ballistic/lifting atmospheric entryspacecraft.ssl.umd.edu/academics/791s12/791s12l05... · dv dt =...

Ballistic and Lifting Atmospheric EntryENAE 791 - Launch and Entry Vehicle Design

U N I V E R S I T Y O FMARYLAND

Ballistic/Lifting Atmospheric Entry• Straight-line (no gravity) ballistic entry based on

altitude, rather than density• Planetary entries (at least a start)• Basic equations of planar motion (with li)• Liing equilibrium glide•

1

© 2012 David L. Akin - All rights reservedhttp://spacecraft.ssl.umd.edu



Parametric Deceleration

2

dv

dt= ve exp

�Be−

hhs

� −B

hs

�e−

hhs

�ve sin γ exp

�Be−

hhs

�

dv

dt=−Bv2

e

hssin γ

�e−

hhs

�exp

�2Be−

hhs

�

dv

dt=−v2

e

2hs�β

�e−

hhs

�exp

�1

�β sin γe−

hhs

�

Let nref ≡v2

e

hs, ν ≡ dv/dt

nref, ϕ ≡ he

hs

ν =−12�β

�e−ϕ h

he

�exp

�1

�β sin γe−ϕ h

he

�



Nondimensional Deceleration, γ=-90°

3

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15 0.2

Deceleratio

Alt

itu

de R

ati

o h

/h

e

0.03 0.1 0.3 1 3

ν�β



Deceleration Equations

4

ν =−12�β

�e−ϕ h

he

�exp

�1

�β sin γe−ϕ h

he

�

n =−ρoV 2

e

2β

�e−

hhs

�exp

�ρohs

β sin γe−

hhs

�

Nondimensional Form

Dimensional Form

Note that these equations result in values <0 (reflecting deceleration) - graphs are absolute values of deceleration for clarity.



Dimensional Deceleration, γ=-90°

5

0

20

40

60

80

100

120

0 500 1000 1500 2000

Deceleration

Alt

itu

de

277 922 2767 9224 27673

(km

)

� m

sec2

�

β

�kg

m2

�



Altitude of Maximum Deceleration

6

ν = −B sin γ�e−

hhs

�exp

�2Be−

hhs

�Returning to shorthand notation for deceleration

ν = −B sin γ�e−η

�exp

�2Be−η

�Let η ≡ h

hs

dν

dη= −B sin γ

�d

dη

�e−η

�exp

�2Be−η

�+

�e−η

� d

dηexp

�2Be−η

��

dν

dη= −B sin γ

�−

�e−η

�exp

�2Be−η

�+

�e−η

� �−2Be−η

�exp

�2Be−η

��

dν

dη= B sin γe−η exp

�2Be−η

� �1 +

�2Be−η

��= 0




7

1 +�2Be−η

�= 0 ⇒ eη = −2B

ηnmax = ln (−2B)

ηnmax = ln�

−1�β sin γ

�

hnmax = hs ln�−ρohs

β sin γ

�

Altitude of maximum deceleration is independent of entry velocity!

Converting from parametric to dimensional form gives




8

0

2

4

6

8

10

12

0 0.2 0.4 0.6 0.8 1

Ballistic Coefficient

Alt

itu

de o

f M

ax D

ece

l h

/h

s

-15 -30 -45 -60 -90

�βγ



Magnitude of Maximum Deceleration

9

and insert the value of η at the point of maximum deceleration

ηnmax = ln�

−1�β sin γ

�⇒ e−η = −�β sin γ

νnmax =−12�β

�−�β sin γ

�exp

�−�β sin γ�β sin γ

�⇒ νnmax =

sin γ

2e

Start with the equation for acceleration -

ν =−12�β

e−η exp�

1�β sin γ

e−η

�

nmax =v2

e

hs

sin γ

2e

Maximum deceleration is not a function of ballistic coefficient!



Peak Ballistic Deceleration for Earth Entry

10

0

1000

2000

3000

4000

5000

6000

0 5 10 15

Entry Velocity (km/sec)

Peak D

ece

lera

tio

n (

m/

sec^

2)

-15 -30 -45 -60 -90γ



Velocity at Maximum Deceleration

11

Start with the equation for velocity

and insert the value of η at the point of maximum deceleration

ηnmax = ln�

−1�β sin γ

�⇒ e−η = −�β sin γ

v

ve= exp

�1

2�β sin γe−η

�

v

ve= exp

�−�β sin γ

2�β sin γ

�⇒ vnmax =

ve√e∼= 0.606ve

Velocity at maximum deceleration is independent of everything except ve



Planetary Entry - Physical Data

12

Radius(km) (km3/sec2) (kg/m3)

hs(km)

vesc(km/sec)

Earth 6378 398,604 1.225 7.524 11.18

Mars 3393 42,840 0.0993 27.70 5.025

Venus 6052 325,600 16.02 6.227 10.37

µ ρo



Comparison of Planetary Atmospheres

13

1E-20

1E-18

1E-16

1E-14

1E-12

1E-10

1E-08

1E-06

1E-04

0.01

1

100

0 50 100 150 200 250 300

Altitude (km)

Atm

osp

heri

c D

en

sity

(kg

/m

3)

Earth Mars Venus



Planetary Entry Profiles

14

V

Ve

γ = −15o

ve = 10km

sec

β = 300kg

m2



Planetary Entry Deceleration Comparison

15

γ = −15o

ve = 10km

sec

β = 300kg

m2



Check on Approximation Formulas

16

γ = −15o

ve = 10km

sec

β = 300kg

m2



Free-Body Diagram

17

D

mg

horizontal

v, s

γL

mv2

r



Dynamics of Lifting Entry

18

Along the velocity vector,

mdv

dt= m

v2

rsin γ −mg sin γ −D

Perpendicular to the velocity vector,

mvdγ

dt= L+m

v2

rcos γ −mg cos γ

(unbalanced lift rotates flight path angle)



