Materi Matematika Kimia Bab 3 The Slide Rule
CHAPTER 3
THE SLIDE
RULE
Wc saw in Chaptcr 2 thc mcthod by which a tablc of logarirhms can
bc uscd to multiply, dividc.,.raisc !9.a Pqwcr.-or c.(tracl a roo(- Thcsc
opcrari6iis can bc car.ricd out mori ripidly.. with somc loss of 4ccuracy, by
mcans of a slidc nilc. A simplg.tO-iosh rulc, sclling.lor about 52. is adc{rrt. io1 ,noit of thc calculaiions thit wc. ciiry out in gcncral chcrniitry.
qll.culaiions. and cxamina(ions
Thc tiric iavcd on'hgmcwoqli, laboraroiy
.
is wcll woiih ihc coit ofihc rulc. ''
'.lt is i*pon"n. to undcrstand rhc principlc upon which thc slidc rulc
is bascd. Whcri wc cairy out the proccsscs o[ mukiplication or division on
a:lidc rulc, what wc arc rcally.doing is adding or subtracting logarirhms.
To undcrstand ho* rht co'mcs aboui, lct us cxaminc rhc sc.rlcs on the slidc
rulc labclcd C and D, which arc idcntical cxbept that thc C scalc is localcd
on ahc movable slidc and rhc D scalc is on thc s(ationary pan o[ thc rulc.
On thcsc scalcs, thc positiom of thc numbcrs arc dctermincd by thc values.
of thcir logarirhms. For cxamplc, thc nrrmbcr 2 is locatcd about J0 pcr
ccnt o[ thc way from rhe lcft inler ( I ) to the right inla.r ( t0). This reflects
rhcfact that the logarirhm of 2 (0.1010) lics 10.10 percent of thc.way bctwccn tht logarithm of I (0.0000) and that of l0 (1.0000). Similarly. thc
numbcr 3 (log 3 - 0.4771) falls at a point a littlc lcss than 50 pcr ccnt o[
thc distancc bctwecn I and 10.
Considcr now thc slidc rulc sctting shown in Figurc J. I . Hcrc. r hc lcft
indcxof thc Cscalc.is placcd dircctlyabovc thc numbcr 2 on thc D scalc.
If wc movc across the scalcs, wc notc that J on thc C scalc is dircctly abovc
6 on thc D scalc. Clearly. by rhis simplc proccss wc have multiplied 2 x 3
to obtain an answcr of 6. What wc havc rcally doneis to add (wo distanccs.
Onc of rhcsc (3.01 inchcs) is proportional to log 2 (0.3010); thi orhcr
(1.?Tinchcs) is proponional to log 3 (0.1771). The total distancc. 7.78
inchcs, is proportional to thc logof 6,0.7781. In othcr words, wc multi27
CHAPIER 3
flGUfE t.f.
Iltc poritioar ol aoobco
o
r$c C ond D
bgorithar-
rdo orc dcrcoiocd by ttcir
girtorel 2 - !.0tin.
bg2;oJOt
O'ttt Dirtoocgl - 3 - 1'77:m'
W'
lca6 - O.7t8 Didoncl - 6 - 778ia.
plicd two numbcrs by. in cffccr, adding thcir logarithms. tn an cntircly
analogous manner, ir is possible to carry out divisions by subtracring dislances proportional to the logarithms o[thc numbcrs involvcd.
3.I
TOCATING NUMBERS
, To ill-u.s1rarc how npr.nbc5s c4n.hq locared on th.c virious scalcs of rhcslidc.rulc. lct us ciiminc tlic-D scalc. . The largc digirs (t.L3,{..:)
rcprescnr rlls inrcgers bctwccri I a;d 10. Norc.thar thc spaccs bcrwccn.
thcsc inrcgcrs bccomi smitlcr.as wc moi,I liom tcfr 19 righr. This gridual
comprcssion o[ihc scalc is a conscquencc of its logarithmic niture...T-trc
distancc bclwccnthe numbcrs I and 2 ii about 3 inchcs (log 2 ; log i'0.3010), rr,hilc rhat'betwccn 9 and l0 is less than half an inch (tog t0 log 9 - 0.0458).
. Wc shall now considcr how ro 6nd thrcc-digir numbcrl in yarious portions of the D scalc (Examplc,3.l). Thc positions of rhese numbcrs arc
indicated by dortcd vcrrical lincs iri Figurc 3.2.
3-42
6.53
i
'6:?
FIGURE
3.2.
Locotion o[ thc
mb.6 ig,3-l2,
cod 653 on thc O (or C] rcolc.
.
IH€ SI,'O€ RUTE
Eromplc 3.1 .Locatc thc following numbers on the D scalc:
a- l-14
b.
3.42
c.
6.51
Soluiion
a.
'
.
.
Focusing our attcntion on thc Po(ion of rhc D scalc be'
twccn the targc digits I and 2, wc recognizc tcn major divisions, cach labclcd with a small digit (t. r. l. r) corrcsponding to thc numbcrs l'1. l'2' l'l' l'4' ' " Bctrvc6n i1r''
numbcrs l.l and 1.4, rhcrc arc tcn small divisions' Ttrc
numbcr l-34 must then lall four small divisions to thc rigtrt
of l-3.
b. Hcrc. wc notc that thc tcn major divisions bctrvccn thc
.
'
largc digirs 3 and 4 arc not numbcrcd- To find thc numbcr
3.4. wc movc four major divisions to thc righ( of thc digit 3.
To locatc 3.42, wc notc that thcrc arc 6vc small divisions
bctwccn J.4 and 3.5. corrcsponding to
l'42' 3'44' 3'46,
numbir J.42 must thcn bc locatcd one
3.48. and 1.50. Thi
smalt divisioh to ihc right o[ 3-'t.
6,5. 6vc major divisions to the
Bctwccn 6.5 and 6.6. thcrc arc two small divisions, which must corrcspond to 6.55 and 6'60' Thc nurn'
ber 6.51 is locarcd a littlc morc rhan half way (viz'; ]3 of
thc way) bc(rvecn 6.50 and 6.55.
.' c. lVc first locati thi numbcr
'
right of
6.
EXERC'5ES
l-
Scr thc tollowing numbcrs on thc D scalc:
d. 3.69
a. 2.02
b. t.68
c. ).42
2-
l-
c. 5.t5
t
5.74
Locatc the numbcrs in pan (l ) on thc A scalc, bctwccn thc lcft and
ccn(cr indcx. Notc that thc A scalc consis(s of two scgmcnt!, onc
running lrom I ro l0 and thc othcr from l0 to 100.
Sct 8 on (hc C scalc dircctly'abovc 4 on thc D scalc' Rcad thc
quotient,2, on rhe C scalc dircctly abovc thc lelr indcx of thc D
scalc. tVhy docs rhis Proccdurc cnablc you to carry ou( thc Process
of division? Explain in tcrms o[ the logarithmic naturc o[ thc
scalcs.
