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Andrew!
Glen Wallace (mailto:gcwallace@xxxxxxxx) published a very smart code on this
a few months ago, October last I think. If he does not respond to you then I
will dig it out from my archive.
Best regards,
Yngvi
hardy@xxxxxxxxxxxxx
-----Original Message-----
From: owner-metastock@xxxxxxxxxxxxx [mailto:owner-metastock@xxxxxxxxxxxxx]On
Behalf Of Kornberg Family
Sent: 1. febrúar 2000 12:57
To: Metastock List
Subject: Formula Question
Hi,
There has been soem discussion on an Australian tech analysis forum about a
formula for an ATR Chandelier stop. Essentially Tharp and Le Beau give this
exit a big wrap.
Someone (Randy aka yankeetrader) on the forum has a function for a
chandelier stop/exit as follows:
atrperiods:=Input("Enter ATR periods",1,39,18);
multiplier:=Input("Enter ATR multiplier",1,10,3);
(HIGH-ATR(atrperiods)*multiplier)
This is OK but as the highs are used to "hang" the Chandelier as the highs
fall then the line drawn using the formula also fall. As we never move our
stops down in a long trade, we must always remember that the highest
Chandelier stop is our stop. The way around this is to do one of two
things - both which have me stumped.
1. Only plot the Chandelier stop from the entry date. Thus the above
formula should prompt for a date and the line should be plotted from that
date with the Stop calculated as exit at the highest high since entry minus
3 ATR's.
I have used a variety of dayofmonth(), month() year() functions to no avail.
Also it appears that a line like this cannot be plotted from a particular
date.
2. Plot the Chandelier Stop after prompting for the entry with the Stop
calculated as exit at the highest high since entry minus 3 ATR's. Thus, the
stop line would never be lowered.
Can the bar of the entry date be painted another color automatically after
prompting?
Any help would be greatly appreciated.
Andrew
PS: A chart base on a template Randy posted is shown to demonstrate how the
Chandelier stop moves up and down.
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