Electronics and Your Slide Rule, Part IV — Combined Operations with Electronic Applications (CIE Electronics Slide Rule No. 515 / Pickett N-515-T)
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CLEVELAND INSTITUTE OF etectRontcs / CLEVELAND, OHIO
Hiectronics and;
Your Slide Rule =
Part IV
An AUTO-PROGRAMMED. lesson
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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:
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Electronics and Your
Slide Rule
Combined Operations with
Electronic Applications
By DARRELL L. GEIGER
Senior Project Director
Cleveland Institute of Electronics
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Tn this lesson you will learn...
37. Solving the Right Triangle When the Hypotenuse c
and One Acute Angle, A or B, Are Known... Page 84
38, Solving the Right Triangle when One Leg, a or b,
and the Opposite Angle Are Know... Page 88
39. Solving the Right Triangle When One Leg, a orb,
and the Adjacent Angle Are Known... Page 90
40, Solving the Right Triangle When the Hypotenuse c
and One Other Side, a orb, Are Given... Page 93
41. Solving the Right Triangle When the Two Legs, a
and b, Are Given... Page 95
42. Finding Trigonometric Functions of Small Angles... Page 97
43, Solving Circuit Problems By Triangle Rules... Page 99
44, Using Scale L... Page 103
Summary... Page 104
Appendix...
Examination...
© Copyright 1966, 1965, 1962, 1958 Cleveland Institute of Electronics.
All Rights Reserved/Printed in the United States of America,
FOURTH EDITION /Fifth Revised Printing/October, 1966.
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Pp
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A chat with your instructor
The Ln scale is used in place of the LL scales foundon log-log type
slide rules. The scale provides a method for solving engineering
problems involving exponents, which are frequently encountered in
electronics. Some examples of the type of practical problems solv-
ed with the Ln scale are given on page 16 of this Appendix.
Specifically, the In scale is a seale of natural logarithms, in con-
trast tothe L scale which isa scaleofcommon logarithms. Common
logarithms, areexponents tobase 10. For example, inthe expression
107 8 = 346, 10 iscalled the base, andthe exponent 2. 539 is called
the logarithm of 346, so that the expression can be written equally
well: log,346 = 2.539.
Any number could be used as the base for a system of logarithms.
However, the only base besides 10 that is actually used for practi-
cal purposes is 2.71828, represented by the symbol e. When e is
usedas the base, the logarithmsare called natural logarithms. The
name comes from the fact that the value e frequently comes up in
problems reiating to natural phenorrena. For example, if you roll
a snow ball down a hill, it will gather additional snow rather slowly
at first, but faster and fasteras theball becomes larger. The weight
of the ball at any time can be roughly figured by the formula, M=
ke" , where e is 2,71828, t is the time that theball has been rolling
and n and k are constants to make the weight come out correctly in
pounds, The derivation of the value 2.71828 is beyond the scope of
this training manual, but it can be found in any calculus text,
Working with natural logarithms follows the same general princi-
ples as withcommon logarithms. However, the characteristic of a
natural logarithm is found in a different way than for common log-
arithms, andhas a rather different meaning. The methodand mean-
ing are explained in the pages that follow. ances con Sit
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Electronics and
Your Slide Rule
Part IV
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SOLUTIONS OF
THE RIGHT TRIANGLE
The general right triangle having hypotenuse c, legs a and b, and
angles A, B, and C = 90° is again picturedin Fig. 31. When any
one side and one acute angle are known, or when any two sides are
known, the remaining parts of the triangle can be found with your
slide rule. This process of finding all the sides and angles of the
triangle is called "solving" the triangle.
37 SOLVING THE RIGHT TRIANGLE WHEN THE HYPOTENUSE ec
AND ONE ACUTE ANGLE, A OR B, ARE KNOWN...The opera-
tions required in solving the right triangle when one side and one
angle are known are just the combined operations of multiplication
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A c+90*
Fig. 31. The general right triangle.
and division with trigonometric functions, opérations with which
you are already familiar. For example, when the hypotenuse c
and the angle A are known, to find the legs of the triangle, a and b,
we use the formulas b=c cos A anda=csin A. Thus in Fig, 31,
ifc = 28.0 and A= 27°, we have b = 28.0 cos 27° and a = 28.0 sin
27°. The combined multiplication operation, 28.0 cos 27°, is per-
formed as usual, by setting the right index of the slide over 28.0
on the D scale and the hairline over 27 on the red numbered 8
scale, then reading b = 28.0 cos 27° = 24.95 onthe D scale. The
slide is in the correct position to perform the operation 28,0 sin
27° and only the hairline has to be moved. Without moving the
slide, perform this operation by moving the hairline over 27 on
the black numbered S scale and read the value of a on the D scale
under the hairline; a = 28.0 sin 27°= 12.7. The angle B can be
found by subtracting 27° from 90°; B= 90° - 27° = 63°.
