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static_stability
Compute static stability.
Available in version 6.3.0 and later.
Prototype
load "$NCARG_ROOT/lib/ncarg/nclscripts/csm/contributed.ncl" ; This library is automatically loaded
; from NCL V6.2.0 onward.
; No need for user to explicitly load.
function static_stability (
p : numeric, ; float, double, integer only
t : numeric,
dim [1] : integer,
sopt [1] : integer
)
return_val [dimsizes(t)] : float or double
Arguments
pArray containing pressure levels (Pa).
tArray containing temperatures (K).
dimThe dimension of t which corresponds to p.
sopt- sopt=0, Return static stability only
- sopt=1, Return static stability, theta, dthdp as type list
Return value
A multi-dimensional array of the same size and shape as t. The output will be double if t is of type double.
Description
Young (2003): "Static stability measures the gravitational resistance of an atmosphere to vertical displacements. It results from fundamental buoyant adjustments, and so it is determined by the vertical stratification of density or potential temperature. It influences the dynamics of many kinds of atmospheric motions, which in turn are responsible for determining its variations."
The formulation used to derive the static stability is from Bluestein (1992): pg 197, eqn 4.3.8. Specifically:
s = -T*d[log(theta)]/dp = -(T/theta)*d(theta)/dpReferences:Bluestein, H.B. (1992): Synoptic-Dynamic Meteorology in Midlatitudes Volume 1: Principles of Kinematics and Dynamics Young, J.A. (2003): Static StabilitySee Also
Examples
Example 1: Return static stability only.
; PRESSURE p1 =(/ 1008.,1000.,950.,900.,850.,800.,750.,700.,650.,600., \ 550.,500.,450.,400.,350.,300.,250.,200., \ 175.,150.,125.,100., 80., 70., 60., 50., \ 40., 30., 25., 20. /)*100 p1@units = "Pa" ; TEMPERATURE (C) t1 =(/ 29.3,28.1,23.5,20.9,18.4,15.9,13.1,10.1, 6.7, 3.1, \ -0.5,-4.5,-9.0,-14.8,-21.5,-29.7,-40.0,-52.4, \ -59.2,-66.5,-74.1,-78.5,-76.0,-71.6,-66.7,-61.3, \ -56.3,-51.7,-50.7,-47.5 /) t1 = t1 + 273.15 ; change to Kelvin t1@units = "K" s1 = static_stability (p1, t1, 0, 0) ; dim=0, opt=0 printVarSummary(s1) print(p1+" "+t1+" "+s1)would yieldVariable: s1 Type: float Total Size: 120 bytes 30 values Number of Dimensions: 1 Dimensions and sizes: [30] Coordinates: Number Of Attributes: 2 long_name : static stability units : K/Pa p(Pa) t(K) s(K/Pa) (0) 100800 302.45 -0.00064084 (1) 100000 301.25 -0.000125448 (2) 95000 296.65 0.000176932 (3) 90000 294.05 0.00042571 (4) 85000 291.55 0.000482328 (5) 80000 289.05 0.000504421 (6) 75000 286.25 0.000513097 (7) 70000 283.25 0.000518761 (8) 65000 279.85 0.000533717 (9) 60000 276.25 0.000600369 (10) 55000 272.65 0.00066135 (11) 50000 268.65 0.00069125 (12) 45000 264.15 0.000652553 (13) 40000 258.35 0.000604632 (14) 35000 251.65 0.000575345 (15) 30000 243.45 0.00048334 (16) 25000 233.15 0.000421649 (17) 20000 220.75 0.000472445 (18) 17500 213.95 0.000697178 (19) 15000 206.65 0.000996432 (20) 12500 199.05 0.00226535 (21) 10000 194.65 0.00519965 (22) 8000 197.15 0.00892678 (23) 7000 201.55 0.0129963 (24) 6000 206.45 0.0151663 (25) 5000 211.85 0.0175898 (26) 4000 216.85 0.0208443 (27) 3000 221.45 0.0230898 (28) 2500 222.45 0.0303216 (29) 2000 225.65 0.0339061Example 2
