Pressure melting

The velocity of sliding caused by pressure melting is found from a calculation of the change in hydrostatic pressure from one side of an obstacle to another. Because the melting temperature of ice varies with pressure, there is a temperature difference across the obstacle which gives rise to the flow of heat, the melting of ice, and the freezing of water. The difference in temperature Δ T is equal to C Δ P, where C is a constant equal to 7.4×10⁻⁹°C. cm.²/dyn. and Δ P is the difference in hydrostatic pressure.

There is some ambiguity connected with the pressure difference Δ P. In the water layer existing between the ice and rock the pressure difference will be Δ P = τ L′²/L _h L _p. In the ice itself, however, the difference may be only τ L′²/3L _h L _p. The factor of

in this latter expression arises because a uniaxial compression or tension produces a hydrostatic pressure of only the magnitude of the compressive or tensile stress. (That is, through a rotation of the coordinate system, a uniaxial stress in one coordinate system can be changed in another coordinate system to a stress system containing only pure shear stresses and a hydrostatic stress. The value of the hydrostatic stress turns out to be the value of the original uniaxial compressive or tensile stress.) In our previous papers on sliding we assumed that the factor should appear in the equations. On the other hand, Reference Kamb and LaChapelleKamb and LaChapelle (1964) consider that it should not. This is a question best answered by experiment, and Kamb and LaChapelle’s (limited) experimental data do seem to support their viewpoint. Therefore in what follows we shall consider that the temperature difference across an obstacle is C τ L′²/L _h L _p rather than the smaller value we used previously. The temperature gradient across the obstacle is C τ L′²/L _h L _p L _d. If all the heat flow is through the obstacle a volume of ice is melted in a unit time which equals the temperature gradient (C τ L′²/L _h L _p L _d) times the area of an obstacle (L _h L _p) times the coefficient of thermal conductivity D of the rock divided by H _ρ , where H is the heat of fusion of ice (80 cal./g.) and p is the density of ice. The velocity of sliding is equal to this volume of melted ice divided by the area of an obstacle. The following equation is obtained for the velocity of the sliding S. which results from pressure melting:

where L is the average dimension of an obstacle (L ³=L _h L _p L _d). Except for the absence of a factor of

equation (1a) is identical to the equation previously obtained for the pressure-melting mechanism. Equation (1a) is derived with the assumption that all the heat flows through the obstacle and none goes through the surrounding ice. To take this latter heat flow into account we rewrite equation (1a) as follows:

where α is a constant whose value is 1 if all of the heat flow is confined to the obstacle and is somewhat larger than 1 if additional heat flows through the ice. The value of α could be determined by laboratory experiments.

Stress concentrations

The velocity of the sliding which results from the creep-rate enhancement caused by stress concentrations is found by noting that the volume of ice which is subjected to the concentrated stress is of the order of the volume of the obstacle itself. The creep rate

of ice is given by Glen’s creep law. It is

where σ is the stress, n is a constant equal to 3 or 4, and B is another constant which is equal to 0.017 bar⁻ⁿ /yr. when the stress is uniaxial tensile or compressive. The compressive stress σ on the up-stream side of an obstacle is equal to τ L′²/β L _h L _p, where β = 2 if the hydrostatic pressure is sufficiently large to prevent cavity formation and β = 1 if it is not. The sliding velocity S ₂ caused by the stress concentration is equal to the creep rate

times the distance in the direction of motion over which this creep rate is effective. Thus

where, as before, L is the average dimension of an obstacle. This sliding velocity is almost the same as that derived previously. The sliding velocity given by equation (3a) is, of course, only a rough estimate of the sliding velocity S ₂ since we do not know exactly the distance over which the creep rate

is effective. A more exact expression for the sliding velocity is

where b is a constant of the order of 1. The exact value of b could be determined from laboratory experiments.