{-
It is reasonablc to suppose lhal thc cnor which a bcginncr mahcl
in sctting a numbcr on the slidc ruic is equivalcnt lo thc smallcst
t
CHAPIER 3
30
division on thc D scalc, which could bc thc distancc bctwccn t'99
(Note that
and 2.00or, alternatively. rhar bctwccn 9.95 and t0.00.
pcrccntagc
of
What
sizc.)
thcse divisioni arc o[ about thc samc
crror docs this rcPrcscnt?
3.2 MULTIPTICATION AND DIVISION
wc
To carry our opcrations involving only multiplication and division,
in
arc
illustratcd
simplc'muttipticatioJrs
Two
us< rhc C and D scalcs.
Examplc 3.2'
Exomplc 3.2 MulriPlY:
a. 2.12 x 4.3>
b.
Solurion
a.
.
b.
5.05
x
4.33 (scc Fig. 1.3)
lVc 6rst scr thc t -ar rhc lcfi of thc C scalc (rhc lcft indcx)
dirccrly abovc 2.12 on rhc D scalc. Now. movc thc lransparcnt plastic curtor sothat iu hairtinc lines up cxactly with
l.f ; on rhc C scalc- Rcad thc ariswcr' aPProximatcly 9'22,
ar rhc hairlinc on thc D scalc.
lf wc atrcmPr ro reP€ar thc proccdurc of pan (a), wc 6nd
rhar rhc numbtr 4Jl on thc C scalc talls beyond the cnd
of thc D scalc- To gcr around this difficrrlty, wc sct thc
ight indct of thc 'C scalc abovc 5.05 on thc D scalc' Now'
movc thc cu6or so that irs hairline coincidcs with 4'33 on
thc C scalc. Dircctly bcncarh' on rhc D scalc, wc hnd that
rhe hairtinc [alls abour half way across thc spacc bctwccn
2. t8 anC 210. Wc might csrimarc its position to bc 2'19'
t(6swing thar rhc pro'dua must bc appro:timatcty 20 (5 x
4 = 20), *c:akc 2l-9 ro bc our answer'
tn gcncral, ro multiply ooc nuribcr by a.,orh.r, -c:
-Set
the.pp-pti".. iodcr of thc C scalc (lcfi or right) direcdy
l.
abovc the 6rsr auobcr oa tbc D scelc2. Movc thc curror so thlt its heirlinc fetls on thc sccond numbcr, on lhc C scalc. 3. Read the answer dircctly bcaearh thc hairlinc on the D scalc'
Thc manipulation involvtd in E:ample 32 pan (b) illustrarcs rhe fact
rhat rhc dccimal poinr in rhc errt*cr inusr bt stt indcpcndcntly of the slidc
L.-
fHE SLIOE RUI.E
UNE (433 on C rcale)
D scale
HAIR LINE (219 on D scate)
flGURE3.3.
Atulliplicotrioor 5.05
x a.3l -
?1.9.
rulc opcration. Locating thc dccimal point, or thc appropriare powcr o
t0, isfacilitatcd by cxprcssing thc numbcrs in cxp6ncntial notation. Sup
posc. for cxamplc. wc wish to .multiply 0.02t2 by 4350. Scrring thc lcf
indcxof thc C scalc abovc 2.12 on the D scalc and moving thc cursor unri
its hairlinc coincidcs with a.35 on thc C scalc, wc rcad thc ansrvcr bcncarl
thc hairlinc on th; D scalc as 9.22. To obtain thc powcr of l0 in tht
aru;wcr! wc writc thc numbcrs in cxponcntial notation:
0.02!2
x
4350
x l0-'
-
2.12
-
(2.12..x a.l5)
x,l-15
x
Wc dcducc that thc p.oa,r., must bc 9-22
0.212
x
43-5
x l0'
(10-2
x lot) = 9 -
x llt , or 92.2. Simita rly:
= 2.12 x tO-r x 4.li x l0r =
ZiZ.x O.OO+I; -2-12
x
lOt x 4.15
l0'
9.22
x l0:r = 9.22 x l0-'
-O.9Zz
Notc that in cach case the slidc rulc opcrarion: arc idcntical. Wc usc rhc
rulcsof cxpoocnts to dctcrminc thc powcro[ 10, or thc dccimal poinr, in
thc answcr.
Sincc thc proccss-of division is thc invcrsc of multiplicarion, ir can bc
accomplishcd on thc slidc rulc by pcrforming in rcvcrse ordcr the manipu-
lations involvcd in mukiplication. Spccifically, ro dividc onc numbcr by
anothcr, wc:
t.
Movc thc cursor so thar its hairlinc falls directly ovcr lhc
numcrator on thc D scalc.
2 Movc thc slidc until the denominaaor, on thc C scalc,falls bcncath thc hairlinc of thc cursor.
3. Rcad tfie rnswer on tlrc D scalcl dircctly [clow llrc iddct o(
thc G rcaic.
r'
CHAPTER 3
Exomplc
l
3.3
Dividc 6'05
x t0'by
9.50
x l0-'.
Solution Usingthc cursor, we linc up 6-05 on thc D scalc with
scalc. Dirccrly bclow thc right indcx of thc C scalc.
To dccidc upon rhc
9.50 on rhc C
wc rcad thc numbcr "6.37" on thc D scalcpropcr powcr of I 0, wc notc that:
I
fr
6.05
9-50
x to' - Lo5 x to' =
9.50
x t0-'
0.6
l0', or, in standard
tt follows tha( our answcr musr. bc 0.637
..porr.n,irt nolalionr 6-37
x lo,
x l0''
{
FIGURE
3.4.
Divirion: 6'O51950
-
0'637'
Many timcs, in working problcms' wc arc rcquircd to carry out a
sl.ccssirc multiplicarions and divisions. In doing this, considcrablc rimc can bc savcd by altcrnating the proccsscs, 6rst dividing, thcn
scrics of
muhiplying. thcn dividing. and so on.
?, M DI" - .
EIO
x t0'l{!4i_"_]!-'_}Q.!jl
(l.24xl0r)(5.10x10')
(6.02
_,
us first dividc 6.02 by 1.24, then multiply by 1.88'
dividc by 5.10, and, 6nally, multiply by 3.1a.
a. To dividc 6.02 by 1.24, usc thc cursor to linc up 6.02 on thc
. D scalc with 1.24 on thc C scalc. The quoticnt, which
nccd not bc rccordcd, can bc found on thc D scalc bclow
the left index of the C scale. Note that the rulc is now in
Solulion
lrt
position lo carry out a multiplication-
rHE .st'OE
33
RUI.E
b. To multiply by l-88' movc thc cursor so thar its hairlinc
fallsatl.Ssonthccscalc.Thcproductwillappcaronthc
D scalc, bcncath thc hairlioc'
dividc by 5'10. lcavc thc cursor whcrc
it is and movc thc
with
thc hairlinc'
up
lincs
Cscalc
thc
5.l0on
until
slidc
thc hairlinc
that
cursor-so
d. To multiply by 3'la' movc thc
D scalc'
ihc
on
"5'62"
Rcad
scalc'
falls at l- 14 onthc C
c. To
wc approximatc
To obtain rhc propcr cxPoncnt in thc answcr'
as follows:
(6 x loDX2 x lorxl) 16 x l.q:=7 x lor
. (txlo')(ixlo') - 5xlo"
thc answer musl bc 5'62 x l0''
Clcarly,
pcrirl1 carrying orrt all thc multiplications 6rst and thcn
manipulacxtra
scvccal
thar
will
6nd
You
forming thc diriions!
thc rccomtions a-rc involrci- This should convince you thar
with
divisions
mukiplications
mcndcd proccdurc o[ altcrnating
mcrit.
has considcrablc
EXERC'SES
Pcrform thc indicatcd oPcrations on thc slidc rulc'
l.
