Modeling Edging Forces in Skiing using Merchant's Theory for Metal Cutting

1 / 23

# Modeling Edging Forces in Skiing using Merchant's Theory for Metal Cutting - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Modeling Edging Forces in Skiing using Merchant's Theory for Metal Cutting' - vanessa

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 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
• Angulation
• difference between edge and lean angles
lean angle

mv²/r

lean angle

mg cos 

edge angle

edge

angle

lean angle vs. turn radius for 5 slopes

V= const 20m/s

90

75

lean angle (deg)

60

50°

45

10°

30

0

10

20

30

40

50

60

lean angle vs. turn radius for 5 speeds

Slope= const 15 deg.

90

75

35m/s

60

30m/s

lean angle (deg)

15m/s

20m/s

25m/s

45

30

15

0

10

20

30

40

50

60

r

Length (L)

Cd

waist

ski

edge angle 

sidecut

snow

Cd

Type

Model

Length (m)

Sidecut (m)

Rossignol

SL

95 Pro

1.631

0.00921

36

GS

1.641

0.00978

34

Volkl

SL

P 40

1.576

0.01238

24

GS

P 40

1.746

0.01122

32

SG

P 30

1.906

0.00938

48

DH

P 20

1.936

0.00702

66

K2

GS

Biaxial

1.670

0.00850

40

edge angle vs. turn radius for different skis

90

80

70

60

Volkl DH

50

edge angle (deg)

40

Volkl SG

Volkl SL

30

20

Volkl GS

K2 GS

10

Rossignol GS

Rossignol SL

0

0

10

20

30

40

50

60

angulation = edge - lean

angulation angle

lean angle

edge angle

speed=20m/s slope=15°

5

-5

angulation (deg)

-15

Volkl DH

VolklSL

-25

Volkl SG

Volkl GS

-35

K2 GS

Rossignol SL

Rossignol GS

-45

40

10

30

20

50

60

70

0

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)
(negative rake)

 EDGE ANGLE

(90+rake)

Ft

SKI

(tool)

M

Fr

SIDE WALL

(relief face)

SPRAY

(chip)

Shear Angle

ø

Fc

p

SHEAR PLANE

Critical Angle

F

from Brown and Outwater 1989

from Brown and Outwater 1989

On the skiability of snow,

• 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
Force relationships

Ski

Snow

p

Fs

Fc

Fn

F

R

--

Ft

N

-

edge angle

shear angle

Forces

Fc = centrifugal

(cutting)

Ft = thrust

Fs = shear

Fn = normal to

shear plane

F = friction on ski

N = normal to ski

ski

snow

p

Fs

Fc

Fn

F

R

--

Ft

N

-

Merchant solution modified for edge angle

Fc = Fs cos  + Fn sin 

Fn = Fs / tan(--)

Fc = Fs(cos  + sin  / tan(--))

 = (-)/2

Merchant’s solution

predicts where the snow will fail when

skidding starts - essential for the solution

Fc tan(--)

p >

 Ls (cos  tan(--) + sin )

Conditions for carving

Fs =  As As = Ls p / sin 

As: area of the shear plane

p: edge penetration

Ls: length of the edge in the snow

: shear strength of the snow

Fc < p  Ls / (cos  + (sin  / tan(--)))

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 of

WPI for help in preparation and equation checking.

Thanks to Dan Mote for explaining that skiing is

machining.

Thanks to Branny von Turkovich for teaching me

machining.