Double-valued nature of the sliding velocity due to stress concentrations

It should be emphasized that in a certain range of values for the overburden pressure the sliding velocity S ₂ actually is a double-valued function of the shear stress τ. The reason for this multiplicity of value is the fact that β may take on one of two possible values. We have noted that if the hydrostatic pressure is very large β = 2 but if it is small β = 1. Suppose however that the hydrostatic pressure has an intermediate value. Let P equal the hydrostatic pressure, where P = ρ g h, ρ is the density of ice, g is the gravitational acceleration, and h is the thickness of the glacier. Let T represent the tensile stress on the down-stream side of an obstacle. This tensile stress is exerted in a direction parallel to the bed. Further, let θ represent the maximum angle between the slope of the obstacle and the average slope of the bed. Thus θ is 90° for an obstacle having part of its surface perpendicular to the average slope of the bed. The maximum tensile stress exerted normal to an ice—rock surface is T sin² θ. In order for a tensile stress to exist on the down-stream side of an obstacle the ice must remain in contact with the rock and thus the hydrostatic pressure must be larger than T sin² θ, where T = τ L′²/2L _h L _p

If the ice loes contact with the down-stream surface of an obstacle only a compressive stress exists on the up-stream side of the obstacle. Its magnitude is τ L′²/L _h L _p. A cavity will form on the down-stream side. The hydrostatic pressure will tend to close the cavity. The rate of closure (Reference NyeNye, 1953) is proportional to P ⁿ . However the cavity is being opened up by ice flow around the obstacle. If P is greater than τ L′²/L _h L _p a cavity will not be able to form. On the other hand if P is less than τ L′²/L _h L _p a cavity can remain open. The value of P can be smaller than τ L′²/L _h L _p and yet be larger than τ L′² sin²θ/2L ^h L ^p, the stress which it must exceed to prevent an ice–rock interface surface from separating and a cavity being formed. Therefore at any value of P = ρ g h in the range

a cavity which is formed behind an obstacle will remain open but if the cavity is not already in existence it will not form. Therefore β can equal either 1 or 2 in this range of values of P. As a result the sliding velocity is double valued.

The glacier sliding velocity

In our previous papers on sliding it was found that the actual sliding velocity of a glacier whose bed contains a full spectrum of obstacle sizes is determined by those obstacles for which S ₁ = S ₂. It was argued that the pressure-melting mechanism enables ice to flow easily around smaller obstacles because this mechanism gives a sliding velocity which is larger the smaller the obstacle. It was argued further that the stress-concentration mechanism enables ice to flow around the larger obstacles without any trouble since for this mechanism the larger the obstacle the greater is the sliding velocity.

We still hold to this viewpoint but we wish to refine the calculation of the actual sliding velocity. In the previous paper it was assumed implicitly that only the “controlling” obstacles, that is, those obstacles whose size is such that S ₁ and S ₂ of equations (1b) and (3) are equal, hinder the ice flow. The resistance offered by obstacles both smaller and larger than the controlling obstacles was considered to be negligibly small. This assumption obviously is only an approximation. The smaller and the larger obstacles also hinder the ice motion although not to the same degree as the controlling protuberances.

We wish now to calculate the sliding velocity of a glacier when the effect of obstacles both smaller and larger than the controlling protuberances is taken into account.

It is postulated that the glacier bed is made up of obstacles of various sizes. One simplification will be made in the distribution of obstacle sizes. It will be assumed that instead of being continuous the spectrum of sizes is discrete. If λ is regarded as the average dimension of the smallest obstacle the next largest obstacle will be taken to have the average dimension 10λ, the next largest rooA, the next 100λ and so on to the largest sized obstacles (which could be of the order of

of the thickness of the glacier).