(6.41
x l0r) x
2. 8.54 x
(l-39
x l0-')
2.90
^
7.12 x l0'
6-10 x l0-'
4.
1.2815.90
-
(6.19
x l0') x
(l-20)
, (1.86) x (l-e5) -- ---L'n
(2-87) x (a'le)
(5-82 x lo') x (4'19 x lo') x (l'14)
"'' --{-'8&;.TT ;G'r2 x ro')
3.3
SQUARES AND SQUARE ROOTS
wc makc usc o[ tlrc
To squarc a nurnbcr or cxtract its squarc rool'
madc up of twq rangesr
is
scatc
A
thc
that
will
note
Yo,,
A and D sc"l.s.
convcnicn( to think of thc index at thc cencich running lrom I ro l. [t is
tcr of thc A icalc as rcprcscnting thc numbcr l0 and thar at tht far riqhr
CHAPTER 3
34
rhis point-of vicw' thc rangc at thc
rr rcrrrcscntinsthc numbcr100' From
runs from l0 ro 100-
,oio,.t it.,t.r at rhc righr vcry simptc proccdurc' All
tit tfiat ruli it "
""
abovc thc numbcr on thc D
*. do is to sct thc tttitfi""'oi titA scalc'
""tt"t
:Thus' wc find (Figurc 3'5) that
rcetc end rcad its,qt'"" ot' ;t
1, -,,3r - 9.(2.14)r - f-i8.
[J'J;ili
Squaring a numbcr
14
FIGURE
To6nd (2.1a
!.5. Sq*'.. ol n'-b""'
x l0-t)r.
(2.t4
22
- l:
(2'l{2 - r58;
!2-9.
(Chaptcr
wc use rhc rulcsofcxponents
x lo-')r = (2'14)2 x (10-t)2 = 4'58 x
l)'
l0-r2
is a slightly morS cometilatla
Extracting thc squarc root of a numbcr
is.to rcvcrsc thc proccss tndo
ro
p'in"iptt'all wc havc
cursor
-r;;i;;;."i,,
is' wc sd thc hairline of thc
volvcd in squaring tttt nu*itt That
root on thc D
ebovc thc number o" ttt]i'tt"f"
rcalc.. ln this way, wc dcducc rhat t
rnd rcad thc squarc
r, = ,
(12.0)'/2 = 3.46
(50.0)'/r =
:
7-07
to
whosc squarc root wc arc askcd
A di6culty ariscs if thc numbcr
numbcr
thc
In lhis casc' we rcwritc
frnd docs nor lalt bctwccnl "na 100'
bctwccn I and
in cxponcntiel form so titt
100 and thc cxPoncnt
tftt *tmcicnt is a rrumbcr
poivcr (t'g" -2'0' 2' 4 ' ' : )'
J;;-t;:"
rHE SUO€
35
RUTE
Exomplc 3.5 Find:
a. (12,600)r'r
. '
b. (1.86 x lor)ul
Sotulion
--'-"-'irz,ooo)tn
(1.26 x l0')r/r' (l'26)'/r x l0r
scalc' we
S.,if"itt.t"i.tii.of rhc cursor at t'26 onl0r'thcorAI lJ'
x
is
l'13
.""ai.ii i"" ,t c D scalc' Our answcf
en cvcn numb. Hcrc, wc musa nrst convcrt thc cxponcnt to
dividc lot
and
l0
by
bcr. To do this, *l-rnultiply r'At
. by l0:
(1.S6
-
x lot)irt -
(18.;
x l0'1r^ - (t8'6)t^ x
l0r
thc hairlinc of the
To 6nd thc squarc root of 18'6' wc movc
cursortols.6onthcAscalc(notcthatlS.6falls.bcyondl0'onour
,t*igt, half of thc A scalc)' ^Wc rcad 4'll on thc D scalc;
;;;must thcnbc4.3l x l0r'
EXERC'SES
'
Find
l.
Z. (6.12
l.
4.
3.4
5.
(1.69)r
(2.02
6. (3.86
7. (2;98
x l0-')r
(1.60)r/r
x lor)ur
x lot ) r/2
x l0-r)r/r
(87.0)r/2
CUBES AND CUBE ROOTS
wc usc thc K scalc in much
To cubc a numbcr or cxtract its cube root.
squarcs or squarc roots'
obrain
to
uscd
is
thc same way that thc A scalc
which mav bc
scgments.
rhrce
inro
dividcd
is
scilc
K
Noticc rhar ihc
and lO0 ro
(ccntcr)'
100
to
,t""Sii"fas runningfrom I to l0 (lc[t), l0
of the
hairline
lhc
wc
sct
numbe:'
a
iOoO"t;St,t. To 6nJ thc cubc of
ensw€r undcr lhc
the
rcad
lnd
scalc
D
thc
on
numbcr
that
arrroa o'"aa
hairlinc on rhc K scalc. Thus. *'c find that:
- 8: (l.o)t = 27; (8.00)' = 512
(l.O I lo-t)'- (l-o)t x (10-r)'= 27 x l0-'
2r
.
CubC f0015 of numbcrs Sslwccn
I and 1000 arc rcadily lound
by
thc numbcr on thc K scale rnd
sctting the hairlinc o[the cursor ebovc
36
rc.ding
CHAPIEI 3
the enswcr bcncath thc
hairlinc on
thi D scalc'
Wc suggcsr
(30b)'/! = 6'69
To cxtracr the cubc root o[ a numbcr which docs not fall bctwccn I
and l0oo, ir is ncccssary lo rcwritc it in cxponcntial form so that thc cocfficicnt is a numbcr bctwccn t and t000 and thc cxPoncnt is cvcnly
divisiblc by thrcc.
Exornplc 3.6 Find:
a. (0.00190)ur
b. (4.68 x l0-')r/r
Solution
f-.. r,.'10;oor90)rl,'- (l-90 x ,g"l1r/r.- {f,9{i)trr
21 1g-'t
Dclos'l.90on thc K scalc, wc rcad l-24 on the D scalc' Our
x l0-r, or0-124.
b. Wc must 6rst rcwrite lhe numbcr so that thc cxponcnt,
upon division by 3, will givc'a wholc numbcr' To obiain
answcr musl bc 1.24
tlic dcsircd form, wc multiply thc cocfficicnt by t0 and
dividc thc cxponcnt bY 10.
(4.68
x l0-')r/'! -
(46.8
x l0-')r/! - (46.$)t/'1
1g-t
Sctting 16.8 on the K scale (bctwccn thc second 4 and 5), *'c read
its c,rbc root on thc D scalc as 3-60. Hcncc, our answer must bc
3.60
x
l0-2.
