Re: Symbolic equation solving in Lua

[Gavin Wraith <gavin@...>]

In message <CAGmA6cGxgim9d1GQB0+r6nb4KOkBcmVxBAYo5QEntokkQpbNEg@mail.gmail.com>
          Hur Butun <hurbutun@gmail.com> wrote:

> I don't really understand what could be the different meanings of solve in
> this case. As I mentioned in my previous message, I would like to obtain
> the value of the friction factor using Colebrook equation.

Sorry - I meant "solve for which variable?". If that is "k" then
what you need is

local x = math.sqrt(lambda)
(exp(-1/(2*x))-2*51/(R*x))/(3.72*d)

On the other hand if you want lambda in terms of R, the Reynold's number, k and
d (hydraulic diameter) then you need

lambda = 1/(math.sqrt (fixpoint (f)))

where

f = function (x)
  local a = ((2.51/R)*x + k/(3.72*d))
  return -2*math.log(a)
  end

where fixpoint(f) solves f(x) == x. To do this you need an initial guess, say 1,
a choice of accuracy epsilon, say 10^(-7) and the Newton-Raphson method:

local df = function (x) -- derivative of f
           local c = -2*(2.51)/R
           local y = (2.51*x)/R + k/(3.72*d))
           return c/y
           end

fixpoint = function (f)
           local g = guess
           while math.abs(g-f(g)) < epsilon do
              g = g - f(g)/df(g)
           end
           return g
           end

This is totally off the cuff, so E&OE.
--
Gavin Wraith (gavin@wra1th.plus.com)
Home page: http://www.wra1th.plus.com/