okay, anyone know of a function (or combination of funcions) that have an output like: it doesn''t have to be exactly like that (ph33r 4h3 m5p41n7 5k1llz), but as long as for about the first 1/3 it stays roughly the same then declines dramatically afterwards. also, a way of modifying when it begins the steep decline would be good too. *sigh* all this to implement view- and aim-independant firing . thanks in advance

How ''bout 1 - x^n? The larger the value of n, the flatter the top and the steeper the drop-off.

HTH

Y = -2^x + 2 seems to work correctly.

Edit: Well, the only problem is that the function starts to dip at about x = -4.5. But other than that, it should roughly model what you're looking for.

[edited by - Apocalypse_Demon on January 16, 2003 8:53:44 PM]

i agree exponential function seems right

Depending on how you interpret that curve, it could be anything. For example, you could draw a straight line between (0,1) and (1,0) and have a curve that at least starts and ends at the right place. Also, you could draw a quarter of a circle of radius 1 centered at the origin and get something close.

Another simple approach that approximate the desired curve would be to have two line segments, one from (0,1) to (1/3,1) and another from (1/3,1) to (1,0). You wouldn''t have the nice smooth transition, but evaluation of the curve would be cheaper than the exponential functions.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

I wonder if the happy smile in the origo indicates something of the nature of this post.... (-?

But yeah, 1-x^n gives good results.

- Mikko Kauppila

How about 2 line-segments and a bezier curve?



[grrrrrr...can''t seem to get this to post! server problems...grrrr]

quote:

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Original post by bob_the_third
How about 2 line-segments and a bezier curve?

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That''d give you something smooth. Not sure if it''d be any cheaper to calc than the exponent.

quote:

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Original post by bob_the_third [grrrrrr...can''t seem to get this to post! server problems...grrrr]

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Yeah, there are problems today.



Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

uutee: i was wondering when someone would pick up on that , unfortunatly the question was actually a serious one!

anyway, thanks for all the replies. 1 - x^n works rather nicely actually (thanks AP).

i''m sure you''ll be seeing some more of my drawings in the near future (this weekend hopefully)...

thanks again