EXERC'SES
.
r.
(6.08)r
2. (1.40
x l0r)t
x l0r)r
3. (5.95
4. (t2-9)v'
3.5
5. (268)ttt
6. (t.96)r/'
7. (1.3? x lOf)'D
8.'(5.92 x I0-')ur
TOGARITHMS
Thc only lincar scalc on thc slidc rulc, whcrc thc numbers arc spaced
o[ thc
at eguat in,.rotr, is rhc L scale Tiris scalc givcs thc logdrithms
If,€
37
SUDE RUI.T
numbcc bctwccn I and t0. To find rhc logarithm o[ a numbcr in this
Erngc, wc sct thc hairlinc ofrhc cursor abovc thc numbcr on thi D scale
and rtad its logarithm on thc L scatc. In this way. wc 6nd that thc logarithm of 2 is, :o thrcc digits. 0.301, that of J is 0.477. and so on.
1l
Dll'rl
2
'
.flGURE
r.6.
togorirhar of nunbco: bE 2-
- 0.301;
lo9
3 ' 0'177.
lf thc numbcr whosc logarithm wc rcquire is lcss than I or grcatcr
than 10, wc procccd as dcscribcd in Chaprcr 2- That is. wc writc lhc numbcrinstandardcxponcotial notation, usc lhe L scalc to 6nd rhc mantissa.
and takc thc cxponcnt to bc thc charactcristic-
Exomplc
3-7
Find thc logarirhm of 3.70
x l0-'.
Solution tog(3.70 x l0-r) = log3.70 + log l0-t - tog3.70 - 5
Sctting 3.70 on the D scalc, wc rced its logarithm. on ihc L scalc.
tobc0-568. Hcncc,log(3.70
x lO-t) -
0.568
- i =
-4.432
Wc scc from this cxamplc that thc L scalc is cquivalcnr to a thrccplacc tablc of logarithms' All of thc opcrations rvhich rvcrc dcscribcd in
Chaptcr 2 in conncction with a tablc of logarirhms can be carricd oul on
jr alidc rulc. using thc L and D scalcs. Wc can' for cxamplc. usc it to 6nd
antilogarithms
Eromple
'
a.
3.8
Find the numbcrs whosc logarithms arc:
0.380
b. 3-r00
c. -2.900
Solution
a. Sct thc hairlinc of thc cursor ovcr 0.180 on thc L scale;
rcad thc numbcr. 2.{0, on thc D scalc.
b. Sct thc hairlinc oicr thc mantissa. 0.100. on thc L scalc.
Ihc numbcr whosc log is 0. 100 is rcad from rhc D sciilc to
,
CHAPIER 3.
The number whosc log is 3 is
antilogJ.l00
c.
-
l0'
l-26 x
in Chaptcr
First, rewritc the number (recall rhe discussion
2) as:
I
EXERC'SES
Find, using rhe slidc rulc:
t
.
log 6.14
log (1.29
2.
3. log (5.84
4. ln 2.82
x l0t)
x l0-')
5. antilog 0.462
6. anrilog 1.250
?.
antilog (-5.400)
3.6 OTHER usFS OF THE SLIDE RULE
chcm'
' The slidc rulc opcrations pcrformc{ most frcqucntly in gencral on
thc
Dcpcnding
3'5'
through
3'2
in
Scctions
ir,ry;;r. ;..n d.r.rib.d
'
numbcrofscalcsavailabtc(andthcPricc!),itispossiblcrouscthcslidc
slide-rule
. ,".i.,y oGth"'optrati6ns' Evcn thc simptcsr
functlr.*r.. (iabcllcd S "ni T; whieh allow us to.obtain oftrigonomctricwill
rulc
slidc
rhe
;;;l;;;., .*.",1- a disc"ssion-oi this applicarion
rulc to carry or,
bc [ound in ChaPrer 8'
6nd thc rcciprocal o[ a
From timc to timc. wc rnaY havc occasion to
as l/r)' This can' o[
numbcr (thc rcciprocal of ''nurnbt" r'' is dcfincd
and traking usc of thc
division
oidinary
courscr bc donc by trcating it as an
is rcguircd' i1 i5- 5s63s'hat
C and D scalcs- Whcrc a""'it' of rcciprocals
Noticc that rhc Cl
scalcs'
morc convcnicn, ,o rn,rt *! of thc C and Cl
Cl scalc stan at
on
thc
numbcrs
thc
scalqamounts to.an invcrsc C scalc;
rhc rcciprocal
find
To
right'
far
the
I
on
ro
l0 on the far lclt and dccrcasc
numbcr on the C
thal
ovcr
cursor
oflhc
hairlinc
wc.sct
thc
of a numbcr,
on thc Cl scalc' In this way'
scalc and rcad its rcciprocat at rhc hairlinc
you can verifl' that:
t/2 = o.'>@; l/6 = 0'16?; l/3't6 -
0'316
to carry
with the D scalc
Thc Cl scale can also bt uscd in conjunction
In both cascst thc ma'
out thc proccslcs of multiplication ind ii'itiott'
IH€ SUOE
39
NUIE
nipulation is somcwhat morc rapid than is thc casc whcn thc C
and D
scalcs arc uscd (Scction 3.2)To multiply 9 by 12, using
lhc Cl and D scalcs' usc thc cursor to linc
up 9 on thc Cl scalc with 12 on thc D scatc. Rcad thc product' 108, on thc
D scatc, dircctly bclow thc indcx of thc Cl (or C) scalc' Notc that rrhcn
multiplications arc carricd out in lhis way, it is ncvcr ncccssary to shift
indiccs to bring the answcr onto thc scalc.
To divide 7.0 by any dcsircd numbcr (c.g.. 5-0, l-5. Z-0)' sct ihc i ndcx
of thc Cl (or C) scale to 7 on thc D scalc- Movc rhc cu6or to thc nu rnbcr
by which you wkh to dividc. on thc Cl scatc- Rcad 166 3ljwcr at thc hair'
linc on rhc D scalc. In this way. wc 6nd rhat:
7.0/2.0 7.0/r.4
5.0
3.5
7.0/1.3 = 2.0
Notc thc advantage of this mcthod of pcrforming divisions
askcd to dividc a lircn numbcr. such as 7.0, by a scrics of
whcn you arc
numbcrs (l'4,
2.0.3.5..-).
Thc following problcms arc dcsigncd (o rcvicw matcrial on cxPoncnts
(Chaptcr l) and logarithms (Chapter 2), in addition to improving yotrr
spced and accuracy in using thc slidc rulc.
Makc usc of thc stidc rulc ro obtain numcrical answcrs to cach protr'
lcm.
