April 1932 Radio News
Wax nostalgic about and learn from the history of early
electronics. See articles from
Radio & Television News, published 1919-1959. All copyrights hereby
Using charts and nomographs to solve calculations is not as
common in today's world of portable high-powered computers as
it was when computers used vacuum tubes or even mechanical gears.
One exception might be the Smith chart, especially in the lab
or field. You can plug numbers into a programmable calculator
or tablet app, but having the solution presented in the form
of a number gives you only that - a number. That's usually good
enough, but if you are doing troubleshooting or tweaking a design,
being able to see how the value got to be what it is by seeing
what's around it can be very helpful. The Smith chart is a particularly
good example when watching the complex impedance point move
around. The more experience you have, the less you need such
devices, but newbies can really benefit from charts. This article
presents a chart (nomograph) that facilitates calculating two
resistors or inductors in parallel, or two capacitors in series
according to the 1/XT = 1/X1 + 1/ X2.
Modern Radio Practice in Using Graphs and Charts
Part Four (see
Calculations in radio design work
usually can be reduced to formulas represented as charts which
permit the solution of mathematical problems without mental
effort. This series of articles presents a number of useful
charts and explains how others can be made
By John M.
A chart for the calculation of the total resistance
of two branches or the total capacity of a series is so simple,
convenient and easy to make that the author has never been able
to understand why it is not used more. In contrast to the chart
which was described last month there is nothing difficult about
this one. Ordinary decimal scales are used and the calibrations
as well as the angles are easily found.
for the solution of our problems are
The similarity of these two equations makes the chart suitable
for the solution of either one of them. All we say in regard
to one equation or its solution is equally applicable to the
In the February, 1932, issue of RADIO NEWS
it was shown that this type of equation could be solved with
the aid of an alignment chart consisting of three scales converging
at one point. If the two angles formed at this point are equal
and of 60 degrees, then the same size unit can be used throughout.
It is not always convenient to make the angles of 60 degrees
for a rather wide sheet of paper becomes necessary.
Fortunately, the two angles can be "anything." If the angle
is a, then the modulus of the slanting scales should be
where Mx is
the modulus of the center scale.
paper is used, the divisions are already present; it is only
necessary to draw the three lines.
It will be seen that
in the case of a large resistor, shunted by a small one, the
ruler, laid across the chart for the reading of the total, will
make a very sharp angle with all three scales and consequently
the accuracy is not so good. This can be obviated by making
another chart with unequal angles for this kind of solution.
The relation still holds when the angles are unequal. This is
proven at the end of this article.
The two types of
charts have been united and you find them both in Figure 1.
The scales marked A belong to the equal-angle chart and are
intended for the solution of the average problem. The scales
marked B form the scale of the unequal-angle type. When the
two resistors in the problem are widely different, this is the
chart to use.
If the range is not large enough for a
particular problem, all values on the three scales can be multiplied
or divided by any number.
Problem 1. Let it be required
to find what is the resistance equivalent to 100 ohms and 150
ohms in parallel. A transparent ruler laid across the chart
so that it connects 100 on one slanting scale and 150 on the
other, crosses the center line at 60. This is the resistance
Problem 2. Suppose there were three resistors
in parallel, for instance, 150, 100 and 40 ohms. In this case,
first determine the resistance of two branches in parallel,
say 100 and 150; our problem No.1. Then the equivalent resistance,
60 ohms, is in parallel with 40 ohms and the chart is used a
second time. Draw a line from 60 on one slanting scale to 40
on the other and find, in the center, the intersection at 24.
This is the resistance of the three branches in parallel.
Problem 3. What is the capacity of 120 and 8 micro-microfarads
in series? Since here the two condensers are so widely different,
it is best to use chart B. A line drawn from 120 on the right
slanting scale to 8 on the extreme left scale crosses B2 at
7.6 micro-microfarads approximately. This is the capacity of
the combination-smaller than the smallest.
Chart Solves Resistance and Capacity Problems
Figure 1. This chart and a ruler are all that you need
for the solution of total resistance of resistor branches and
for the solution of total capacity for condensers in series
is of course necessary to read the same units on all scales.
They should be all micro-microfarads or all micro-farads for
one particular problem.
Some equations, frequently used
in optics, are of the same form as those here treated and therefore
can be solved with the aid of the same chart.
formula, giving the relation between focal length of a lens
and the distances of image and object, is
where f is the focal distance, i the image distance and
o the object distance. Read f on the middle scales: A2 or B2.
A second one is the formula for the focal length of
a system of two or more lenses.
where F is the focal length of the system and f1
f2are the focal
lengths of the individual lenses making up the system. In this
case F is to be read on the middle scale.
In Figure 2 are shown three scales converging at one point
and with the angles p and q unequal. It is required to find
the relation between the segments OA, OX and OB when A, X and
B are on a straight line.
The solution is found in the
same way as in previous examples, in the March and February
issues of RADIO NEWS.
Draw the two rectangular triangles
ACX and XDB. Since these triangles are simi-lar, we can write
OB.OX sin q
- OA.OB cos p sin q
= OA.OB sin p cos q - OA.OX sin p
Solving for OX:
OX (OA sin
p + OB sin q)
= OA.OB (cos q sin p + cos p sin q)
From trigonometry: cos q sin p + cos p sin q = sin (p+q).
Substituting this in the equation (1) and writing it in the
When the modulus on the three scales is also taken
in consideration, OX, OA and OB must be replaced by their respective
aMa and bMb
The formula then becomes
Figure 3 shows the principle of constructing the symmetrical
chart. A constructional line drawn through division 50 on the
center scale and perpendicular to it, intersects both slanting
scales at division 100. All other lines, parallel to it, pass
through divisions on the slanting scales which indicate twice
that of the one in the center.
In the case of a non-symmetrical scale, the construction
is made as in Figure 4. The constructional lines form parallelograms
and the numbers are the same at three angles.
constructions are easy to understand, for in each case they
are really sample calculations as we described last month, under
the subhead, "The Automatic Method."
Nomographs Available on RF Cafe:
Voltage and Power Level Nomograph
- Voltage, Current, Resistance,
and Power Nomograph
- Earth Curvature Nomograph
- Coil Design Nomograph
Coil Inductance Nomograph
- Antenna Gain Nomograph
Posted October 4, 2013