The Friction of tilted Skates on Ice

The friction felt by a speed skater is calculated as function of the velocity and tilt angle of the skate. This calculation is an extension of the more common theory of friction of upright skates. Not only in rounding a curve the skate has to be tilted, but also in straightforward skating small tilt angles occur. The tilt increases the friction substanstially and even for small tilts the increase is relevant. The increase of the friction with the velocity, which is very slow for the upright skate, becomes more pronounced for large tilts. keywords: solid friction, fluid mechanics, lubrication.


Introduction
Skating is an intriguing sport from the physics viewpoint, as ice seems to be the only substance that allows skating in a remarkable range of temperatures, velocities and skater weights. The physical problem is twofold: the ice should allow to push oneself forward and the friction should be low enough to glide. This is achieved by the form of the skate which has sharp edges and a thin profile, leading to a small friction in the forward direction and a large friction in the sideways direction. It is mainly the small friction in the forward direction that begs for an explication. Indeed the measured friction forces are orders of magnitude lower than the friction between steel and another solid medium. In spite of the fact that skating has been around for centuries, there is still no consensus on the friction mechanism. One school of thought [1,2,3] holds that the key to skating is the structure of the surface of ice that is "wet" within the temperature range from 0 to -30 centigrade. "Wet" means that the surface layers of ice are very mobile. The similar Arrhenius behaviour of the surface mobility and the friction supports this explanation.
The more conventional school of thought [4,5,6] explains the low friction by the formation of a liquid layer between skate and ice. The friction provides the heat to melt a layer and the layer co-determines the friction. Important is an additional friction component as the skate ploughs through the ice. But also in this school there are differences of opinion in the precise mechanism of friction, in particular regarding the relation between the pressure in the water layer and the reaction rate of the ice to that pressure. These studies deal with the friction of an upright skate. However the upright position is rather rare in skating. Even in straightforward skating the skate mostly has a small tilt angle. In curves the tilt angle may be very large i.e. substantial more than 45 0 . There are few measurements of the friction in real skating. The only measurements to date are of de Koning et al. [7], which indicate that the upright position gives the lowest friction, but detailed values of the friction as function of the tilt angle are lacking. The upright position has obtained the most attention as it is easier to treat because of the left-right symmetry and the way the ice is touched. In this note we study the friction of a skate under an arbitrary tilt angle.
The tilt angle may result from two reasons: one is that of the beginner in skating, seeking stability from the large transverse friction. An experienced skater on the other hand has the skate permanently in line with the legs, such that the system of body and skate can be considered as rigid. We are interested in the latter case. Fig. 1 shows a speed skater in a curve with a rather large tilt angle. (Short trackers experience in the curve of the track, which has a shorter radius of curvature, even larger tilt angles.) The body needs for stability a tilt angle parallel to the resultant of the gravitational force M g and the centrifugal force M V 2 /R c , where M is the mass of the skater, V the velocity and R c the radius of the curve. These forces act on the center of mass of the skater and must be compensated by an equal and opposite force from the ice exerting on the skate. From this equilibrium we can calculate the tilt angle ψ of the skater Although the skate is perfectly in line with the standing leg, the tilt angle of the skate is not the same as that of the body (defined by the line from the skate to the center-of-mass of the skater). For the forces on the skate we need the tilt angle φ of the skate. As the Fig. 1 shows, ψ will be somewhat smaller than φ. We will calculate the difference. This implies that not only the basis of the skate feels a pressure, but to a lesser extend, also the side. Actually the way in which the skate is pushed against the ice is complicated, as it can be varied by the muscles in the foot of the skater, without changing the overall forces as weight and centrifugal force. The pressure can be shifted from front to rear and from basis to side. Moreover the equilibrium to which we alluded in Eq. (1) does not need to be realised as the skater can shift weight from one foot to the other. Here we leave all these nuances aside and concentrate on the friction of a skate which makes a tilt angle φ with the normal and we do not consider force components in the forward direction other than the friction between the skate and the ice.
We study the friction of a skate which is mainly pushed in the ice in the direction of the tilt (see fig. 2). Our study is an extension of the treatment given earlier for the upright skate [6], which appears here as limit of zero tilt angle. There are many parameters such as temperature, weight, speed, mass and type of skate, which all can be varied, leading to myriad of cases. In order to avoid this, we focus on "standard" skating conditions: the skater's mass M = 72 kg, the velocity V = 10 m/s, the curvature of the skate blade R = 22 m, the width of the blade w = 1.1 mm and the temperature T a few degrees below freezing, which we acknowledge by the choice for the hardness of ice p h = 10 MPa. The hardness is the maximum pressure that one can exert on ice while remaining in the elastic deformation regime. Unfortunately the data in the literature on the hardness of ice show a large variation [9,10,2]. The choice p h = 10 MPa is a compromise.
Our main interest is the tilt angle dependence of the friction. After introducing the useful coordinate systems, the pressure in the water layer is derived from the hydrodynamic equations. A hydrodynamic treatment is relevant since the water layer has a thickness of the order of a µm. Important for the solution are the boundary conditions, for which we derive expressions. The calculation of the thickness of the water layer proceeds along similar lines as in the case of an upright skate [6]. The water layer at the basis and the side have to be discussed separately, due to different boundary conditions. The paper closes with a presentation of the results and a discussion of the main features of the solution.

