Electronics and Your Slide Rule, Part II — Combined Operations with Electronic Applications (CIE Electronics Slide Rule No. 515 / Pickett N-515-T)
electronics
CLEVELAND INSTITUTE OF evectnonics / CLEVELA ND, OHIO
Electronies and
Youn Slide Rule =
Part {I
An AUTO-PROGRAMMED Lesson
Processediby Reicale
Provided | bysoe)H
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
ABOUT THE AUTHOR
Through over 15 years experience in helping students learn through home
study, Mr. Geiger has obtained an intimate understanding of the prob-
lems facing home-study students. He has used this knowledge to make
many improvements in our teaching methods. Mr. Geiger knows that
students learn fastest when they actively participate in the lesson, rather
than just reading it. Accordingly, you will find many “What Have You
Learned?" sections in this lesson, to assist you in getting a firm grasp of
each topic.
Mr. Geiger edits much of our new lesson material, polishing up the manu-
scripts we receive from subject-matter experts so that they are easily read-
able, contain only training useful to the student in practical work, and are
written so as to teach, rather than merely presenting information.
Mr. Geiger's book, Successful Preparation for FCC License Examinations
(published by Prentice-Hall), was chosen by the American Institute of
Graphic Arts as one of the outstanding text books of the year.
) Lesson exclusive with the
Cleveland Institute of Electronics, It is your assur-
ance of the finest in technical education and training.
bed ere 4
goitED Sc, 2
£ % Accredited by the Accrediting Commission zo
z +, of the National Home Study Council ge
3 3 Tix Goeretting Somntechn ben ten erent ee E>
4% E Office of Education as a “r antaed acereditinge gs
“Ones ot 3:
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Electronics and Your
Slide Rule
Combined Operations with
r Electronic Applications
By DARRELL L. GEIGER
Senior Project Director
Cleveland Institute of Electronics
icon ae
‘joy RF Catel
wen
»
ded by
A:
z
&
is 3
es
oo
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
15. Reciprocals... Page 24
16. Combined Multiplication and Division... Page 25
17, Resetting the Slide... Page 30
18, Handling Several Factors with CI Scale... Page 32
19, Solving Practical Problems With the Slide Rule... Page 35
20, Repeated Operations With One Factor Constant, .. Page 37
21. Squares und Square Roots... Page 40
22, Evaluating Expressions Containing Squares... Page 44
23. Evaluating Expressions Containing Square Roots... Page 48
24. Proportion... Page 53
Examination... Page 55a
3
x oe
© Copyright 1966, 1965, 1962, 1958 Cleveland Institute of Electronics. : ;
All Rights Reserved/Printed in the United States of America. se
FOURTH EDITION/Fifth Revised Printing/October, 1966, 3:
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
our rmstructor
Some people think they know how touse aslide rule if they can multi-
ply or divide two numbers with it. While a slide rule can be helpful
even if that is allone knows about it, the real potential of aslide rule
as a time saving device is realized in problems involving several
steps. Rather thanworkingsucha problem step-by-step and reading
off theanswer to each before starting the following step, the expert
combines the steps into one blended operation, The only reading he
generally makes is the final answer. This lesson will teach you the
way of the expert.
In the last lesson you learned the basic use of the C and D scales.
In this lesson you will be introduced to three additional scales, the
Cl, the Aand the B. While the CI scale is basically one for finding
reciprocals, you will learn how it is used in conjunction with other
scales to simplify many problems. The A and B scales are basic-
ally for squares and square roots. By combining these scales with
other scales, you can solve problems quickly that would involve sey-
eral steps if worked by pencil.
We have emphasized before that you need pay no attention to the dec-
imal point in the number while manipulating the slide rule. There
is one very important exception. That is problems where square
roots are involved. While studying this lesson pay particular atten-
tion to how the decimal! point location affects the method of taking
square root.
You Will also learn in this lesson the use of the slide rule for prob-
lems in proportion. Proportional problems are particularly easy to
understand and work by the slide rule method, and such problems
are widely found in electronics.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Processed by RF Cafe
Provided by Joe H.
Electronics and
Your Slide Rule
Part II
1 RECIPROCALS...The reciprocal of a number is that number
divided into 1. Thus the reciprocal of 245 is equal to 1/245,
which by regular division on the slide rule is found to be
0.00408. However, since reciprocals are so widely used,
the CI scale is placedon the slide rule so that reciprocal values
can be read directly. To find the reciprocal of 245 using the
Cl scale, set the hairline of the indicator over 245 on scale C
and then read the digits of the answer, 408 under the hairline
on the CI scale. See Fig. 14. The slide need not be moved in
finding reciprocals. To locate the decimal point, notice that
1/200 is equal to 0.005. Hence, the correct answer must be
0. 00408,
To fing the reciproce! of 245
set howling over 245 of scale C
Lees aneeencsenescressm: Oe
1
ee a ef es ret or ame — —7
245
then reod 408 on. scale Ci under hoirline
Fig. 14. Reading reciprocal of 245, which equals 0.00408, on slide rule.
