# Nomograms

## Nomograms

These are graphical devices for evaluating a formula. A slide-rule can be considered as a basic form of nomogram, but is in fact more of a general-purpose device. Nomograms on the other hand are designed usually for the evaluation of one particular formula. In their simplest form, there are two outer scales, on which you locate the values of two quantities involved in the calculation. You draw a line between these two points, or use the edge of a ruler to join them (as you would not want to mark the paper). Ideally, a transparent ruler with a thin line running down its centre provides the best accuracy. Where the line intersects the central scale is the result of the calculation. Some nomograms are so complex that you wonder how anybody ever came to invent them. Let’s try designing one to evaluate a simple formula, for example one for the volume of a cone:

V = phd2/12,

where V is the volume, h is the height of the cone, and d is the diameter of the base.

Fig.1. First stages in designing a nomogram for the volume of a cone.

We first draw two vertical scales, a convenient distance apart, the left one showing h and the right one showing d. Fig. 1 shows the steps in this construction. Since the formula involves multiplications rather than additions, we can anticipate that these two scales will be of the same form as those on a slide-rule – so-called logarithmic scales. We can start by letting both scales run from 1 to 10. Now, we don’t know what numbers to put on the "result" scale – we don’t even know where it will be, so let’s try to locate a particular result, e.g. a volume of 1 unit. This will correspond to a mark on the result scale, labelled with the number "1". If the diameter d is one unit, then V will be 1 when h is 12/p = 3.8197. Therefore, we find (as close as possible) the value 3.8197 on the "h" scale, and join this to the "1" on the "d" scale with a straight line. Now a volume of 1 could also arise from the combination d = 1.5, h = 16/(3p) = 1.6977, and we join these two points with another straight line. These two lines are shown in red in Fig. 1. The result V = 1 must lie on both of these lines, and therefore is at the point where they cross.

At this stage, we could try one or two more combinations that give the same result V = 1. All the lines added to the diagram should pass through the same point. (If they don’t, then we know that there is something wrong with our assumptions – perhaps the scales should not be logarithmic, for example. If they do all meet, then that doesn’t prove that our assumptions are all correct, but it gives some confidence!)

Next we look for combinations of d and h that lead to the result V = 10. Possible choices are d = 2 with h = 10×12/(4p) = 9.5493, and d = 4 with h = 2.3873. Drawing these lines (shown in green in Fig. 1), we find the point where they cross – that will give the result V = 10. Finally, we draw a straight line through the two results V = 1 and V = 10, and fill in a scale, running from 1 to 10, with logarithmic spacing. This line is shorter than the two outer scales, so it is useful to extend it in both directions, so that any points on the h and d scales can be joined without the result going off the V scale. That gives the finished nomogram, as shown in Fig. 2. This is able to give the volume of a cone of any dimensions, provided that we can correct for factors of 10 mentally. For example, h = 3 and d = 3 predict that the volume is about 7.07 (7.0686 by calculator). If the dimensions are h = 30 and d = 3, then the volume is 10 times larger, i.e. 70.7, while if the dimensions are h = 3 and d = 30, the volume is 100 times larger, i.e. 707. Of course, it is possible to use the nomogram inversely – for example if you are given the height and the volume, then you can find the diameter, but you have to be even more careful about factors of 10.

Fig. 2. Finished nomogram for the volume of a cone. In practice the diagram would be drawn much larger, with a larger number of graduations.

Other formulas involving the multiplication or division of two input variables can be treated the same way except that, if the formula has a division by some quantity x (or x2, etc), then the x scale should be inverted – i.e. the values on it should increase from top to bottom.

One rather clever nomogram, shown in Fig. 3, evaluates the formula

1/x + 1/y = 1/f.

This gives the relationship between the positions of an object and its image when viewed using a lens with focal length f. Here x is the distance of the object from the lens and y the distance of the image.

Fig. 3. The lens nomogram, for converging lenses, and real objects and images.

The striking features about this nomogram are that it uses non-parallel scales, and solves a non-linear problem (i.e. one involving more complicated operations than simple addition) but using uniformly-divided scales.

For a converging lens (like a magnifying glass), the focal length f is assigned a positive value, and is the distance, from the lens, that the image is formed when the object is at infinity. It is where the hot spot is formed if you focus light from the sun (which is appropriate, as focus is the Latin word for hearth). Now the x scale doesn’t go as far as infinity but, if it did, the line from the infinity point, through the given value of f, would be horizontal, and you can see that the corresponding value of y is exactly equal to f, in accordance with the definition of f. As another general example, we see that when x = 2f, then y = 2f also. A line has been drawn (in red) on the diagram showing that, if the lens has a focal length 2 cm, and the object is placed at 3 cm, then the image is formed at 6 cm.

If x is less than f, then the constructed line (such as the green line in Fig. 3) does not meet the y scale, so where is the image? We can extend the y scale downward to negative values, as is done in Fig. 4, and find a result. For example, if f = 10 cm and x = 5 cm, then the green line gives the result y = –10 cm. The negative sign means that the image is not formed by light that has passed through the lens, but is on the same side as the object. The light emerging from the lens appears to come from a point 10 cm behind the lens. This is known as a virtual image. This is precisely the way that a magnifying glass is used (i.e. you look “through the lens” at the object) and, in this example, the image would appear twice as large as the object.

Fig. 4. The lens nomogram of Fig. 3 extended to include diverging lenses, and virtual objects and images.

We can also extend the x scale to negative values. If the object is at –10 cm, for example, then there isn’t actually an object at all, but the light hitting the lens is already converging towards a virtual object at a point 10 cm on the other side. (It doesn’t pass through this point, because the lens gets in the way!)

Finally, we can extend the 45° line to represent lenses with a negative focal length, or diverging lenses. With a real object, these always form a virtual image. For example if the lens has focal length –8 cm and the object is 8 cm behind the lens, then the blue line in Fig. 4 shows that the image is 4 cm behind the lens. If the incoming light is already converging strongly enough to give a virtual object with |x| < |f|, then a real image can be formed.

The nomogram in Fig. 3 can also be used for combining two electrical resistors in parallel (because the same formula applies), and here of course only the positive values are meaningful. Searching for "nomogram" on the Internet will give you hundreds of different examples.

C.J.Evans, Dec.2008