Cosine interpolation - GameCreators Forum

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David T

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Posted: 22nd Jun 2004 19:56

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Hi,

I was wondering if anybody could figure this out. I'm trying to get simple cosine interpolation working between two points (x1,y2)(x2,y2).

This is what I have:

+ Code Snippet

x1 = 100
y1 = 100
x2 = 200
y2 = 200

ink rgb(255,0,0),0
line x1,y1,x2,y2

ink rgb(255,255,255),0

for t = 1 to 10
inc mu#,0.1
ix = int(InterpolateX(x1,x2,mu#))
iy = int(CosineInterpolateY(y1,y2,mu#))
dot ix,iy
next t



sync
wait key

function InterpolateY(y1,y2,mu#)
returnval# = (y1*(1-mu#)+(y2*mu#))
endfunction returnval#

function InterpolateX(x1,x2,mu#)
diff = x2 - x1
returnval# = (x1*1.0) + (diff * mu#)
endfunction returnval#

function CosineInterpolateY(y1,y2,mu#)
   mu2# = (1-cos(mu#*(3.1412)))/2;
   returnval# = InterpolateY(y1,y2,mu2#)
endfunction returnval#

Which basically boils down to my equation for interpolation being

mu# = [value between 0.0 and 1.0]
mu2# = (1-cos(mu#*(3.1412)))/2
interpolatedy# = (y1*(1-mu2#)+(y2*mu2#))

Yet I get returned the original y1 value. Anybody help (doubt it, bit of a niche this)?

Two strings walk into a bar. I'll have a pint says the first$%ASLDJ09920D"$"$D. Excuse my friend says the second, he isn't null terminated.

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Phaelax

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Posted: 22nd Jun 2004 20:16

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Try this. Function interpolates from A to B based on mu#. (0-1)

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function interpolateX#(a#, b#, mu#)

   returnVal# = a# + ((b# - a#) * mu#)

endfunction returnVal#

As for cosine interpolation, I have no idea what that is.

"eureka" - Archimedes

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David T

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Posted: 22nd Jun 2004 20:24

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Quote: "As for cosine interpolation, I have no idea what that is."

It's basically interpolating but to gain a smoother curve between points. It's based on, apparently, chaning the value of mu# to this:

mu2# = (1-cos(mu#*(3.1412)))/2

And then running a normal interpolation.

Two strings walk into a bar. I'll have a pint says the first$%ASLDJ09920D"$"$D. Excuse my friend says the second, he isn't null terminated.

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las6

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Posted: 22nd Jun 2004 21:02

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I think I did cosine interpolation once. that one isn't really good for just about anything. In 2d, it behaves somewhat normally (not good or predictable) - but when used alone in 3d it's just like linear interpolation with a slow movement at the start and the end of the line... :\

You're much better of with real splines. Beziers, preferably. Hermite or Cubic Interpolations are good...

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David T

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Posted: 22nd Jun 2004 21:09

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Quote: "I think I did cosine interpolation once. that one isn't really good for just about anything. In 2d, it behaves somewhat normally (not good or predictable) - but when used alone in 3d it's just like linear interpolation with a slow movement at the start and the end of the line... :\"

I'm interested in 2d at the moment

I remember now, you did that path making app. Do you still have the code you used?

Two strings walk into a bar. I'll have a pint says the first$%ASLDJ09920D"$"$D. Excuse my friend says the second, he isn't null terminated.

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las6

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Posted: 22nd Jun 2004 21:46 Edited at: 22nd Jun 2004 21:50

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here's what I found:
and like for linear interpolation, you'll have to do these separately for x and y.
(dunno why I have used "y#" as the generic variable... could have used a# b# c# d# like for the linear one...