Because of the existence of the shear stress τ, a glacier transmits over an area A a total force τA to the bed of the glacier. This force is transmitted through the obstacles and thus each size group will transmit some fraction of the total force. The obstacles in the controlling size group will, of course, transmit the major portion of the force. Suppose we let τ _i A represent that force which the obstacles of size (10) ⁱ λ transmit to the bed. We must have the condition that Στ _i A = τ A, or

The stress τ _i can be regarded as the effective shear stress which causes the flow of ice around obstacles of the size (10) ⁱ λ. Since the sliding velocity must be the same for every sized obstacle it follows that

where S is the actual sliding velocity of the glacier and the subscripts refer to the values of L, L′, etc., for the particular size of obstacle.Footnote *

Equation (6) gives τ _i in terms of the sliding velocity S. If all of the values of τ _i are substituted into the summation term of equation (5) an equation is obtained which contains only the sliding velocity S and the applied shear stress τ as the independent variables. The sliding velocity has been determined as a function of the shear stress acting at the bed of a glacier. By inspection it is obvious that this equation is a complicated one. A good approximation to this more exact equation can be found by observing that in equation (6) the first term on the right-hand side is predominant for obstacles smaller than the controlling sizeFootnote † whereas the last term predominates for obstacles larger than the controlling size. Thus we can make the approximation that S = (S ₁) _i for the smaller obstacles, that S = (S ₂) _i for the larger, and that S = (S ₁+S ₂) = 2S ₁ = 2S ₂ for the controlling obstacle size.

Now let Λ represent the average overall dimension and Λ _d the dimension in the direction of motion of the controlling obstacles. Let us assume that the “roughness” of the bed is not a function of the size of obstacles. Therefore L′ _i /L _i = r is constant for all obstacle sizes. The term r is a measure of the roughness of the bed. Also let

be a constant independent of obstacle size. Let τ _Λ represent the effective shear stress acting on the controlling obstacles. Since the sliding velocity is identical for all obstacles the effective shear stresses acting on obstacles smaller than the controlling size are τ Λ/5, τ _Λ /50, τ _Λ /500, etc. Similarly, the effective shear stresses acting on the larger obstacles are τ _Λ /(β/β _Λ )/5^1/n , τ _Λ /(β/β _Λ )/50^1/n , τ _Λ /(β/β _Λ )/500^1/n etc., where β _Λ is the value of β for the controlling obstacles.

With these values for the stresses τ _i equation (5) reduces to

when the value of β for all the larger obstacles is identical to β _Λ of the controlling obstacles. Since n ≈ 3 this last equation reduces to

If β _Λ = 1 for the controlling obstacle size but β of the larger obstacles is equal to 2, equation (7b) becomes

Equations (7b) and (7c) represent two limiting cases.

It can be seen that the effective shear stress acting on the controlling protuberances is only about

to of the applied shear stress. Thus an appreciable part of the resistance to sliding comes from obstacles other than the controlling obstacles. This additional resistance is provided principally by the obstacles which are larger than the controlling size.

If we substitute equations (7) into equation (6) and set L _i = Λ and S ₁ = S ₂, we obtain the following expression for the sliding velocity of a glacier

where k represents the constant term of equations (7) and has a value which lies between. 2.314 and 3.405. An expression for Λ is derived from the condition that S ₁ = S ₂ at the controlling obstacle size:

The substitution of this expression into (8a) results in the equation

This equation gives the sliding velocity of a glacier as a function of the shear stress acting at the bed of a glacier. Apart from a constant factor it is identical to the sliding velocity calculated previously. Both the sliding velocity and the controlling obstacle size derived previously (Reference WeertmanWeertman, 1957, Reference Weertman1962) can be obtained from these last two equations by setting k = b = γ = 1,

and β _Λ = 2, and by dividing the right-hand side of equation (8b) by 2.

