# Re: intp2

From: Dennis Shea <shea_at_nyahnyahspammersnyahnyah>
Date: Mon May 30 2011 - 08:18:13 MDT

It appears that the documentation is not correct.
The data must start at the 'surface' . This will be loked at.

You can use the ::-1 syntax to reorder.

linlog = 0 ;
;linlog = 2 ; ln(p) interpolation

pi =(/ 1000.,925.,850.,700.,600.,500.,400.,300.,250., \
200.,150.,100.,70.,50.,30.,20.,10. /)

xi =(/ 28., 23., 18., 10., 2., -4., -15.,-30.,-40., \
-52.,-67.,-78.,-72.,-61.,-52.,-48.,-46. /)

po =(/ 1000.,950.,900.,850.,800.,750.,700.,600.,500., \
425.,400.,300.,250.,200.,100.,85.,70.,50.,40.,\
30.,25.,20.,15.,10. /)

print(" --------------------------------------- ")
xo_a = int2p (pi,xi,po,linlog) ; descending (pi, po)
xo_b = int2p (pi(::-1),xi(::-1),po(::-1),linlog) ; ascending (pi, po)
xo_c = int2p (pi,xi,po(::-1),linlog) ; pin: descending , po: ascending
print(xo_a+" "+xo_b+" "+xo_c)
; xo_a will contain (/ 28.,24.71,21.37,18. ,...., -48.,-47.17,-46./)

print(" --------------------------------------- ")
XO_A = int2p_n (pi,xi,po,linlog,0) ; descending (pi, po)
XO_B = int2p_n (pi(::-1),xi(::-1),po(::-1),linlog,0); ascending (pi, po)
XO_C = int2p_n (pi,xi,po(::-1),linlog,0) ; pin: descending , po:
ascending
print(XO_A+" "+XO_B+" "+XO_C)
; XO_A will contain (/ 28.,24.71,21.37,18. ,...., -48.,-47.17,-46./).

On 5/29/11 7:34 PM, Agnes Lim wrote:
> Hi
>
>
> For int2p, does it matter if pin is monotonically decreasing whereas
> pout is monotonically increasing?
> From the help, it seems that pin and pout need not be in the same trend.
>
>
> Agnes
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Received on Mon May 30 08:18:20 2011

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