The Geometry of the Indentation
There is ample evidence that a skate deforms the ice plastically. The skater leaves a visible trail behind in virgin ice and a skating rink has to be mopped up regularly in order to improve the skating conditions. Recently Th. Boudewijn [8] has carefully measured the indentation that a skate (in the upright position) leaves behind after it has been pushed into the ice. He finds a trough with sharp walls at the position of the edges of the skate. So the indentation is not elastic but plastic. The skate ploughs a furrow in the ice, which is as deep as the skate penetrates. Thus we ignore possible elastic deformations.
The force drives the skate into the ice mainly in the direction of the tilt. For a sufficiently large tilt the skate touches the ice with one side and with the basis of the blade. The other side remains up in the air. A picture in the transverse direction of the skate is drawn in fig. 2. We have to treat two different surfaces between skate and ice: the side surface, which hardly exerts pressure on the ice and the basis surface which is driven into the ice by a force. In order to describe them it is convenient to use a rotated coordinate system, with coordinates (ζ, ξ) instead of the standard reference system with coordinate z in the vertical direction and y in the transverse direction. The forward coordinate x is the same in both systems. The (ζ, ξ) system is rotated over the tilt angle φ in the (z, y) plane. The x axis is along the line where the skate meets the ice at the side. The length over which the skate makes contact with the ice the along x axis is the The boundaries of the side segment are the line ζ = 0 where it meets the ice surface and the locus of the edge is given by the functions The boundaries of the basis segment at the ice surface is given by the edge ξ = 0 and the line where it meets the ice surface given by the functions We clearly have d = −ζ e (0) and f = ξ s (0). A point on the edge of the skate is denoted by the coordinate x e . It can equally well be labelled by the corresponding ζ e or ξ e . The friction is the sum of the friction in the two surfaces. Both have a friction due to the water layer which forms between ice and skate and the basis piece has in addition a ploughing friction, due to the generation of the dent into the ice by the skate. We can calculate the frictions in a similar way as the in the upright position with as main difference the geometric shape of the contact surfaces.
All parameters can be expressed in the contact length, which is determined by fitting the applied force.

