[Sidenote: ⵜ
ⵜ
ⵏ
ⵏ]
Set 27 on C to 43·5 on D, and as with this _division_ the slide is to
the right, make the first ⵏ mark. Bring cursor to 29·4 on C, and as in
this _multiplication_ the slide is to the right, make the first − mark,
cancelling as shown. Setting 3·83 on C to the cursor, requires the
second ⵏ mark, which, however, is cancelled in turn by the
multiplication by 51. The division by 10·5 requires the third ⵏ mark,
and after multiplying by 32 (requiring no mark) the final division by
1·31 requires the fourth ⵏ mark. Then, as there are 8 numerator digits,
6 denominator, and 2 uncancelled memo-marks (which, being 1, are
additive) we have
Number of digits in result = 8 − 6 + 2 = 4.
Had the uncancelled marks been − in character, the number of digits
would have been 8 − 6 − 2 = 0.
For quantities less than 0·1 the digit place numbers will be _negative_.
The troublesome addition of these may be avoided by transferring them to
the opposite side and treating them as positive.
_2_ _4_
0·00356 × 27·1 × 0·08375
Thus:— ───────────────────────── = 288
0·1426 × 9·85 × 0·00002
_2_ _1_ _1_
The first numerator, 0·00356, has −2 digits. Note this by placing 2
_below the lower line_ as shown. 27·1 has 2 digits; place 2 over it.
0·08375 has −1 digit; hence place 1 _below the lower line_. The first
denominator has no digits; the second, 9·85, has 1 digit; hence place 1
under it. 0·00002 has −4 digits; place 4 _above the upper line_. The sum
of the top series is 2 + 4 = 6; of the bottom series 2 + 1 + 1 = 4.
Subtracting the bottom from the top, we have 6 − 4 = 2 digits, to which
1 has to be added for an uncancelled memo-mark, and the result is read
as 288.
Moving the decimal point often facilitates matters. Thus, (32·4 × 0·98 ×
432 × 0·0217)/(4·71 × 0·175 × 0·00000621 × 412000) is much more
conveniently dealt with when re-arranged as (32·4 × 9·8 × 432 ×
2·17)/(4·71 × 17·5 × 6·21 × 4·12) = 141.
To determine the number of figures in the result by rough cancelling and
mental calculation, we note that 4·71 enters 432 about 100 times; 9·8
enters 17·5 about 2; 6·21 into 32·4 about 5; and 2·17 into 4·12 about 2.
This gives (500)/(4) = 125, showing that the result contains 3 digits.
From the slide rule we read 141, which is therefore the result sought.
The occasional traversing of the slide through the rule, to interchange
the indices—a contingency which the use of the C and D scales always
involves—may often be avoided by a very simple expedient. Such an
example as (6·19 × 31·2 × 422)/(1120 × 8·86 × 2.09) = 3·93 is sometimes
cited as a particularly difficult case. Working through the expression
as given, two traversings of the slide are necessary; but by taking the
factors in the slightly different order, (6·19 × 31·2 × 422)/(8·86 ×
2·09 × 1120), _so that the significant figures of each pair are more
nearly alike_, we not only avoid any traversing the slide, but we also
reduce the extent to which the slide is moved to effect the several
divisions.
Such cases as (_a_ × _b_)/(_c_ × _d_ × _e_ × _f_ × _g_) or (_a_ × _b_ ×
_c_ × _d_ × _e_)/(_f_ × _g_) really resolve themselves into (_a_ × _b_ ×
1 × 1 × 1)/(_c_ × _d_ × _e_ × _f_ × _g_) and (_a_ × _b_ × _c_ × _d_ ×
_e_)/(_f_ × _g_ × 1 × 1 × 1), but, of course, if rules are used to
locate the decimal point, the 1’s so (mentally) introduced are not to be
counted as additional figures in the factors.