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Cauldron subsidence and subglacial floods

Published online by Cambridge University Press:  14 September 2017

G.W. Evatt
Affiliation:
Mathematical Institute, University of Oxford, 24–29 St Giles’ Way, Oxford OX1 3LB, UK E-mail: evatt@maths.ox.ac.uk
A.C. Fowler
Affiliation:
Mathematical Institute, University of Oxford, 24–29 St Giles’ Way, Oxford OX1 3LB, UK E-mail: evatt@maths.ox.ac.uk
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Abstract

Ice cauldrons are depressions which form at the surface of ice sheets when an underlying subglacial lake empties, in particular when subglacial volcanic eruptions occur. Notable examples of such cauldrons occur on the surface of the Vatnajökull ice cap in Iceland. More generally, cauldrons will form when a subglacial lake empties during a jökulhlaup. The rate of subsidence of the ice surface is related to the rate at which the subglacial water empties from the lake. We use a viscous version of classical beam theory applied to the ice sheet to determine the relation between the subsidence rate and flood discharge. We use the results to make inferences concerning ring fracture spacings in cauldrons, the consequent effect on flood discharge dynamics and the likely nature of subsidence events in the Antarctic Ice Sheet.

Information

Type
Research Article
Copyright
Copyright © The Author(s) [year] 2007 
Figure 0

Fig. 1. An ice cauldron forming after the Gjálp eruption under Vatnajökull in 1996. The cauldron is about 2 km in diameter and hundreds of metres deep. Ring shear fractures can be seen, indicating yield of the ice. The subsidence rate of the ice surface was initially about 12 mh–1; for further information see Guðmundsson and others (2004). Photograph courtesy of M.T. Guðmundsson.

Figure 1

Fig. 2. Schematic diagram of cauldron geometry.

Figure 2

Fig. 3. Solution of Equation (35) (beginning at t = 0 days) at times t = 350 days (a), t = 371 days (b), t = 382 days (c) and t = 393 days (d) using Equation (40) for Nþ, with t = 1, c = 2.5 10–3, α = 3, γβ/ν2 = 15.8, and a time scale of t0 = 416 days. The units of vertical subsidence are in metres, and the units of horizontal distance are km. The results in the figure correspond to a choice of μd = N0ig = 3 m, and thus N0 = 0.3 bar. The results are plotted for a lake of width 20 km, in which case if d = 4 km and γβ = 0.2 γβ = 0.63, corresponding to a value of A = 0.48 bar. No surface fracture is expected in this case.