Figure 1 shows plots of sliding velocity at constant stress versus the roughness factor r of the bed. (The larger is r the smoother is the bed.) Also shown in these plots is Λ, the controlling obstacle size, versus r. In the calculations for these plots it was assumed that α = 1, n = 3, γ = 1, b = 1, C = 7.4×10^−3°C./bar, B = 0.017 bar⁻³/yr., D = 0.005 cal./°C. sec. (a typical value for rocks), and τ = 1.0 bar. In the figure one set of curves was found for β _Λ = 1 and k = 2.31 and another set for β _Λ = 2 and k = 2.31. These values of β _Λ and k represent limiting cases. The set of values β _Λ = 1 and k = 3.405 correspond to curves which are intermediate to those plotted.

Fig. 1. Plot of sliding velocity S and size of controlling obstacles Λ as a function of the roughness factor r for a shear stress τ = 1 bar, k = 2.31 and β_Λ = 1 or 2. (The case of k = 3.4 and β_Λ = 1 falls between the curves shown)

In these plots the controlling obstacle size at a particular sliding velocity can be found by drawing a vertical line from the sliding velocity curve to the obstacle size curve. This is done in Figure 1 for a sliding velocity of 80 m./yr., which is of the order of the sliding velocity of glaciers having a shear stress around 1 bar at their beds. The roughness factor of a glacier which slides with the velocity of 80 m./yr. under a stress of 1 bar must be of the order of r = 15 to 20. Also shown in Figure 1 are the obstacle sizes and roughness factors connected with sliding velocities of 10 m./yr. and 1.0 m./yr. under a 1 bar shear stress. A somewhat different method of presenting the results of equations (8) and (9) is shown in Figures 2 and 3. Here we find plotted the sliding velocity and controlling obstacle size as a function of the shear stress for various values of the bed roughness.

Fig. 2. Plot of sliding velocity S and controlling obstacle size Λ as a function of the shear stress τ for various values of the roughness factor r and k = 2.31 and β_Λ = 2

Fig. 3. Plot of sliding velocity S and controlling obstacle size Λ as a function of the shear stress τ for various values of the roughness factor r and k = 2.31 and β_Λ = 1

Reference Kamb and LaChapelleKamb and LaChapelle (1964) measured a sliding rate of 5.8 m. /yr. at the end of their ice tunnel. They estimated that the roughness factor of the bed was equal to 9 and the shear stress acting on the bed was 0.7 bar. For these values of the roughness and the shear stress Figure 2 predicts a sliding rate of 2 m./yr., a value close to the observed rate. The controlling obstacle size obtained from this plot is about 4 cm. This size is comparable to the maximum thickness (about 3 cm.) of the observed regelation layer. If the maximum thickness of the regelation layer is taken as an approximate measure of the controlling obstacle size, as was suggested by J. W. Glen at the Berkeley symposium, it would appear that theory and observation agree reasonably well. Nevertheless, Kamb and LaChapelle feel that the controlling obstacle size is much greater because the regelation layer behind the obstacles they examined appeared to be undeformed by creep. However, I pointed out ([Union Géodésique et Géophysique Internationale], in press) the improbability of finding an obstacle of a size exactly equal to the controlling obstacle size. It is only for this special size that the regelation and creep flow contribute comparable amounts to the total slippage. A given obstacle is more likely to be either smaller or larger than the controlling size. If it is smaller, flow of ice around it occurs predominately by regelation; if it is larger the flow takes place largely through creep deformation.

Glen’s condition

The values of the sliding velocity and the controlling obstacle size depend upon whether or not Glen’s condition is satisfied. Glen’s condition is satisfied if the hydrostatic pressure ρgh is sufficiently large.

The effective stress on the controlling obstacles is τr ²/β _Λ k, where β _Λ = 2 when Glen’s condition holds and β _Λ = 1 when it does not. If the hydrostatic pressure ρgh is greater than τ r ²/k, Glen’s condition always is valid; if ρgh is less then τ r ² sin² θ/2k, Glen’s condition never is valid (θ is the angle between the maximum slope of an obstacle and the average slope of the bed) ; hut in the region where τr ² sin² θ/2k <ρgh < tr ²/k Glen’s condition may or may not he valid. In this region the sliding velocity is double-valued. (It is interesting to note that Meier (private communication) has some field observations which indicate, through an indirect calculation, the occurrence of a double value in the sliding velocity of Nisqually Glacier.)

