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Gary Fritz wrote:
I am taking the sine of 180* times the index value which is from 0 to
1 instead of taking sine of the angle. This index is a linear
function,
for starters then, note that the sine function is very non-linear, so
why involve it in some kind of an index calculation that you want to
have linear behavior.
As Gary says below, perhaps if you would better describe what you are
trying to accomplish graphically, then others could help with solutions.
You don't have to describe your trading methodology if you don't want
to, just what the info it is you have available at the time you want to
start drawing and what it is you want to see pictorially based on that info.
Rod G.
and 180* this = angle at 0*, 90*, etc. but not in between.
Problem is I don't know how to convert the linear index into the
appropriate angle. Any ideas?
I'm not real clear on what you're doing. Are you saying your
index goes from 0 at the point on the right side of the circle,
0.5 at the circle center, and 1 at the left side? So 180* this
index gets you the correct angle at 0, 90, and 180?
If so, that's basically just luck. That linear index has no
direct relationship to the angle.
Let's see, to map that index to angle... first I would map the
index to -1 .. +1, corresponding to the cosine of the angle. So
let's say NewIndex = 1-2*OldIndex. Take the arc cosine of the
NewIndex and you'll have the angle. Take the sine of that and
you have the vertical offset for your circle at that point. (No
square roots!!) Actually you divide the sine by 2 because the
radius of your circle is 0.5, not 1, since it goes from 0 on one
side to 1 on the other.
So:
NewIndex = 1-2*OldIndex { cos of the angle at that x position }
Angle = arccos(NewIndex) { the angle }
Yoffset = sine(Angle)/2
plot1(Centerprice + Yoffset);
plot2(Centerprice - Yoffset);
That gets you a lovely circle. At least in Excel.
Problem: TS doesn't seem to have arccos. Maybe it's just too
early on a Sunday morning to think that hard, but I can't see how
to calculate arccos using arctan or other available functions.
You could approximate it with a Taylor series, but I don't know
if you want to work that hard. :-) Maybe somebody else can help
on that one.
Gary
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