origin and nature of lunar soil • soil mechanics...1 2 k p cos2 1 tan n q = exp ⇥ 3⇡ 2 tan ⇤...

Terramechanics ENAE 788X - Planetary Surface Robotics

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

Terramechanics 1

• Origin and nature of lunar soil • Soil mechanics • Wheel-soil interactions

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

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)

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

Particle Size Distribution in Regolith

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

Soil Characterization – Direct Shear

cross-section A

WShear Stress ⌧ =

Normal Stress � =

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

P = applied pressure

z = compression depth

k, n = heuristic parametersP

Effects of Soil Mechanics

00 2 4 6 8 10

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

Scaled PressureP

kSoilPenetrationDepthz

Lunokhod 1 Penetrometer Data

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”

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

wheel width b

area of compressed soil A = bd

distance rolled d

Displacement EnergyE

�Pdz

� zo

0Pdz =

� zo

0kzndz = k

Rolling Resistance

Total Energy

Abd = k

Given a force resisting rolling ⌘ R,

the energy required to roll a distance d is

= Edisplacement

) Rd =E

Rolling Resistance

For n = 1 : P = kz;E

For n =

: P 2= k2z;

kbz32o

For n = 0 : P = k;E

;R = kbzo

Generic case: P = kzn;E

n+ 1;R = kb

Soil Displacement Calculations

Wwheel width b

R�Z ✓

0dF sin ✓ = 0

�W +

Z ✓o

0dF cos ✓ = 0

dF cos ✓ = �Pb dx

dF sin ✓ = Pb dz

Z ✓o

0Pb dz W = �

Z ✓o

0Pb dx

In general, P = kx

W = �Z z

dF = Pbr d✓

wheel width b

2� (z

� z)

� AB

2� (z

� z)

�✓D

� z)� (zo

� z)2

2 = [D � (zo

� z)](zo

� z)

Soil Compression Calculations

But D � zo

2 ⇡ D(zo

� z) ) 2x dx = �D dz

so from W = �Z z

ndx we get W = �

W = �bk

✓�D

2pDpzo

W = bk

Define zo

� z ⌘ t2 ) dz = �2tdt

W = bkpD

� t2)ndt

W ⇡ bkpDz

(3� n)

for n = 1 ) W =

for n =

for n = 0 ) W = bkpDz

Taylor Series expansion (zo

� t2)n ⇠=

� nzn�1o

t2 + · · ·

Rolling Resistance as f(W)

for n = 0 ) W = bkpDz

R = kbzo

) R =kb

(kb)2W 2

D) R =

for n =

pD ) z

) R =2

◆ 32

◆ 32 W

R = 0.876W

Rolling Resistance as f(W)

for n = 1 ) W =

bkz32o

pD ) z2

◆ 43

) R =1

◆ 43

◆ 43✓

◆ 13

R = 0.859

✓W 4

◆ 13

Rolling Resistance as f(W) (Generic)

W =bkpDz

(3� n) =bkpD

3zn+ 1

(3� n)

3� n

◆ n+1

3� n

◆ 2(n+1)2n+1

n+ 1zn+1o

3� n

◆ 2(n+1)2n+1

3� n

◆ 2(n+1)2n+1

◆ 12n+1

More Detailed Soil Compression Equation

b+ k�

P =�

b+ k�

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

Soil Characteristics

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

Terzaghi Soil Bearing Capacity Factors

N� =

✓Kp�

2 �� 1

◆tan�

⇥�3⇡2 � �

�tan�

⇡4 +

Kp� =�8�2 � 4�+ 3.8

�tan2

✓⇡

Nc = cot�

⇥�3⇡2 � �

�tan�

⇡4 +

⌘ � 1

; = cot� (Nq � 1)

� = Angle of internal resistance of soil

Parameter Definition

Kc = (Nc � tan�) cos2 �

K� =

tan�+ 1

�cos

↵ ⌘ Angle of approach of wheel to soil

↵ = cos

✓1� 2z

� ⌘ Weight density of soil

Kc ⌘ Cohesive modulus of soil deformation

K� ⌘ Frictional modulus of soil deformation

Soil Bearing Limit

Safe weight on the soil

Ws = A

✓cNc + �zNq +

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 =�

⇥zn+1

(3� n)bkpd

� 22n+1

Soil Compression – Reece Formulation

kc units�< N/m(n+1) >

k� units�< N/m(n+2) >

Problem is that kc and k� have variable dimensions,

✓kcb

+ k�

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(kc + bk�)

�12n+1

�3Ww

(3� n)⇥

⇥ 2(n+1)2n+1

Rc =12

(kc + bk�)�13

�3Ww

⇥ 43

Lunar Regolith Soil Properties

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)

⇥ 23

= 2.03 cm

Rc =12

(0.14 + 17.4 � 0.827)�13

�3 � 2532⇥

⇥ 43

= 29.8 N

�N�1/3

cm�2/3

⇥ �N4/3

⇥= N

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

Bulldozing Resistance

Rb =b sin (↵+ �)

2 sin↵ cos�

�2zcKc + �z2K�

�General case:

For tracked vehicles, only the first term applies:

Rb =b sin (↵+ �)

2 sin↵ cos�

= z tan2✓⇡

4� �

⇣⇡2� �

⌘+ c`2

1 + tan

✓⇡

◆�

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

�1� 2z

All angles in radians!