3.1
(5.49 x lOrX6.O2
4.91 x t0-:
6.84 x l0'
(3.08 x l0')r
log(2.41 x l0t;
x l0-r)
3-3
3.{
3.5 11.28 x t0-')t/'
3.6 antilog 0.649
3.7 (5.00 x lQ:1ul
3.8 (t.29 x lo:u)'
3.e l/(t.gS x l0-')
- -^ (t.28 x ro')(3.14)
3-10
------r6;98 x l0'
3.1I 2.7.1 (3.08 x lo-r)r/r
THE SLIDE
RULE
Wc saw in Chaptcr 2 thc mcthod by which a tablc of logarirhms can
bc uscd to multiply, dividc.,.raisc !9.a Pqwcr.-or c.(tracl a roo(- Thcsc
opcrari6iis can bc car.ricd out mori ripidly.. with somc loss of 4ccuracy, by
mcans of a slidc nilc. A simplg.tO-iosh rulc, sclling.lor about 52. is adc{rrt. io1 ,noit of thc calculaiions thit wc. ciiry out in gcncral chcrniitry.
qll.culaiions. and cxamina(ions
Thc tiric iavcd on'hgmcwoqli, laboraroiy
.
is wcll woiih ihc coit ofihc rulc. ''
'.lt is i*pon"n. to undcrstand rhc principlc upon which thc slidc rulc
is bascd. Whcri wc cairy out the proccsscs o[ mukiplication or division on
a:lidc rulc, what wc arc rcally.doing is adding or subtracting logarirhms.
To undcrstand ho* rht co'mcs aboui, lct us cxaminc rhc sc.rlcs on the slidc
rulc labclcd C and D, which arc idcntical cxbept that thc C scalc is localcd
on ahc movable slidc and rhc D scalc is on thc s(ationary pan o[ thc rulc.
On thcsc scalcs, thc positiom of thc numbcrs arc dctermincd by thc values.
of thcir logarirhms. For cxamplc, thc nrrmbcr 2 is locatcd about J0 pcr
ccnt o[ thc way from rhe lcft inler ( I ) to the right inla.r ( t0). This reflects
rhcfact that the logarirhm of 2 (0.1010) lics 10.10 percent of thc.way bctwccn tht logarithm of I (0.0000) and that of l0 (1.0000). Similarly. thc
numbcr 3 (log 3 - 0.4771) falls at a point a littlc lcss than 50 pcr ccnt o[
thc distancc bctwecn I and 10.
Considcr now thc slidc rulc sctting shown in Figurc J. I . Hcrc. r hc lcft
indcxof thc Cscalc.is placcd dircctlyabovc thc numbcr 2 on thc D scalc.
If wc movc across the scalcs, wc notc that J on thc C scalc is dircctly abovc
6 on thc D scalc. Clearly. by rhis simplc proccss wc have multiplied 2 x 3
to obtain an answcr of 6. What wc havc rcally doneis to add (wo distanccs.
Onc of rhcsc (3.01 inchcs) is proportional to log 2 (0.3010); thi orhcr
(1.?Tinchcs) is proponional to log 3 (0.1771). The total distancc. 7.78
inchcs, is proportional to thc logof 6,0.7781. In othcr words, wc multi27
CHAPIER 3
flGUfE t.f.
Iltc poritioar ol aoobco
o
r$c C ond D
bgorithar-
rdo orc dcrcoiocd by ttcir
girtorel 2 - !.0tin.
bg2;oJOt
O'ttt Dirtoocgl - 3 - 1'77:m'
W'
lca6 - O.7t8 Didoncl - 6 - 778ia.
plicd two numbcrs by. in cffccr, adding thcir logarithms. tn an cntircly
analogous manner, ir is possible to carry out divisions by subtracring dislances proportional to the logarithms o[thc numbcrs involvcd.
3.I
TOCATING NUMBERS
, To ill-u.s1rarc how npr.nbc5s c4n.hq locared on th.c virious scalcs of rhcslidc.rulc. lct us ciiminc tlic-D scalc. . The largc digirs (t.L3,{..:)
rcprescnr rlls inrcgers bctwccri I a;d 10. Norc.thar thc spaccs bcrwccn.
thcsc inrcgcrs bccomi smitlcr.as wc moi,I liom tcfr 19 righr. This gridual
comprcssion o[ihc scalc is a conscquencc of its logarithmic niture...T-trc
distancc bclwccnthe numbcrs I and 2 ii about 3 inchcs (log 2 ; log i'0.3010), rr,hilc rhat'betwccn 9 and l0 is less than half an inch (tog t0 log 9 - 0.0458).
. Wc shall now considcr how ro 6nd thrcc-digir numbcrl in yarious portions of the D scalc (Examplc,3.l). Thc positions of rhese numbcrs arc
indicated by dortcd vcrrical lincs iri Figurc 3.2.
3-42
6.53
i
'6:?
FIGURE
3.2.
Locotion o[ thc
mb.6 ig,3-l2,
cod 653 on thc O (or C] rcolc.
.
IH€ SI,'O€ RUTE
Eromplc 3.1 .Locatc thc following numbers on the D scalc:
a- l-14
b.
3.42
c.
6.51
Soluiion
a.
'
.
.
Focusing our attcntion on thc Po(ion of rhc D scalc be'
twccn the targc digits I and 2, wc recognizc tcn major divisions, cach labclcd with a small digit (t. r. l. r) corrcsponding to thc numbcrs l'1. l'2' l'l' l'4' ' " Bctrvc6n i1r''
numbcrs l.l and 1.4, rhcrc arc tcn small divisions' Ttrc
numbcr l-34 must then lall four small divisions to thc rigtrt
of l-3.
b. Hcrc. wc notc that thc tcn major divisions bctrvccn thc
.
'
largc digirs 3 and 4 arc not numbcrcd- To find thc numbcr
3.4. wc movc four major divisions to thc righ( of thc digit 3.
To locatc 3.42, wc notc that thcrc arc 6vc small divisions
bctwccn J.4 and 3.5. corrcsponding to
l'42' 3'44' 3'46,
numbir J.42 must thcn bc locatcd one
3.48. and 1.50. Thi
smalt divisioh to ihc right o[ 3-'t.
6,5. 6vc major divisions to the
Bctwccn 6.5 and 6.6. thcrc arc two small divisions, which must corrcspond to 6.55 and 6'60' Thc nurn'
ber 6.51 is locarcd a littlc morc rhan half way (viz'; ]3 of
thc way) bc(rvecn 6.50 and 6.55.
.' c. lVc first locati thi numbcr
'
right of
6.
EXERC'5ES
l-
Scr thc tollowing numbcrs on thc D scalc:
d. 3.69
a. 2.02
b. t.68
c. ).42
2-
l-
c. 5.t5
t
5.74
Locatc the numbcrs in pan (l ) on thc A scalc, bctwccn thc lcft and
ccn(cr indcx. Notc that thc A scalc consis(s of two scgmcnt!, onc
running lrom I ro l0 and thc othcr from l0 to 100.
Sct 8 on (hc C scalc dircctly'abovc 4 on thc D scalc' Rcad thc
quotient,2, on rhe C scalc dircctly abovc thc lelr indcx of thc D
scalc. tVhy docs rhis Proccdurc cnablc you to carry ou( thc Process
of division? Explain in tcrms o[ the logarithmic naturc o[ thc
scalcs.
