terramechanics - university of maryland...terramechanics 1 enae 788x - planetary surface robotics u...

Terramechanics 1 ENAE 788X - Planetary Surface Robotics

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

Terramechanics

• Origin and nature of lunar soil • Soil mechanics • Wheel-soil interactions • Tandem wheel calculations • Grousers

1

© 2016 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu

http://spacecraft.ssl.umd.edu



Lunar Regolith• Broken down from larger pieces over time • Major constituents

– Rock fragments – Mineral fragments – Glassy particles

• Local environment – 10-12 torr (= 1.22x10 -10 Pa = 1.93x10 -14 psi) – Meteorites at velocities >105 m/sec – Galactic cosmic rays, solar particles – Temperature range +250°F – -250°F

2



Regolith Creation Process

• Only “weathering” phenomenon on the moon is meteoritic impact!

• Weathering processes – Comminution: breaking rocks and minerals into smaller

particles – Agglutination: welding fragments together with molten

glass formed by impact energy – Solar wind spallation and implantation (miniscule) – Fire fountaining (dormant)

3



JSC-1 Simulant

• Ash vented from Merriam Crater in San Francisco volcano field near Flagstaff, AZ

• K-Ar dated at 150,000 years old ± 30,000 • Major constituents SiO2, TiO2, Al2O3, Fe2O3, FeO,

MgO, CaO, Na2O, other <1% • Represents low-Ti regolith from lunar mare • MLS-1 simulant (U.Minn.) preferred for simulation

of highland material • BP-1 (Flagstaff, AZ) is ground basaltic lava -

higher fidelity because of angular grain shapes

4



Particle Size Distribution in Regolith

5

Rahmatian and Metzger, “Soil Test Apparatus for Lunar Surfaces” Earth and Space 2010, ASCE



Soil Testing Apparatus

Bevameter (force vs. displacement)

Internal friction angle ϕ

Shear deformation modulus K

6



Soil Characterization – Direct Shear

7

Soil

cross-section A

T

WShear Stress ⌧ =

T

A

Normal Stress � =

W

A

c = Cohesion

� = angle of internal friction

Normal Stress �

Shear Stress ⌧



Modeling Soil Reaction to a Wheel

Assume soil reaction is like a (nonlinear) spring

P = kzn

8

P = applied pressure

z = compression depth

k, n = heuristic parametersP

z



Effects of Soil Mechanics

9

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10

n=0n=0.5n=1.0n=1.5

Scaled PressureP

kSoilPenetrationDepthz



Lunokhod 1 Penetrometer Data

10

Mitchell et.al., Mechanical Properties of Lunar Soil: Density, Porosity, Cohesion, and Angle of Internal Friction”Proceedings of the Third Lunar Science Conference, Vol. 3, MIT Press, 1972



Wheel-Soil Interaction

Wheel rolling over soil does work – Compression – “Bulldozing”

11

from Gibbesch and Schafer, “Advanced and Simulation Methods of Planetary Rover Mobility on Soft Terrain” 8th ESA Workshop on Advanced Space Technologies for Robotics and Automation, Noordwijk, The Netherlands, November, 2004



Wheel-Soil Interactions

12

zo

R

W

wheel width b

area of compressed soil A = bd

distance rolled d

Displacement EnergyE

A=

�F

Adz =

�Pdz

E

A=

� zo

0Pdz =

� zo

0kzndz = k

zn+1o

n + 1



Rolling Resistance

13

Total Energy

E

AA =

E

Abd = k

zn+1o

n+ 1

bd

Given a force resisting rolling ⌘ R,

the energy required to roll a distance d is

Eroll

= Rd

Eroll

= Edisplacement

) Rd =E

Abd



Rolling Resistance

14

For n = 1 : P = kz;E

A= k

z2o

2

;R =

1

2

kbz2o

For n =

1

2

: P 2= k2z;

E

A=

2

3

kz32o

;R =

2

3

kbz32o

For n = 0 : P = k;E

A= kz

o

;R = kbzo

Generic case: P = kzn;E

A= k

zn+1o

n+ 1;R = kb

zn+1o

n+ 1



Soil Displacement Calculations

15

zo

R

Wwheel width b

✓

✓o

dF

Fx

Fz

R�Z ✓

o

0dF sin ✓ = 0

�W +

Z ✓o

0dF cos ✓ = 0

dF cos ✓ = �Pb dx

dF sin ✓ = Pb dz

R =

Z ✓o

0Pb dz W = �

Z ✓o

0Pb dx

In general, P = kx

n

W = �Z z

o

0bkz

ndx

dF = Pbr d✓




16

zo

wheel width b

✓

A

Bz

D

2

AB =D

2� (z

o

� z)

x

x

2 =

✓D

2

◆2

� AB

2=

✓D

2

◆2

�D

2� (z

o

� z)