Figures 4 and 5 show regions of validity and non-validity of Glen’s condition. In these figures glacier thickness h is plotted versus shear stress for various values of the roughness factor r. The intermediate areas of these plots indicate the region within which β _Λ can have either one of two values. Here the sliding velocity is double valued.

Fig. 4. Diagram of glacier thickness h and shear stress τ showing the regions in which β_Λ = 1, β_Λ = 2 and β_Λ = 1 or 2. The roughness factor r = 10 and k = 2.31. The ratio of maximum angle θ between slope of the surface of the obstacle and the average slope of the bed is assumed to be 30°

Fig. 5. Same diagram as Figure 4 expect that r = 5 and r = 15

When Glen’s condition does not hold, cavities form behind the controlling obstacles and also possibly behind obstacles larger than those of the controlling size. Reference LliboutryLliboutry (1959) first predicted the existence of cavities at the bed of a glacier. His prediction was based on a model of a glacier bed, his washboard model, which is rather different than the one we have employed. Nevertheless the physical reason for the appearance of cavities in our model of a glacier bed is much the same as in his.

The length of the cavity which is formed behind an obstacle can be estimated as follows. The stress which causes sliding around an obstacle is τ r ²/k, where β = 1. The sliding velocity is proportional to (τ r ²/k) ⁿ . Since the overburden pressure is ρ g h the velocity with which the cavity is closed is proportional to (ρ g h) ⁿ . Therefore the cavity behind an obstacle of size Λ will be closed off at a distance from the obstacle which is of the order of Λ(τ r ²/k) ⁿ /(ρ g h) ⁿ . The cross-sectional area of the cavity approaches zero at this distance. At the head of the cavity the cross-sectional area is that of the obstacle itself. The probability that another obstacle of size Λ is directly down-stream from the obstacle increases the further down-stream one goes. A simple calculation shows that the probability is of the order of 1 at the distance Λr ². Therefore the length of the cavity cannot exceed Λr ².

It is evident that when the length of the cavities becomes as long as Λr ² the bottom of the glacier touches the bed only at the tops of the obstacles. This is the condition fundamental to Lliboutry’s theory of sliding (1959). I presented an argument (Reference WeertmanWeertman, 1962) that this condition can occur only for extremely rapid sliding velocities such as would exist during the avalanching of thin ice slabs. The argument that we presented still is valid. If an appreciable fraction of a glacier loses contact with the bed through the formation of cavities the effective hydrostatic pressure at the bottom of the glacier is increased over the value ρ g h by the factor μ (the area of the bed divided by the area of ice in contact with the bed). This increase in the hydrostatic pressure increases the rate of closure of a cavity. The length of a cavity now is of the order of Λ(τ r ²/k) ⁿ /(μ ρ g h) ⁿ .

The value of μ may be estimated easily. An area (r Λ)² of the bed contains one obstacle of size Λ. The area of the glacier bed underneath the cavity behind this obstacle is approximately Λ ²(τ r ²/k) ⁿ /(μ ρ g h) ⁿ . From these two areas the following equation is found for the ratio μ:

Figure 6 shows a plot of µ versus h for various values of r. It is assumed that the shear stress τ = 1 bar and n = 3. (According to Figures 4 and 5 it is possible for β _Λ to equal 2 and to equal 1 for any ice thickness to the right of the vertical hatches in Figure 6.) If the ice is riding on top of the obstacles, as it is pictured doing in Lliboutry’s theory, µ is approximately equal to r ². In order for µ to have this value under a stress of τ = 1 bar and a roughness of r = 15, 10, or 5, the ice thickness must be less than 10 m. This conclusion is in harmony with my previous discussion (Reference WeertmanWeertman, 1962) of Lliboutry’s paper. It can be seen from Figure 6 that an appreciable separation of ice from rock can occur, although not to the extent envisaged by Lliboutry, for glacier thicknesses of the order of 100 m.

