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Tempo ramps - demystify comments a bit.

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nick_m 2016-02-26 13:46:06 +11:00
parent 895bbbba11
commit d8549c6ff5
1 changed files with 23 additions and 12 deletions
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libs/ardour/tempo.cc
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@ -261,6 +261,7 @@ TempoSection::frame_at_beat (double beat, framecnt_t frame_rate) const
{
return frame_at_tick (beat * BBT_Time::ticks_per_beat, frame_rate);
}
/*
Ramp Overview
@ -273,7 +274,7 @@ Ta----|--------------|* |
| * | |
T0----|* | |
* | | |
_______________|___|_____
_______________|___|____
time a t (next tempo)
[ c ] defines c
@ -281,32 +282,42 @@ Duration in beats at time a is the integral of some Tempo function.
In our case, the Tempo function (Tempo at time t) is
T(t) = T0(e^(ct))
where c is the function constant:
where c is the function constant
c = log(Ta/T0)/a
and
so
a = log(Ta/T0)/c
Given the function constant, the integral of the Tempo function (the beat function, which is the duration in beats at some time t) is:
The integral over t of our Tempo function (the beat function, which is the duration in beats at some time t) is:
b(t) = T0(e^(ct) - 1) / c
To find the time t at any beat b, we use the inverse function of the beat function (the time function) which can be shown to be :
To find the time t at beat duration b, we use the inverse function of the beat function (the time function) which can be shown to be:
t(b) = log((cb / T0) + 1) / c
We define c for a tempo ramp by placing a new tempo at some distance t away from our existing one.
The problem is that we usually don't know t.
We do know the duration in beats where the next tempo section lies.
Where t = a (when we define c), we can solve a in terms of beat duration b and the two relevant tempos using the beat function:
a = b log (Ta / T0) / (T0 (e^(log (Ta / T0)) - 1))
The time t at which Tempo T occurs is a as above:
t(T) = log(T / T0) / c
We then use a to set the function constant c (above).
Solving a and setting c in one operation allows us to further reduce the problem to:
We define c for this tempo ramp by placing a new tempo section at some time t after this one.
Our problem is that we usually don't know t.
We usually do know the duration in beats between this and the next tempo section.
Where t = a (i.e. when a is equal to the time of the next tempo section), we can solve t in terms of
beat duration and our two tempos.
A bit of scribbling with the beat function gives us:
t = b log (Ta / T0) / (T0 (e^(log (Ta / T0)) - 1))
By substituting our expanded t as a in the c function above, we see that our problem is reduced to:
c = T0 (e^(log (Ta / T0)) - 1) / b
We can now evaluate and store c for use in beat, time and tempo calculations until the following tempo section
(the one that defines c in conjunction with this one) is changed or moved.
Most of this stuff is taken from this paper:
WHERE’S THE BEAT?
TOOLS FOR DYNAMIC TEMPO CALCULATIONS
Jan C. Schacher
Martin Neukom
Zurich University of Arts
Institute for Computer Music and Sound Technology
https://www.zhdk.ch/fileadmin/data_subsites/data_icst/Downloads/Timegrid/ICST_Tempopolyphony_ICMC07.pdf