Simple analytical models of glacier-climate interactions - by Prof. J ...
Simple analytical models of glacier-climate interactions - by Prof. J ... Simple analytical models of glacier-climate interactions - by Prof. J ...
H(r) = 8/5 3/8 b R /(3 R A*) 1/8 R 5/3 - r 5/3 3/8 (P 2.8) 3500 3000 2500 b=const b=b R r/R 2000 H (m) 1500 1000 500 0 -500 0 500 1000 1500 2000 r (km) 30
Problem West Antarcic ice sheet (P 3) The mass budget equation is obtained by setting the total accumulation equal to the total flux across the grounding line. This flux equals the ice velocity times the outlet cross section. Therefore π a R = 2 π R f * ρ w ρ i d 2 = 2 π R f d 2 (P 3.1) The water depth at the grounding line equals b 0 - s R, so we have d 2 = b 0 2 + s 2 R 2 - 2 b0 s R (P 3.2) Combining yields 2 f s 2 R 2 - (4 b 0 f s + a) R + 2 f b 0 2 = 0 (P 3.3) Special case: b 0 = 0 (a purely marine ice sheet). So the highest point of the continent is just at sea level. Eq. (P 2.3) reduces to 2 f s 2 R 2 - a R = 0 (P 3.4) → R = a 2 f s 2 (P 3.5) Application to the WAIS (R = 600 km, f = 1 yr -1 , a = 0.25 m yr -1 ): 1/2 → s = a = 0.25 1/2 = 0.00046 (P 3.6) 2 f R 2 x 1 x 600,000 Sensitivity to changes in accumulation rate: ∂R ∂a = 1 2 f s 2 = 2.37x106 = 23.7 km/% (P 3.7) 31
- Page 1 and 2: SIMPLE ANALYTICAL MODELS OF GLACIER
- Page 3 and 4: 1. A mass-balance model The process
- Page 5 and 6: (iii) A 0 > A 1 (continuous ablatio
- Page 7 and 8: 2. Ice deformation: perfect plastic
- Page 9 and 10: = ρ i h ρ i - ρ m = -δ h . (2.7
- Page 11 and 12: 3. A simple glacier model We consid
- Page 13 and 14: d L d T a = ∂ L ∂ E d E d T a =
- Page 15 and 16: For L < L ub the solution is given
- Page 17 and 18: 5. A volume time scale for valley g
- Page 19 and 20: Problem: • The time scale derived
- Page 21 and 22: Note that values of L for which N <
- Page 23 and 24: 7. Steady state ice-sheet profiles
- Page 25 and 26: Fig. 7.2 3500 3000 2500 h (m) 2000
- Page 27 and 28: Problem HMB-feedback and response t
- Page 29: Problem ice-sheet profile (P 2) For
H(r) = 8/5 3/8 b R /(3 R A*) 1/8 R 5/3 - r 5/3 3/8 (P 2.8)<br />
3500<br />
3000<br />
2500<br />
b=const<br />
b=b R<br />
r/R<br />
2000<br />
H (m)<br />
1500<br />
1000<br />
500<br />
0<br />
-500<br />
0 500 1000 1500 2000<br />
r (km)<br />
30
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