What algorithm solves this problem? Non-linear measuring tape 
A measuring tape is marked at 0, 5, 15 and 40. The distances between each mark are marked on top. At what distances should I mark 1 through 4, as well as 6-14 and 16-39? My math knowledge does not cover this.
Obviously 0-1 is the largest distance, then 1-2, 2-3 and so on.
Which algorithm is appropriate here? I do not even know what to search for. I tried some online regression calculators (logarithmic) but I failed to find a solution for this.
Also, English is not my first language. I'm sorry if anything is confusing above.
 A: I don't know of a particular algorithm, but you can try fitting the points to different curves. You basically want to find $y(x)$, where you have:
$$y(0)=0,\ y(5)=62,\ y(15)=87,\ y(40)=108$$
By looking at the curve you can see that the slope decreases sharply though the function is increasing. Thus, assuming the function is monotonic, either $y(x)=Ax^b, 0<b<1$ or $y(x)=A\log(bx)$ are candidate models, though of course, infinitely many more complex functions may fit at least as well.
Using a least squares algorithm, I found:
$$y(x)\approx 0.301 \times e^{0.0452 x}$$
Which although it gives $0$ at $x\approx 0.3$ instead of zero, gives a nice overall fit.
Using this function you can fill in the blanks.
A: Fitting any empirical model on so little data is indeed a random guesswork. (Anyway, I would certainly not think of a logarithmic or exponential model, as the first data point is $(0,0)$).
To get any sensible answer, it is better that you explain how this data is generated.
With no better information, natural cubic spline interpolation should be appropriate.
A: An nbubis answered, many models could fit the data and the problem here is that you have very very few data points.
Using as a model  $y=a x^b$ and minimizing the sum of  the squares of absolute errors, I obtained  $a=41.8586$ and $b=0.259482$; for these parameters, we obtain values of $0.00$, $63.55$, $84.52$ and $109.02$.
Minimizing the sum of  the squares of relative errors, I obtained  $a=40.8547$ and $b=0.267670$; for these parameters, we obtain values of $0.00$, $62.85$, $84.34$ and $109.66$.
It could be very good if you could have more data points.
