Slide rule 2

We knew that slide rule can do multiplication. This post is focus on the applications.

Ratio:
That is the simplest function I can think of. For example, exchange rate. Say, 1 USD = 7.78 HKD. We simply set the D row be USD, C row be HKD. Align the C(7.78) on D(1), then done! If you want to check 4.5 USD = ? HKD, look at the D(4.5) and find out the number in C row.
10 HKD = ? USD, look at the C(10) and find the number on D row.

Given the c and a, find out the b in a^2+b^2= c^2 :
since b = Sqrt(c^2-a^2) = Sqrt( (c+a) (c-a) ) = Sqrt(c+a) Sqrt(c-a)
Thus, we have to use the A and B row, which is Sqrt root.
First, locate A(c+a) and align B(1) on it, check the B(c-a) and read the D row.

Square of x = x^2 :
using the fact that A(x) = D ( Sqrt(x) ) <-> A(x^2) = D(x), so simple read the A row of D(x).

Division x / y :
Locate D(x) and align C(y) on it, read the D row on the C(1). This is a reverse process of multiplication! We really do thing in backward sequence!
another way to do this is using n / m = n x 1/m, thus, Locate D(n) on Cl(1) and read the D row on Cl(m).

another use of Cl row :
For ruler shape user, you will find out 2x6 will be out range, the ruler is just long enough. So, you are going to use the Cl row. The principle is, n x m = n / (1/m), a multiplication is equal 2 times division. The step is: align D(n) on Cl(m), than read the D row on Cl(1). You will find that the answer is 10 times lesser. This out-range problem can be overcome by a circlur slide rule.

10^x :
we have to use the L row. locate the L(x) and read the corresponding C row or D row, no need to slide.


log(x) :
if x <10 , use D(x) and read L row if 100>x >10, use D(x) and read L row, but add 1.
if 1000>x>100, add 2 on the L row.


e^x :
from the fact that, if LL2(x) = D( ln(x) <-> LL2( e^x) = D(x). so, read the LL2 row on the D(x). and if it is out range, use LL3.
e.g.

a^x :
this times, we have to use the fact that a^x = e ^ ( x ln(a) ). so, we align the C(1) on LL2(a), then the LL3 or LL2 row on C(x), the decision on whether LL2 or LL3 is on the range of the result. The reason for this is, LL2(a) + C(x) = log( ln^(a) ) + log (x) = log ( x ln(a) ) = D( x ln(a) ) = LL2( e^ (x ln(a) ) = LL2 (a^x).
e.g.
2^1.1 : LL2(2) + C(1.1) = LL2(2.14)
2^2 : LL2(2) + C(2) = LL3(4)
3^n : LL3(3) + C(n) = LL3( 3^n)


log(x, base a) :
log(x, base a) = log(x) / log(a). this is quit complicate as there is only 1 L row.


x + y :
L(x) + L(y) = D( 10^x) + D (10^y) = D (10^(x+y)) = L (x + y)
align D(1) with L(x), read the D row of L(y) = D(z), remember, and read the L row on D(z).

by applying the above methods, one can calculate everything. The master of slide rule skill require deep understanding of log function, and inverse function. Thus, this is a very good tool to train mathematical skill.

for example,

Sqrt( x^2 + y^2 ) :
use square method, and apply addition method, then use square root method.