The boundary conditions for the layers
The counter forces of the ice are mediated to the skate by the water layers. Therefore we have to know the pressure distribution in the water layers. In Appendix A we summarise the standard theory for the velocity field v and the pressure p in a layer of water of thickness h, which is sheared on top by a skate with velocity V in the x direction and sticks to the ice at the bottom. There is a force on the layer in the downward direction and due to the resulting pressure, the water also flows in the transverse y direction, with a much smaller velocity v y . The water layer is pushed down at the top with a velocity v sk and the ice yields with a rate v ice at the bottom. The flow and the pressure are characterised by parameters a, b and c. We here only need the pressure at the top and bottom of the layer, of which the dependence on the transverse y direction is given by with η = 1.737 · 10 −3 Pas. This expression is applied both to the basis water layer and the side water layer. The parameters of the side layer are indexed as a s , b s , c s with the coordinate y replaced by ζ and z by ξ. Those at the bottom are a b , b b , c b with coordinate y replaced by ξ and z by ζ. Thus we have to determine six parameters. Five of them follow from the boundary conditions on the pressure, the parameter a b will be determined from the rate at which the water layer is squeezed out at the sides.
In previous calculations [4,6] the pressure was assumed to vanish at the edges of the skate in the upright case, in the idea that it should equal the outside air pressure, which is virtually zero as compared to the MPa pressures inside the layer. However, measurements [8] of the track left behind by the skate, indicate sharp walls in the ice caused by the edges. So at the edge the pressure must have been equal or larger than the hardness in order to give these plastic deformations. Therefore we expect the pressure in the basis water layer to exceed the hardness p h and take it equal at p h , at the point where the side layer meets the surface of the ice (ζ = 0). Fortunately a recalculation of the friction of the upright skate with the new boundary conditions, does not give a large difference for the friction. For completeness we give in Appendix B the changes in the equation with respect to the previous calculation [6], due to the changes in the boundary conditions.
On the other hand the formation of the side layer does not involve a (plastic) deformation and therefore the pressure in the side water layer will be below the hardness p h . Continuity of the pressure then implies that the pressure at the edge, sandwiched by the two layers, will be equal to p h . At the point where the side layer meets open air, we assume the pressure to vanish.
We first discuss the side layer and observe that a s = 0. It follows from Eq. (17) since both velocities vanish: v sk = 0, because the skate does not move in the direction ζ of the layer thickness and v ice = 0, as the pressure in the layer will not exceed the hardness. At the edge the variable ζ in the side surface assumes the value ζ e (x). So the condition that the pressure equals p h at the edge implies For ζ = 0 the pressure vanishes, which gives the second condition c s = 0. Therefore b s has the value leading to the expression for the pressure in the side water layer .
For the basis layer the coordinate ξ = 0 at the edge. This gives the condition At the other side of the basis layer, at the value ξ = ξ e , we have again the pressure p h , leading to the condition Combining these relations we can write the pressure Eq. (5), in the basis layer as Anticipating that we can find a b from relation (17), once we have solved the thickness h b , we have found an expression for all the parameters. But we still have to find a proper boundary condition for the integration of the thickness of the water layer at the side surface. Take a point x e , the value of x where it starts from the edge. The thickness h s (x e ) is the value of the layer on the side at the edge. Let the corresponding layer thickness on the basis at the edge be h b (x e ). Then we require the integrated flow towards the edge in the basis layer to be equal to the flow in the side layer away from the edge to the outside.
The minus sign results from the fact that the flow in the basis layer is in the −ξ direction and in the side layer in the +ζ direction. Later on we will see that we do not need to calculate the h b (x e ) in order to find h s (x e ).

Force and Friction at the basis
We will now use the layer equation for the generation of the water layer derived in a previous paper [6] reading It describes the growth of the thickness h downward along the skate. The first term on the right hand side gives the growth due to melting. The second term accounts for the compression due to the rate v sk at which the ice comes down and rate v ice at which ice gives in due to the pressure in the water layer. k is a small length (∼ 10 −11 m) with ρ is the density of ice ρ = 916.8 kg/m 3 and L h the latent heat of melting L h = 0.334 · 10 6 J/kg. The expression for v sk reads As long as the pressure in the water layer is below the hardness p h of ice v ice = 0. For higher pressures we assume the relation (like a Bingham solid) In the calculations we set γp h = 2 mm/s, which seemed a reasonable guess [6]. The last equation that we use, follows from the incompressibility of water, leading to the relation For the layer equation of the basis water layer we first have to solve for the constant a b . From Eq. (16) and Eq. (11), we get for v ice at the bottom of the layer v ice = γηa b ξ(ξ e − ξ).
Inserting this into Eq. (17) and using Eq. (15) for v sk yields for a b Using this expression in Eq. (17) gives for the layer equation of the basis the form with ξ e (x) given by Eq. (4). The equation has to be integrated from x = x s (ξ), where h b (x s , ξ) = 0, downwards to x = 0. Note that at the edge, where ξ = 0 the equation simplifies to The pressure follows from Eq. (11) and Eq. (19) as The friction, due to the water layer is given by the integral Apart from this friction we have also the ploughing friction as a result from making the indentation. It is given by the integral The fraction x/R gives the component of the force that has to be exerted in the forward direction. The normal force exerted by the basis surface on the skate equals