WHAT HAVE YOU LEARNED?
Find the reciprocals of the following values:
1...0.178 5... 0.00064
2 -2...2,849 6. ..32, 000
z2 0 3...87 7...33.6
i 4...0.4444 8...4.15 x 107%
Ee 24
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
9... The conductance of acircuit is 0.72 mhos. What is its 25
resistance?
10,..Four resistors of 24, 130, 45, and 73 ohms are connected
in parallel. What is the combined resistance?
11...Three resistors of 18, 32, and 26 ohms are connected in
parallel, What is the combined resistance?
ANSWERS
1. 5.78 2. 0.351 3. 0.01149 4. 2.25 5. 1562
6. 0.0000312 7. 0.0298 8. 2.41 x10’ 9, 1.389 ohms
10. 11.73 chms 11. 7.98 ohms
16 COMBINED MULTIPLICATION AND DIVISION... By handling
Provided by Joe H,
Processed by RF Cafe
problems where both multiplication and division are involved
in the correct manner, the number of steps can be reduced,
saving labor and reducing the possibility of making a mistake.
ee re
Example...5a
Find the value of 4saA * 8.51
91.5
Solution...
For the simplest method, first divide 7.24 by 91.5 and then
multiply by 8.51, as follows:
(1) Push haifline tu 724 on scale D.
(2) Move slide so that 915 on scale C is under the
hairline,
(3) Push hairline to 851 on scale C.
(4) Under hairline on scale D read, 673, the digits
of the answer.
To locate decimal point: =* = 0.56. Hence, the answer
will be 0. 673.
Explanation. . .
Fig. 15 illustrates this problem, The dashed vertical line
indicates the first position of the hairline, in which 724 on
scale D is opposite 915 on scale C. The solid vertical line
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Provided by Joe H.
124 X85!
To find cee
set 913 00 scole C opposite 724 on scole D
then opposite @5/ on scale C read 673 on scole D0
Fig. 15. Finding ™-24-X 4.51 = 0.673
represents the second position of the hairline, where it is
placed over 851 on scale Cand reads the answer 673 on scale D.
Tounderstand how the steps of the solution result in the answer
for us x 851, first divide 724 by 915 in the regular manner,
writing down the answer 791. To get the answer to the prob-
lem, 791 must now be multiplied by 851. To do this the right
index of scale C is placed opposite 791 on scale D. Providing
the slide was not moved alter writing down the value 791, it
willbe found that the right index of scale C is already opposite
791 on scale D. Thus, no resetting of the slide is necessary,
it being already in the proper position. Hence, the step where
the index of the.C.scale is set opposite the quotient of 724/915
on the D scale can be entirely eliminated, the stidé @tready
being in its proper position. For the same reason it is not
necessary to read or to consider the quotient of 724/915. With
this unnecessary work eliminated the problem reduces to the
four steps given under the solution above.
In order for you to see the importance of working problems in
the proper order where several factors are involved, you
should rework Example 5 above by first multiplying 724 by
851, and then dividing by 915, You will find that the slide
must be set twice when the problem is worked in this manner,
thus making considerably more work, and increasing the
chances of making a mistake. Also, the accuracy is reduced
Processed by RF Cafe
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
to some extent, because the more times the slide must be set
the less the final accuracy. That is because any setting of the
slide is subject to a certain amount of error, so that the more
the slide is set the greater the total error. However, even
with many settings this error should be small with proper care.
By the method of Example 5(a) the combined resistance of two
resistors in parallel can be found quickly by using the formula,
_R, xR,
Rt =k +R,
Example... 5b
What is the combined resistance of 23 ohms and 30 ohmsin
parallel?
Solution... See Fig. 16
(1) Set hairline over 30 on scale D,
(2) Mentally add 23 to 30 giving 53. Move slide
so that 53 on scale C is under the hairline,
(3) Move hairline over 23 on scale C.
(4) Read the figures of the result, 1302, under
the hairline on scale D.
(5) Tolocate the decimal point we make use of the
rule that the combined resistance of two re-
sistors in parallel is always less than that of
the smaller of the two resistors, but never
less than half the resistance of the smaller
resistor.
Hence, in this case the resistance must be less than 23 ohms
and greater than 11.5 ohms. The answer must then be 13, 02
ohms or 13 ohms for all practical purposes.