{-
It is reasonablc to suppose lhal thc cnor which a bcginncr mahcl
in sctting a numbcr on the slidc ruic is equivalcnt lo thc smallcst
t
CHAPIER 3
30
division on thc D scalc, which could bc thc distancc bctwccn t'99
(Note that
and 2.00or, alternatively. rhar bctwccn 9.95 and t0.00.
pcrccntagc
of
What
sizc.)
thcse divisioni arc o[ about thc samc
crror docs this rcPrcscnt?
3.2 MULTIPTICATION AND DIVISION
wc
To carry our opcrations involving only multiplication and division,
in
arc
illustratcd
simplc'muttipticatioJrs
Two
us< rhc C and D scalcs.
Examplc 3.2'
Exomplc 3.2 MulriPlY:
a. 2.12 x 4.3>
b.
Solurion
a.
.
b.
5.05
x
4.33 (scc Fig. 1.3)
lVc 6rst scr thc t -ar rhc lcfi of thc C scalc (rhc lcft indcx)
dirccrly abovc 2.12 on rhc D scalc. Now. movc thc lransparcnt plastic curtor sothat iu hairtinc lines up cxactly with
l.f ; on rhc C scalc- Rcad thc ariswcr' aPProximatcly 9'22,
ar rhc hairlinc on thc D scalc.
lf wc atrcmPr ro reP€ar thc proccdurc of pan (a), wc 6nd
rhar rhc numbtr 4Jl on thc C scalc talls beyond the cnd
of thc D scalc- To gcr around this difficrrlty, wc sct thc
ight indct of thc 'C scalc abovc 5.05 on thc D scalc' Now'
movc thc cu6or so that irs hairline coincidcs with 4'33 on
thc C scalc. Dircctly bcncarh' on rhc D scalc, wc hnd that
rhe hairtinc [alls abour half way across thc spacc bctwccn
2. t8 anC 210. Wc might csrimarc its position to bc 2'19'
t(6swing thar rhc pro'dua must bc appro:timatcty 20 (5 x
4 = 20), *c:akc 2l-9 ro bc our answer'
tn gcncral, ro multiply ooc nuribcr by a.,orh.r, -c:
-Set
the.pp-pti".. iodcr of thc C scalc (lcfi or right) direcdy
l.
abovc the 6rsr auobcr oa tbc D scelc2. Movc thc curror so thlt its heirlinc fetls on thc sccond numbcr, on lhc C scalc. 3. Read the answer dircctly bcaearh thc hairlinc on the D scalc'
Thc manipulation involvtd in E:ample 32 pan (b) illustrarcs rhe fact
rhat rhc dccimal poinr in rhc errt*cr inusr bt stt indcpcndcntly of the slidc
L.-
fHE SLIOE RUI.E
UNE (433 on C rcale)
D scale
HAIR LINE (219 on D scate)
flGURE3.3.
Atulliplicotrioor 5.05
x a.3l -
?1.9.
rulc opcration. Locating thc dccimal point, or thc appropriare powcr o
t0, isfacilitatcd by cxprcssing thc numbcrs in cxp6ncntial notation. Sup
posc. for cxamplc. wc wish to .multiply 0.02t2 by 4350. Scrring thc lcf
indcxof thc C scalc abovc 2.12 on the D scalc and moving thc cursor unri
its hairlinc coincidcs with a.35 on thc C scalc, wc rcad thc ansrvcr bcncarl
thc hairlinc on th; D scalc as 9.22. To obtain thc powcr of l0 in tht
aru;wcr! wc writc thc numbcrs in cxponcntial notation:
0.02!2
x
4350
x l0-'
-
2.12
-
(2.12..x a.l5)
x,l-15
x
Wc dcducc that thc p.oa,r., must bc 9-22
0.212
x
43-5
x l0'
(10-2
x lot) = 9 -
x llt , or 92.2. Simita rly:
= 2.12 x tO-r x 4.li x l0r =
ZiZ.x O.OO+I; -2-12
x
lOt x 4.15
l0'
9.22
x l0:r = 9.22 x l0-'
-O.9Zz
Notc that in cach case the slidc rulc opcrarion: arc idcntical. Wc usc rhc
rulcsof cxpoocnts to dctcrminc thc powcro[ 10, or thc dccimal poinr, in
thc answcr.
Sincc thc proccss-of division is thc invcrsc of multiplicarion, ir can bc
accomplishcd on thc slidc rulc by pcrforming in rcvcrse ordcr the manipu-
lations involvcd in mukiplication. Spccifically, ro dividc onc numbcr by
anothcr, wc:
t.
Movc thc cursor so thar its hairlinc falls directly ovcr lhc
numcrator on thc D scalc.
2 Movc thc slidc until the denominaaor, on thc C scalc,falls bcncath thc hairlinc of thc cursor.
3. Rcad tfie rnswer on tlrc D scalcl dircctly [clow llrc iddct o(
thc G rcaic.
r'
CHAPTER 3
Exomplc
l
3.3
Dividc 6'05
x t0'by
9.50
x l0-'.
Solution Usingthc cursor, we linc up 6-05 on thc D scalc with
scalc. Dirccrly bclow thc right indcx of thc C scalc.
To dccidc upon rhc
9.50 on rhc C
wc rcad thc numbcr "6.37" on thc D scalcpropcr powcr of I 0, wc notc that:
I
fr
6.05
9-50
x to' - Lo5 x to' =
9.50
x t0-'
0.6
l0', or, in standard
tt follows tha( our answcr musr. bc 0.637
..porr.n,irt nolalionr 6-37
x lo,
x l0''
{
FIGURE
3.4.
Divirion: 6'O51950
-
0'637'
Many timcs, in working problcms' wc arc rcquircd to carry out a
sl.ccssirc multiplicarions and divisions. In doing this, considcrablc rimc can bc savcd by altcrnating the proccsscs, 6rst dividing, thcn
scrics of
muhiplying. thcn dividing. and so on.
?, M DI" - .
EIO
x t0'l{!4i_"_]!-'_}Q.!jl
(l.24xl0r)(5.10x10')
(6.02
_,
us first dividc 6.02 by 1.24, then multiply by 1.88'
dividc by 5.10, and, 6nally, multiply by 3.1a.
a. To dividc 6.02 by 1.24, usc thc cursor to linc up 6.02 on thc
. D scalc with 1.24 on thc C scalc. The quoticnt, which
nccd not bc rccordcd, can bc found on thc D scalc bclow
the left index of the C scale. Note that the rulc is now in
Solulion
lrt
position lo carry out a multiplication-
rHE .st'OE
33
RUI.E
b. To multiply by l-88' movc thc cursor so thar its hairlinc
fallsatl.Ssonthccscalc.Thcproductwillappcaronthc
D scalc, bcncath thc hairlioc'
dividc by 5'10. lcavc thc cursor whcrc
it is and movc thc
with
thc hairlinc'
up
lincs
Cscalc
thc
5.l0on
until
slidc
thc hairlinc
that
cursor-so
d. To multiply by 3'la' movc thc
D scalc'
ihc
on
"5'62"
Rcad
scalc'
falls at l- 14 onthc C
c. To
wc approximatc
To obtain rhc propcr cxPoncnt in thc answcr'
as follows:
(6 x loDX2 x lorxl) 16 x l.q:=7 x lor
. (txlo')(ixlo') - 5xlo"
thc answer musl bc 5'62 x l0''
Clcarly,
pcrirl1 carrying orrt all thc multiplications 6rst and thcn
manipulacxtra
scvccal
thar
will
6nd
You
forming thc diriions!
thc rccomtions a-rc involrci- This should convince you thar
with
divisions
mukiplications
mcndcd proccdurc o[ altcrnating
mcrit.
has considcrablc
EXERC'SES
Pcrform thc indicatcd oPcrations on thc slidc rulc'
l.
