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To: B. Thomson who wrote (11693)3/16/2001 10:56:48 PM
From: MechanicalMethod  Read Replies (1) | Respond to of 12039
 
You'll need to convert to TC2000 programing syntax but this may give you an idea.

if c >= c[60] * 1.5 then alert = true; {150% rise}
if c <= c[60] * .5 then alert = true; {50% decline}

Depending on your definition of rise and decline you might try this:
if H >= c[60] * 1.5 then alert = true; {150% rise}
if L <= c[60] * .5 then alert = true; {50% decline}

or this:
if H >= H[60] * 1.5 then alert = true; {150% rise}
if L <= L[60] * .5 then alert = true; {50% decline}

If you post some code we can probably figure out how TC2000 inputs, variables and alert calls work so you can get your scans running.



To: B. Thomson who wrote (11693)3/20/2001 1:03:14 PM
From: MechanicalMethod  Read Replies (1) | Respond to of 12039
 
Here's an alternate approach that returns true if the conditions met anytime during the last 3 months. You'll need to convert it to TC2000.

var: decline(false), rise(false), both(false);

if lowest(L,60)[1] < c[60]*.5 then decline=true;
if highest(H,60)[1] > c[60]*1.5 then rise=true;
if lowest(L,60)[1] < c[60]*.5 and highest(H,60)[1] > c[60]*1.5 then both=true;