Example 5(a) involves only three terms. When more than
three are involved in a combined multiplication and division
problem, the problem is worked by an extension of the method
above. The rule for working such problems with least labor
and minimum settings of the slide is to divide first and then to
alternately multiply and divide.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
27
Provided by Joe H,
Processed by RF Cafe
Provided by Joe H.
To find the combined resistence of 23. «© poralie! with 30
set 23 + 30+ 53 on scale C opposite JO on scale D
Sees re —— re
|
2 3 eee ee
1 of e213) a 3 er isd 3
v4 bones
bee c=
Ss
then opposite 23 on scole C recd /3 on scole D
Fig. 16. Finding that the combined resistance of 23 ohms in parallel with
30 ohms is 13 ohms.
Example... 6
Find the value ot 51.7 x 23.1 x 3.27
42 x 7.53
Solution. ..
First divide 517 by 42. Then multiply by 231. Follow this by
dividing by 753, and then multiplying by 327. The problem
could, of course, be worked equally well in some other order,
just so you divideas the first step and then follow by alternate-
ly multiplying and dividing.
The work is as follows:
(1) Set hairline over 517 on scale D.
(2) Push slide so that 42 on scale C is under the
hairline,
(3) Push hairline so that it is over 231 on scaleC.
(4) Push slide so that 753 on scale C is under the
hairline.
(5) Push hairline so that it is over 327 on scaleC.
(6) Read the digits of the answer, 1235, under
hairline on scale D.
To locate decimal points:
56 x 20 x 3 _ 300 _ ox, 308. 19
ap x7 26 “PPYOX. “og
Hence, the answer to the problem is 12.35.
Processed by RF Cafe
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Example 6 could just as well have been worked in some other
order, just so a division is made first and that followed by
alternately multiplying and dividing. As anexercise you should
work the problem in some other order, Asa further exercise,
work the problem by first multiplying together the numerator
factors, 517 x 231 x 327, then multiplying the denominator
factors, 42 x 753, and then dividing the numerator product by
the denominator product. You will find the labor by this meth-
od to be many times that of the proper method explained
above. Also, you are much more apt to make a mistake. Yet
you will find many technicians using their slide rules in this
cumbersome manner.
Example. ..'7
Find the value of lH ER.
72 x 38.2
Solution. ..
First divide 87 by 72, then multiply by 461, and then divide by
382, as follows:
(1) Push hairline so that it is over 87 on scale D.
(2) Push slide so that 72 on scale C is under the
hairline.
(3) Pushhairlineso that it is over 461 on scale C.
(4) Push slide so that 382 on scale C is under the
hairline.
(5) Read the digits of the answer, 1458, opposite
the left index of scale C
To place decimal! point:
76° 46 28
where 2” means a value less than 2. Hence, the answer to the
problem is 1, 458.
Explanation...
The difference between this problem and Examples 5(a) and 6
is that the last step in this problem is one of division. Hence,
you read the answer opposite the index of C, the same as you
would for any ordinary division problem.
eee een eee ee ee ee ed
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Provided by Joe H,
Processed by RF Cafe
1
Provided by Joe H.,
Processed by RF Cate
RESETTING THE SLIDE...In working any combination prob-
lem it sometimes happens that an operation cannot be per-
formed because the part of the slide that must be used is pro-
jecting beyond the body of the rule. This difficulty is over-
come by a process called resetting the slide, This is done by
pushing the hairline so that it is over the C-scale index that
is within the body of the rule, in order to mark the position of
that index.
Then the slide is pushed so that the C-scale index that was pro-
jecting beyond the body of the rule is now brought under the
hairline; that is, it takes the place originally occupied by the
other index. For example, in Fig. 17(a) the left index of
scale C is opposite 6 on scale D. To reset the slide, push
the slide so that the right index of scale C is opposite 6, as
in Fig. 17(b).
Resetting the slide will never invalidate the answer to any
problem. Therefore it may be used without hesitation in a
problem whenever necessary in order to proceed with the fol-
lowing steps of the problem.
eee eee eee eee eee eee eee eee eee ee ee
Example... 8
Find the value of eos
(1) Set hairline over 629 on scale D,
(2) Push slide so that 35 on scale C is under the
hairline. The next step would normally be to
push the hairline so that it is over 74 on scale
C. However, this is impossible with the slide
in its present position. Therefore the slide
must be reset.
(3) Reset slide, as follows:
(a) Push hairline over left index of scale C.
under hairline.
(4) Push hairline over 74 on scale C,
(5) Read the digits of the answer, 1330, onscaleD.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
(b) Push slide so that right index of scale C is
To place decimal point: 00 xf = 120. Hence, the answer to
the problem is 133, 0.
