (See Later Description) 28

29 Pflanz s Method (3) A = 2 M V ( C D S) 0 ρ 1 V 1 τ infl Ribbon/Ringslot à h=1 Solid, Elliptical, Flat àh=2 Extended Skirt, Reefed àh=1/2 τ infl = n D 0 V 1 k n Canopy Fill Constant k = Decelleration Exponent Solid, Elliptical Chute n 4 k 0.85 h= 1 h =2 h = 1/2 29

30 Pflanz s Method (4) η = 2 Load Reduction Factor, X 1 η =1 η =1/ 2 η =1/ 2 η =1 η =1/ 2 η = 2 η =1 η = 2 A = 2 M V ( C D S) 0 ρ 1 V 1 τ infl 30

31 Pflanz s Method Example Drogue deploy 31

32 Pflanz s Method Example (2) Desired Terminal Velocity = m/sec Get Nominal Parachute Size V terminal = S 0 parachute = 2M vehicle g ρ parachute vehicle ( C D S) 0 + ( CD S) ref M vehicle g 1 2 ρ V 2 terminal vehicle ( C D S) ref C D0 parachute 0.85 X = = M 2 D 0 parachute = cm 32

33 Subsonic Inflation time Pflanz s Method Example (3) τ infl = n D 0 V 1 k n Canopy Fill Constant k = Decelleration Exponent D 0 parachute = cm Solid, Elliptical Chute n 4 k 0.85 V 1 = m/sec τ infl = = sec A = 2 M V ( C D S) 0 ρ 1 V 1 τ infl = =

34 Pflanz s Method Example (5) η = 2 = η =1 η =1/ 2 η =1/ 2 η =1 η =1/ 2 η = 2 η =1 η = 2 A = 2 M V ( C D S) 0 ρ 1 V 1 τ infl = X 1 =

35 F peak = q 1 ( C D S) 0 C x X 1 Pflanz s Method Example (5) Pa q 1 Dynamic Deployment 1.8 C x Shock Load Factor X 1 Opening Force Reduction Factor ( C D S) 0 Nominal Drag Full Inflation m 2 = = N X 1 = Near infinite mass 35

36 Pflanz s Method Example 2 (2) 36

37 F P = q C (t tsi ) ( S D 0 ) C x Inflation Curve Methods Ignores parachute mass (conservative) t t si t inf τ infl t si η n Since direct Sim accounts for deceleration No X 1 used in this method Direct Simulation Verification Tool τ infl η M V M V 37

38 Inflation Curve Method (2) Inflation Data from Doherr Ribbon/Ringslot à h=1 Solid, Elliptical, Flat àh=2 Extended Skirt, Reefed àh=1/2 ( C D S 0 ) t ( C D S 0 ) steady η =1/ 2 η =1 t t si t inf t si τ infl 38

39 Inflation Curve Method (3)

40 Inflation Curve Method (4) Direct Simulation Response, compare peak load to Pflanz Method = N

41 Questions?? 41

Differential Equations and the Parachute Problem

Differential Equations and the Parachute Problem Ronald Phoebus and Cole Reilly May 10, 2004 Abstract The parachute problem is a classical first semester differential equations problem often introduced