The publication by Mercanton and Renaud Reference Mercanton and Renaud ^{1 ,} Reference Mercanton ² of the long profile of the Unteraar Glacier measured by seismic sounding makes possible a rough comparison between theory and observation. The measured profile based on 750 soundings over the glacier surface is reproduced from reference Reference Mercanton2 in Fig. 1.

Fig. 1. Long profile of the Unteraar Glacier measured by seismic sounding, reported by Mercanton and Renaud. Vertical scale exaggerated by a factor of two. The observed curve is compared with one calculated by assuming a constant slope of the bed, β = 0.0213 radians

Suppose a wedge-shaped block of ice, whose surface slope is α, as shown in Fig. 2b (below), rests on an inclined rough plane of slope β, where α and β are both small angles. Taking the x axis parallel to the bed and pointing uphill we may consider the equilibrium of the section ABCD, of unit thickness perpendicular to the diagram, bounded by two planes drawn perpendicular to the bed separated by a distance δx. Let us assume that the normal pressure on AB is given to the first approximation by the hydrostatic head, which increases from zero at A to approximately ρgh at B, where AB = h, ρ is the density, assumed constant, and g is the acceleration due to gravity. The mean pressure is , and so the normal force on AB is . The normal force on DC is, therefore, . The other forces acting parallel to the bed are τδx, directed uphill, where τ is the shear force per unit area exerted by the bed on the ice, and hδx. ρg sin β, the component of the weight of the section, directed downhill.

Fig. 2.

Thus, to the first approximation,

Hence, carrying out the differentiation and dividing by ρgh δx, we have

But is the slope of the upper surface, α. Hence,

To this approximation, therefore, the shear stress at the bed is numerically the same as if the bed were of slope α and the block were parallel-sided rather than wedge-shaped. It is essentially the slope of the surface rather than that of the bed which, together with the local depth, decides the shear stress on the bed. A derivation of formula (1) starting from a different assumption is given in reference Reference Nye3 (equation 16).

If it is assumed that τ is constant over the bed and that β is also constant (Fig. 2 a), so that we have a block of ice of varying thickness resting on a plane slope, equation (1) may be integrated to give

where h ₀ is written in place of τ/ρg, and h=0 when x=0. This is the equation of the surface profile of the block, It does not hold near the end of the block where h is small, because there dh/dx, and hence α, become numerically large.

The calculation may now be modified so as to take account of the sides of the glacier valley. Let Fig. 2 a and b now represent the central longitudinal section of a symmetrical valley. The cross-section perpendicular to the bed at B is shown by EBF in Fig. 2c . EF is assumed to be a straight horizontal line. The cross-section at C is shown by GCH, drawn so that GH coincides with EF. Let it be assumed that the cross-section of the valley remains the same at all relevant places ; in the figure, then, EBF is at a constant vertical distance δh from GCH. The forces acting parallel to the bed on the ice between these two sections are due, as before, to the pressure difference on the sections, the weight, and the drag of the bed. The force due to the pressure difference is the force acting on the strip shown in Fig. 2c between GCH and EBF. This is

where t is the depth at a distance z from AB. The difference between taking the limits of integration from E to F and G to H is small. The force is thus

where A is the area of the section EBF.

The downhill force due to the weight is

The uphill force due to the drag of the bed is

where p is the perimeter EBF and τ is the average shear traction over the perimeter. Hence equating the forces, to the first approximation,

where R=A/p, the hydraulic radius of the section.

It will be noted that

Thus once again it is α, the slope of the surface, which, together with R, gives the average shear stress on the bed. In fact formulae (2) and (5) are both special cases of the general relation

This relation is exactly true for a glacier of constant cross-section and slope α, as may be proved by simple resolution of forces. ⁴ It is also approximately true in a glacier of slowly varying cross-section if the current values of R and the surface slope α arc used. This may be proved as follows. Formula (6) gives, approximately, the average shear stress acting on an imaginary cylindrical surface whose cross-section is the same as that of the glacier bed but whose generators are of slope α. This surface makes everywhere only a small angle with the actual bed; a small rotation only of the stress tensor is therefore needed at each point to give the shear stress on the bed. The formula is thus still a valid approximation. Formula (6) applies equally well to an ice sheet and a glacier and there is no restriction on the value of α. The only proviso is that the area of cross-section should not change rapidly.

This reasoning shows that the derivation of formulae (4) and (5) illustrated in Fig. 2 b and c is unnecessarily restrictive in that the cross-section of the valley was assumed to remain constant down its length. A derivation on the same lines in which the cross-section is allowed to change slowly would be slightly more complicated because an extra term would enter the equations due to the longitudinal component of the normal pressure exerted by the valley on the ice.

In principle, if the thickness of the glacier at any one cross-section is known, if the shape of the valley is known and if τ is assumed to be constant, equation (4) can be integrated numerically, step by step, to give the complete surface profile of the glacier. For the present application it is reasonable, as shown below, to simplify the procedure by assuming that β is constant and that R is proportional to h over the range to be considered. Thus

where c is a constant. Then equation (4) becomes

where τ′=τ/c.

Equation (7) is the same as equation (1) with the effective value of τ changed. For constant β it therefore integrates in the same way to give

where h′₀=τ′/ρg and h=0 when x=0.

To apply this equation to the Unteraar Glacier we first need to know the value of c. If the cross-section of the valley is approximated by a parabola we find that using Koechlin’s figures Reference Koechlin ⁵ the best shape of parabola is given by

Further calculation shows that R=220 m. when h=400 m. i.e. c = 0.55,

The mean value of c in this range is thus

The bed of the Unteraar as drawn in Fig. 1. may be approximated reasonably well by a straight line of slope 0.0213 radians, labelled “assumed bed.” With this value for β, equation (8) gives the best fit with the observed profile when h′₀=15.0 m. This gives τ′=1.32 bars (1 bar=10⁶dynes/cm.²) and hence τ=0.585 × 1.32=0.77 bars. The curve so calculated is drawn in Fig. 1. and labelled “theoretical surface”, so that the agreement may be judged.

As a check on the value of τ used, it may be mentioned that starting with Mercanton’s figuresReference Mercanton ⁶ for the width, depth and surface slope at three different places on the Unteraar and assuming parabolic sections in order to calculate R, the values of the average shear stress on the bed at these three places come out to be 0.566, 0.557 and 0.485 bars. The mean value is 0.54 bars, which differs from 0.77 bars by 30 per cent. Another check on the value is the fact that shear stresses of about these magnitudes have been found by GlenReference Glen ⁷ in laboratory experiments on ice to produce the rates of shear observed in glaciers ⁴