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Fast W'bal integration

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Dave Waterworth has fixed my implementation of W'bal to
remove my "optimisations" and fix the math.

The W'bal is now computed in a single pass and is just
as fast as the differential form from Andy Froncioni but
has the benefit that is still uses Tau.

Many thanks to Dave, and I'll write this up shortly.

Conflicts:
	src/DBAccess.cpp
	src/WPrime.cpp
This commit is contained in:
Mark Liversedge
2014-10-11 12:40:05 +01:00
parent 1f00790069
commit 4dcde2a8ca
1 changed files with 9 additions and 29 deletions
Download Patch File Download Diff File Expand all files Collapse all files

38
src/WPrime.cpp
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@@ -219,19 +219,15 @@ WPrime::setRide(RideFile *input)
int stop = last / 2;
WPrimeIntegrator a(inputArray, 0, stop, TAU);
WPrimeIntegrator b(inputArray, stop+1, last, TAU);
WPrimeIntegrator a(inputArray, 0, last, TAU);
a.start();
b.start();
a.wait();
b.wait();
// sum values
for (int t=0; t<=last; t++) {
values[t] = a.output[t] + b.output[t];
xvalues[t] = double(t) / 60.00f;
values[t] = a.output[t];
xvalues[t] = t / 60.00f;
}
// now subtract WPRIME and work out minimum etc
@@ -428,19 +424,14 @@ WPrime::setErg(ErgFile *input)
int stop = last / 2;
WPrimeIntegrator a(inputArray, 0, stop, TAU);
WPrimeIntegrator b(inputArray, stop+1, last, TAU);
WPrimeIntegrator a(inputArray, 0, last, TAU);
a.start();
b.start();
a.wait();
b.wait();
// sum values
for (int t=0; t<=last; t++) {
values[t] = a.output[t] + b.output[t];
xvalues[t] = t * 1000;
values[t] = a.output[t];
}
// now subtract WPRIME and work out minimum etc
@@ -549,22 +540,11 @@ void
WPrimeIntegrator::run()
{
// run from start to stop adding decay to end
for (int t=begin; t<end; t++) {
double I = 0.00f;
for (int t=0; t<=end; t++) {
if (source[t] <= 0) continue;
// start at 1, since the actual value shouldn't be adjusted with itself !
for (int i=1; i < (TAU*3) /*WPrimeDecayPeriod*/ && t+i < source.size(); i++) {
double value = source[t] * pow(E, -(double(i)/TAU));
// diminishing returns - we're dealing in kJ, so 10J is nothing !
// this saves about 20% in calculation time typically
if (value < 10.00f) break;
// integrate
output[t+i] += value;
}
I += exp(((double)(t) / TAU)) * source[t];
output[t] = exp(-((double)(t) / TAU)) * I;
}
}