Resetting of the slide can often be avoided by changing the
order of working theproblem. Take, for example, the problem
194 x 16.5 x 7.4
4.71 x 39
divided by 471 and then multiplied by 165, However, it is
found that the slide must be reset in order to multiply by 165.
However, we can multiply by 74 without resetting, and this
should be done, saving the factor 165 until later. That is,
divide 194 by 471 and then multiply by 74. Follow this by
dividing by 39 and then multiplying by 165, giving 129.0 as
the answer. Thus, no resetting of the slide is necessary in
this problem.
Working in the usual manner, 194 is first
~ cabs
{ . 1 6 ee ws
o0 ft 5 4 $ 6 * @ $ ©
fe) origine! pesition of slide
{b) the slide reset
Fig. 17. When the port of the slide to be used for the next operation lies
beyond the body of the rule, the slide is reset.
WHAT HAVE YOU LEARNED?
In order that you can check to see if you are working the fol-
lowing problems in the most efficient manner, the minimum
number of slide settings for each problem is given in the
answers.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
31
Processed by RF Cafe
Provided by Joe H,
bt
OC
Processed by RF Cafe
Provided by Joe H,
534 x 6.1 3400 x 9.77 x 16
1.-.-—“Syg—Ci ? 1200“ asex gee
8.66 x 420 _ 0.001622 x 84.4 x 63
3s ? 65: rE =?
4.38 «x 212 1,818 =~ 0.079 x 26.6
aes GS 6 : 9... "35-4 x 0.52
0.237 x 1621 _ 6830 7.5 _
mnie 43.2 li 10...599x 0.169 °°
284 x 0.861 11. 274 x 3,53
6... 5030 ? Ai... 0.0382 x 320 u
19 x 44°x 35.2 66.7 x 1490
6...“ xag«" ? «+12... Soo x 204,600
ANSWERS
Answer Minimum slide Answer Minimum sli
settings sett
1. 4460 1 7. 761 2
2. 388 1 8. 0.00451 2
3. 14.84 1 9. 0.314 2
4. 8.89 2 10. 101,400 3
5, 0.0486 1 ll. 3,25 2
6. 43.8 2 12. 7.14x1075% 2
HANDLING SEVERAL FACTORS WITH CI SCALE...We have
found that the most efficient way to work problems involving
more than two factors is first to divide and then to alternately
multiply and divide. This process cannot be carried all the
way through the problem unless the number of factors in the
denominator is equal to or is one less than the number of
factors in the numerator. However, when this is not the case,
it is a simple matter to change the problem so that the factors
are properly balanced between numerator and denominator for
most efficient handling.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
32
26 x 57°
cannot be done efficiently by the method so far explained, be-
cause it would require two divisions to be done consecutively.
32 x 57
26 .
Dividing by a number is the same as multiplying by the recip-
rocal of that number, Hence, rather than divide by 57, we can
Suppose, for example, that we wish to evaluate This
4
' However, we can rewrite the problem to read,
multiply by = Since the CI scale is a scale of reciprocals,
to set up = we set up 57 on the CI scale. To work this prob-
lem then, start in the regular manner by dividing 32 by 26.
Then push the hairline so that it is over 57 on scale CI. Under
the hairline read the answer, 0.0216, on scale D. The follow-
ing rule takes care of all situations involving several factors:
—>— Rule... Inproblems in multiplication and division involving
several factors, a division should be performed first, and this
followed by alternate multiplication and division. If when a
multiplication is required by this rule and no unused factor is
left in the numerator, then a factor from the denominator may
be used by setting its value on the CI scale. Or ifa division
is required by this rule and no unused factor is left in the de-
nominator, a factor in the numerator may be used on the Cl
scale.
ee | ee i eT ee ee ne
Example... 9
Find the value of 716 x 0,203 x 6.7
Solution. ..
The method is to divide 716 by the reciprocal of 203 (which
gives the same result as multiplying by 203) and then multiply
by 67,
(1) Set the hairline over 716 on scale D.
(2) Push the slide so that 203 on scale CI is under
the hairline,
(3) Push the hairline so that it is over 67 on scale C.
(4) Read the answer, 974, under the hairline on
scale D.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Provided by Joe H,
Processed by RF Cafe
34 Example... 10
Find the value of =
64 x 6.1 x 24 x 1.26
Solution. ..
(1) Set the hairline over 17 on scale D.
(2) Push the slide until 64 on scale C is under the
hairline. We have now divided 17 by 64. Lf we
try to multiply by the reciprocal of 6.1 as the
next step we find that the slide must be reset
because 6.1 on scale CI projects beyond the
body of the rule. To avoid resetting the slide,
we multiply by the reciprocal of 24 instead.
