how about (phase-/phase)%lineReactance
note that the diagonal elements of phase-/phase will be 0 but also note
that with line reactances of 0 the 0/0 terms will be 0
however- the problems may well be with the non zero phase differences
and 0 reactances. so you need to eliminate the non-zero/zero items.
As I indicated before- nodal methods have an advantage
a)fewer equations in general
b)( phase(i)-phase(j))*B(ij) isn't threatened by divide by 0 as B(ij )
for non-connected busses will be 0
possibly a discussion of the approach that you are using rather than the
programming, might be useful.
I may be able to help as power system analysis is something that I am
familiar with. Feel free to contact me
Don Kelly
On 21/05/2013 9:49 AM, P T wrote:
I am almost close to finishing an initial version. But, I don't know how to
write a verb P(ij) = (phase(i) - phase(j))/lineReactance(ij) where
*lineReactance*
0 0.005 0.00625 0 0 0 0
0 0 0 0
0.005 0 0.04 0.00333333 0.01 0 0
0 0 0 0
0.00625 0.04 0 0 0 0 0 0.01
0.025 0 0
0 0.00333333 0 0 0.005 0.005 0.00333333
0 0 0 0
0 0.01 0 0.005 0 0 0.005
0 0 0 0
0 0 0 0.005 0 0 0.005
0 0 0 0.00666667
0 0 0 0.00333333 0.005 0.005 0
0 0 0 0.00833333
0 0 0.01 0 0 0 0 0
0.01 0.00625 0
0 0 0.025 0 0 0 0
0.01 0 0.02 0
0 0 0 0 0 0 0 0.00625
0.02 0 0.02
0 0 0 0 0 0.00666667 0.00833333
0 0 0.02 0
*phase *
9.49428 5.92433 7.70672 4.02707 4.03065 2.58708 3.0874 5.96236 6.03134
4.82902 0
I can compute one by one like
((0{phase)-phase)% 0{lineReactance
0 713.99 286.01 _ _ _ _ _ _ _ _
((1{phase)-phase)% 1{lineReactance
_713.99 0 _44.5596 569.181 189.368 _ _ __ __ _ _
But, how do I write this so that it computes for all items in the matrix?
Thanks,
PT
How do I write a verb that computes
On Tue, May 21, 2013 at 8:40 AM, Linda Alvord <[email protected]>wrote:
odd=:2 & |
halve=:-:
veco =: 3 : '(c*1+3*y) + (halve y) * (1-c =. odd y)'
data =: 1 + i. 10000
veco3=:'((odd y)*1+3*y) + (halve y) * (1-odd y)'(13 : )
(veco3 data) -: (veco data)
1
veco3
(odd * 1 + 3 * ]) + halve * 1 - odd
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Linda
Alvord
Sent: Tuesday, May 21, 2013 9:21 AM
To: [email protected]
Subject: Re: [Jprogramming] newbie help: how to avoid division by zero
'(c*1+3*y) + (halve y) * (1-c =: odd y)'(13 : )
3 : '(c*1+3*y) + (halve y) * (1-c =: odd y)'
'(c*1+3*y) + (halve y) * (1-c =. odd y)'(13 : ) halve (* + [ * 1 - ])
odd
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Linda
Alvord
Sent: Tuesday, May 21, 2013 7:13 AM
To: [email protected]
Subject: Re: [Jprogramming] newbie help: how to avoid division by zero
I didn't realize that this definition for veco2 did not work.
data =: 1 + i. 10000
veco =: 3 : '(c*1+3*y) + (halve y) * (1-c =. odd y)'
'(c*1+3*y) + (halve y) * (1-c =. odd y)'(13 : ) halve (* + [ * 1 - ])
odd
veco2=:halve (* + [ * 1 - ]) odd
(veco2 data) -: (veco data)
0
So, although your substitution works, it doesn't help.
veco3=:halve (* -.) odd
(veco3 data) -: (veco data)
I'll be more careful about testing the results of: ' '(13 : )
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of David Ward
Lambert
Sent: Monday, May 20, 2013 12:24 PM
To: programming
Subject: Re: [Jprogramming] newbie help: how to avoid division by zero
I'd replace ([ * 1 - ]) with (* -.) .
As previously observed, 13 :'quoted expression' won't write hooks.
(([ * 1 - ]) -: (* -.)) -:i:8 NB. monad
1
(([ * 1 - ])/ -: (* -.)/)~ -:i:8 NB. dyad
1
Date: Sun, 19 May 2013 11:43:52 -0400
From: "Linda Alvord" <[email protected]>
To: <[email protected]>
Subject: Re: [Jprogramming] newbie help: how to avoid division by zero
Message-ID: <000101ce54a7$aa2d4c10$fe87e430$@net>
Content-Type: text/plain; charset=US-ASCII
It is worth noting further simplification by J:
veco=: 13 :'(c*1+3*y) + (halve y) * (1-c =. odd y)'
veco
halve (* + [ * 1 - ]) odd
Linda
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