From: Josef 'Jupp' SCHUGT <jupp@...>
Date: 2004-02-03T08:30:51+09:00
Subject: Re: Ruby math: Newton's Law of Cooling

Hi!

* Theodore Knab:
> I was wondering how I would compose the following "Newton's Law of
> Cooling" into a Ruby calculation:
> 
> http://scienceworld.wolfram.com/physics/NewtonsLawofCooling.html
> 
> t = time
> T = changing temp of object
> T_s = temp of surrounding environment
> T_0 = initial temperature of the object
> K = an experimental constant that has to do with water and surface area
> 
> T(t) = T_s + (T_0 - T_s)e^(-Kt)

Perhaps it is helpful to explain the physics behind that formula. It
will be very elementary. What do we know?

First of all we know that we start with some initial temperature of
the object. We call it T_0.

Then we know that if we wait long enough the temperature will become
that of the surrounding environment.

Experiments show that the temperature change is proportional to the
difference between the temperature of the object and the temperature
of the environment and the timespan you wait (given that the
temperature does not change significantly during that timespan) So
for a short timespan dt the following holds:

    T(t + dt) - T(t) =  K * [T(t) - T(s)] * dt

Now let us take a look at K. This constant describes how fast the
heat exchange takes place. This depends on several factors (in no
particular order):

- The surface through which the exchange takes place. If the surface
  is large the echange is faster.
- The amount of matter that has to be cooled down. Dinosaurs were de
  jur cold-blooded but de facto warm-blooded.
- The heat capacity of the material that is cooled down.
- Who likely the energy is transported away (depends on isolation,
  humidity of the environment, etc.)

Anyway. In the limit of dt versus zero the above difference equation
becomes Newtons Law of Cooling. Now let's come to the problem of how
to use the equation. For this you have to assume that the temperature
is the only enviromental influence that is subject to change.
Otherwise things become *very* complicated.

The first task is finding out K. On a day where wether is very
constant so that you need not bother about a changing environment
temperature you measure the temperature of the object at fixed time
intervals and write down temperature and time. Now let's modifiy the
equation:

T(t) = T_s + (T_0 - T_s)e^(-K*t)
T(t) - T_s = (T_0 - T_s) * e^(-K*t)
(T(t) - T_s) / (T_0 - T_s) = e^(-K*t)
ln (T(t) - T_s) - ln( T_0 - T_s) = -K*t
ln (T(t) - T_s) =  ln( T_0 - T_s) - K*t

Let's use the abbreviations x(t) = ln (T(t) - T_s) and
x_0 = ln (T_0 - T_s):

x(t) = x_0 - K * t

This means: If you plot x(t) on the vertical axis and t on the
horizontal axis the slope is -K.

Does that answer the question?

Josef 'Jupp' SCHUGT
-- 
http://oss.erdfunkstelle.de/ruby/    -     German comp.lang.ruby-FAQ
http://rubyforge.org/users/jupp/     -     Ruby projects at Rubyforge
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