� = density of soil

LRV Bulldozing Example (1)

� = 33o = 0.576 rad

↵ = cos

✓1� 2z

◆= cos

✓1� 2(1.812)

◆= 17.18o = 0.2999 rad

e(1.5⇡��) tan�

⇡4 +

⌘= 32.23

= z1 tan2

✓⇡

2� �

◆= 0.5341

Nc =Nq � 1

tan�= 48.09

N� =2(Nq + 1) tan�

1 + 0.4 sin 4�= 33.27

Kc = (Nc � tan�) cos2 � = 33.37

K� =

✓2N�

tan�+ 1

◆cos

2 � = 72.77

� = 0.002595N

cm3; c

= 0.017N

✓1� 2z

◆= 12.18 cm

LRV Terzhagi Parameters

Red lines represent values calculated for this example

Rb =b sin (↵+ �)

2 sin↵ cos�

⇣⇡2� �

⌘+ c`2

1 + tan

✓⇡

◆�

hRbi = cm

cm3cm2

◆+ cm3 N

cm2cm2

Rb = 94.98 + 0.000131 + 0.014 = 95.00 N per leading wheel

Rb,total

= 190.0 N

Tandem Wheels

✓0✓1 ✓02

z1z0 z2

z0 = z1 + z2

Assume n =1

2=) P = k

P1 = kpz1 P2 = k

pz1 + z2

Soil Weight Bearing Analysis

In general,

Z ✓0

0dF cos ✓ =

Z ✓0

0Pb ds cos ✓

Z ✓0

0bkpz cos ✓ ds

Z ✓0

pr(cos ✓ � cos ✓0) cos ✓ d ✓

ds = r d ✓ z = r(cos ✓ � cos ✓0)

Generic Wheel Soil Suspension

Assuming small sinkage,

z ! small, ✓ ! small,

cos ✓ ⇡ 1

cos ✓ d ✓⇡ d✓

cos ✓ ⇡ 1� ✓2

+ (higher order terms)

W =bkr3/2p

Z ✓0

q✓20 � ✓2 d✓

W =bkr3/2p

✓20 sin

✓✓

◆+ ✓

q✓20 � ✓2

�✓0

Weight on the Front Wheel

W =⇡bkr3/2

Front wheel: z1 = r1 � r1 cos ✓0

z1 = r1 � r1

✓1� ✓20

2+ · · ·

◆=) ✓20 ⇠= 2

W1⇠=

⇡bkz1pr1

Weight on Back Wheel

Change to limits of integration:

W2 =bkr3/22p

Z ✓0

q✓202 � ✓2 d✓

0 �! ✓0, r �! r2

q✓202 � ✓2 ⇠= ✓02

✓1� 1

✓202+ · · ·

Z ✓0

q✓202 � ✓2 d✓ ⇠= ✓02✓1 �

✓31✓02

= ✓02✓1

✓1� 1

✓21✓202

◆⇠= ✓02✓1

Weight on Back Wheel

W2 =bkr3/2p

2✓02✓1

z0 ⇠= r2✓2022

z2 ⇠= r2✓212

✓02 ⇠=r2z0r2

✓1 ⇠=r2z2r2

W2 = bkp2r2

Track Depth of Tandem Wheels

Front: z1 =

⇡bkpr1

z0 = z1 + z2 =2p2W1

⇡bkpr1

(bk)22r2

Back: z2 =

bkp2r2

Much algebra then ensues...

z20 � 2p2W1

⇡bkpr1

z0 +W 2

(bk)22r2= 0

Rolling Resistance of Tandem Wheels

Solve the quadratic equation to get

⇡pr1

⇡2r1+

This was all done for n =

=) R =

bkz3/20

⇡pr1

⇡2r1+

⇡pD1

⇡2D1+

Nondimensional Forms

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

For W1 = W2 =

=) a = 1

1 + aW W2 =

1 + aW

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

⇡pD1

⇡2D1+

Define wheel diameter ratio ⇢ ⌘ D1

Nondimensional Forms

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

⇡p⇢+

1 +4a2

⇡2⇢

Let ⇠ ⌘ 2

(a+ 1)3/2

⇡p⇢+

1 +4a2

⇡2⇢

R =⇠pbk

Simple Example Case

Consider ⇢ = 1 (D1 = D2 = D)

✓W1 = W2 =

For tandem wheels, R =

0.580pbk

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

0.876pbk

Tandem wheels reduce rolling resistance by 34%

Dual Wheels

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

R =0.876p

Rdual =0.619p

Dual wheel rolling resistance 29% less than single,

7% higher than tandem

origin and nature of lunar soil • soil mechanics...1 2 k p cos2 1 tan n q = exp ⇥ 3⇡ 2 tan ⇤...

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