Modeling Edging Forces in Skiing using Merchant's Theory

for Metal Cutting

Christopher A. Brown

Mechanical Engineering Department

Worcester Polytechnic Institute

Worcester, Massachusetts, USA

outline

• Lean and edge angle– speed, radius, side cut and angulation

• Ski-snow forces– Merchant theory– friction, edge angle and penetration

Lean and edge angle

• Lean angle and balancing centrifugal forces– changes with speed and slope

• Edge angle and geometric turning– considering side cut radius

• Angulation– difference between edge and lean angles

lean angle

mv²/r

mg cos

lean angle

)cos(**)tan(

2

gr

vLeanAngle

edgeangle

edge angle

lean angle vs. turn radius for 5 slopesV= const 20m/s

30

45

60

75

90

0 10 20 30 40 50 60

turn radius (m)

lea

n a

ng

le (

de

g)

50°

10°

lean angle vs. turn radius for 5 speedsSlope= const 15 deg.

15

30

45

60

75

90

0 10 20 30 40 50 60turn radius (m)

lea

n a

ng

le (

de

g)

15m/s 20m/s30m/s

25m/s

35m/s

Length (L)

r

Cd

rCd

LCd

*24

22

waist

edge angle

Cd

sidecut

snow

ski

cos

sidecutCd

Rossignol

Volkl

K2

SL

Type

SG

GS

Model

DH

P 40

95 Pro

Length (m)

Biaxial

GS

SL

GS

P 40

P 20

P 30

0.00921

Sidecut (m)

0.01238

0.00978

0.00938

0.01122

0.00702

0.00850

1.631

1.641

1.670

1.906

1.746

1.576

1.936

max. radius (m)

36

24

32

48

66

40

34

edge angle vs. turn radius for different skis

0

10

20

30

40

50

60

70

80

90

0 10 20 30 40 50 60

turn radius (m)

ed

ge

an

gle

(d

eg

)

Volkl SG

Volkl GS

Volkl DH

Volkl SL

Rossignol SLRossignol GSK2 GS

angulation angle

edge angle

lean angle

angulation = edge - lean

angulation vs. radius

-45

-35

-25

-15

-5

5

0 10 20 30 40 50 60 70turn radius (m)

an

gu

lati

on

(d

eg)

speed=20m/s slope=15°

Volkl SL

Rossignol GS

Volkl GS

Rossignol SL

K2 GS

Volkl SG

Volkl DH

Ski snow forces -Machining analogy

• Tool = Ski

• Workpiece = Snow

• Cutting = Skidding • limiting condition on carving

• Cutting force = Turning force

• Rake angle = Edge angle (+90 deg)

M

Fr

Fc

Ft

SIDE WALL(relief face)

SKI(tool)

(negative rake)

EDGE ANGLE(90+rake)

p

SHEAR PLANE

Shear Angleø

SPRAY

(chip)

F

Critical Angle

from Brown and Outwater 1989

from Brown and Outwater 1989 On the skiability of snow,

Objectives of machining calculations

- minimum conditions for carving

• Turning force from mass, speed and radius

• Edge penetration – as a function of edge angle and friction

• Thrust force (normal to the snow)– can be influenced by body movements

FcFn

FtR

N

F

--

-

pFs

SnowSki

Force relationships

ForcesFc = centrifugal

(cutting)Ft = thrustFs = shearFn = normal to

shear planeF = friction on skiN = normal to ski

shear angleedge angle

FcFn

FtR

N

F

--

-

pFs

snowski

Fc = Fs cos + Fn sin

Fn = Fs / tan(--)

Fc = Fs(cos + sin / tan(--))

for min Fc: = (-)/2

- predicts where the snow will fail whenskidding starts - essential for the solution

Merchant solution modified for edge angle

Conditions for carving

Fs = As As = Ls p / sin

As: area of the shear planep: edge penetration

Ls: length of the edge in the snow: shear strength of the snow

Fc < p Ls / (cos + (sin / tan(--)))

p > Fc sin tan(--)

Ls (cos tan(--) + sin )

Edge Angle vs. Snow Penetration

0

2

4

6

8

10

12

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90

Edge Angle, Theta (deg)

Sn

ow

Pe

ne

tra

tio

n (

mm

)

velocity = 20 m/sradius = 20 mmass = 90 kgslope = 15 deg.snow strength = 0.06 mPa

Increasing the friction coefficient

= 0.10

= 0.02

Edge Angle vs. Thrust Force

0

1000

2000

3000

4000

5000

6000

7000

8000

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90

Edge Angle, Theta (deg)

Th

rus

t F

orc

e (

N)

Increasing the friction coefficient

= 0.10

= 0.02

velocity = 20 m/sradius = 20 mmass = 90 kgslope = 15 deg.snow strength = 0.06 MPa

Edge Angle vs. Axial Force

0

1000

2000

3000

4000

5000

6000

7000

8000

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90

Edge Angle, Theta (deg)

Ax

ial F

orc

e (

N)

velocity = 20 m/sradius = 20 mmass = 90 kgslope = 15 deg.snow strength = 0.06 MPa

Increasing the friction coefficient

= 0.10

= 0.02

discussion

• Negative now angulation predominates

• Edge roundness, penetration and length– shorter skis should hold better

• Penetration can be a function of snow strength

• Leg strength should put a lower limit on edge angle

acknowledgements

Thanks to Chris Hamel and Mike Malchiodi ofWPI for help in preparation and equation checking.

Thanks to Dan Mote for explaining that skiing is machining.

Thanks to Branny von Turkovich for teaching memachining.