(6.41
x l0r) x
2. 8.54 x
(l-39
x l0-')
2.90
^
7.12 x l0'
6-10 x l0-'
4.
1.2815.90
-
(6.19
x l0') x
(l-20)
, (1.86) x (l-e5) -- ---L'n
(2-87) x (a'le)
(5-82 x lo') x (4'19 x lo') x (l'14)
"'' --{-'8&;.TT ;G'r2 x ro')
3.3
SQUARES AND SQUARE ROOTS
wc makc usc o[ tlrc
To squarc a nurnbcr or cxtract its squarc rool'
madc up of twq rangesr
is
scatc
A
thc
that
will
note
Yo,,
A and D sc"l.s.
convcnicn( to think of thc index at thc cencich running lrom I ro l. [t is
tcr of thc A icalc as rcprcscnting thc numbcr l0 and thar at tht far riqhr
CHAPTER 3
34
rhis point-of vicw' thc rangc at thc
rr rcrrrcscntinsthc numbcr100' From
runs from l0 ro 100-
,oio,.t it.,t.r at rhc righr vcry simptc proccdurc' All
tit tfiat ruli it "
""
abovc thc numbcr on thc D
*. do is to sct thc tttitfi""'oi titA scalc'
""tt"t
:Thus' wc find (Figurc 3'5) that
rcetc end rcad its,qt'"" ot' ;t
1, -,,3r - 9.(2.14)r - f-i8.
[J'J;ili
Squaring a numbcr
14
FIGURE
To6nd (2.1a
!.5. Sq*'.. ol n'-b""'
x l0-t)r.
(2.t4
22
- l:
(2'l{2 - r58;
!2-9.
(Chaptcr
wc use rhc rulcsofcxponents
x lo-')r = (2'14)2 x (10-t)2 = 4'58 x
l)'
l0-r2
is a slightly morS cometilatla
Extracting thc squarc root of a numbcr
is.to rcvcrsc thc proccss tndo
ro
p'in"iptt'all wc havc
cursor
-r;;i;;;."i,,
is' wc sd thc hairline of thc
volvcd in squaring tttt nu*itt That
root on thc D
ebovc thc number o" ttt]i'tt"f"
rcalc.. ln this way, wc dcducc rhat t
rnd rcad thc squarc
r, = ,
(12.0)'/2 = 3.46
(50.0)'/r =
:
7-07
to
whosc squarc root wc arc askcd
A di6culty ariscs if thc numbcr
numbcr
thc
In lhis casc' we rcwritc
frnd docs nor lalt bctwccnl "na 100'
bctwccn I and
in cxponcntiel form so titt
100 and thc cxPoncnt
tftt *tmcicnt is a rrumbcr
poivcr (t'g" -2'0' 2' 4 ' ' : )'
J;;-t;:"
rHE SUO€
35
RUTE
Exomplc 3.5 Find:
a. (12,600)r'r
. '
b. (1.86 x lor)ul
Sotulion
--'-"-'irz,ooo)tn
(1.26 x l0')r/r' (l'26)'/r x l0r
scalc' we
S.,if"itt.t"i.tii.of rhc cursor at t'26 onl0r'thcorAI lJ'
x
is
l'13
.""ai.ii i"" ,t c D scalc' Our answcf
en cvcn numb. Hcrc, wc musa nrst convcrt thc cxponcnt to
dividc lot
and
l0
by
bcr. To do this, *l-rnultiply r'At
. by l0:
(1.S6
-
x lot)irt -
(18.;
x l0'1r^ - (t8'6)t^ x
l0r
thc hairlinc of the
To 6nd thc squarc root of 18'6' wc movc
cursortols.6onthcAscalc(notcthatlS.6falls.bcyondl0'onour
,t*igt, half of thc A scalc)' ^Wc rcad 4'll on thc D scalc;
;;;must thcnbc4.3l x l0r'
EXERC'SES
'
Find
l.
Z. (6.12
l.
4.
3.4
5.
(1.69)r
(2.02
6. (3.86
7. (2;98
x l0-')r
(1.60)r/r
x lor)ur
x lot ) r/2
x l0-r)r/r
(87.0)r/2
CUBES AND CUBE ROOTS
wc usc thc K scalc in much
To cubc a numbcr or cxtract its cube root.
squarcs or squarc roots'
obrain
to
uscd
is
thc same way that thc A scalc
which mav bc
scgments.
rhrce
inro
dividcd
is
scilc
K
Noticc rhar ihc
and lO0 ro
(ccntcr)'
100
to
,t""Sii"fas runningfrom I to l0 (lc[t), l0
of the
hairline
lhc
wc
sct
numbe:'
a
iOoO"t;St,t. To 6nJ thc cubc of
ensw€r undcr lhc
the
rcad
lnd
scalc
D
thc
on
numbcr
that
arrroa o'"aa
hairlinc on rhc K scalc. Thus. *'c find that:
- 8: (l.o)t = 27; (8.00)' = 512
(l.O I lo-t)'- (l-o)t x (10-r)'= 27 x l0-'
2r
.
CubC f0015 of numbcrs Sslwccn
I and 1000 arc rcadily lound
by
thc numbcr on thc K scale rnd
sctting the hairlinc o[the cursor ebovc
36
rc.ding
CHAPIEI 3
the enswcr bcncath thc
hairlinc on
thi D scalc'
Wc suggcsr
(30b)'/! = 6'69
To cxtracr the cubc root o[ a numbcr which docs not fall bctwccn I
and l0oo, ir is ncccssary lo rcwritc it in cxponcntial form so that thc cocfficicnt is a numbcr bctwccn t and t000 and thc cxPoncnt is cvcnly
divisiblc by thrcc.
Exornplc 3.6 Find:
a. (0.00190)ur
b. (4.68 x l0-')r/r
Solution
f-.. r,.'10;oor90)rl,'- (l-90 x ,g"l1r/r.- {f,9{i)trr
21 1g-'t
Dclos'l.90on thc K scalc, wc rcad l-24 on the D scalc' Our
x l0-r, or0-124.
b. Wc must 6rst rcwrite lhe numbcr so that thc cxponcnt,
upon division by 3, will givc'a wholc numbcr' To obiain
answcr musl bc 1.24
tlic dcsircd form, wc multiply thc cocfficicnt by t0 and
dividc thc cxponcnt bY 10.
(4.68
x l0-')r/'! -
(46.8
x l0-')r/! - (46.$)t/'1
1g-t
Sctting 16.8 on the K scale (bctwccn thc second 4 and 5), *'c read
its c,rbc root on thc D scalc as 3-60. Hcncc, our answer must bc
3.60
x
l0-2.
EXERC'SES
.
r.