Fig. 6. Plot of the ratio µ (ratio of area of bed to area of ice in contact with bed) versus the thickness h of a glacier when τ = 1 bar, k = 2.31 and β_Λ = 1

In Lliboutry’s theory the ice separation at the bed profoundly influences the sliding velocity. It is to be emphasized strongly that the separations occuring in the present analysis do not have this strong influence on the velocity except when µ ≈ r ². In fact so long as µ < r ² the sliding velocity is not influenced by separation.Footnote *

Our analysis of cavities was based on the assumption that the cavities are free of water. Lliboutry also considered cavities filled with water. His treatment was criticized by the present author ( [Union Géodésique et Géophysique Internationale], 1963, p. 67) as being incomplete. There is no way to estimate the magnitude of the water pressure in a cavity from his analysis, yet a knowledge of this pressure is important since the sliding velocity he derived depends sensitively upon it.

I should like now to give an argument in favor of the idea of Lliboutry that cavities formed by obstacles normally are filled with water. Consider the cavity whose cross-sectional and top views are shown in Figure 7. If the cavity does not contain water the pressure P ^* of ice against rock at the periphery of the cavity will be smaller than the pressure P at the ice-rock interface a distance away from the cavity. This conclusion can be demonstrated quantitatively from Nye’s theory (1953) of the closing of tunnels for the case when n of Glen’s creep equation is equal to 3. (If n were smaller than about 2, P ^* would be larger than P according to Nye’s theory.) The water at the bottom of a glacier which exists in the ice-rock interface will always flow down a pressure gradient. Since the pressure gradient is towards the cavity a water-free cavity will become filled with water.

Fig. 7. Water-free cavity behind an obstacle (a) cross-sectional view, (b) looking down from a point directly above the cavity. The pressure P is the average hydrostatic pressure at the bed and the pressure P^* is that exerted near the cavity (P^* < P)

Suppose that an isolated cavity is filled with water. Assume that for a constant sliding velocity the cavity is in a steady-state condition. It is growing neither smaller nor larger. The water pressure at the bottom of the cavity must be identical with that under the ice. Otherwise water would flow into or out of the cavity. At the top of the cavity the water pressure could differ from the pressure in the ice by an amount which at most is of the order of (ρ _w –ρ) gΛ, where ρ _w and ρ are the densities of water and ice respectively. For an obstacle 1 cm. in size this pressure difference is of the order of 10⁻⁴ bar. Therefore only a negligibly small pressure difference is available for causing the closure of the cavity. (In a steady-state condition the closure is exactly balanced by the sliding process which opens up the cavity.) Contrary to our assumption the isolated cavity cannot be in a steady-state condition. If the cavity remains completely filled with water it must become longer and longer since there is insufficient pressure available to close it off. Eventually it will connect with other cavities. When enough cavities interconnect the water within them can drain off. Once this happens the average diameter of a cavity will decrease through the creep flow of ice until the flow of water is so restricted that the cavity again can be completely filled with water. The cavity would have a tapered profile in the direction of ice flow similar to Figure 7 except that it would not be completely closed off at its down-stream end. The length of the tapered cavity would depend not only on the size of the obstacle, the roughness, and the shear stress acting across the bed, but also on the amount of water flowing at the bottom of a glacier. If the amount of water flow is large the average diameter of interconnecting cavities must be large in order to accommodate it. If it is small the average diameter will be small. In the latter situation the total area of the water-filled cavities at the glacier bed would be about the same as for the case in which the cavities were assumed to be water free. The velocity of sliding in this situation would be the same regardless of whether the cavities were water-free or water-filled.

