how is interpolate used

Dubbsta

New member
Joined
Jul 13, 2017
Messages
186
i put values in one of the methods and the object just keep going
ex. ball.x += InterpolateBounceInOut( 0.1, 0.5, 0.1)
so im pretty sure im not using it right. assistance please thx :)
 

Phil7

Administrator
CX Code Contributor
3rd Party Tool Dev
Joined
Jun 26, 2017
Messages
535
The thing is that these functions are pretty low level mathematic ones.
with the first two parameters you set the starting and the end values you want to have. They are usualy fix values for one ease.
The last parameter is the variable one between 0 and 1 that states how far away you are from the starting (0) or the end point (1)

To get a visual impression you could do something like
Local relativeX:Float = Float(MouseX()) / Float(DeviceWidth()) ' This gives you a value between 0.0 and 1.0
ball.y = Interpolate_YOURTYPE(100, 300, relativeX )
If you move the mouse horizontally you see the ball moving vertically according to the interpolation function from 100 to 300 in y coordinates.

Here is a simple example to show the effect. Note that the y values are flipped vertically because y is pointing downwards.

code_language.cerberus:
Strict

Import mojo
Import interpolate


Class myApp Extends App
  
    Method OnCreate:Int()
        SetUpdateRate 60
        Return 0
    End
  
  
    Method OnRender:Int()
        Cls 0,0,0
        Local relativeX:Float = Float(MouseX()) / Float(DeviceWidth()) ' This gives you a value between 0.0 and 1.0
        Local ballY:Float = InterpolateBounceEaseInOut(100, 300, relativeX )
        DrawCircle(MouseX(), ballY, 20)
        Return 0
    End
  
End


Function Main:Int()
    New myApp
    Return 0
End
 
Last edited:

Dubbsta

New member
Joined
Jul 13, 2017
Messages
186
thx alot phil7 i actually used second param as variable before. thx a ton
 
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