Function Linear_Interpolate#(a#,b#,step#)
   t# = a# * (1.0-step#) + b# * step#
Endfunction t#

function Cosine_Interpolate#( y1#, y2#, k1# )
    k2# = (1 - cos( k1#*180 ))/2.0
    newY# = (y1#*(1-k2#)+y2#*k2#)
endfunction newY#

Function CubicInterpolate(y0#, y1#, y2#, y3#, mu#)

mu2# = mu# * mu#
   a0#  = y3# - y2# - y0# + y1#
   a1#  = y0# - y1# - a0#
   a2#  = y2# - y0#
   a3#  = y1#

result# = a0# * mu# * mu2# + a1# * mu2# + a2# * mu# + a3#

endfunction result#

Function HermiteInterpolate(y0#, y1#, y2#, y3#, mu#, tension#, bias#)

mu2# = mu# * mu#
   mu3# = mu2# * mu#

m0# =        (y1#-y0#)*(1.0+bias#)*(1.0-tension#)/2.0
   m0# = m0# + ((y2#-y1#)*(1.0-bias#)*(1.0-tension#)/2.0)
   m1# =        (y2#-y1#)*(1.0+bias#)*(1.0-tension#)/2.0
   m1# = m1# + ((y3#-y2#)*(1.0-bias#)*(1.0-tension#)/2.0)
   a0# =  2.0 * mu3# - 3.0 * mu2# + 1.0
   a1# =        mu3# - 2.0 * mu2# + mu#
   a2# =        mu3# -       mu2#
   a3# = -2.0 * mu3# + 3.0 * mu2#

result# = a0# * y1# + a1# * m0# + a2# * m1# + a3# * y2#
Endfunction result#

+ Code Snippet

Function Linear_Interpolate#(a#,b#,step#)
   t# = a# * (1.0-step#) + b# * step#
Endfunction t#

function Cosine_Interpolate#( y1#, y2#, k1# )
    k2# = (1 - cos( k1#*180 ))/2.0
    newY# = (y1#*(1-k2#)+y2#*k2#)
endfunction newY#

Function CubicInterpolate(y0#, y1#, y2#, y3#, mu#)

   mu2# = mu# * mu#
   a0#  = y3# - y2# - y0# + y1#
   a1#  = y0# - y1# - a0#
   a2#  = y2# - y0#
   a3#  = y1#

   result# = a0# * mu# * mu2# + a1# * mu2# + a2# * mu# + a3#

endfunction result#


Function HermiteInterpolate(y0#, y1#, y2#, y3#, mu#, tension#, bias#)

   mu2# = mu# * mu#
   mu3# = mu2# * mu#

   m0# =        (y1#-y0#)*(1.0+bias#)*(1.0-tension#)/2.0
   m0# = m0# + ((y2#-y1#)*(1.0-bias#)*(1.0-tension#)/2.0)
   m1# =        (y2#-y1#)*(1.0+bias#)*(1.0-tension#)/2.0
   m1# = m1# + ((y3#-y2#)*(1.0-bias#)*(1.0-tension#)/2.0)
   a0# =  2.0 * mu3# - 3.0 * mu2# + 1.0
   a1# =        mu3# - 2.0 * mu2# + mu#
   a2# =        mu3# -       mu2#
   a3# = -2.0 * mu3# + 3.0 * mu2#

   result# = a0# * y1# + a1# * m0# + a2# * m1# + a3# * y2#
Endfunction result#

and treat that y1# as the start point and y2# as the end point in cubic and hermite interpolations... y0# is the point before that or just a separate control point you can use to change the curve behaviour. y3# is the point after the end point or a control point just like y0#.

there's a easier way now. Just store all the data in vectors and use the interpolation from the vector commands. (there's a working hermite right in there - very easy to use)

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las6

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Posted: 22nd Jun 2004 21:51

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here's the better alternative, I think:
http://www.dannywartnaby.co.uk/rgt/index.php?board=8;action=display;threadid=3408

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David T

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Posted: 22nd Jun 2004 22:29

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Cheers

Two strings walk into a bar. I'll have a pint says the first$%ASLDJ09920D"$"$D. Excuse my friend says the second, he isn't null terminated.

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