In some cases it is not possible to perform the second operation
without moving the slide. For instance, if in the preceeding ex-
ample A was 20° instead of 27°, the slide wauld not be in the cor-
rect position to perform the operation b = 28.0 sin 20°, since 20°
on the black portion of the S scale would be out of the range of the
D scale. In these cases the slide must be moved so that the other
index of the slide covers the value of the hypotenuse on the D
scale, Hence, to find b= 28.0 sin 20°, you must move the slide
so that the left index rests over 28.0 on the D scale.
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To solve the triangle when the value of the hypotenuse and one
acute angle A or B, is given, set the index of the slide over the
value of the hypotenuse on the D scale. To find the side opposite
the given angle (side a if angle A is given, side b if angle B is
given), set the hairline over the value of the angle on the black S
scale and read the value of the opposite side under the hairline of
the D scale. Then set the hairline over the value of the angle on
the redS scale. (This may or may not require the resetting of
the slide.) The value of the remaining side, which is adjacent to
the given angle, can then be read under the hairline on the D
scale. The decimal is placed by noting that the two legs of the
triangle are never greater than the hypotenuse and never less than
one tenth the hypotenuse for the angles within the range of the S
and T scales, The remaining angle of the triangle can be found by
subtracting the given angle from 90°,
Example...47
In Fig. 32 the right triangle is shown for c = 20, B= 25°. Solve
this triangle.
Solution...
1...Set the left index of the slide opposite 20 on the D
scale,
2...Here side b is the side opposite the known angle B
=25°, To find b set the hairline over 25° on the
black S scale and read 446 on the L scale, Since
the value of b must lie between 20 and 2.0, b=
8.46.
B-25°
. o
=
o
© 5
s A . c
3
; Fig. 32. The triangle to be used in Example 47.
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$...To find the value of the adjacent side a the slide
must be reset. Move the slide so that the right
index is opposite 20 on the D scale and move the
hairline over 25 on the red numbered S scale.
Read the value of a on the D scale; a = 18.13.
4... Find angle A by subtracting 25° from 90°; A = 90°
~ 25° = 65°,
ee
WHAT HAVE YOU LEARNED?
Solve the right triangle of Fig. 31 for the following cases:
1...¢ = 14.6, A= 18° 7...C€= 249, A= 28.6°
2...¢7 17.2, B= 32. 6° 8...¢= 68.1, A = 32.7°
3...¢ = 39.7, B= 53. 2° 9...c = 43.5, B = 46.5°
4...¢= 108, A= 64,1° 10...c = 50.3, B= 55°
5...C = 8.61, B= 24.3° 11...¢= 72.1, B= 19.7°
6...c = 1.75, A= 36. 4° 12...c = 432, A= 25. 5°
ANSWERS
1,..B = 90° — 18° = 72° §...B = 57.3*
a = 14.6 sin 18° = 4.51 a = 36.8
b = 14.6 cos 18° = 13. 89 b = 57.3
2...A=57.4° 9,..A = 43, 5°
a =14,5 a = 29,9
b = 9,27 b = 31.55
3...A = 36. 8° 6...B = 53. 6° 10...A = 35°
a = 23.8 a =1.038 a = 28.85
b = 31.8 b=1.41 b = 41,2
4...B = 25,9° 7...B=61.4° 11...A = 70. 3°
a =97.1 a=119 a = 67.9
b = 47.2 b = 218.6 b = 24.3
5...A = 65. 7° 12...B = 64, 5°
a =7.84 a = 186
b = 3.54 b = 389
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3 SOLVING THE RIGHT TRIANGLE WHEN ONE LEG, a OR b, AND
THE OPPOSITE ANGLE ARE KNOWN,..In Fig. 31, when the side
a and the angle A opposite this side are known, the hypotenuse c
can be found by the formula, c = a/sin A, Once c has been found,
the other leg of the triangle, b, can be obtained by the formula,
b=c cos A,
For the case where a= 25 and A = 35°, we have c = 25/sin 35° and
b=c cos 35°. First find c by setting the hairline over 25 on the
D scale and moving the slide so that the hairline also covers 35 on
the black numbered S scale. Read c = 43.6 on the D scale under
the right index of the slide. Note that the slide is in the correct
position to find b = 43.6 cos 35° and need not be moved in this
case. However, in some cases the slide must be reset to perform
this second operation. Set the huirline over 36 on the redS seale
and read b = 35.7 under the hairline on the Dscale. Find angle
B by subtraction; B = 90° - 35° = 65°,
In general, when one acute angle and the side opposite this angle
is known, set the hairline over the value of the side on the D
scale. Move the slide so that the hairline covers the value of the
angle on the black numbered S scale, Read the value of the hy-
potenuse on the D scale under the index of the slide. This value
will be no less than the value of the known side and no greater
than 10 times it. Then move the hairline so that it covers the
value of the known angle on the red numberedS scale. (This may
or may not require a resetting of the slide so that the other index
covers the value-af the hypotenuse on the D scale.) Read the value
of the remaining side, which is adjacent to the known angle, under
the hairline on the D scale. The unknown angle {s found by sub-
tracting the known angle from 90°.