(3) Push the hairline so it is over 24 on scale CI.
(4) Push the slide so that 61 on scale C is under
the hairline.
(5) Push the hairline over 126 on scale Cl.
(6) Read the answer, 0.00144, on scale D.
Explanation. ..
The actual form of the problem as the slide rule works it is
as follows:
2 ae
17 x 24 x 126
64 x 61
The method is the same as in Example 6 and similar problems
except that reciprocal values are set on the CI scale.
Example...11 :
Find the value of 56.3 x a7
Solution. ..
The method is to divide 1 by 55.3 and then multiply by the re-
ciprocal of 27.
(1) Set hairline to 1 on acale 7D (that will be the
right index of scale D).
: (2) Pushslide so that 55.3 on scale C is under the
3 M4 hairline.
ze (3) Push hairline over 27 on scale Cl.
33 (4) Read the answer 0, 000670 on scale D under the
; g hairline,
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
_
ce)
Provided by Joe H.
Processed by RF Cate
Explanation... 3S
In step 1 the hairline could have been set to either the right or
the left index of scale D, since both represent the number 1,
However, if the left index was used it would be necessary to
reset the slide to obtain the answer to the problem. It is dif-
ficult for a beginner to know which index to select for the
shortest solution. However, with some experience it is usual-
ly easy to tell which index is best.
WHAT HAVE YOU LEARNED?
1...2.71 * 6.42% §.21=? 6...84.4 x 0.00701 x 9.89 x 8.9 = ?
23.1
2..-551 x 3a01 =? 7, 0:53 x 4.5 x 50.9 =?
13.2
473
a 1 -
2.75 x 8.5 8...7-9 x 0.0214 =?
432
4...>>>— =? 1
SB % Sr 9.--0 O47 x 25.1% 0.56a.*
24 x 68 x 57 x 49
5... =? 45 x 46 x 57
32.7 10
"tex 49x 50x51
ANSWERS
1, 90.6 2, 0,00278 3, 20,2 4, 57.6 6, 139,400
6. 52.1 7. 9.20 8 9.74 9. 1.506 10. 0,01967
SOLVING PRACTICAL PROBLEMS WITH THESLIDE RULE...
In working a practical problem with the slide rule, the various
operations that must be performed to obtain the answer should
be written down before any calculations are made. Then the
required operations can be performed as a single problem on
the slide rule, thus making most efficient use of the rule. You
have already learned how the slide rule saves the most labor
in problems that involve a number of operations.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Provided by Joe H.
Example... 12
How many feet of wire will be required to wind 168 turns on a
form 3.8 inches in diameter?
Solution...
The first thing to do is to find the length of one turn. Since the
circumference of a circle is equal to the diameter multiplied
by s, the length of one turn will be 3.8 x 3. 14 inches.
To find the total length of the wire the length of one turn
should be multiplied by the number of turns, giving 3.8 x 3.14
x 168 inches as the total length of wire. This value must be
divided by 12 to find the length of wire in feet. Hence, the
complete problem becomes:
3.8 x 3.14 x 168
12
Solving with the slide rule gives 167 feet as the answer to the
problem.
WHAT HAVE YOU LEARNED?
Work the following problems by writing down the complete
problem before making any calculations.
1...A de motor draws 23 amps at 440 volts. What is the
horsepower input to the motor? The reverse side of your
slide rule will tell you the number of watts in a horsepower.
2...The plate input to a vacuum tube is 1.7 amps at 650 volts.
The efficiency of the tube is 55%. What is the power output
from the stage?
3... Two resistors are in series. The voltage across the first
Processed by RF Cafe
voltage across the second resistor?
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
resistor is 78 volts, and the resistance is 425 ohms. The
resistance of the second resistor is 167 ohms. What is the
Processed by RF Cafe
Provided by Joe H.
4,...The voltage across a resistance of 6, 300 ohms is 83 volts.
Find the peak value of the current through this resistor.
ANSWERS
lL P= ners = 13.57 horsepower, ans.
2. P=1.7 x 650 x 0.55 © 608 watts, ans.
3. E,= mee = 30.6 volts, ans.
4. ly = eae = 0.01863 amps, ans.
REPEATED OPERATIONS WITH ONE FACTOR CONSTANT...
It frequently occurs in electronics that a number of values
must be multiplied or divided by some constant factor. The
slide rule is particularly adaptable to this type of work.
ee
Example...13
In a series of measurements on certain equipment the follow-
ing effective voltage values were obtained: 133 volts, 268 volts,
431 volts, 740 volts, and 806 volts. It is desired to know the
peak values of these voltages.
The peak voltages are found by multiplying the effective values
by 1.414.