�2

=

✓D

2

◆2

�✓D

2

◆2

+ 2D

2(z

o

� z)� (zo

� z)2

x

2 = [D � (zo

� z)](zo

� z)



Soil Compression Calculations

17

But D � zo

� z

x

2 ⇡ D(zo

� z) ) 2x dx = �D dz

so from W = �Z z

o

0bkz

ndx we get W = �

Z zo

0bkz

n�D

2x

dz

W = �bk

Zz

o

0zn

✓�D

2pDpzo

� z

◆dz

W = bk

Zz

o

0zn p

Ddz

2pzo

� z

!dz




18

Define zo

� z ⌘ t2 ) dz = �2tdt

W = bkpD

Z pz

o

0(z

o

� t2)ndt

W ⇡ bkpDz

o

3zno

(3� n)

for n = 1 ) W =

2

3

bkzo

pDz

o

for n =

1

2

) W =

5

6

bkzo

pD

for n = 0 ) W = bkpDz

o

Taylor Series expansion (zo

� t2)n ⇠=

zno

� nzn�1o

t2 + · · ·



Rolling Resistance as f(W)

19

for n = 0 ) W = bkp

Dzo

) zo

=

✓W

bk

◆21

D

R = kbzo

) R =kb

(kb)2W 2

D) R =

W 2

kbD

for n =

1

2

) W =

5

6

bkzo

pD ) z

o

=

6

5

W

bkpD

R =2

3kbz

32o

) R =2

3kb

✓6

5

W

kbpD

◆ 32

=2

3

✓6

5

◆ 32 W

32

pkbD

34

R = 0.876W

32

pkbD

34



Rolling Resistance as f(W)

20

for n = 1 ) W =

2

3

bkz32o

pD ) z2

o

=

✓3W

2kbpD

◆ 43

R =1

2kbz2

o

) R =1

2kb

✓3W

2kbpD

◆ 43

=1

2

✓3

2

◆ 43✓

W 4

kbD2

◆ 13

R = 0.859

✓W 4

kbD2

◆ 13



Rolling Resistance as f(W) (Generic)

21

W =bkpDz

o

3zno

(3� n) =bkpD

3zn+ 1

2o

(3� n)

zn+ 1

2o

=3

(3� n)

W

bkpD

zn+1o

=

✓3

3� n

W

bkpD

◆ n+1

n+12=

✓3

3� n

W

bkpD

◆ 2(n+1)2n+1

R =bk

n+ 1zn+1o

=bk

n+ 1

✓3

3� n

W

bkpD

◆ 2(n+1)2n+1

R =1

n+ 1

✓3

3� n

WpD

◆ 2(n+1)2n+1

✓1

bk

◆ 12n+1



More Detailed Soil Compression Equation

k =kc

b+ k�

P =�

kc

b+ k�

⇥zn

b = wheel width

kc units�< N/m(n+1) >

k� units�< N/m(n+2) >

kc = modulus of cohesion of soil deformation

k� = modulus of friction of soil deformation

22



Soil Characteristics

23

soil type n kc � N

mn+1⇥ k� � N

mn+2⇥

Dry Sand 1.1 990 1,528,000

Lunar Regolith 1 1400 820,000

Sandy Loam 0.7 5270 1,515,000

Sandy Loam (MER-B) 1 28,000 7,600,000

Slope Soil(MER-B) 0.8 6800 210,000

Clay (Earth) 0.5 13,190 692,200



Terzaghi Analysis of Soil Deformation

24



Terzaghi Soil Bearing Capacity Factors

25

N� =

1

2

✓Kp�

cos

2 �� 1

◆tan�

Nq =

exp

⇥�3⇡2 � �

�tan�

⇤

2 cos

2⇣

⇡4 +

�2

⌘

Kp� =�8�2 � 4�+ 3.8

�tan2

✓⇡

3+

�

2

◆

Nc = cot�

8<

:exp

⇥�3⇡2 � �

�tan�

⇤

2 cos

2⇣

⇡4 +

�2

⌘ � 1

9=

; = cot� (Nq � 1)