SOR REE TEETER EE eee eee eee eset
3
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ze Example...48
HW In Fig. 33 the right triangle is shown for b = 55 and B = 32°.
Be Solve this triangle.
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Fig. 33. The triangle to be used for Example 48.
Solution...
1...Set the hairline over 55 on the D scale,
2,..Move the slide so that the hairline is over 32 on
the black § scale and read 1038 on the D scale un-
der the index of the slide. Since the value of c
must lie between 55 and 550, c = 108, 8.
3... Move the hairline so that it covers 32 on the red
S scale, this can be done without resetting the slide
in this case, Read the value of the remaining side,
a, below the hairline on the D scale. The value of
a must lie between 10,4 and 104, hence a = 88,0.
4... Find angle A by subtraction; A = 90° - 32° = 58°,
WHAT HAVE YOU LEARNED?
Solve the right triangle of Fig. 31 in the following cases:
l...a = 14,2, A= 24° 1...b= 42.3, B= 28°
2...b=138, B= 39,5° 8...b= 28.5, B=63.1°
3...a= 65.1, A= 44,2° 9...a=278, A= 75°
4..,a=1,39, A= 22. 4° 10...a= 72.5, A=16.1° 2
5...b= 34,5, B= 13° ll...b= 11.1, B= 35.2° #
6...b= 654, B= 44,1° 12...a= 709, A= 21.3° i
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90 ANSWERS
1...B = 66° 7...& = 62°
c = 34.9 c = 90.1
d = 31.9 a= 79.5
2...A=50.5° 8...A= 26.9°
c = 217 c = 31.95
a = 167.4 a = 14,45
3...B= 45, 8° 9...B = 16°
c = 93.5 c = 288
b = 67.0 2
4...B= 67.6° ate
c = 3.65 c = 261.5
b'=3.37 b = 251
5...A= 77° 11...A = 54.8°
c = 153.4 c = 19.26
= 149 a = 15.73
6...A = 45.9° 12...B = 68, 7°
c = 939 c = 1951
a = 674 b = 1818
39 SOLVING THE RIGHT TRIANGLE WHEN ONE LEG, a OR b, AND
THE ADJACENT ANGLE ARE KNOWN... With reference to Fig.
31, when side a and the angle B adjacent to this side are known,
the hypotenuse © can be found by the formula c — a/cos B. Qnoe o
has been found, the other leg of the triangle can be found by the
formula b=c sin B. Note that these formulas are similar to
those of the previous section where side a and angle A were given.
For this reason the operations are quite similar. In fact, you
will find the operations are the same except the roles of the red
and black numbered § scales are interchanged.
To see this similarity, consider the example given in the beginning
of the previous section, Here a= 25 and B= 55°. Find c by set-
ting the hairline over 25 on the D scale and this time move the
slide so that the hairline also covers 55 on the redS scale. The
value c = 43.6 is read under the right index of the slide on the D
scale, In the previous section this operation was performed by
setting the slide such that the hairline covered the value of A= 35°
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on the black § scale----you can see this gives the same slide rule
settings. The slide is now in the correct position to perform the
operation b = 43.6 sin 55° and need not be reset. Set the hairline
over 55 on the black numbered S scale and read the value of b on
the D scale under the hairline; b = 35.7. Here again the settings
are identical to the settings for the operation b = 43.6 cos 35°.
When one acute angle and the side adjacent to this angle are
known, set the hairline over the value of the known side on the D
scale, Move the slide so that the hairline covers the value of the
angle on the red S scale. Read the value of the hypotenuse on the
D scale under the index of the slide. Then move the hairline so
that it covers the value of the known angle on the black § scale,
(This may or may not require a resetting of the slide.) Read the
value of the remaining side under the hairline on the D scale, Find
the unknown angle by subtracting the known angle from 90°,
Example...49
In Fig. 34 the right triangle is shown for b = 30 and A = 60°.