Force and Friction on the side
The layer equation for the water layer at the side simplifies since both v sk = 0 and v ice = 0. The former since the ice is not pushed down at the side and the latter since the pressure stays below p h . So the layer equation becomes for side layer h s The integration starts from x e at the edge, with the solution The initial condition h s (x e ) has to be determined from Eq. (12). The thickness depends implicitly on the value of ζ corresponding to the value of x e on the edge. To make this dependence explicit we first rearrange the expression for h s (x e ). The force and friction on the side can be disentangled from that of the basis, although they seem connected by the flow Eq. (12). At the edge, where ξ = 0 we have, due to Eq. (19), for a b the value which implies for b b (x), with the aid of Eq. (10), Then Eq. (12) gives for the boundary value h s (x e ), with the value Eq.
As one sees this is independent of h b (x). We rewrite the expression a bit with the aid of Eq. (14) as Using the values of ρ, L h and p h = 10 MPa for ice, one has for the dimensionless combination λ λ = ρL h p h 30.
We use the result (31) for the solution of the layer h s . The integration of the thickness of the side water layer proceeds by fixing a value of ζ in the region −d ≤ ζ ≤ 0 and starting the integration at x e (ζ) on the edge. Writing ξ e also as function of the corresponding ζ via ξ e = −ζ(f /d), the value of h s (ζ) at the edge becomes the initial thickness h 0 (ζ) With this value of h 0 the solution for the side water layer reads more explicitly The friction is then, for a value of ζ, given by .
The integral is elementary and reads The total friction is the integral which has to be performed numerically. The total friction is the sum of three contributions The pressure in the side layer is given by Eq. (8). The total force on the side layer equals 6 The Force Balance The forces F b and F s of the ice on the skate, given by Eqns. (25) and (39), are normal to the surface. The components in the z and y direction are formed as the combinations The component in the z direction balance the weight M g of the skater and the y component balances the centrifugal force Thus we find the relation between the body inclination ψ and the skate tilt φ as Since both F b and F s are positive, ψ < φ, as one observes from Fig. 1. An alternative form of Eq. (42) reads which also follows directly from the balance of forces in the coordinate system of the skates. In order to get an impression of these tilt differences we have plotted in Fig.4 φ − ψ as function of the velocity for a number of tilt angles φ. The difference decreases slowly with the velocity and increases with the tilt angle. For a given tilt angle φ the radius of curvature has to chosen such that Eq. (1) is fulfilled. Velocities below 1 m/s make little sense in rounding a curve.

Results
For the calculation of the friction one has to know the thickness of the water layers. That of the basis follows from the integration of Eq. (20) in the x direction for each ξ in the interval 0 ≤ ξ ≤ f . If f exceeds the width of the skate w (as happens for very small tilt angles), the upper limit of the ξ interval has to be replaced by w. The thickness of the side water layer is given by Eq. (32) for the values of ζ in the interval −d ≤ ζ ≤ 0. In general the side layer contributes a modest amount to the friction, only at large tilt angles and high velocities it starts to count. In Fig. 5 we have plotted the contributions of the friction of the water layers and the ploughing friction, together with the total friction as function of the tilt angle. The chosen velocity is V = 10 m/s. One observes that, while the layer friction is rather insensitive, the ploughing friction increases as function of the tilt angle. As a result the friction in a curve is substantially larger friction than that of the upright skate. This happens already for small tilt angles. One should realize that a tilt of a few degrees easily occurs even for straightforward skating, rendering already some 20 % increase in the total friction.
In Fig. 6 we have plotted the friction as function of the velocity for various tilt angles, which can be translated with Eq. (2) in the radius R c of the curve. We see that the smaller Figure 5: The friction contributions as function of the tilt angle the radius of the curve the larger the friction. Apart from a region of velocities smaller than walking speed, the increase sets in for all curvatures at higher speeds. Note that for tan(φ) = 0.1, which can hardly be called a tilt, the friction already rises from 1.2 at slow speeds to almost 2 for V =10 m/s. This is surprising since the concomittant equilibrium curvature of the stroke is R c = 100 m for V =10 m/s, which is not distinguishable from a straight stroke.
By our choice of the boundary conditions on the pressure we have always a positive contribution to the force exerted by the ice on the side of the skate. This means that the resultant of the force on the basis and the side has a direction (slightly) different from the tilt angle φ of the skate. The resultant points in the direction of the center of mass of the skater as seen from the skate on the ice.

