A four-variable nomogram for Newton's law of cooling
What happens if you put an object with one temperature in an
environment with a different temperature? In idealized conditions,
Newton's law of cooling says that the object will change
temperature at a rate that's proportional to the difference in
temperature between the object and the environment. After a long
time, the temperature of the object and the environment will be
essentially the same.
This nomogram exposes these relations and allows you to make
quantitative calculations: Suppose you know the initial
temperature of an object and the temperature of the
environment. You want to see how the temperature of the object
varies as a function of time.
Find the current time on the vertical axis. The top
of the axis is time \(t=0,\) while later times are lower.
Locate the starting temperature of the object on
the red axis, and the
temperature of the environment on
the blue axis.
Follow the guidelines corresponding to each temperatures
until you reach the height given by the current time.
The horizontal separation between the two guidelines
represents the current temperature of the object. And by varying
the time (vertical position), you can discern exactly how temperature changes with time.
Appendix: Creating nomograms with oriented transparency
This nomogram is an example of nomogram with oriented
transparency. In general, nomograms with oriented
transparencies involve an equation of four variables that can be
expressed as follows:
$$A_1(x_1) = A_{2,3}(x_2, x_3) + A_{2,4}(x_2,x_4).$$
To plot such a nomogram, first choose an orientation for your
\(x_2\) axis. Here, unlike for the cooling law nomogram above,
we'll choose a horizontal \(x_2\) axis which points
rightward; we plot the curve \(\langle x_2, 0\rangle\) ranging
through all values of \(x_2\) that we're interested in, and we'll
regularly mark points on the curve with the value of \(x_2\) that
generated them.
Next, draw curves for the \(A_{2,3}\) field. To do so, choose
several values of \(x_3\) and for each one, plot the curve
\(\langle x_2, A_{2,3}(x_2,x_3)\rangle\) which you get by holding
\(x_3\) constant and varying \(x_2\) through all values you're
interested in. Label each curve with the value of \(x_3\) that
generated it.
Third, draw curves for the \(-A_{2,4}\) field— note the
minus sign. To do so, choose several values of \(x_4\) and for
each one, plot the curve \(\langle x_2, -A_{2,4}(x_2,
x_4)\rangle\). Label each such curve with the value of \(x_4\)
that generated it.
Fourth, prepare a transparency for the variable \(x_1\). In
other words, on a separate, transparent sheet with the same
coordinate system, plot the curve \(\langle 0, A_1(x_1)
\rangle\). To make gradations, regularly label points on the
curve with the value \(x_1\) that generated them. Also take care
to mark the origin \(\langle 0, 0\rangle\), as you'll need it to
align the transparency with the other sheet.
This completes the construction of the nomogram. You can solve
for any of the variables given the other three. For example, to
find \(x_1\), you simply find the guidelines corresponding to
the given values of \(x_3\) and \(x_4\), then follow them until
you reach the horizontal position corresponding to the given
value of \(x_2\). Overlay the transparency so that it is
perpendicular to the \(x_2\) axis and so that its origin
\(\langle 0, 0\rangle\) is on one of the guidelines at the
appropriate height. The \(x_1\) curve will cross the other
guideline at a particular point; this point on the transparency
reveals the desired value of \(x_1\).
Why does this work?
Suppose you are given the values of \(x_2, x_3,\) and \(x_4\).
Based on the way we constructed this nomogram, the guidelines
for \(x_3\) and \(x_4\)
will be at the appropriate height at the coordinates \(\langle
x_2, A_{2,3}(x_2,x_3)\) and \(\langle x_2,
-A_{2,4}(x_2,x_4)\). Their vertical separation is just the
difference in the second coordinates, that is
$$A_{2,3}(x_2,x_3) - (- A_{2,4}(x_2, x_4) ).$$
But the equation we're plotting is
$$A_1(x_1) = A_{2,3}(x_2,x_3) + A_{2,4}(x_2,x_4)$$
which means that their vertical separation is also \(A_1(x_1)\)
for the unknown value of \(x_1\). The transparency allows us to
align an \(A_1\) axis to measure the separation, and because
points on the \(A_1\) axis are labeled with the values \(x_1\)
that generate them, we can then read off the missing value of \(x_1\).
Transforming oriented nomograms
Sometimes, this method
produces a slightly cluttered nomogram. There are several
transformations you can perform to make it more visually
appealing.
First, you can change the scale of the \(x_2\) axis with any
monotonic function \(f\). To make this change, draw the
\(x_2\) axis using \( \langle f(x_2), 0\rangle\) and mark a
few points with the value of \(x_2\) which generated
it. Also change the first coordinate of the fields
\(A_{2,3}\) and \(A_{2,4}\) to \(f(x_2)\) instead of just
\(x_2\). The transparency (with \(x_1\)) will be unaffected.
Second, you can vertically separate the \(A_{2,3}\) and
\(A_{2,4}\) fields by any amount \(\sigma\). To do so,
translate one of the fields vertically by an amount
\(\sigma\), and translate the origin of the
transparency by the same amount in the opposite
direction.
This transformation is valid because $$A_1(x_1) + \sigma =
A_{2,3}(x_2,x_3) + A_{2,4}(x_2,x_4) + \sigma$$
is true whenever the original equation is.
Constructing a nomogram for Newton's cooling law
As noted above, in idealized settings, an object will change
temperature at a rate proportional to the difference between its
current temperature and the temperature of the
environment. Formally, we can express this relation with a
differential equation
$$ \frac{dT}{dt} \propto T(t) - T_{\infty} $$
where \(t\) is the current time, \(T(t)\) is the current
temperature of the object, and \(T_\infty\) is the (assumed
constant) temperature of the environment.
The solution to this differential equation is
$$T(t) = T_0 \cdot \exp(-k t) \quad +\quad T_\infty\cdot
\left[1-\exp(-kt)\right]$$
where \(T_0\) is the temperature of the object initally —
that is at time \(t=0\), and \(k\) is the constant of
proportionality mentioned in the differential equation. If we
subsume \(k\) and \(t\) into a single variable \(u\equiv kt\), this
solution is already in the form of a nomogram with oriented
transparency; we have