Solve this triangle.
Solution...
1...Set the hairline over 30 on the D scale.
2... Move the slide so that the hairline covers 60 on
the red S scale and read 600 under the index of the
slide on the D scale, The value of c must lie be-
tween 30 and 300, hence c = 60.0.
3, .. Move the hairline so that it covers 60 on the black
§ scale. This can be done without reseting the
slide. Read the value of a onthe Dscale. This
value must lie between 6,00 and 60.0, hence a=
52.0.
4... Find angle B by subtraction; B = 90° - 60° = 30°,
WHAT HAVE YOU LEARNED?
Solve the right triangle of Fig, 31 in the following cases:
1,..a = 36.7, B= 52, 3° 3...b= 108, A= 44.1°
2...8 = 22.2, B= 19.5" 4...a= 39.1, B= 14. 6°
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60°
P +
be30 s
Fig. 34. The right triangle with one acute angle and an adjacent side given.
§...b= 1.52, A= 24.7° 9...b= 21,5, A= 62.3°
6...b= 50.2, A = 65,1° 10,..a = 2,87, B= 41.2°
7...0= 1.73, A= 26° 11...b= 109, A= 12.4°
8...a= 15.7, B= 39.1° 12...a = 67.5, B = 24,3°
ANSWERS
1.ccA = 37:7° 7...B= 65°
c = 60 c = 1.909
b = 47.5 a = ,807
2...A=70.5° 8...A=50.9°
c = 23.55 c = 20.23
b = 7.86 b = 12.75
3...B= 45.9° 9,..B= 27.7°
c = 150.4 c = 46.25
a = 104.7 a = 40.95
4...A= 75.4° 10... A = 48, 8°
c = 40.4 c = 3.815
b = 10.19 b =2,51
5...B = 65.3° 11...B = 77.6°
: a = 0.699 a = 23.98
td 6...B = 24.9° 12...A = 65.7°
z= c = 119.2 c = 74.2
i a = 108 b = 30.5
ze
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40 SOLVING THE RIGHT TRIANGLE WHEN THE HY POTENUSE c
AND ONE OTHER SIDE, a OR b, ARE GIVEN... When c and,
say, b are known, the triangle can be solved in the following way.
Since b/c = cos A, cos A is known, hence A can be found. Angle
B is obtained by subtracting A from 90° and a is obtained from the
formula a=c sin A. These operations will now be combined to
give the most efficient way of solving the right triangle when the
hypotenuse and one other side are known.
Consider a triangle in which c = 5 and b= 4. Set the right index
of the slide opposite 5 on the D scale and the hairline over 4 on the
D scale, Notice that the hairline now covers the value 4,5 = 0.80
on the C scale. Here you have actually divided 4 by 5 using the C
and D scales in anew way. Since the index of the slide rests over
5 and the hairline covers 4 on the D scale, the number under the
hairline on the C scale is the number which multiplies 5 to give 4,
or just 4/5, This type of division has the advantage of yielding
the answer on the C scale instead of the Dscale. The ratio 4/5 =
0.80, under the hairline is cos A= b/c, hence A can be read di-
rectly as 36,9° under the hairline on the red numbered portion of
the S scale. Since the black numbers of the S scale are 90° minus
the red numbers, the angle B = 90° - A can be read under the
hairline on the black numbered portion of the S scale; B = 53.1%.
Do not move the slide since it is in the correct position to per-
form the operation a= c sin A=5 sin 36.9°, Move the hairline
over 36.9° on the black numbered S scale and read a = 3.00 on the
D scale under the hairline. In placing the decimal! point we have
again made use of the fact that the legs of the right triangle are
never less than one tenth the hypotenuse and never greater than it
for angles within the range of the S scale.
To solve the right triangle when the hypotenuse and one other side
are given, set the index of the slide over the value of the hypote-
nuse on the D scale and then set the hairline over the value of the
known side on the D scale. Read the value of the angle between
the hypotenuse and this known side on the redS scale. Read the
value of the other acute angle on the black S scale. Now set the
hairline over the value of the first angle you found on the black
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S scale. (This may require a resetting of the slide.) Read the
value of the unknown side under the hairline on the D scale.
Fig. 35. The cose where the hypotenuse ond one side are known.
Example...50
In Fig. 35 the right triangle is shown for c = 25 anda=5. Solve
this triangle.
Solution...
1...Set the index opposite 25 on the D scale, and
the hairline over 5 on the D scale,
2,.,.Here the angle B is the angle between the hy-
potenuseand the known side a. Read the value
of B, 78.48°, on the red S scale,
3...Read angle A= 11,52° on the black S scale.
