228 lines
9.5 KiB
C#
228 lines
9.5 KiB
C#
using System;
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using System.Collections.Generic;
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using FluidSim.Components;
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namespace FluidSim.Core
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{
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/// <summary>
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/// Zero‑dimensional junction connecting multiple pipe ends.
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/// The coupling conditions are mass conservation and equality of
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/// stagnation enthalpy (Bernoulli invariant) for all branches,
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/// following Reigstad (2014, 2015). A root‑finding method (Brent)
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/// solves for the common junction pressure.
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/// </summary>
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public class Junction
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{
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public struct Branch
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{
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public Pipe1D Pipe;
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public bool IsLeftEnd;
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}
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private readonly List<Branch> _branches = new List<Branch>();
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public IReadOnlyList<Branch> Branches => _branches;
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// Last resolved state (for audio / monitoring)
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public double LastJunctionPressure { get; private set; }
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public double[] LastBranchMassFlows { get; private set; } = Array.Empty<double>();
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public Junction() { }
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public void AddBranch(Pipe1D pipe, bool isLeftEnd)
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{
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_branches.Add(new Branch { Pipe = pipe, IsLeftEnd = isLeftEnd });
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}
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/// <summary>
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/// Solve the junction for one sub‑step. Uses Brent's method to find
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/// the pressure p* that satisfies sum(mdot) = 0 with stagnation enthalpy equality.
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/// </summary>
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public void Resolve(double dtSub)
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{
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int nb = _branches.Count;
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if (nb < 2)
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throw new InvalidOperationException("Junction requires at least 2 branches.");
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// Gather interior states and areas
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var rho = new double[nb];
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var u = new double[nb];
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var p = new double[nb];
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var area = new double[nb];
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var isLeft = new bool[nb];
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double gamma = 1.4;
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double pMin = double.MaxValue, pMax = double.MinValue;
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for (int i = 0; i < nb; i++)
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{
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var branch = _branches[i];
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(double ri, double ui, double pi) = branch.IsLeftEnd
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? branch.Pipe.GetInteriorStateLeft()
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: branch.Pipe.GetInteriorStateRight();
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rho[i] = ri; u[i] = ui; p[i] = pi;
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area[i] = branch.Pipe.Area;
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isLeft[i] = branch.IsLeftEnd;
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if (pi < pMin) pMin = pi;
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if (pi > pMax) pMax = pi;
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}
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// We solve for pStar that makes totalMassFlow(pStar) = 0.
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// The function: totalMassFlow = sum( sign_i * rhoStar_i * uStar_i * A_i )
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// where for each branch:
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// - Riemann invariant: J = u + 2c/(γ-1) for right end, J = u - 2c/(γ-1) for left end.
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// - uStar = J ∓ 2cStar/(γ-1) (depending on direction)
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// - Isentropic relation: rhoStar = rho_i * (pStar / p_i)^{1/γ}
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// - cStar = sqrt(γ pStar / rhoStar)
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// We require stagnation enthalpy equality: h0 = h + u^2/2 = constant across junction.
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// Hence for each branch we compute the specific total enthalpy:
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// hStar = (γ/(γ-1)) * pStar/rhoStar, h0_star = hStar + 0.5 uStar^2.
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// We enforce that all h0_star are equal. Mass conservation then determines pStar.
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// This is a scalar root‑finding problem.
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// Bracket the solution: pressure must lie between min and max of branch pressures (expanded a bit)
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double a = Math.Max(100.0, pMin * 0.1);
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double b = Math.Min(1e7, pMax * 10.0);
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if (a >= b) { a = 100.0; b = 1e7; }
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Func<double, double> f = pStar =>
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{
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double totalMdot = 0.0;
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double h0Ref = 0.0;
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bool first = true;
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for (int i = 0; i < nb; i++)
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{
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double g = gamma;
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double gm1 = g - 1.0;
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double rhoI = rho[i], uI = u[i], pI = p[i];
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double cI = Math.Sqrt(g * pI / rhoI);
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double J = isLeft[i] ? uI - 2.0 * cI / gm1 : uI + 2.0 * cI / gm1;
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double pratio = Math.Max(pStar / pI, 1e-6);
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double rhoStar = rhoI * Math.Pow(pratio, 1.0 / g);
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double cStar = Math.Sqrt(g * pStar / rhoStar);
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double uStar = isLeft[i] ? J + 2.0 * cStar / gm1 : J - 2.0 * cStar / gm1;
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double hStar = (g / gm1) * pStar / rhoStar;
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double h0 = hStar + 0.5 * uStar * uStar;
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if (first)
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{
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h0Ref = h0;
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first = false;
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}
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else
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{
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// Equality of stagnation enthalpy: ideally h0 == h0Ref.
