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John,
let me restate what seems to be the core of the problem you want to solve.
If it isn't, please let me know.
On a semi-log chart (regular time scale, logarithmic price scale), let a
straight line be defined by two points P1(x1,y1) and P2(x2,y2). Given a
third point P3(x3,y3) not incident to the original straight line, we are
looking for a point P4 such that, on the semi-log chart, the straight line
P3->P4 will be parallel to the trend line P1->P2.
Solution: P4(x3+x2-x1, y3*y2/y1)
Again, there are infinitely many other possible points P4.
Hope this helps.
Michael Suesserott
-----Ursprüngliche Nachricht-----
Von: john hamon [mailto:jhamon@xxxxxxxxxxx]
Gesendet: Thursday, January 04, 2001 21:00
An: Bob Fulks; Omega_List
Betreff: RE: AW: back to the well again... parallel trend lines
which, bob is my exact problem and which i carelessly forgot to mention. i
will map the trend line in and out of arithmetic space to determine my
deltas. i forgot to mention that.
jh
-----Original Message-----
From: Bob Fulks [mailto:bfulks@xxxxxxxxxxxx]
Sent: Thursday, January 04, 2001 11:48 AM
To: fritz@xxxxxxxx
Cc: omega-list@xxxxxxxxxx
Subject: Re: AW: back to the well again... parallel trend lines
At 12:30 PM -0700 1/4/01, Gary Fritz wrote:
> > Solution: P4 = (x3+x2-x1, y3+y2-y1)
>
>Awww, that's too EASY... :-)
>
>Good point, Michael. I got so caught up in trigonometry that I
>didn't think of the obvious approach.
>
>That will produce a line from P3 to P4 at the same angle as the
>original P1-P2 line. P3 will be one endpoint. If you wanted the
>parallel line centered around P3, you could just add/subtract 1/2 the
>x2-x1 and y2-y1 values to the P3 location.
Unless you are using a log price scale...
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