4,..Set the hairline over 78.48° on the blackS
seale. Read the value of the unknown side,
b = 24.5, on the D scale.
ee ee eee eee ee eee eee eee ee ee
WHAT HAVE YOU LEARNED?
Solve the right triangle of Fig. 31 in the following cases:
1...¢ = 35.0, b= 15.0 5...0 = 16.3, a= 9.12
2...¢ = 128, a= 54.3 6...¢ = 44.5, b = 20.2 aa
3,..¢= 11.1, a = 7.04 7...0 = 112, b= 39.0 H
4...0= 115, b= 24.6 8...¢= 32.1, a= 28.6 H
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4
9...C = 42.6, b=11.4 ll...c = 1.04, b=,.213
10...c = 75.5, a = 38.9 12...c = 8.64, a= 4,72
ANSWERS
1.,.A = 64. 6° 7...A = 69. 6°
B= 25,4° B= 20.4°
a = 31.6 a =105
2...A = 25.1° 8...A=63.1°
B= 64,9° B = 26.9°
b = 115.9 b =14.5
3...A = 39,4° 9...A = 74,47°
B = 50.6° B = 15, 53°
b = 8.58 a =4l
4...A=77.65° 10.,.A = 31.0°
B = 12. 35° B= 59. 0°
a = 112.2 b = 64.7
5...A= 34,0° 11... A= 78.19°
B= 56, 0° B= 11,81°
b=13.5 a=1.019
6...A = 63, 0° 12...A=33.1°
B = 27. 0° B = 56.9°
a = 39.65 b = 7.24
SOLVING THE RIGHT TRIANGLE WHEN THE TWO LEGS, a AND
b, ARE GIVEN ,..If a and b are known, tan A = a/b is known and
A can be found. Then to find the hypotenuse c we make use of the
formula c = a/sin A, and to find B we subtract A from 90°, For
example, consider the triangle of Fig. 31 with a = 3 and b= 6.
Set the right index of the slide over 6 on the D scale and then the
hairline over 3 onthe D scale. The ratio, a’b= tan A= 0.50,
then appears on the C scale. Since this ratio is less than one,
A must be less than 45° and can be read directly under the hair-
line on the black numbered T scale as A = 26.56°, Since the red
numbers of the T scale are 90° minus the black one, angle B can
be read on the red numbered portion as 63.44°. The hairline is in
the correct position to perform the operation c = 3/sin 26. 56° and
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only the slide has to be moved. Move the slide so that the hair-
line covers 26.56° on the black S scale. The value of the hypote-
nuse, 6.71, then appears opposite the index of the slide on the D
scale,
In general, to solve the right triangle when the two legs are given
set the index of the slide over the value of the larger of the two
sides on the D scale. Next, set the hairline over the value of the
smaller of the two sides, also on the D scale. Read the value of
the angle between the larger side and the hypotenuse on the black
T scale. Read the value of the other acute angle on the red T
scale, Without moving the hairline, move the slide so that the
hairline covers the value of the first angle on the black S scale or
the second angle on the redS scale, Read the value of the hypote-
nuse under the index of the slide on the D scale,
Example... 51
In Fig. 31 assume that b= 8 and a= 39. Solve the triangle.
Solution...
1...Set the left index of the slide over 39 on the D
scale.
2...Set the hairline over 8 on the D scale and read the
value of B on the black T scale; B= 11.59". Read
the value of A on the red T scale; A= 78, 41°,
3... Without moving the hairline, move the slide so that
the hairline covers 11,59 on the black scaue.
Read the value of c under the left index of the slide;
c= 39.8.
ee Ce Cr
WHAT HAVE YOU LEARNED?
= Solve the right triangle of Fig. 31 in the following cases:
id
: l...a = 27.2, b= 18.9 3...a= 7,23, b= 25.3
; 2,..a = 108, b= 57.5 4...a=1.73, b= 1.05
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5...8 = 31.4, b= 14.2 9...a = 33,6, b= 98.9
6...a =128, b= 39.1 10.,.a= 16.2, b= 9.12
7...a=.111, b=.789 ll...a= 32.1, b= 49.7
8...a= 75.6, b= 106 12...a= 141, b= 200
ANSWERS
1.,.A = 55, 2° 7...A= 8,01"
