Setting the R.H. index of the slide to 5 on A, it is seen that 1·71 on
D coincides with 1·71 on ᗺ. Then setting the L.H. index to 5 on A,
further coincidences are found at 3·68 and at 7·93, the three values
thus found being the required roots. Note that the first root was
found on that portion of the D scale lying under 1 to 5 on A; the
second root on that portion lying under 5 to 50 on A; and the third
root on that portion of D lying under 50 to 100 on A. In this
connection, therefore, scale A may always be considered to be divided
into three sections—viz., 1 to _n_, _n_ to 10_n_, and 10_n_ to 100.
For all numbers consisting of 1, 1 + 3, 1 + 6, 1 + 9—_i.e._, of 1, 4,
7, 10, or −2, −5, etc., figures—the coincidence under the first
section is the one required. If the number has 2, 5, 8, or −1, −4, −7,
etc., figures, the coincidence under the second section is correct,
while if the number has 3, 6, 9, or 0, −3, etc., figures, the
coincidence under the last section is that required. The number of
digits in the root is determined by marking off the number into
sections, as already explained.
_Cube Root (Pickworth’s Method)._—One of the principal objections to the
two methods described is the difficulty of recollecting which scales are
to be employed and with which index of the slide they are to be used.
With the direct method another objection is that the readings to be
compared are often some distance apart, the maximum distance intervening
being _two-thirds_ of the length of the rule. To carry the eye from one
to another is troublesome and time-taking. With the inverted scale
method the reading of a scale reversed in direction and with the figures
inverted is also objectionable.
With the author’s method these objections are entirely obviated. The
_same scales and index are always used_, and are read in their normal
position. The three roots of _n_, 10_n_ and 100_n_ (_n_ being less than
10 and not less than 1) are given with one setting and appear in their
natural sequence, no traversing of the slide being needed. The readings
to be compared are always close together, the maximum distance between
them being _one-sixth_ of the length of the rule. The setting is always
made in the earlier part of the scales where closer readings can be
obtained, and finally, if desired, the result may be readily verified on
the lower scales by successive multiplication.
For this method two gauge points are required on C. To conveniently
locate these, set 53 on C to 246 on D; join 1 on D to 1 on A with a
straight-edge and with a needle point draw a short fine line on C. Set
246 on C to 53 on D, and repeat the process at the other end of the
rule. The gauge points thus obtained (dividing C into three equal parts)
will be at 2·154 and 4·641, and should be marked ∛(10) and ∛(100)
respectively.[6]
EX.—Find ∛(2·86,) ∛(28·6) and ∛(286).
Set cursor to 2·86 on A and drawing the slide to the right find 1·42
under 1 on C, when 1·42 on B is under the cursor. Then reading under
1, ∛(10) and ∛(100,) we have
∛(2·86) = 1·42; ∛(28·6) = 3·06 and ∛(286) = 6·59.
It will be seen that factorising with powers of 10, we multiply the
initial root by ∛(10) and ∛(100). Obviously the three roots will always
be found on D, in their natural order and at intervals of one-third the
length of the rule. The number of digits in the roots of numbers which
do not lie between 1 and 1000, is found as before explained.
In any method of extracting cube roots in which the slide has to be
adjusted to give equal readings on B and D, the author has found it of
advantage to adopt the following plan:—The cursor being set to, say, 4·8
on A, bring a near _main_ division line on B, as 1·7, to the cursor;
then 1 on C is at 1·68 on D. The difference in the readings is two small
divisions on D, and moving the slide forward by _one-third the space
representing this difference_, we obtain 1·687 as the root required.
With a little practice it is possible to obtain more accurate results by
this method than by comparing the reading on D with that on the less
finely-graded B scale.