� = Angle of internal resistance of soil



Parameter Definition

26

Kc = (Nc � tan�) cos2 �

K� =

2N�

tan�+ 1

�cos

2 �

K� ⌘ Modulus of cohesion of soil deformation

Kc ⌘ Modulus of density of soil deformation

↵ ⌘ Angle of approach of wheel to soil

↵ = cos

�1

✓1� 2z

D

◆

� ⌘ Weight density of soil

✓N

m3

◆



Soil Bearing Limit

27

Safe weight on the soil

Ws = A

✓cNc + �zNq +

1

2�bN�

◆

c ⌘ Soil cohesion (Pa)

b ⌘ Wheel width (m)



Equations for Compression Resistance

Ww = weight on wheel

d = wheel diameter

Rc = compression resistance (per wheel)

Rc =�

bk

n + 1

⇥zn+1

28

z =

3Ww

(3� n)bkpd

� 22n+1



Soil Compression – Reece Formulation

kc units�< N/m(n+1) >

k� units�< N/m(n+2) >

29

Problem is that kc and k� have variable dimensions,

P =

✓kcb

+ k�

◆zn

based on n

c = Cohesion

� = angle of internal friction

Normal Stress �

Shear Stress ⌧



Compression Resistance (Lunar Soil)

kc = 0.14 N/cm2

k� = 0.827 N/cm3

n = 1

Rc =1

n + 1(kc + bk�)

�12n+1

�3Ww

(3� n)⇥

d

⇥ 2(n+1)2n+1

Rc =12

(kc + bk�)�13

�3Ww

2�

d

⇥ 43

30



Compression Resistance (Lunar Soil)

n = 1

Rc =1

n + 1(kc + bk�)

�12n+1

�3Ww

(3� n)⇥

d

⇥ 2(n+1)2n+1

Rc =12

(kc + bk�)�13

�3Ww

2�

d

⇥ 43

31

k� = 0.827 N/cm3

kc = 0.14 N/cm2

Rc = 20.89 N/wheel = 83.57 N (total)



Lunar Regolith Soil Properties

32

Symbol Description Value Used in

n Exponent of sinkage 1.0 EVERYTHING

kc Cohesive modulus of soil deformation 1400 N/m2 Rc, Rb, H

kф Frictional modulus of soil deformation 820,000 N/m3 Rc, Rb, H

θ Slope angle Rg

Cf Coefficient of friction 0.05 Rr

γ Soil density 1.6 g/cm3 Rb

Co Cohesive strength of soil 0.017 N/cm2 H

ф Angle of internal friction 30-40 degrees Rb

K Coefficient of soil Slip 1.8 cm H

s slip 0.02 – 0.05 H



Apollo Lunar Roving Vehicle Example

check units -

z =�

3 � 2532(0.14 + 17.4 � 0.827)

⇥82

⇥ 23

= 2.03 cm

Rc =12

(0.14 + 17.4 � 0.827)�13

�3 � 2532⇥

82

⇥ 43

= 29.8 N

�N�1/3

cm�2/3

⇥ �N4/3

cm2/3

⇥= N

33



Rolling and Gravitation Resistance

• Rolling resistance (tires, bearings, etc.)

• Gravitational resistance

• LRV examples (15° slope)

Rr = Wvcf

Wv = weight of vehicle

Rg = Wv sin �slope

Rg = 262 N

cf = coe�cient of friction (typ. 0.05)

Rr = 51 N

34



Bulldozing Resistance

35

Rb =b sin (↵+ �)

2 sin↵ cos�

�2zcKc + �z2K�

�General case:

For tracked vehicles, only the first term applies:

Rb =b sin (↵+ �)

2 sin↵ cos�

�2zcKc + �z2K�

�

`o

= z tan2✓⇡

4� �

2

◆

+`3o

�

3

⇣⇡2� �

⌘+ c`2

o

1 + tan

✓⇡

4+

�

2

◆�

� = angle of attack of wheel in soil ⇥ cos�1

�1� 2z

D

⇥

� = density of soil�

kgm3

⇥

All angles in radians!



