My school does Orienteering in PE, and basically the grades are from 0 to 20, the is the best,0 is the worst and 10 is 50%, well my teacher calculates the grades this way, if you do the task in 5 minutes you get a 20, if you do in 10 you get a 10, and calculates the rest, with a linear equations(I did it myself and it is: $y= -2x + 30$.

(Not so important to read, my opinion):Well is this really the best model? I don't think so, the effort someone makes to get a 15(knowing he had 10) is the same someone with a 15 makes to get a 20, which is 2 minutes and 30 seconds, obviously 2 minutes and 30 isn't the same for someone everyone, but the effort for someone who has a 18 to get a 19 or even twenty, at least to me, should be mucher higher(less time), than the effort someone makes to get a 13, knowing he has a 10

What kind of model should I choose, so in general between the grades 15 and 20, the difference of time is less than between 15 and 10(and how can I change it, like, if i wanted it to be from 17-20 and 0-17. But still:



f(x)=20 <=> x<=5 I can asure this by making the function recursion

is it a logistic model?


There are many functions that would do what you want for $x\ge 5$, but if you want a logistic $f$ it would be $$f(x)=\frac{40}{1+\exp\frac{x-5}{5/\ln 3}}.$$This implies $f(10)=10,\,f(5)=20,\,f(7.5)=\frac{40}{1+\sqrt{3}}\approx 14.64,\,f(x)=15\implies x=5+\frac{5\ln\frac{5}{3}}{\ln 3}\approx 7.32.$That achieves what you want, albeit only just in that example. You might want to look for a quadratic solution instead; if nothing else, it's easier to explain.

  • $\begingroup$ That's actually so nice! How did you get to that function? I'm not really sure what I'm looking for, but between [5,10] always concave downwards, the first derivative is always decreasing, it means that, if you can improve your result by 1 second, the lower your grade is, the bigger the change in grades is. (Is what I said right? I looks like to me) $\endgroup$ – Nuno Mateus Mar 14 at 15:44
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    $\begingroup$ @NunoMateus I solved for $a/(1+\exp (x-5)/b)$ to get the right values of $f(5),\,f(10)$. $\endgroup$ – J.G. Mar 14 at 16:13

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