[RFC PATCH] Make fake profile average depth come out right

[Linus Torvalds]

On Sun, Nov 3, 2013 at 2:44 PM, Anton Lundin <glance at acc.umu.se> wrote:
>> +/*
>> + * Good fake dive profiles are hard.
>> + *
>> + * "depthtime" is the integral of the dive depth over
>> + * time ("area" of the dive profile). We want that
>> + * area to match the average depth (avg_d*max_t).
>> + *
>> + * To do that, we generate a 6-point profile:
>> + *
>> + *  (0, 0)
>> + *  (t0, max_d)
>> + *  (t1, max_d)
>> + *  (t2, d)
>> + *  (t3, d)
>> + *  (max_t, 0)
>
> Hmm.. These ones should be t1 to t4, not t0 to t3?

Yes.

>> + * That six-point profile gives a total area of:
>> + *
>> + *   (max_d*max_t) - (max_d*t1) - (max_d-d)*(t4-t3)
>
> Why do you care about the max_d*t1 part? Isn't that just the accent
> slope? There should be more factors there if we should care about all
> the accent slopes in that formula.

Heh. Draw it out. Yes, you can make the formula much more complex like
yours is, but by selecting the shapes right, you get the much simpler
one.

The max_t*t1 is twice the area of the triangle generated by the
descent, yes. But because the ascent has the same shape, you can
actually find the *same* triangle on the ascent path. And then the
only remaining shape to subtract from the "big rectangle" is the time
we spend at depth "d" (which is a rectangle of size
"(t4-t3)*(max_d*d)").

> Shouldn't the whole thing look something more like:
> (max_d*max_t) - (max_d*t1*0.5) - (max_d-d)*(t3-t2)*0.5
> - (max_d-d)*(t4-t3) - d*(max_t-t4)*0.5 - (max_d-d)*(max_t-t4)

Simplify it. You should get to the exact same answer. In particular,
the same slope part:

>> + * And the "same ascent/descent rates" requirement
>> + * gives us (time per depth must be same):
>> + *
>> + *   t1 / max_d = (t3-t2) / (max_d-d)
>> + *   t1 / max_d = (max_t-t4) / d

can be used to simplify your complex expression to my simple one.
Except I didn't do it that way, it's easier to do by just partitioning
the shape to begin with and noticing that the (two) triangles that
happen at ascent (first from d_max to d, and then from d to 0) fit
with the triangle at descent.

> The rest looks sane based on those assumptions, but i can't really
> figure out why you care about the (max_d*t1) part.

Your equation had it too, look at it. You just had the first obvious
triangle (the "(max_d*t1*0.5)" in your equation), it's just that your
other terms end up adding up to another triangle, and the sum of those
is my "max_d*t1".

And no, I didn't actually do the simplification, and I didn't check
that your more complex expression is correct.. But I'm pretty sure my
simpler one *is* correct, though.

                Linus