Tandem Wheels

36

W1W2

D2

D1

✓0✓1 ✓02

z1z0 z2

z0 = z1 + z2

Assume n =1

2=) P = k

pz

P1 = kpz1 P2 = k

pz1 + z2



Soil Weight Bearing Analysis

37

In general,

W =

Z ✓0

0dF cos ✓ =

Z ✓0

0Pb ds cos ✓

W =

Z ✓0

0bkpz cos ✓ ds

W =

Z ✓0

0bkr

pr(cos ✓ � cos ✓0) cos ✓ d✓

ds = r d✓ z = r(cos ✓ � cos ✓0)



Generic Wheel Soil Suspension

38

Assuming small sinkage,

z ! small, ✓ ! small,

cos ✓ ⇡ 1

cos ✓ d✓ ⇡ d✓

cos ✓ ⇡ 1� ✓2

2

+ (higher order terms)

W =bkr3/2p

2

Z ✓0

0

q✓20 � ✓2 d✓

W =bkr3/2p

2

1

2

✓20 sin

�1

✓✓

✓0

◆+ ✓

q✓20 � ✓2

�✓0

0



Weight on the Front Wheel

39

W =⇡bkr3/2

4p2

✓20

Front wheel: z1 = r1 � r1 cos ✓0

z1 = r1 � r1

✓1� ✓20

2+ · · ·

◆=) ✓20 ⇠= 2

z1r1

W1⇠=

⇡bkz1pr1

2p2



Weight on Back Wheel

40

Change to limits of integration:

W2 =bkr3/22p

2

Z ✓0

0

q✓202 � ✓2 d✓

0 �! ✓0, r �! r2

q✓202 � ✓2 ⇠= ✓02

✓1� 1

2

✓2

✓202+ · · ·

◆

Z ✓0

0

q✓202 � ✓2 d✓ ⇠= ✓02✓1 �

1

6

✓31✓02

= ✓02✓1

✓1� 1

6

✓21✓202

◆⇠= ✓02✓1



Weight on Back Wheel

41

W2 =bkr3/2p

2✓02✓1

z0 ⇠= r2✓2022

z2 ⇠= r2✓212

✓02 ⇠=r

2z0r2

✓1 ⇠=r

2z2r2

W2 = bkp2r2

pz0z2



Track Depth of Tandem Wheels

42

Front: z1 =

2

p2W1

⇡bkpr1

z0 = z1 + z2 =2p2W1

⇡bkpr1

+W 2

2

(bk)22r2

1

z0

Back: z2 =

W2

bkp2r2

1pz0

�2

Much algebra then ensues...

z20 � 2p2W1

⇡bkpr1

z0 +W 2

2

(bk)22r2= 0



Rolling Resistance of Tandem Wheels

43

Solve the quadratic equation to get

z0 =1

bk

0

@p2W1

⇡pr1

+

s2W 2

1

⇡2r1+

W 22

2r2

1

A

This was all done for n =

1

2

=) R =

2

3

bkz3/20

R =2

3

1pbk

0

@p2W1

⇡pr1

+

s2W 2

1

⇡2r1+

W 22

2r2

1

A3/2

R =2

3

1pbk

0

@ 2W1

⇡pD1

+

s4W 2

1

⇡2D1+

W 22

D2

1

A3/2



Nondimensional Forms

44

Total wheel load W = W1 +W2 Wheel weight ratio a ⌘ W1

W2

For W1 = W2 =

W

2

=) a = 1

W1 =a

1 + aW W2 =

1

1 + aW

R =2

3

1

(a+ 1)3/2W 3/2

pbk

0

@ 2a

⇡pD1

+

s4a2

⇡2D1+

1

D2

1

A3/2

Define wheel diameter ratio ⇢ ⌘ D1

D2



Nondimensional Forms

45

R =2

3

1

(a+ 1)3/2W 3/2

D3/42

pbk

2a

⇡p⇢+

s

1 +4a2

⇡2⇢

!3/2

Let ⇠ ⌘ 2

3

1

(a+ 1)3/2

2a

⇡p⇢+

s

1 +4a2

⇡2⇢

!3/2

R =⇠pbk

W 3/2

D3/42



Simple Example Case

46

Consider ⇢ = 1 (D1 = D2 = D)

a = 1

✓W1 = W2 =

W

2

◆

For tandem wheels, R =

0.580pbk

W 3/2

D3/42

For single wheel (n=1/2), R =

0.876pbk

W 3/2

D3/42

Tandem wheels reduce rolling resistance by 34%



Dual Wheels

47

Equivalent to single wheel case twice as wide b =) 2b

R =0.876p

2

1pbk

W 3/2

D3/42

Rdual =0.619p

bk

W 3/2

D3/42

Dual wheel rolling resistance 29% less than single,

7% higher than tandem

terramechanics - university of maryland...terramechanics 1 enae 788x - planetary surface robotics u...

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