Modern Radio Practice in Using Graphs and Charts
April 1932 Radio News Article
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.
April 1932 Radio News
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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. Borst
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.
The formulas 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 other one.
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.
When cross-section 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
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
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.
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
Chart Solves Resistance and Capacity Problems
It 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.
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.
The well-known 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
Draw the two rectangular triangles ACX and XDB. Since
these triangles are similar, we can write
OB.OX sin q - OA.OB
cos p sin q
= OA.OB sin p cos q - OA.OX sin p
OX (OA sin p + OB sin q)
(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 desired form:
When the modulus on the three scales is also taken in consideration,
OX, OA and OB must be replaced by their respective values
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.
These 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."
Posted October 4, 2013