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  • sx = x starting point
  • ex = x end point
  • sy = y starting point
  • ey = y max

having all the above, plus the current x, how do I calculate with a formula an y which isn't linear but that goes faster at first and slower approaching ey instead? (such as resembling gravity)

I guess I should use some sqr or some log somewhere, but I fail to see how exactly. I know this should be trivial but it looks like my basic math skills vanished, not having used them for too long, I'm sorry.

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What exactly do you want your function to do? What's it for? – J. M. Apr 26 '11 at 12:23
I'll try to clarify later, thanks. – o0'. Apr 26 '11 at 13:05
I suggest you provide a figure showing a sketch of the desired arc, the more exact the better. We then compare it with the arcs in our catalogue and can furnish a suitable formula. – Christian Blatter Apr 26 '11 at 13:22

There are many possible answers. One cannot know what sort of answer is most suitable for your particular application without further information. I will supply one answer that more or less fits your description. In order to avoid confusion with multiplication, I will change your notation slightly, using for example $s_x$ instead of sx, $e_x$ instead of ex, and so on.

Here is a suggested formula.
It is easy to check that this works. When $x=s_x$, the squared term is equal to $1$, and $y=e_y-(e_y-s_y)=s_y$. When $x=e_x$, the squared term is $0$, and $y=e_y$. The fact that I squared ensures that the approach to $e_y$ is rapid at the beginning and slow when $y$ is close to $e_y$. I used the square function as a reaction to your mention of gravity. One can adjust the rate of approach by changing the exponent from $2$ to something else. For a more rapid approach at the beginning, use, for example, $4$ instead of $2$. You can use non-integer exponents, like $1.6$ or $2.7$ instead of $2$ to adjust things so that they fit your application reasonably.

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