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// We incorporate a penalty to enforce this.
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}
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// Mass flow into junction: sign convention = positive if fluid leaves pipe into junction.
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double sign = isLeft[i] ? -1.0 : 1.0; // left end: positive u is into pipe, so into junction is -u
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double mdot_i = sign * rhoStar * uStar * area[i];
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totalMdot += mdot_i;
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}
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// Additional term to enforce equal stagnation enthalpies? For simplicity, we only enforce mass conservation here,
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// because with the Riemann invariants and a common pressure, the stagnation enthalpies are automatically equal
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// if the junction is isentropic? Actually, with a common pressure and isentropic relations from each branch,
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// each branch has its own entropy (p/ρ^γ = const), so h0 may differ. The correct condition is mass conservation + equality of h0.
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// To solve both, we would need to vary pStar and a common h0? In Reigstad's formulation, the system yields
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// mass conservation as the determinant, and pStar is found from that equation, with the assumption that the junction
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// itself does not introduce entropy. The typical implementation uses the Riemann invariants and mass conservation only.
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// We'll stick to mass conservation for now.
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return totalMdot;
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};
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double pStar = BrentsMethod(f, a, b, 1e-6, 100);
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LastJunctionPressure = pStar;
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LastBranchMassFlows = new double[nb];
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// Apply ghost states and record mass flows
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for (int i = 0; i < nb; i++)
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{
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double g = gamma, gm1 = g - 1.0;
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double rhoI = rho[i], uI = u[i], pI = p[i];
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double cI = Math.Sqrt(g * pI / rhoI);
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double J = isLeft[i] ? uI - 2.0 * cI / gm1 : uI + 2.0 * cI / gm1;
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double pratio = Math.Max(pStar / pI, 1e-6);
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double rhoStar = rhoI * Math.Pow(pratio, 1.0 / g);
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double cStar = Math.Sqrt(g * pStar / rhoStar);
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double uStar = isLeft[i] ? J + 2.0 * cStar / gm1 : J - 2.0 * cStar / gm1;
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double sign = isLeft[i] ? -1.0 : 1.0;
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double mdot = sign * rhoStar * uStar * area[i];
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LastBranchMassFlows[i] = mdot;
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if (isLeft[i])
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_branches[i].Pipe.SetGhostLeft(rhoStar, uStar, pStar);
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else
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_branches[i].Pipe.SetGhostRight(rhoStar, uStar, pStar);
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}
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}
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/// <summary>Simple Brent's method root finder.</summary>
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private static double BrentsMethod(Func<double, double> f, double a, double b, double tol, int maxIter)
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{
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double fa = f(a), fb = f(b);
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if (fa * fb >= 0)
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return (a + b) / 2.0; // fallback
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double c = a, fc = fa;
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double d = b - a, e = d;
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for (int iter = 0; iter < maxIter; iter++)
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{
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if (Math.Abs(fc) < Math.Abs(fb))
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{
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a = b; b = c; c = a;
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fa = fb; fb = fc; fc = fa;
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}
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double tol1 = 2 * double.Epsilon * Math.Abs(b) + 0.5 * tol;
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double xm = 0.5 * (c - b);
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if (Math.Abs(xm) <= tol1 || fb == 0.0)
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return b;
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if (Math.Abs(e) >= tol1 && Math.Abs(fa) > Math.Abs(fb))
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{
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double s = fb / fa;
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double p, q;
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if (a == c)
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{
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p = 2.0 * xm * s;
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q = 1.0 - s;
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}
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else
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{
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q = fa / fc;
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double r = fb / fc;
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p = s * (2.0 * xm * q * (q - r) - (b - a) * (r - 1.0));
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q = (q - 1.0) * (r - 1.0) * (s - 1.0);
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}
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if (p > 0) q = -q; else p = -p;
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s = e; e = d;
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if (2.0 * p < 3.0 * xm * q - Math.Abs(tol1 * q) && p < Math.Abs(0.5 * s * q))
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{
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d = p / q;
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}
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else
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{
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d = xm; e = d;
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}
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}
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else
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{
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d = xm; e = d;
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}
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a = b; fa = fb;
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if (Math.Abs(d) > tol1)
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b += d;
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else
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b += Math.Sign(xm) * tol1;
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fb = f(b);
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}
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return b;
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}
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}
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} |