If the average diameter of a cavity remains large because of an abundance of melt-water supply the interconnecting water-filled cavities can be approximated in a limiting case by a sheet of water of uniform thickness at the bed of a glacier. The thickness of such a sheet of water was shown (Reference WeertmanWeertman, 1962) to be determined by the amount of water flowing through any section of the bed. We consider the effect of a water layer on the sliding velocity in the following section.

Effect of the water layer at the bed of a glacier

The field observations of Elliston ([Union Géodésique et Géophysique Internationale], 1963, p. 65–66) on the Gornergletscher show convincingly that water at the bottom of a glacier can change markedly the velocity of glacier movement. He finds that in the winter the glacier velocity is 20–50 per cent slower than the annual mean velocity, and the summer velocity exceeds the average by 20–80 per cent. These changes can be understood if melt water acts as a lubricant at the bottom of a glacier. I have shown (Reference WeertmanWeertman, 1962) that if the water layer at the bottom of a glacier is thicker than the height of the controlling obstacles an increase in the sliding velocity will occur.

In our previous treatment it was found that the water layer has no effect on the sliding rate until the thickness of the water layer is as great as the height of the controlling obstacles. We should like to point out now that a water layer with a thickness an order of magnitude smaller than the controlling obstacle size can cause an appreciable increase in the sliding velocity. The reason that in the present theory a water layer of such small thickness can affect the sliding rate may be seen from equations (5) and (6). No obstacles smaller than the thickness of the water layer can cause a hindrance to the sliding motion. Thus the effective shear stress τ_i in equation (5) acting on these obstacles is zero. As a result the effective stress on the larger obstacles is raised and the sliding velocity is increased.

When the thickness d of the water layer is smaller than Λ, equation (7a) for τ _Λ becomes

The number of terms in the series

… depends on the thickness of the water layer. If the water layer is equal to or larger than Λ/10 no term is retained in the series. If the water layer is equal to or larger than Λ/100 but smaller than Λ/10 the first term only is retained; if the layer is larger than or equal to Λ/1000 but smaller than Λ/100 the first two terms are retained; and so on. If the water layer is somewhat thicker than Λ/10 but not so thick as Λ, equation (11) reduces to

or k = 2.092. This value of k is almost 10 per cent smaller than the previous value of 2.314. Thus the sliding velocity, which is inversely proportional to the square of k, is approximately 20 per cent larger than it would be if the water layer thickness were very much smaller. This is an appreciable increase in the sliding velocity. If the water layer thickness were of the order of Λ /100, k would be 1 per cent smaller and the sliding rate 2 per cent faster, and so on for still smaller thicknesses of the water layer.

If the water layer is thicker than the height of the controlling obstacles, equations (5) and (6) predict that the sliding velocity S ^* will be

where β′ is the value of β for the obstacles which are just bigger than the thickness of the water layer and Λ ′ is the size of these obstacles. The term k is equal to 1.82 if the value of β is the same for all obstacles larger than the thickness of the water layer, and k = 2.04 if β′ = 1 for obstacles of size Λ′ and β = 2 for all larger obstacles.

Figure 8 shows a plot of the ratio of the sliding velocity S ^* when an appreciable water layer is present to the ordinary sliding velocity S versus the thickness of the water layer. (It is assumed that β = 2 for all obstacles.) The plot is a step function because we have employed a discrete rather than continuous distribution function for obstacle sizes. It is expected that the curve of S ^*/S versus water layer thickness for a continuous distribution function approximates the dashed curve drawn in this figure.

Fig. 8. Plot of the ratio S^*/S versus thickness of the water layer at the bed. (S* is the sliding velocity when an appreciable water layer is present and S the sliding velocity when the water layer thickness is extremely small.) The solid curve gives values of S*/S for the bed with the discrete spectrum of obstacle sizes considered in the text. The dashed curve shows a possible variation of S*/S with the thickness of the water layer for a bed with a continuous spectrum of obstacle sizes