(1) Set the left index of the C scale opposite 1.414
on scale D.
(2) Opposite 133 on scale C read 188.1 volts on
scale D, the peak value of 133 volts.
(3) Similarly, opposite 268 on scale C read 379
peak volts on scale D, and opposite 431 on
scale C read 609 on scale D.
(4) To read the peak values corresponding to 740
and 806 volts, reset the slide.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Provided by Joe H,
38 (5) Opposite 740 and 806 on scale C read the cor-
responding peak values, 1046 and 1140 volts.
Example 13 shows that to maltiply several values by a common
factor only one or two movements of the slide is required.
Repeated operations in division involving a common factor
should be changedto a problem in multiplication so as to retain
the advantage of a single (or possibly two) settings ofthe slide.
Example. ..14
The length of five antennas are 31.5 meters, 40.2 meters,
47.1 meters, 50.1 meters, and 52,3 meters. Find the length
of these antennas in feet.
Solution...
The back ofthe slide rule shows that one foot is equal to 0.3048
or about.0.305 meter. Hence, each of the antenna lengths must
be divided by 0. 305 to get their lengths in feet. Dividing in the
regular way it is found that the slide must be changed for each
value to be read, This can be avoided by changing the problem
to one in multiplication.
To do this remember that dividing by 0.305 is the same as
multiplying by the reciprocal of 0.305. By using the C and CI
scale as explained in Topic 15 the reciprocal of 0.305 can be
found to be 3.28, Hence, by setting the right index of the C
scale opposite 3.28 on scale D, we can directly read off tne
required values as in Example 13. However, there isa
quicker way to get the slide in the proper position, as follows:
Find the reciprocal of 0. 305 by dividing 1 by 0,305, To do this:
(a) Move the slide so that 0.305 on scale C is op-
posite 1 (the left index) on scale D.
$ (b) Notice that the right index of scale C is now
£ opposite 3.28, the reciprocal of 0.305, on
= scale D. Hence, no further adjustment of the
‘ slide is necessary.
£ Opposite the values in meters on scale C read the required
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
values in feet on scale D, giving the following: 31.5 meters =
103.3 feet. 40.2 meters = 131.9 feet. 47.1 meters = 154.5
feet, 50.1 meters = 164.3 feet, 52.3 meters = 171.5 feet, the
required answers.
Example...15
Divide 291 consecutively by the values 35.3, 42.3 and 58.4.
Solution. .,
Dividing by these values is the same as multiplying by their
reciprocals, Hence,
(1) Set the left index of scale C opposite 291 on
scale D.
(2) Move hairline so that it is over 35. 3 on scale CI.
(3) Under the hairline on scale D read 8,24 on
scale D. This will be the value of 291 divided
by 35.3.
(4) In the same manner opposite 42.3 and 58.4 on
scale Cl read 6.88 and 4.98 on scale D, Hence,
291 291
ja.3 = 8-88: and 754 = 4,98,
ee) ee eee mew we eee eer ee eee ee +e ee ewe
WHAT HAVE YOU LEARNED?
Do not use mvure than two settings of the slide for any single
computation in any of these problems.
1...Compile a table that will give the circumference of any
circle with a diameter between one and three inches, the di-
ameter being measured to the nearest quarter inch.
2...Change the following fractions to decimal forms:
1/64, 3/64, 9/64, 13/64, 17/64, 21/64, 23/64, 25/64,
3...There are 37 questions in an examination. If all are cor-
rect the grade is 100%. Assuming each question has equal
weight, make a table showing the proper grade for any number
of missed questions up to 10,
Processed by RF Cafe
Provided by Joe H.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
4... The voltage across a parallel circuit is 135volts. Find
the current values for circuit resistances of 220, 330, 440,
550, and 660 ohms.
Diameter
1, 1 in:
11/4 in.
1 1/2 in,
1 3/4 in.
2 in.
21/4 in.
21/2 in.
2 3/4 in.
3 in.
ANSWERS
Circumference
3.14 in.
3.93 in.
4,71 in,
.30 in.
. 28 in.
. 07 in.
. 85 in.
. 64 in,
. 42 in,
omanra nun
2.
1/64 = 0.0156.
3/64 = 0.0469
9/64 = 0.1406
13/64 = 0.203
17/64 = 0. 266
21/64 = 0, 328
23/64 = 0. 359
25/64 = 0.391
3. Questions missed Grade Questions missed Grade
1 97. 3% 6 83. 8
2 94.6% 7 81.1%
3 91.9% 8 78.4%
4 89. 2% 9 75.7%
5 86.5% 10 73.0%
4. Resistance Current
220 ohms 614 milliamperes
830 ohms 409 milliamperes
440 ohms 307 milliamperes
550 ohms 245 milliamperes
660 ohms 205 milliamperes
wy)
_
Provided by Joe H.
