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

[Anton Lundin]

On 04 November, 2013 - Anton Lundin wrote:

> On 03 November, 2013 - Linus Torvalds wrote:
> 
> > 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)").
> > 
> 
> Ah. It was the tricky part about using the same accent and decent speed
> that i didn't get.
> 
> Then everything makes sense again.
> 
> But i still thinks that your formula is missing a (max_d-d)*(max_t-t4)
> factor. Its not a big one, but its still one.

Stupid me. Its there, it was just i who missed it.

Ok, then i'm happy with the math in this one.


We should fix the comment with the t0..t3 vs. t1..t4. We should probably
add a explanatory comment that explains just points out that with the 
same accent/decent speed this is true:


(max_d*t1) == (max_d*t1*0.5) + (max_d-d)*(t3-t2)*0.5 +
              d*(max_t-t4)*0.5 + (max_d-d)*(max_t-t4)
And
t1 = t3-t2 + max_t-t4

So that no one else needs to get confused by this =)


So for what its worth:

Reviewed-by: Anton Lundin <glance at acc.umu.se>

//Anton

> 
> > > 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
> 
> -- 
> Anton Lundin	+46702-161604
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-- 
Anton Lundin	+46702-161604