B= 34, 8* B= 61,99°
ec = 33,1 c =.797
2...A = 62.0° 8...A# 35. 5°
B = 28.0° B= 54.5°
c = 122.5 c = 130.2
3...A = 15.92" 9..,A = 18,76°
B = 74. 08° B = 71. 24°
c = 26.3 c = 104.5
4,..A = 58.75° 10...A = 60. 64°
B= 31, 25° B = 29, 36°
ec = 2,025 c = 18.6
5...A = 65, 68° 11...A = 32, 9°
B = 24. 32° B = 57.1°
ec = 34.5 c = 69.2
6,..A = 73, 01° 12,..A = 35.19°
B = 16, 99° B = 54.81°
c = 133.8 c = 245
4 FINDING TRIGONOMETRIC FUNCTIONS OF SMALL ANGLES...
The sine and tangent of small angles (below the range of scales S
and T) are approximately equal and can be found approximately by
multiplying the size of the angles in degrees by the factor 0, 01745,
This is the method used for finding the functions of angles too
small to be found on the S or T scales. It is not necessary to
remember the constant value 0.01745 because it is permanently
marked on the C, D, and Cl scales of your slide rule and identi-
fied by a small degree mark (°).
To find the sine of 0.7°, place the right index of scale C opposite
7 on scale D, and then opposite the degree mark between 1.7 and
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1.8 on scale C, read 122 on scale D. Thus the sine of 0.7° is
0.0122. This is also the tangent of 0. 7°,
The sine and tangent values for angles below the range of the S and
T scales have a 0 between the decimal point and the first signifi-
cant figure, If the angle size was less than 0. 573° there would be
more than one zero between the decimal point and the first figure,
but it is unlikely that you will be working with angles that small.
As another example with small angles, suppose we wish to find
the angle whose tangent is 0.0838. Since there is a zero between
the decimal point and the first figure of the value, we know that
the angle required is below the range of the S or T scale. To find
the tangent of a small angle we multiply the angle size by 0.01745.
Conversely, to find the angle corresponding to a given tangent
value, we would divide that value by 0.01745, Thus, we set the
degree mark on scale C opposite 838 on scale D and then read
4.8°, the answer, opposite the C index on scale D.
WHAT HAVE YOU LEARNED?
1... Find the sine and tangent of each of the following angles:
(a) 1.72° (e) 3.82°
(b) 3.59° (f) 2.21°
(c) 4. 86° (g) 1.68°
(d) 1.53° (h) 5.31°
2... Find the angle A in each of the following problems:
(a) tan A= .0437 (e) sin A= .0296
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(b) sin A=.0252 (f) tan A= .0341
(c) tan A=.0621 (g) sin A= .0557
(d) tan A= .0843 (h) tan A= .0201
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ANSWERS
1... (a) .0300 (e) . 0666
(b) . 0626 (f) . 03856
(c) .0848 (g) . 0293
(d) .0267 (h) 0926
2... (a) 2.505" (e) 1. 695°
(b) 1.445° (f) 1.955°
(c) 3.56° (g) 3..19°
(d) 4.83” (h) 1. 152°
4 SOLVING CIRCUIT PROBLEMS BY TRIANGLE RULES... The
trigonometric scales of the slide rule are cxtré mély useful in
working a-c circuit problems. As an example, consider the series
RLC circuit. From a-c circuit theory it is known that a phasor
impedance triangle can be drawn for such a circuit. This triangle
and its circuit are shown in Fig. 36. The resistance R and the
total reactance X = X,_ - Xo make up the two legs of the triangle.
The impedance of the circuit, Z, is the hypotenuse, and the angle
@ between R and Z represents the angle by which the current |
leads the voltage E. The solutions to the right triangle discussed
in the previous sections can be directly applied to this Impedance
triangle,
4
ge Example.,,.52
Be In the circuit of Fig. 36 let R= 509, xX, = 80 2, Xe =6592, and
i I= .5lamp. Find the impedance Z, the phase angle 6 and the
ce voltage E.
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100 Solution...
Here the two legs of the impedance triangle are known; R
50 2 and X = 80 - 65= 15 Q, the hypotenuse Z and one acute an-
gle, 6, must be found. With reference to the previous sections
HOKE Ke
=
Fig. 36. The series circuit and its impedance triangle.
the triangle is solved as follows:
(a) Set the right index of the slide over the value R =
50 on the D scale and the hairline over X = l5also
on the D scale.
(b) The angle © is the angle between R, (which is the
larger of the two legs) and Z. Hence this angle is
read on the black T scale as 0 = 16.7°,
(c) Without moving the hairline, set the slide so that
the hairline covers 16.7 on the black S$ scale and
read the value of Z on the D scale; Z = 52.32.
(d) Do not move the slide since it is in the correct
position to find the voltage: E=1Z. Set the hair-
line over 1=.51l0n the C scale and read E = 26.6
volts (52.3 x 0.51). on the D scale.