Equations of Planar Lifting Entry

19

dv

dt=

�v2

r− g

�sin γ − D

m

vdγ

dt=

L

m+

�v2

r− g

�cos γ



Equilibrium Glide• Forces perpendicular to velocity vector are balanced

• Typically very shallow glide

20

=⇒ dγ

dt= 0; γ = constant

=⇒ assume γ → 0; sin(γ) → 0; cos(γ) → 1



Equilibrium Glide Equations

21

dv

dt= −D

m

0 =L

m+

�v2

r− g

�

v2

r− g = −1

2ρov

2 cLA

me−

hhs

dv

dt= −1

2ρov

2 cDA

me−

hhs

D =1

2ρv2cDA

ρ = ρoe− h

hs



Dynamics Perpendicular to Velocity

L/D set by vehicle aerodynamics, flight velocity, and angle of attack (assumed constant)

22

cLcD

=L

D

v2

r− g = −1

2ρov

2 L

D

cDA

me−

hhs

v2

r= −1

2ρov

2L/D

βe−

hhs + g

v2 = −1

2ρorv

2L/D

βe−

hhs + gr



More Lift-Direction Dynamics

23

Let e−hhs ≡ σ

�= density ratio ≡ ρ

ρo

�

v2 +1

2ρorv

2L/D

βσ = gr

v2�1 +

1

2ρor

L/D

βσ

�= gr

v =

�gr

1 + ρorσ(L/D)2β



Velocity during Entry

24

v

vco=

�1

1 + ρoroσ(L/D)2β

vco =

�µ

ro=

√goro

�µ = gor

2o

�

vco ∼= ve (within 1-2% for Earth)

v

ve=

�1 +

ρoro2β

L

De−

hhs

�− 12

Entry trajectory as a function of altitude, ratio

�β

L/D

�



Equilibrium Glide Velocity Trends

25



Trends with “Lifting Ballistic Coefficient”

26



Nondimensional Form of Equation

27

v

ve=

�1 +

ρoro2β

L

De−

hhs

�− 12

v

ve=

�1 +

ρohs

2β

rohs

L

De−

hhs

�− 12

Remember �β ≡ β

ρohs

v

ve=

�1 +

1

2�βrohs

L

De−

hhe

hehs

�− 12



Deceleration

28

L

m= g − v2

r= g − gv2

gr= g

�1−

�v

vco

�2�

dv

dt= −D

m= − L

L/D

1

m= − 1

L/D

L

m

dv

dt= − g

L/D

�1−

�v

vco

�2�



Deceleration

29

Let n ≡ 1

g

dv

dt(deceleration in g’s)

n = − 1

L/D

�1−

�v

vco

�2�

n = − 1

L/D

1− 1

1 + ρoro2β

LD e−

hhs

�Note: 1− 1

1 +K=

1 +K

1 +K− 1

1 +K=

K

1 +K

�



Deceleration

30

n = − 1

L/D

ρoro2β

LD e−

hhs

1 + ρoro2β

LD e−

hhs

n = −ρoro2β e−

hhs

1 + ρoro2β

LD e−

hhs

n =−1

2βρoro

e+hhs + L

D

n monotonically increases with decreasing altitude



Deceleration Trends with Altitude

31



Limiting Deceleration

32

nlimit =−1

2βρoro

+ LD

β ∼ O(103); ρo ∼ O(1); ro ∼ O(10

6) =⇒ 2β

ρoro∼ O(10

−3)

nlimit∼=

−1

L/D

Li significantly moderates peak g’s on entry (L/D=0.25 --> nlimit=4 g‘s)



Time for Entry

33

dv

dt= − g

L/D

�1−

�v

vco

�2�

dt =−(L/D)dv

g

�1−

�v

vco

�2�

� t

0dt =

� 0

ve

−(L/D)dv

g

�1−

�v

vco

�2�



Time for Entry

34

∆t =1

2

�rogo

L

Dln

1 +�

vvco

�2

1−�

vvco

�2

Time for entry ∝ L

D

Not a function of β



Time From Entry Interface

35



Distance Along Flight Path

36

For shallow entry,ds

dt∼= v ds = v dt

ds =−(L/D)v dv

g

�1−

�v

vco

�2�

Let u ≡ 1−�

v

vco

�2

du =−2v

v2codv v dv =

−v2co2

du

�ds =

L/D

2g

�v2codu

u



Distance Along Flight Path

37

∆s =(L/D)v2co

2glnu =

(L/D)goro2go

ln

�1−

�v

vco

�2�

∆s =ro2

L

Dln

�1−

�v

vco

�2�

Not a function of β or g



Distance from Entry

38



Bank Angle

39

D

mg

horizontal

Lφ

L cos (φ)

φ ≡ Bank Angle

Level turn: L cosφ = mg

(g’s you pull in the banked turn)

L

mg=

1

cosφ≡ nbank



Crossrange (without derivations)

40

• Use of li to deviate laterally from planar groundtrack

• Optimum bank angle to maximize crossrange

• Maximum achievable crossrange

φopt∼= cot−1

�

1 + 0.106

�L

D

�2

ymax∼=

ro5.2

�L

D

�2 1�1 + 0.106

�LD

�2



Crossrange Based on L/D

41



Early Shuttle Design Configurations• Delta wing configuration

– High hypersonic li– High landing velocity– High crossrange

• Straight-wing configuration– High subsonic li– Low(er) landing velocity– Low crossrange

42

ballistic/lifting atmospheric entryspacecraft.ssl.umd.edu/academics/791s12/791s12l05... · dv dt =...

Documents