Processed by RF Cafe
SQUARES AND SQUARE ROOTS...TheA and B scalesare used
in conjunction with the C and D scales for working problem:
involving squares and square roots. One way to find the squars
of a number, say of 1.6 is to multiply 1.6 by 1.6 in the regula)
way using the Cand D scales. A quicker way is to move the
hairline until it is over 1.6 on scale D and to then read th
Square, 2.56, under the hairline on scale A. See Fig. 18.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Since the A scale cannot be read with as great an accuracy as
the C or D scales, this method is not used when the greatest
possible accuracy is needed. However, it is accurate enough
for most requirements.
To fing (1.6)*:
set hairline over 1.6 on seole o
a) (a _ 6
then reed 256 on scole A
Fig. 18. Finding thet (1.6)? = 2.56 (or finding thot \/2.56 — 1.6).
Finding the square root is the reverse of finding the square of
a value. In finding the square root the hairline is placed over
the value of which the root is wantedonscaleA, and then under
the hairline on scale D the value of the root is read. For
example to find the square root of 9, slide the hairline until it
is over 9 on scale A, and then under the hairline on scale D
read 3, the value of the square root of 9.
Inall pfoblems up this time the location of the decimal point
has been found by inspection after all slide rule manipulations
were completed, In extracting the square root the position of
the decimal point must always be considered before any work
is done with the rule.
Since the decimal point must be considered, the slide rule is
only directly suitable for finding square roots of numbers
which are within the range of the A or B scale, which you will
notice is from 1 to 100. However, it is very simple to adapt
numbers below 1 or above 100 so that their roots may be ob-
tained on the slide rule.
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
41
Provided by Joe H,
Processed by RF Cate
ee
Example...16
Find the square root of 85. 4.
Solution. ..
Since this value is between 1 and 100, it can be set up directly
on the A scale.
(1) Move the hairline until it is over 85.4 on scale
A. This value is near the right end of the A
scale.
(2) Under the hairline on scale D read 9.24, ans.
Explanation...
The hairline must be set to 85.4 on the A scale, and not on
8.54. 85.4 is between the marks 80 and 90 near the right of
the A scale. 8.54 would be between 8 and 9 near the center
of the A scale. As an exercise you should extract the root of
8.54 on your slide rule, showing that the answer is 2.92.
re Oe
When the number of which the root is wanted does not lie be-
tween 1 and 100, the number should first be divided off into
periods of two digits, starting at the decimal point, exactly
the same as when finding the square root arithmetically. Then
extract the root on the slide rule considering the decimal! point
as being after the first period which is not zero in value.
ee eee eee ree eee nee
Example... 17
Find the square root of 71, 400.
Solution... ’
(1) Divide into periods: 7'14'00. (Remember to
start at the decimal point for this operation.)
2 (2) Consider the decimal point as being at the end
ze of the first period; that is, after the 7.
ze (3) Find the square root of 7.14: Move hairline
33 sothatit is over 7.14 on scale A and then read
gE the root digits, 267, on scale D,
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
(4) To locate the decimal point use the rule that
for numbers greater than 1 there will be as
many digits to the left of the decimal point In
the root as there are periods to the left of the
decimal point. So the square root will have
three digits to the left of the decimal point.
Hence, 267 is the answer.
Example. ..18
Find V0. 0000015
Solution...
(1) Dividing into periods: 0.00'00'31'5
(2) Consider the decimal point as being at the end
of the first period not zero in value; that is,
after 31.
(3) Find the digits of the square root of 31.5 to be
561,
(4) To locate the decimal point use the rule that
for numbers less than 1 there will be one zero
immediately to the right of the decimal point
in the answer for each period of zeros im-
mediately to the right of the deciinal point in
the number. The number 0.00'00'31'5 has
two periods of zero value immediately to the
right of the decimal point.
Hence, 0.00561 is the root of this number.
Seereeeerreererereerereeeraretrereeetererrert ee eeeeeertreneeeensieee
WHAT HAVE YOU LEARNED?
1... (0.873)? = ? 6...¥5 = ?
2... (46.5)? = ? 7...V50 = ?
3...(6. 25)? = ? 8...v4 = ?
4...(0.001234)? = ? 9...V40 = ?
5...v4.235 = ? 10...¥33.34 = ?
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
43
Provided by Joe H.
Processed by RF Cafe
Provided by Joe H.
4 11.../0501T = ? 14... V0. 000383 = ?
12...¥754569 = ? 15...V¥0.06= ?
13...¥250,000,000 = ? 16...V846 = ?