In the circuit of Fig. 36, the power which the circuit dissipates is
given by the formula P = El cos 6, The power factor of the cir-
cuit is cos 6 = frand represents the ratio of the actual power sup-
plied to the circuit and the product of the impressed voltage and
current, The essential a-c circuit formulas involving circuit
power are permanently marked on the back of your slide rule un-
ready familiar.
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der the heading "Ohm's Law Formulas, AC Circuits". By means
of these formulas, problems involving circuit power can also be
handled with your slide rule by methods with which you are al-
Example...63
The series circuit of Fig. 36 is connected to a supply of E = 110
volts. If R = 1,100 2 and X = 1,500 2, what is the power factor
of this circuit? How much power will the resistance R dissipate ?
Solution, ..
The power factor is cos @ and can be found from X and R, The
impedance Z can also be found from X and R, and then the formu-
2
la P= E’ oes Siwhich can be found on the back of the rule) can be
used to find the power dissipated.
(a) Set the right index of the slide over 1, 500 on the D
scale and the hairline over 1,100 on the D scale.
Read © = 53,7° on the red T scale,
(b) Set the slide so that the hairline covers 53.7 on the
redS scale, Read the value of Z on the D scale as
Z=1,8602.
(c) Notice that since the hairline now lies over the
value of 6 on the red S scale, the value of cos 0
must lie under the hairline on the C scale; cos @ =
.592. The power factor is then 59.2%.
(d) The circuit power is the power which R dissipates
and can now be calculated:
p = EX.cos @ ~ (110)? (. 592)
Z (1, 860)
In the preceeding examples the trigonometric scales were used
with a minimum number of settings. In problems such as this,
where a number of combined operations are involved, there is no
general rule which will give the minimum number of settings.
¥ou should, however, be alert at «i! times to the posttton of the
slide and hairline and look for the most direct means of obtaining
the desired result.
= 3.85 watts.
WHAT HAVE YOU LEARNED?
1...In the series circuit of Fig. 36, R = 25 2, and X = 80 Q,
What is the impedance Z? What is the power factor of this
circuit?
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101
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102 2...1f the voltage supply E of a circuit similar to Problem 1 is
220 volts, the current I is 1 ampere and the power factor is
0.72, what is the value of Z. R, and X? What is the power
delivered to the circuit?
3... The circuit of Fig. 36 is to have an impedance of 1, 200 2
when R is 800 2. What must the value of X be? What is the
power factor?
4...The circuit of Fig. 36 is to have an impedance of 625 2 when
X= 1579. What must the value of R be? If the circuit pow-
er is 200 watts, what is the value of E and I?
5...In Fig, 36, if the impedance is 836 2 and the power factor is
0.350, what is the value of R, X, and 0?
6...How much power is delivered to the circuit of Fig. 36 if1 =
.52 amperes, R = 78.6 2, and X = 28.92? What is the im-
pedance and phase angle of this circuit?
7...A 110 volt electric heater has a resistance of 8.3 9 and there-
fore draws 13,25 amperes when in operation. It is desired to
operate this heater from a 220 volt line. What reactance must
be put in series with this heater so that it only draws 13.25
amperes from the 220 volt line?
ANSWERS
1...2= 83.82, Pe= .2982
2...Z = 220 2, P= 158.4 watts, R= 198.42, X = 152.89
3...P,=.667, X= 895 2
4...R = 605 2, E = 359.5 volts, I = .575 amperes
5...R = 292.52, 6 = 69,5", X= 782.52
6...2= 83.79, O= 20,2*, P= 21,2
7...X%= 14.42
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44 USING SCALE L...Scale L is used in conjunction with scale D to 103
find the mantissa of logarithms to the base 10, and conversely to
find antilogarithms, To find the mantissa of the logarithm of a
number, place the hairline over that number on scale D and then
read the mantissa of the logarithm under the hairline on scale L.
Thus, using this method we find the mantissa of the logarithm of
64.5 to be 0.809. The characteristic will be 1 in accordance with
the usual rules for finding the characteristic. Hence, log 64.5 =
1,809,
Finding antilogarithms is, of course, just the reverse of finding
logarithms. To find the number whose logarithm is 4.4152 we set
the hairline over the mantissa, 0.415, on scale L and then read
the digits of the logarithm, 260, on scale D, Locating the deci-
mal point in accordance with the rules of logarithms gives us an-
tilog 4.415 = 26, 000.
Example, ..564
The output of an amplifier is 12 watts. The input to the amplifier
is 0.032 watts. Find the gain of the amplifier in decibels.
Solution, .. P 19
(1) db = 10 log 5. = 10 log KY
(2) Set the left index of scale C opposite 12 on scale D.
