# 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
 $$A_1(T) = T$$ $$A_{2,3}(u,T_0) = T_0\cdot \exp(-u)$$ $$A_{2,4}(u,T_\infty) = T_\infty\cdot \left[ 1 - \exp(-u)\right]$$