ANSWERS
1. 0.762 5. 2.06 9. 6,32 13. 1,581 x 10
2. 2160 6, 2.24 10. 5.77 14, 1,957 x 10
3. 39.1 7. 7.07 11. 0.708 15. 0.245
4. 1.523x 10° 8. 2.00 12. 869 16, 29.1
y+) EVALUATING EXPRESSIONS CONTAINING SQUARES. .. Wor
ing problems on the slide rule where power or roots are i
volved often require no more work than if the power or roc
were not required. The following examples show how to hanc
problems involving powers and the next topic will cover prc
lems involving roots,
Example. ..19
Find the value of (19.4 = 4.21)?
Solution. ..
Multiply 19. 4 by 4.21 m me usual manner, but Insteadof rer
ing the answer under the hairline on scale D, read thedigits
the answer, 667, on scale A instead. Reading on scale
would give the product of 19.4 x 4.21, while reading on sc
A will give (19.4 x 4.21)’. That is because the square of :
value on scale D is given directly above on scale A. To.
cate the decimal point, note that 20 x 4 = 80, and 80’ = 6,4
5 Hence, the answer will be 6670.
&
= Example... 20
i Find the value of (2:4)
g 4.21
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Solution...
Divide 19.4 by 4. 21 in the usual manner, but instead of read-
ing the answer on scale D opposite the right index of scale C,
read the answer, 21.2, on scale A opposite the right index of
scale B. The principle here is the same as for Example 19.
We want the square of the result obtained by division; there-
fore, we read from scale A rather than from scale D.
The following example, as well as many other problems to
follow, makes use of the fact that the A and B scales can be
used for ordinary multiplying and dividing, just as can the C
and D scales. An A or B scale is nothing more than a C or
D scale reduced to half-size, so that it can be repeated twice
on the rule.
Example... 21
Find the value of 4,4 x (1, 34)’,
Solution. ..
(1) Set the left index of the C scale opposite 1.34
on scale D.
(2) Push the hairline so that it is over 44 or 4.4
on scale B.
(3) Under the hairline on scale A read the answer
digits, 790.
(4) Locaung the decimal point gives 7.90 as the
answer.
Explanation. .,
In step (1) by setting the left index of C opposite 1.34 on scale
D, the left index uf B is placed opposite 1.34? on scale A. We
now use the A and B scales to multiply together the two values
4.4 and 1347. Since the left index of B is already opposite-
134? it is only necessary to set the hairline to 44 on scale B
and read the answer on scale A.
In step (2) notice that there are two different positions on scale
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
Provided by Joe H,
Processed by RF Cafe
Provided by Joe H.
B where the hairline is over 44, one on the left half on the B
scale and the other on the right half of the scale. Either posi-
tion may be used. It is only in extracting the square root that
the proper half of the A or B scale must be chosen.
Example..
Find the a ig of ee
Solution...
(1) Set hairline over 82 on scale D.
(2) Move slide so that 314 on either half of scale
B is under the hairline,
(3) Read digits of answer, 214, opposite left index
of B on scale A,
Explanation...
Setting the hairline over 82 on scale D at the same time places
it over 82? on scale A. We now divide 82? by 3. 14 in the usual
manner, except that we use the A and B scale for the purpose,
and the value 82 has already been set on scale A. The answer
is 21, 4,
Example... 23 ave
Find the value of (3.14)
Solution...
(1) Set the hairline over either 82 on scale A.
(2) Move slide until 314 on scale C is under the
harrline.
(3) Read answer digits, 832, opposite an index o!
scale B on scale A.
(4) Locating decimal point the answer is 0.832.
Explanation. ..
This is another problem in dividing using the Aand B scales.
In step (2) we want (314)? under the hairline on the B scale.
By placing 314 on the C scale under the hairline, weat once
have 314? on scale B under the hairline.
Processed by RF Cafe
ee ee **
This manual was scanned by Mr. Joe Hernandez (KN2A) and then processed by RF Cafe (KB3UON).
Please visit www.RFCafe.com
WHAT HAVE YOU LEARNED?
1...(7.1 x 0.734)? = ? 2...(0.0239 x 560)? = ?
3... (41)? x (0.56)? = ? [This expression is equivalent to
(41 x 0. 56)*)
148.5\ 2 0.616 \ *
4., (+46:5) = Beas (Sa5 =?
6 (66.9) = ? [This expression is equivalent to ($2) -
*** (74.6) ) = 14.6
_ (0.97)? _
7... 9.26 x (4,48)? = ? 10...95 Sag 7?
8... (4,93)? x 0,86 = ? 11 sie =?
+. (4, ) . “"" (9, 18)
(5.33)? _ 8.41 _
ar