(3) Move the indicator so that the hairline is over 32
on scale Cl,
(4) Under scale L read 0.574. This is the mantissa
of the logarithm.
(5) Since 12/0, 032 is roughly 400, the characteristic
of the logarithm will be 2.
(6) 10 x 2.574= 25.74 db, the answer.
Explanation...
In step (1) the proper formula to use is obtained from the back of
the Electronics slide rule. In steps (2) and (3) 12 is divided by
0, 032 by multiplying 12 by the reciprocal of 0. 032 (see Topic 18),
& using the CI scale. The reason for multiplying in this manner is
He: so that the hairline is over the quotient on the D scale, and con-
ze sequently over the logarithm of the quotient on the L scale. In
° : this manner the required mantissa can be read directly on the L
gé scale without the necessity of first reading the quotient on scale D.
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104 WHAT HAVE YOU LEARNED?
Find the logarithm of:
1...427.6 3... 25, 000 §.5.05218
2...4.276 4...0.00915 CEs TF
Find the antilog of:
Riigw Be 43T 8...0.815 9,..4.377 10...1.444
11...The output from an amplifier is 25 watts, and the input to
the amplifier is 0.025 watts. What is the gain of the amplifier in
db?
12.., Find the cube root of 365.2. (Hint: Divide the logarithm of
365. 2 by 3 and take the antilogarithm of this result.)
ANSWERS
1. 2.631 5. 1.338 9. 0.000238
2. 0.631 6. 0.845 10. 0.278
3. 4,398 7. 273.6 ll. 30 db
4. 3.961 8. 6,531 12, 7,148
SUMMARY
The following table summarizes the application of the opera~
tional methods explained in the manual to typical formulas in
electronics. The text pages referred to in the table give de-
tailed explanations of the procedures and also examples or
practice problems.
The method given in the table for solving each formula ig one
of the shortest methods, In general there are many methods
of working any problem on a slide rule, but some methods
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are much more convenient than others. Every time a quantity 105
is set up on a slide rule, read from the rule, or transferred to
another scale, an error can possibly be made. Hence, the use
of two operations where one is sufficient is not the best practice.
A formula not listed can be handled efficiently by locating in the
table one equivalent in form and then solving by the method for
the equivalent formula,
In using the following methods it is assumed that the slide will
be reset (see page 29) if this operation should become neces-
Sary in order to continue with the method.
SUMMARY OF SETTINGS FOR COMMON FORMULAS
Explained in
Forniula How to Solve
G= i (1) Opposite R on scale C, read Topic 15
R G on scale Cl.
R,R, (1) Opposite R, on scale D, set the Topic 16
Rr™ R, +R; sum, R,+ R, (found mentally),
on scale C.
(2) Opposite R, on scale C, read R
on scale D.
P=PFR (1) Set index of slide opposite I on Topic.22
scale D.
(2) Opposite the resistance R on
scale B, read P on scale A.
1. ye (1) Set P on scale A opposite Ron Topic 23
R scale B.
2 (2) Opposite the slide index, read I
xz?
gk on scale D. If answer is un~
z= reasonable, repeat step (1) ex-
33 cept use the other half of scale
Be B for setting R.
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106
Formula How to Solve Explained in
r-2 (1) Set slide index opposite 1 on Example 23
r scale D.
(2) Opposite P on scale A, read R
on scale B.
p= E? (1) Opposite E on scale D, set Ron Example 22
"R scale B.
(2) Read P opposite the slide index
on scale A.
E = VPR (1) Set index of scale B opposite P Example 24
on scale A,
(2) Opposite R on scale B, read E
on scale D, If result is unrea-
sonable, repeat step (1) except
use other half of scale A for
setting P.
{- P (1) Opposite P on scale D, set E on Topic 18
ae scale C.
(2) Opposite I on scale Cl, read pf
on scale D.
te 1 (1) Determine approximate value of Topic 29
2aVLC { from Decimal Point Locator
scales.
(2) Opposite L on scale H, set C on
scale B.
(3) Opposite the slide index, read f
on scale D. If aot in agreement
with approximate value, repeat
step (2) except use the other half
of scale B for setting C.
1 (1) Set index of scale C opposite fon Topic 29
(2xf)2C scale D.
(2) then opposite Con scale B, read
L on scale H.
(3) Find decimal point from Decimal!
Point Locator scales.
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Formula
1
C" Gem L
XL = 20fL
ee
2e
XL
Lf St
1
XC * Sic
|
22CXc
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How to Solve
(1)
(2)
(3)
(1)
(2)
(3)
(1)
(2)
(3)
(1)
(2)
(1)
(2)
(3)
(1)
(2)
(3)
Set index of scale C opposite f on
scale D.