Math equation to calculate different speeds for fade animation

Question

I'm trying to add a fade effect to my form by manually changing the opacity of the form but I'm having some trouble calculating the correct value to increment by the Opacity value of the form.

I know I could use the AnimateWindow API but it's showing some unexpected behavior and I'd rather do it manually anyways as to avoid any p/invoke so I could use it in Mono later on.

My application supports speeds ranging from 1 to 10. And I've manually calculated that for a speed of 1 (slowest) I should increment the opacity by 0.005 and for a speed of 10 (fastest) I should increment by 0.1. As for the speeds between 1 and 10, I used the following expression to calculate the correct value:

double opSpeed = (((0.1 - 0.005) * (10 - X)) / (1 - 10)) + 0.1; // X = [1, 10]

I though this could give me a linear value and that that would be OK. However, for X equal 4 and above, it's already too fast. More than it should be. I mean, speeds between 7 and 10, I barely see a difference and the animation speed with these values should be a little more spaced

Note that I still want the fastest increment to be 0.1 and the slowest 0.005. But I need all the others to be linear between them.

What I'm doing wrong?

EDIT: It actually makes sense why it works like this, for instance, for a fixed interval between increments, say a few milliseconds, and with the equation above, if X = 10, then opSpeed = 0.1 and if X = 5, then opSpeed = 0.47. If we think about this, a value of 0.1 will loop 10 times and a value of 0.47 will loop just the double. For such a small interval of just a few milliseconds, the difference between these values is not that much as to differentiate speeds from 5 to 10.

Answer 1

Here's how I'd program that linear relationship. But first I'd like to make clear what I think you're doing.

You want the rate of change in opacity to be a linear function of speed:

o(v) = o1*N1(v) + o2*N2(v) so that 0 <= v <=1 and o(v1) = o1 and o(v2) = o2.

If we choose N1(v) to equal 1-v and N2(v) = v we end up with what you want:

o(v) = o1*(1-v) + o2*v

So, plugging in your values:

v = (u-1)/(10-1) = (u-1)/9

o1 = 0.005 and o2 = 0.1

So the function should look like this:

1. o(u) = 0.005*{1-(u-1)/9} + 0.1*(u-1)/9
2. o(u) = 0.005*{(9-u+1)/9} + 0.1*(u-1)/9
3. o(u) = 0.005*{(10-u)/9} + 0.1(u-1)/9

You can simplify this until you get a simple formula for o(u) where 1 <= u <= 10. Should work fine.

Answer 2

If I understand what you're after, you want the equation of a line which passes through these two points in the plane: (1, 0.005) and (10, 0.1). The general equation for such a line (as long as it is not vertical) is y = ax+b. Plug the two points into this equation and solve the resulting set of two linear equations to get

 a = (0.1 - 0.005) / 9 = 0.010555555555...
 b = 0.005 - a        = -0.00555555555...

Then, for each integer x = 1, 2, 3, ..., 10, plug x into y = ax+b to compute y, the value you want.

Answer 3

Your problem stems from the fact that the human eye is not linear in its response; to be precise, the eye does not register the difference between a luminosity of 0.05 to 0.10 to be the same as the luminosity difference between 0.80 and 0.85. The whole topic is complicated; you may want to search for the phrase "gamma correction" for some additional information. In general, you'll probably want to find an equation which effectively "gamma corrects" for human ocular response, and use that as your fading function.