(6.08)r
2. (1.40
x l0r)t
x l0r)r
3. (5.95
4. (t2-9)v'
3.5
5. (268)ttt
6. (t.96)r/'
7. (1.3? x lOf)'D
8.'(5.92 x I0-')ur
TOGARITHMS
Thc only lincar scalc on thc slidc rulc, whcrc thc numbers arc spaced
o[ thc
at eguat in,.rotr, is rhc L scale Tiris scalc givcs thc logdrithms
If,€
37
SUDE RUI.T
numbcc bctwccn I and t0. To find rhc logarithm o[ a numbcr in this
Erngc, wc sct thc hairlinc ofrhc cursor abovc thc numbcr on thi D scale
and rtad its logarithm on thc L scatc. In this way. wc 6nd that thc logarithm of 2 is, :o thrcc digits. 0.301, that of J is 0.477. and so on.
1l
Dll'rl
2
'
.flGURE
r.6.
togorirhar of nunbco: bE 2-
- 0.301;
lo9
3 ' 0'177.
lf thc numbcr whosc logarithm wc rcquire is lcss than I or grcatcr
than 10, wc procccd as dcscribcd in Chaprcr 2- That is. wc writc lhc numbcrinstandardcxponcotial notation, usc lhe L scalc to 6nd rhc mantissa.
and takc thc cxponcnt to bc thc charactcristic-
Exomplc
3-7
Find thc logarirhm of 3.70
x l0-'.
Solution tog(3.70 x l0-r) = log3.70 + log l0-t - tog3.70 - 5
Sctting 3.70 on the D scalc, wc rced its logarithm. on ihc L scalc.
tobc0-568. Hcncc,log(3.70
x lO-t) -
0.568
- i =
-4.432
Wc scc from this cxamplc that thc L scalc is cquivalcnr to a thrccplacc tablc of logarithms' All of thc opcrations rvhich rvcrc dcscribcd in
Chaptcr 2 in conncction with a tablc of logarirhms can be carricd oul on
jr alidc rulc. using thc L and D scalcs. Wc can' for cxamplc. usc it to 6nd
antilogarithms
Eromple
'
a.
3.8
Find the numbcrs whosc logarithms arc:
0.380
b. 3-r00
c. -2.900
Solution
a. Sct thc hairlinc of thc cursor ovcr 0.180 on thc L scale;
rcad thc numbcr. 2.{0, on thc D scalc.
b. Sct thc hairlinc oicr thc mantissa. 0.100. on thc L scalc.
Ihc numbcr whosc log is 0. 100 is rcad from rhc D sciilc to
,
CHAPIER 3.
The number whosc log is 3 is
antilogJ.l00
c.
-
l0'
l-26 x
in Chaptcr
First, rewritc the number (recall rhe discussion
2) as:
I
EXERC'SES
Find, using rhe slidc rulc:
t
.
log 6.14
log (1.29
2.
3. log (5.84
4. ln 2.82
x l0t)
x l0-')
5. antilog 0.462
6. anrilog 1.250
?.
antilog (-5.400)
3.6 OTHER usFS OF THE SLIDE RULE
chcm'
' The slidc rulc opcrations pcrformc{ most frcqucntly in gencral on
thc
Dcpcnding
3'5'
through
3'2
in
Scctions
ir,ry;;r. ;..n d.r.rib.d
'
numbcrofscalcsavailabtc(andthcPricc!),itispossiblcrouscthcslidc
slide-rule
. ,".i.,y oGth"'optrati6ns' Evcn thc simptcsr
functlr.*r.. (iabcllcd S "ni T; whieh allow us to.obtain oftrigonomctricwill
rulc
slidc
rhe
;;;l;;;., .*.",1- a disc"ssion-oi this applicarion
rulc to carry or,
bc [ound in ChaPrer 8'
6nd thc rcciprocal o[ a
From timc to timc. wc rnaY havc occasion to
as l/r)' This can' o[
numbcr (thc rcciprocal of ''nurnbt" r'' is dcfincd
and traking usc of thc
division
oidinary
courscr bc donc by trcating it as an
is rcguircd' i1 i5- 5s63s'hat
C and D scalcs- Whcrc a""'it' of rcciprocals
Noticc that rhc Cl
scalcs'
morc convcnicn, ,o rn,rt *! of thc C and Cl
Cl scalc stan at
on
thc
numbcrs
thc
scalqamounts to.an invcrsc C scalc;
rhc rcciprocal
find
To
right'
far
the
I
on
ro
l0 on the far lclt and dccrcasc
numbcr on the C
thal
ovcr
cursor
oflhc
hairlinc
wc.sct
thc
of a numbcr,
on thc Cl scalc' In this way'
scalc and rcad its rcciprocat at rhc hairlinc
you can verifl' that:
t/2 = o.'>@; l/6 = 0'16?; l/3't6 -
0'316
to carry
with the D scalc
Thc Cl scale can also bt uscd in conjunction
In both cascst thc ma'
out thc proccslcs of multiplication ind ii'itiott'
IH€ SUOE
39
NUIE
nipulation is somcwhat morc rapid than is thc casc whcn thc C
and D
scalcs arc uscd (Scction 3.2)To multiply 9 by 12, using
lhc Cl and D scalcs' usc thc cursor to linc
up 9 on thc Cl scalc with 12 on thc D scatc. Rcad thc product' 108, on thc
D scatc, dircctly bclow thc indcx of thc Cl (or C) scalc' Notc that rrhcn
multiplications arc carricd out in lhis way, it is ncvcr ncccssary to shift
indiccs to bring the answcr onto thc scalc.
To divide 7.0 by any dcsircd numbcr (c.g.. 5-0, l-5. Z-0)' sct ihc i ndcx
of thc Cl (or C) scale to 7 on thc D scalc- Movc rhc cu6or to thc nu rnbcr
by which you wkh to dividc. on thc Cl scatc- Rcad 166 3ljwcr at thc hair'
linc on rhc D scalc. In this way. wc 6nd rhat:
7.0/2.0 7.0/r.4
5.0
3.5
7.0/1.3 = 2.0
Notc thc advantage of this mcthod of pcrforming divisions
askcd to dividc a lircn numbcr. such as 7.0, by a scrics of
whcn you arc
numbcrs (l'4,
2.0.3.5..-).
Thc following problcms arc dcsigncd (o rcvicw matcrial on cxPoncnts
(Chaptcr l) and logarithms (Chapter 2), in addition to improving yotrr
spced and accuracy in using thc slidc rulc.
Makc usc of thc stidc rulc ro obtain numcrical answcrs to cach protr'
lcm.
3.1
(5.49 x lOrX6.O2
4.91 x t0-:
6.84 x l0'
(3.08 x l0')r
log(2.41 x l0t;
x l0-r)
3-3
3.{
3.5 11.28 x t0-')t/'
3.6 antilog 0.649
3.7 (5.00 x lQ:1ul
3.8 (t.29 x lo:u)'
3.e l/(t.gS x l0-')
- -^ (t.28 x ro')(3.14)
3-10
------r6;98 x l0'
3.1I 2.7.1 (3.08 x lo-r)r/r