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Step One

I need to find the equation that will describe a cartesian path that looks like the following red line:

enter image description here

Step Two

Given a constant start point $A(0, 0)$, a variable end point $B(x, y)$ and a variable percentage, I need to find a point $C(x, y)$ that is percentage distance on that line:

enter image description here

Let me know if I can provide anything to make the question more clear. Thanks for your help!

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Do you require a single equation for the curve $A,C,B$? I mean could it be e.g. 3 equations, two of which would be $y=y_A=0$ (horizontal line) and $x=x_B$ (vertical line)? Why the (linear-algebra) tag? – Américo Tavares Jun 2 '12 at 0:16
I think that one equation will make it much easier to find the point C based on a percentage distance. Sorry about the 'linear-algebra' tag. My options were limited since this is my first question. Thanks! – Wigz Jun 4 '12 at 17:00
up vote 1 down vote accepted

You can try curves of the form $$y = \dfrac{x^{n}}{(\alpha-x)^m}, \text{ where $x \in [0,\alpha)$}$$ The parameters $m$ and $n$ controls how steep you want your growth to be and the parameter $\alpha$ controls where you want your curve to go to infinity. Below are some such curves for different values of $m$ and $n$ with $\alpha$ fixed as $5$. enter image description here

Now given $B$ on the curve, compute the length of the curve from $A$ to $B$, the length of the curve is given by $$L_B = \int_{x_A}^{x_B} \sqrt{1 + \left(\dfrac{dy}{dx} \right)^2} dx$$ Similarly, compute the length of the curve till $C$ which is given as $$L_C = \int_{x_A}^{x_C} \sqrt{1 + \left(\dfrac{dy}{dx} \right)^2} dx$$

Say you want $C$ to be at a distance such that $\dfrac{L_C}{L_B} = \dfrac{\beta}{100}$ where $\beta$ is the $\%$ distance of $C$ on the curve from $A$ to $B$. Once you compute $L_c$ from above, find $x_c$ such that $$L_C = \displaystyle \int_{x_A}^{x_C} \sqrt{1 + \left(\dfrac{dy}{dx} \right)^2} dx$$

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I think this will work what I need, thank you. It's not imperative that the path goes directly vertical after the curve and the trade-off for having one, elegant equation is definitely worth it. Now, I just need to take this over to JavaScript and get it working for what I need. I'll post the result. Thanks for your help! – Wigz Jun 4 '12 at 17:02

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