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

▶

Jul 31, 2023 317 likes •581 views

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

SLIDE 1

The Parachute Problem

Ethan Retherford

College of the Redwoods

May 15, 2015

SLIDE 2

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, x0, and falling towards Earth under the influence of gravity.

SLIDE 3

SLIDE 4

SLIDE 5

The Basic Model

We start with analyzing the parachutist’s free-fall from, x0.This is done by applying Newton’s Second Law of Motion, ΣF = ma.

SLIDE 6

Free-fall and parachute deployed stages of the jump

x0 Fg = −mg Fd = −k1v Free − fall x0 Fg = −mg Fd = −k2v Parachute Deployed Assume the parachute deploys instantaneously at td

SLIDE 7

Deriving the differential equations k =

0 ≤ t < td k2, t ≥ td ΣF = ma Fg + Fd = ma −mg − kv = ma

ma + kv = −mg

SLIDE 8

Deriving the differential equations

First-order system for position and velocity x′ = v, x(0) = x0 v ′ = −g − k mv, v(0) = 0.

SLIDE 9

Free-fall During free-fall k = k1 for t < td v ′ + k1 mv = −g, v(0) = 0 Solving for v(t), we get the velocity while in free-fall v(t) = mg k1

e−k1t/m − 1

SLIDE 10

Free-fall Now, to obtain the position x(t) we can sub v(t) into the position equation x′ = v x′ = mg k1

e−k1t/m − 1
,

x(0) = x0 Solving x(t) = x0 − mg k1 t − m2g k2

e−k1t/m − 1

SLIDE 11

Parachute is deployed

v ′ + k2 m v = −g, v(t−

d ) = v(t+ d )

Obtaining the general solution and using the initial condition mg k1

e−k1td/m − 1
= −mg

k2 + Ce−k2td/m C = mg k2 ek2td/m + mg k1 ek2td/m e−k1td/m − 1

Substituting C back into v(t), t ≥ td

v(t) = mg k1 e−k2(t−td)/m e−k1td/m − 1

+ mg

kd

e−k2(t−td)/m − 1

SLIDE 12

Parachute is deployed

Use Same method for Position x(t) = x0 − mg k1 td −

e−k1td/m − 1

m2g k2

+ m2g k1k2

e−k2(t−td)/m − 1
− m2g

k2

e−k2(t−td)/m − 1
+ mgtd

k2 − mgt k2

SLIDE 13

Position and Velocity for Entire Jump

x(t) =                x0 − mg

k1 t − m2g k2

e−k1t/m − 1
,

t < td x0 − mg

k1 td −

e−k1td/m − 1
m2g

1 + m2g

k1k2

e−k2(t−td)/m − 1
− m2g

e−k2(t−td)/m − 1
+ mgtd

− mgt

k2 ,

t ≥ td v(t) =       

mg k1

e−k1t/m − 1
,

t < td

mg k1 e−k2(t−td)/m

e−k1td/m − 1

+ mg

e−k2(t−td)/m − 1
,

t ≥ td

SLIDE 14

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

SLIDE 15

Analysis of Basic Model

k1 = 14 kg/s, k2 = 160 kg/s

Position (x/200) Velocity (v/10) Acceleration (a/g)

10 20 30 40 time (s)

5 10

SLIDE 16

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 (≈ 3Gs 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.

SLIDE 17

Deployment stage Added

x0 Fg = −mg Fd = −k1v Free − fall x0 Fg = −mg Fd = −kdv Deployment x0 Fg = −mg Fd = −k2v Parachute Deployed

SLIDE 18

SLIDE 19

Connecting the constant states in a continuous manner

k(t) =      k1, 0 ≤ t < td kd, td ≤ t < td + τd k2, t ≥ td + τd.

SLIDE 20

Linear Interpolation

20 23.5 t 14 160 k(t)

(td, k1) (td+τd, k2)

kd(t) = k1 + k2 − k1 τ (t − td)

SLIDE 21

Plots with Linear Interpolation

Position (x/1000) Velocity (v/10) Acceleration (a/g)

20 40 60 80 100 time (s)

SLIDE 22

Jerk

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

SLIDE 23

Linear Interpolation with Jerk

Position (x/1000) Velocity (v/10) Acceleration (a/g) Jerk (j/g)

20 22 24 26 time (s)

2 4

SLIDE 24

Cubic Interpolation

20 23.5 t 14 160 k(t)

(td, k1) (td+τd, k2)

kd(t) = (−2k2 + 2k1) t − td τd 3 + (3k2 − 3k1) t − td τd 2 + k1.

SLIDE 25

Plots with Cubic Interpolation

Position (x/1000) Velocity (v/10) Acceleration (a/g) Jerk (j/g)

20 22 24 26 time (s)

2 4