Same as Example 1 but opt=1 so three variables are returned: static stability, potential temperature and d(theta)dp.opt = 1 ; return 3 varaibles as part of a list dim = 0 S1 = static_stability (p1, t1, opt, dim) printVarSummary(S1) S1_s = S1[0] ; explicitly extract each variable from the list S1_pt = S1[1] ; not necessary but clearer S1_dthdp = S1[2] printVarSummary(S1_s) printVarSummary(S1_pt) printVarSummary(S1_dthdp)would yieldVariable: S1 Type: listTotal items: 3 Variable: S1_s Type: float Total Size: 120 bytes 30 values Number of Dimensions: 1 Dimensions and sizes: [30] Coordinates: Number Of Attributes: 2 long_name : static stability units : K/Pa Variable: S1_pt Type: float Total Size: 120 bytes 30 values Number of Dimensions: 1 Dimensions and sizes: [30] Coordinates: Number Of Attributes: 2 units : K long_name : potential temperature Variable: S1_dthdp Type: float Total Size: 120 bytes 30 values Number of Dimensions: 1 Dimensions and sizes: [30] Coordinates: Number Of Attributes: 2 long_name : vertical derivative of theta with pressure units : K/Pa p(Pa) t(K) s(K/Pa) Pot(K) d(theta)dp (0) 100800 302.45 -0.00064084 301.762 0.000639381 (1) 100000 301.25 -0.00012544 301.25 0.000125448 (2) 95000 296.65 0.000176932 301.034 -0.000179547 (3) 90000 294.05 0.00042571 303.045 -0.000438733 (4) 85000 291.55 0.000482328 305.421 -0.000505276 (5) 80000 289.05 0.000504421 308.098 -0.000537662 (6) 75000 286.25 0.000513097 310.798 -0.000557098 (7) 70000 283.25 0.000518761 313.669 -0.000574472 (8) 65000 279.85 0.000533717 316.543 -0.000603696 (9) 60000 276.25 0.000600369 319.706 -0.000694812 (10) 55000 272.65 0.00066135 323.491 -0.000784671 (11) 50000 268.65 0.00069125 327.553 -0.00084281 (12) 45000 264.15 0.000652553 331.919 -0.000819968 (13) 40000 258.35 0.000604632 335.753 -0.000785782 (14) 35000 251.65 0.000575345 339.777 -0.000776828 (15) 30000 243.45 0.00048334 343.521 -0.000682019 (16) 25000 233.15 0.000421649 346.597 -0.000626816 (17) 20000 220.75 0.000472445 349.789 -0.000748612 (18) 17500 213.95 0.000697178 352.211 -0.00114772 (19) 15000 206.65 0.000996432 355.528 -0.00171429 (20) 12500 199.05 0.00226535 360.783 -0.00410601 (21) 10000 194.65 0.00519965 376.058 -0.0100456 (22) 8000 197.15 0.00892678 405.988 -0.0183828 (23) 7000 201.55 0.0129963 431.206 -0.0278049 (24) 6000 206.45 0.0151663 461.598 -0.0339099 (25) 5000 211.85 0.0175898 499.026 -0.0414339 (26) 4000 216.85 0.0208443 544.465 -0.0523356 (27) 3000 221.45 0.0230898 603.697 -0.0629453 (28) 2500 222.45 0.0303216 638.883 -0.0870846 (29) 2000 225.65 0.0339061 690.782 -0.103797 Example 3: Calculate the static stability for the following.
Let p1 be a one-dimensional array containing isobaric levels: p1(klev); Let p2 be a two-dimensional array containing isobaric levels: p2(klev,ncol); Let p3 be a three-dimensional array containing pressure levels: p3(klev,nlat,mlon); Let p4 be a four-dimensional array containing pressure levels: p4(ntim,klev,nlat,mlon); Let p5 be a five-dimensional array containing pressure levels: p5(nens,ntim,klev,nlat,mlon); Let t1 be a one-dimensional array containing temperature: t1(klev); (0) Let t2 be a two-dimensional array containing temperature: t2(klev,ncol); (0,1) let t3 be a three-dimensional (klev,nlat,mlon); (0,1,2) let t4 be a four-dimensional (ntim,klev,nlat,mlon); (0,1,2,3) let t5 be a five-dimensional (nens,ntim,klev,nlat,mlon) (0,1,2,3,4) s1 = static_stability(p1,t1, 0, sopt) s3 = static_stability(p1,t2, 0, sopt) s3 = static_stability(p1,t3, 0, sopt) s4 = static_stability(p1,t4, 1, sopt) s5 = static_stability(p1,t5, 2, sopt) s33 = static_stability(p3,t3, 0, sopt) s44 = static_stability(p4,t4, 1, sopt)