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RE: [Metastockusers] RE:Re: Bar Count since certain date



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<FONT face=Arial 
size=2>Robin:
<FONT face=Arial 
size=2> 
Can you direct me to 
the portion of the cited web site where it says that 3 points are sufficient to 
define a parabola? Three points would be sufficient to define a parabola if you 
know the semi latus rectum (factor F below).
<FONT face=Arial 
size=2> 
Please bear with me. 
While both the circle and the parabola are conic sections their equations are 
different.
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size=2> 
Equation of Circle : 
X**2 + Y**2 = R**2    R is the radius
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Equation of 
parabola: Y**2 = 
4FX            F is the 
semi latus rectum
<FONT face=Arial 
size=2> 
Here is a little 
exercise you can do.  All you need are two pieces of paper, a 
pencil, a compass, a straight edge (a ruler will do) and maybe an 
eraser.
 
Place 3 points on a 
piece of paper, they must not be on a straight line. Call them A, B, and C. Draw 
the lines A-B and B-C. Bisect these 2 lines and draw the perpendicular 
bisectors. the bisectors will meet at the center of the circle. Next select any 
other 3 points on the circle and repeat this little exercise.  These 
bisectors will also meet at the center of the circle.  Before the days 
of electronic navigation, the basis of this method was used by 
navigators near shore or in narrow channels to locate the position of 
the vessel.
<FONT face=Arial 
size=2> 
If you draw a 
parabola and select two or more sets of three points on the parabola 
and repeat the exercise, the bisectors will not meet at the same 
place.
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size=2> 
If you have any 
further questions please write me privately
<FONT face=Arial 
size=2> 
<FONT face=Arial 
size=2>Lionel
<FONT face=Arial 
size=2> 
<FONT face=Arial 
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<FONT face=Arial 
size=2> 
 

<FONT face=Tahoma 
size=2>-----Original Message-----From: Robin Hood 
[mailto:robinhood@xxxxxxxx] Sent: Sunday, January 05, 2003 4:51 
AMTo: Metastockusers@xxxxxxxxxxxxxxxSubject: Re: 
[Metastockusers] RE:Re: Bar Count since certain date
Hi Lionel
 
Three points are sufficient as parabola is made by tho points equidistanced 
(? in english ?) form another one.
<A 
href="http://www.xahlee.org/SpecialPlaneCurves_dir/Parabola_dir/parabola.html";>http://www.xahlee.org/SpecialPlaneCurves_dir/Parabola_dir/parabola.html for 
examples (by google).
 
Here is a chart form an Italian site which I don't name because of 
netiquette (I think...). They build a parabola from three points, in effect. 
It's the same effect I'd like to reproduce with Metastock... but the trouble are 
the input dates (x coordinates) :-(
 
 
 
    robin
 
 
<BLOCKQUOTE 
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
  ----- Original Message ----- 
  <DIV 
  style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black">From: 
  Lionel 
  Issen 
  To: <A 
  title=Metastockusers@xxxxxxxxxxxxxxx 
  href="mailto:Metastockusers@xxxxxxxxxxxxxxx";>Metastockusers@xxxxxxxxxxxxxxx 
  
  Sent: Saturday, January 04, 2003 4:57 
  PM
  Subject: RE: [Metastockusers] RE:Re: Bar 
  Count since certain date
  
  Three points 
  always define a circle. I think that you need at least 4 points to define 
  a parabola. When curve fitting a parabola you need more than 4 points as data 
  points always have some noise, they don't fall on a nice smooth 
  curve.
   
   
  
  
  <FONT 
  face=Tahoma size=2>-----Original Message-----From: Robin Hood 
  [mailto:robinhood@xxxxxxxx] Sent: Saturday, January 04, 2003 8:25 
  AMTo: Metastockusers@xxxxxxxxxxxxxxxSubject: Re: 
  [Metastockusers] RE:Re: Bar Count since certain date
  Hi Group
   
  Thanks to Spyros for his formula.
   
   
  In order to build a parabolic trend-line, I need three points (geometry 
  requires it). It means three x,y (date, price) coordinates.
   
  The problem is that metastock accepts only six inputs (in oder to define 
  dates I need 3+3+3=9 inputs :-( ).
   
  Is there another way to make this?
   
  Now, the formula inputs are as follows, but I'd like to change the 
  numbers with dates since I don't understand very well and immediately their 
  sequence
   
  x1:=Input("point 1 : x1",-100000000,100000000,0);y1:=Input(" 
  y1",-100000000,100000000,0);x2:=Input("point 2 : 
  x2",-100000000,100000000,0);y2:=Input(" 
  y2",-100000000,100000000,0);x3:=Input("Point 3: 
  x3",-100000000,100000000,0);y3:=Input(" y3",-100000000,100000000,0);
   
   
   
  Thanks in advance, Robin
   
   
  <BLOCKQUOTE 
  style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
    ----- Original Message ----- 
    <DIV 
    style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black">From: 
    SR 
    To: <A 
    title=Metastockusers@xxxxxxxxxxxxxxx 
    href="mailto:Metastockusers@xxxxxxxxxxxxxxx";>Metastockusers@xxxxxxxxxxxxxxx 
    
    Sent: Saturday, January 04, 2003 2:24 
    PM
    Subject: [Metastockusers] RE:Re: Bar 
    Count since certain date
    bruneski came up with a variation of my approach which is 
    corrects somefailures of my version.So I suggest trying this 
    one:{BarsSinceDate}d:=Input("Day",1,31,1);m:=Input("Month",1,12,1);y:=Input("Year",1920,2030,2002);mydate:= 
    10000*y + 100*m + d;date:= 10000*Year() + 100*Month() + 
    DayOfMonth();count:=BarsSince(mydate<=date); {use '<' if first day 
    is not to becounted}count:=count + 
    Cum(If(count=0,1,0));countTo unsubscribe from 
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