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Pumi says

Hey igisoftware, thats great explanation … I know understand it, and why he told us that we can never solve it :)

Thanks everyone for trying to help me.

3 years ago in thread Any math expert here ??
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Pumi says

igisoftware, i think we did learn most of them.

I dont know serbian but my parents do and i can also use google translator, so ok you can explain it in serbian if you wish.

Thanks

3 years ago in thread Any math expert here ??
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Pumi says
Y is a value that is always equal itself in any direction. No matter how many times, y1, y2, or even y2, the result of y? will always be Y = 25. And Y x Y Will always result in birth.

y1 = 25.

y2 = 50.

a12 will always equal the value simplified as A = 1.

a22 = 4.

(2a = 27 / (27 -1) = 1) + (1a = 1)

1 + (32 – 16) = 17.

25 – 9 = 17.

17 – 9 = I.

23 = 2+3 = 5. 5 is a number that perpetuates its equivalent. Like 5 times its value is equal to 5×5 = 25. 23 chromasomes. Mayans believed that the world was created 5 times before. F is failure, F is imperfect. F is death. But F is not the end. P is the prophecy. B is birth. P – B = N. N = 1+4 = E. E is a value that perpetuates its value. BENJ . PALWS – 10 = PALMS . Fingers, (starting from the left hand and alternating the value from left to right): 16, 27, 38, 49, 510. 16(P), 27(A), 38(L), 49(W), 510(S). S – J = I. I + (1-(0)) = 10. W = J = M. PALM .

Wow, you are a genius :P

3 years ago in thread Any math expert here ??
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Pumi says

Hello,

Our Physics proffesor told us that for us is amost impossible to solve a problem, and who will solve it will get a 5 mark. So here is it:

y1 = a1·cos(wt-p1)
y2 = a2·cos(wt-p2)

I = y1 + y2 = ?

and he gave us the result which is :

I = a1² + a2² + 2·a1·a2·cos( p2-p1 ) .

So is there anyone who can give me a step by step solution of this ??

3 years ago in thread Any math expert here ??
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Pumi says

lol :D

3 years ago in thread Looking for a menu
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Pumi says
for blur filter, decrease blur quality (3 = cpu killer) and use multiples of 2 for better performances.
don’t draw all the background since you only need a dWidth x dHeight rectangle.

instead look at .draw() method description for how to use a box, this will also lead to more optimization:
– no need to create temp myBd_
– no need to use myBd_.copyPixels

don’t call applyFilter with the same bitmapData as source and destination -> unoptimized since a temp buffer has to be created each time

instead, create a buffer once and use that for applyFilter stuff.

also important is to avoid vectorial if you want speed.
so get rid of that mcGlass stuff and just use the bitmap as the window background
you can just use fillRect do draw window borders

Thanks bitfade, here is the optimized code… I decreased blur quality, and used .Draw() clipRect. I just couldnt understand you about applyFilter. And mcGlass is the movieClip where i will put my content.

You really are the master of BitmapData, love your effects :)

3 years ago in thread BitmapData help
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Pumi says

No better and faster way ?!

3 years ago in thread BitmapData help
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Pumi says

LuizZak I only need to blur the area where window is, not entire background.

trimax I think it is for AS3 , isnt it ?? Site is in AS2 .

In my current code, i am using 2 bitmapData, one to draw mc in to, and second is copying needed area from the first one, and then drawing it on screen using beginBitmapFill() ... I am sure there is a better ( and faster ) way of doing it.

Thanks, Pumi

3 years ago in thread BitmapData help
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Pumi says

ohh… forgot to mention, background is animated. So thats why i need it to be done with bitmapData so it can update.

3 years ago in thread BitmapData help
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Pumi says

Hello all,

I making a window, that will open in my flash site, which will have a blurred transparent background. I did it using BitmapData, but i think the way ive done it is pretty CPU heavy. Here is the code im using to do it. I dont know much about BitmapData so please help me.

Please tell me if there is any other better way of doing this. :)

Thanks, Pumi

EDIT : Opps double Thread… Srry !!

3 years ago in thread BitmapData help
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