Finding the Equation to Place RGB=255 128 0 on a Long Thin Shape

  • Context: High School 
  • Thread starter Thread starter Wes Turner
  • Start date Start date
  • Tags Tags
    Shape
Wes Turner
Messages
68
Reaction score
15
I have a long thin shape (rope, bar). The color of the shape changes smoothly from red (RGB=255 0 0) at one end to yellow (RGB=255 255 0) at the other. I need an equation that will allow me to adjust the G value from 0 to 255 in a way that I can choose where on the shape it has the value 127.5 (equal parts red and yellow, RGB = 255 128 0), if the 127.5 is rounded.

The linear equation y = 255x + 0 places that point at the midpoint (0.5) of the shape. The rounded values at 0.1 intervals are:
Color shading.jpg


I would like an equation that will enable me to place the G=127.5 value anywhere on that shape from 0.0 to 1.0. I'm not sure if a quadratic equation is the best choice or some type of exponential. For the quadratic, I would have three points ((0.0), (k,127.5), (1,255)) and could solve 3 equations in 3 unknowns.

Is there a better way?

Thanks
 

Attachments

  • Color shading.jpg
    Color shading.jpg
    12.8 KB · Views: 597
Mathematics news on Phys.org
In general a quadratic equation will go to negative values if your midpoint is too close to the 255 side, and above 255 if it is tooclose to the right side. You could use two linear relations left and right of it, or use a spline.

An exponential distribution with an offset (##a+e^{bx+c}##) would work as well.
 
  • Like
Likes   Reactions: Wes Turner
mfb said:
In general a quadratic equation will go to negative values if your midpoint is too close to the 255 side, and above 255 if it is too close to the right side.
Yeah, after some testing, I discovered that the quadratic is not a good choice.

You could use two linear relations left and right of it, or use a spline.
I'm not sure what you mean by two linear relations, but a spline might be a good choice. I didn't think of that. Thanks.

An exponential distribution with an offset (##a+e^{bx+c}##) would work as well.
Another good suggestion. Thanks.
 
Wes Turner said:
I'm not sure what you mean by two linear relations
G(x)=c*x up to the fixed point in the middle, and G(x)=a+b*x from there to the end (where a+b=255 to satisfy G(1)=255).
 
  • Like
Likes   Reactions: Wes Turner
mfb said:
G(x)=c*x up to the fixed point in the middle, and G(x)=a+b*x from there to the end (where a+b=255 to satisfy G(1)=255).
Got it, thanks.