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
lean angle
mg cos 
v2
tan(LeanAngle) 
r * g * cos( )
edge angle
edge
angle
lean angle vs. turn radius for 5 slopes
V= const 20m/s
lean angle (deg)
90
75
60
50°
45
10°
30
0
10
20
30
40
turn radius (m)
50
60
lean angle vs. turn radius for 5 speeds
Slope= const 15 deg.
90
lean angle (deg)
75
35m/s
60
30m/s
15m/s
20m/s
25m/s
45
30
15
0
10
20
30
40
turn radius (m)
50
60
2
r
L
Cd 
4 r
2 * Cd
2
Length (L)
Cd
edge angle 
sidecut
snow
sidecut
Cd 
cos
Cd
Type Model Length (m) Sidecut (m) max. radius (m)
Rossignol
SL
95 Pro
GS
1.631
0.00921
36
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
Biaxial
1.670
0.00850
40
K2
GS
edge angle vs. turn radius for different skis
90
edge angle (deg)
80
70
60
Volkl DH
50
40
Volkl SG
Volkl SL
30
20
Volkl GS
10
0
Rossignol GS
0
10
20
30
K2 GS
Rossignol SL
40
turn radius (m)
50
60
angulation
angle
angulation = edge - lean
lean
angle
edge
angle
angulation vs. radius
speed=20m/s slope=15°
angulation (deg)
5
-5
-15
Volkl DH
Volkl SL
-25
Volkl SG
Volkl GS
-35
K2 GS
Rossignol GS Rossignol SL
-45
0
10
20
40
30
turn radius (m)
50
60
70
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)
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,
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
Force relationships

edge angle

Ski
shear angle
Snow
p
Fs
Fc
Fn
F

R
N
--
Ft

-
Forces
Fc = centrifugal
(cutting)
Ft = thrust
Fs = shear
Fn = normal to
shear plane
F = friction on ski
N = normal to ski
Merchant solution modified for edge angle


ski
snow
p
Fs
Fc
Fn
F

R
N
--
Ft

Fc = Fs cos  + Fn sin 
Fn = Fs / tan(--)
-
Fc = Fs(cos  + sin  / tan(--))
 for min Fc:  = (-)/2
- predicts where the snow will fail when
skidding starts - essential for the solution
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(--)))
Fc sin  tan(--)
p>
 Ls (cos  tan(--) + sin )
Edge Angle vs. Snow Penetration
12
velocity = 20 m/s
radius = 20 m
mass = 90 kg
slope = 15 deg.
snow strength = 0.06 mPa
Snow Penetration (mm)
10
8

= 0.02
6
Increasing the
friction coefficient
4
 = 0.10
2
0
5
10
15
20
25
30
35
40
45
50
55
60
Edge Angle, Theta (deg)
65
70
75
80
85
90
Edge Angle vs. Thrust Force
8000
7000
velocity = 20 m/s
radius = 20 m
mass = 90 kg
slope = 15 deg.
snow strength = 0.06 MPa
Thrust Force (N)
6000
5000
 = 0.10
4000

= 0.02
3000
Increasing the
friction coefficient
2000
1000
0
5
10
15
20
25
30
35
40
45
50
55
60
Edge Angle, Theta (deg)
65
70
75
80
85
90
Edge Angle vs. Axial Force
8000
velocity = 20 m/s
radius = 20 m
mass = 90 kg
slope = 15 deg.
snow strength = 0.06 MPa
7000
Axial Force (N)
6000
5000
 = 0.10
4000

= 0.02
3000
Increasing the
friction coefficient
2000
1000
0
5
10
15
20
25
30
35
40
45
50
55
60
Edge Angle, Theta (deg)
65
70
75
80
85
90
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.