Normalize $X$ to $0$ to $10$ scale with asymptotes at either end

Question

I am trying to find a scaling function that mimics the gas gauge in a car. I would like to map a value to a $0$ to $10$ scale, on which I have two known points. For example:

$X_1 = 2$, $Y_1 = 2.5$, $X_2 = 4$, $Y_2 = 7.5$, $X_3 =$ __

The known points, $X_1$ and $X_2$ should always correspond to $Y$ values of $2.5$ and $7.5.$ A simple linear equation is not sufficient, since its possible for the $X_3$ value to generate a $Y$ that is greater than $10$, or less than $0.$

Can you think of a suitable function that matches the known points, but will approach a limit at $0$ and $10$? Thanks for any guidance!

Answer 1

You can use the arctangent function, which has horizontal asymptotes going to $\pm \infty$. As it ranges from $\frac {-\pi}2$ to $\frac \pi 2$ we need to rescale to get your range of $(0,10)$, so we want $10\left(\frac 1\pi \arctan X_{scaled} + \frac 12\right)$. Now all we have to do is figure out how to scale $X$. This is not too hard as $X_{scaled}=\pm 1$ gives $2.5, 7.5$, so $X_{scaled}=\frac {2(X-(X_1+X_2)/2)}{X_2-X_1}$. The final answer is $Y=10\left(\frac 1\pi \arctan \frac {2(X-(X_1+X_2)/2)}{X_2-X_1} + \frac 12\right)$

Answer 2

Another option is to use a function of the form $$ \frac{10}{1+ae^{-bx}} $$ which is bound by 0 and 10 as required.

With the requirements of $(2,2.5),(4,7.5)$ we have

$$ a e^{-2b} =3 \qquad \text{ and } ae^{-4b}=\frac{1}{3}$$ Dividing to eliminiate the $a$, we have $e^{2b}=9, b =\frac{\ln(9)}{2}$ and $a = 27 $. This reduces to

$$\frac{10}{1+27(3^{-x})}$$

This has the benefit of being fairly linear through your data points as shown by the dashed line in the figure.