Conclusion
Using that skates trace a furrow in the ice by a plastic deformation, we have derived a set of equations from which the forces on the skate can be calculated by (numerical) integration of the layer equations for the basis and the side of the skate. We get the forces as function of the contact length. For a giving weight of the skate one must find the contact length by an iterative procedure matching the normal force component on the ice with the weight of the skater.
The problem requires boundary conditions on the pressure in the water layer, resulting from melting due to the frictional heat. The plastic deformation implies that the pressure exceeds the hardness in the basis layer and stays below the hardness p h in the side layer. Continuity of the pressure then gives the value p h at the edge of the skate. As a result we find the a small force on the side of the skate. In the equilibrium situation where the centrifugal force and the normal force are both balanced by the generated forces of the ice on the skate, we find the tilt angle ψ of the body as function of the tilt angle φ of the skate.
Our calculation is a simplified version of the equations proposed for the upright position [4,6] where heat generation and flows in the bulk are taken into account. In general these extensions are of minor influence except for more special circumstances as very low temperatures. We have focused on the influence of the tilt angle and find that the friction increases substantial with the tilt angle. In particular, for the unavoidable small tilts occurring in skating, the influence is large as shown in Fig. 5. Note that the friction increases slowly with the velocity at fixed tilt angle, a feature which is also found in the upright skate. The mechanism is the same: due to the increasing pressure in the basis water layer the skate is lifted (aqua planing) and therefore the skate makes less contact with the ice, which lowers the friction.
Although the calculated friction for tilts is larger than in the upright position, the values are still about half those found in real skating experiments [7]. Part of it can be explained by our idealisation of the circumstances: perfectly polished ice and skate surfaces. The above mentioned influence of the heat generation and flow in the bulk ice will also enlarge the friction. By changing parameters as the coefficient γ, about which little is known and the hardness p h , which is poorly known, one could bring theory and experiment closer together, but in view of the risk of unrealistic values, there is as yet no point in fitting the parameters to the experiment. .
Due to the pressure water is also squeezed out sideways in the y direction as a Poisseuille flow v y = −(ay + b)z(z + h).
At the top and bottom of the layer v y = 0. The velocity in the z direction has the profile v z = a z 3 3 This component is dictated by the requirement of incompressible flow and the condition that v z = 0 for z = −h. The pressure which causes the flow in the y direction has the form with η = 1.737 · 10 −3 Pas, the viscosity of water. On verifies that the hydrodynamic equations ∇p = η∆v, are fulfilled. The constants a, b and c have to be obtained from the boundary conditions.

B The upright skate
In the upright position the skate touches the ice the the interval −w/2 < y < w/2, where w = 1.1mm is the width of the skate. We find for the pressure, at the top or bottom of the water layer, the expression p(x, y) = p h + ηa(x) w 2 4 − y 2 The constant p h raises the pressure at the edges equal to the hardness, which was missing in the previous treatment [6]. It was shown there that to a good approximation we may connect v ice with the average v ice = γ w/2 −w/2 dw w (p(x, y) − p h ) = γηa(x)w 2 /6.
Combining this with the incompressability of water, as expressed in Eq. (17), gives the relation for a(x) .
Then we can write the layer growth equation explicitly as and the pressure expression as p(x) = p h + ηV xw 2 R(h 3 (x) + γηw 2 ) .
Integration of Eq. (53) yields the layer thickness h(x) and Eq. (54) gives the pressure profile.