More Slide Rule Stuff
Seems I’m not quite done with slide rule posts. I am going to attempt to show some operations concerning electronic formulae in this dissertation. I may not cover everything as the post would be too large!
I think I will use the Post Versalog 1460 slide rule this time. I recently acquired a couple of them from E-Bay, one full sized 1460 and the 6” model 1461, which is identical to the big brother 1460.
I think, to keep this short, I will only address Ohm’s law, parallel resistance, inductive and capacitive reactances, impedance, and perhaps polar and rectangular conversions.
Ohm’s Law
Ohm’s law is primarily performed by simple multiplication and division. To keep it simple, I will use standard notation for the calculations, as it makes decimal location easier. The basic formula for voltage, current and resistance is straightforward,
\[\frac{E}{I \times R}\]
The scales primarily used are the C and D scales, and possibly the CI scale for ease of calculation, so I won’t go into further detail as that is pretty basic stuff.
Parallel Resistance
On the other hand, doing parallel resistance is more interesting. To start, we look at two resistors in parallel to determine total resistance. The formula is the product of the two resistors divided by their sum.
\[R_{x} = \frac{1}{\frac{1}{R1} + \frac{1}{R2}} \quad or \quad \frac{R1 \times R2}{R1 + R2}\]
Also the total resistance is always less than the smaller R value. So, let’s use some values of 20 and 30 ohms for the two resistors.
\[R_{x} = \frac{20 \times 30}{20 + 30} = \frac{600}{50}\]
As 50 will go into 600 about 12 times, that is the answer. However, using the slide rule, we can set 50 (C scale) over 20 (D scale), then move the hairline (HL) over 30 (C scale), and read 12 ohms (D scale). Basically combined operations. That was a rather simplistic example, able to be worked in your head.
Let’s try a more rigorous example, with R values of 900 k\(\Omega\) and 3.5 M\(\Omega\).
\[R_{x} = \frac{0.9 \times 3.5}{0.9 + 3.5}\]
First we place both in the same units, M\(\Omega\), meaning 0.9 and 3.5, so the answer on the D scale would also be in M\(\Omega\). Next, set 4.4 (C scale) over 0.9 (D scale), move HL to 3.5 (C scale) and read 0.716 M\(\Omega\) on D scale.
Some value combinations result in the slide projecting far out one side or the other. To prevent this, use the A and B scales instead.1
Inductive and Capacitive Reactances
The formula for inductive reactance, \(X_{L} = 2 \pi f L\), is basically a continued-product operation. So, if we have inductance of 25 uH at 10 kHz, we multiply \(6.28 \times 2.5e^{-5} \times 1e^{4}\), giving 15.7, having estimated \(3 \times 6 = 18\) (2 digits). For multiplication, we add the exponents (-5 + 4) = -1, so we move the decimal one left, for XL = 1.57.
The formula for capacitive reactance, \(X_{C} = \frac{1}{2 \pi f C}\) is best performed using the CIF scale, an inverted CF scale for reciprocal functions. But we can simplify this formula, by finding the reciprocal, \(\frac{1}{2 \pi}\). Placing the HL over \(\pi\) (D scale), move the CIF scale index (near center of slide) under the HL, moving the HL to 2 (CIF) scale, read 0.159 under the HL on C scale. This transforms the XC formula to,
\[ X_{C} = \frac{0.159}{f C},\]
another continued operations formula. Now, if we have a capacitance of 12 pF at a frequency of 3 MHz, we have the formula, \(X_{C} = \frac{1.59e^{-1}}{1.2e^{-11} \times 3e^{6}}\). Setting HL at 1.59, moving 1.2 (C scale) under HL, sliding HL to 3 (CI scale) gives 4.42. For division decimal placement we subtract, (-1) - (-11) - 6 = 4. So the answer has 4 digits for 4420.
Impedance
The impedance of a series circuit containing a resistance and inductance is given by Pythagoras’ Theorem as,
\[Z = \sqrt{R^{2} + X_{L}^{\; 2}}\]
If we wished to use a short-cut for this, we would have to use the R scales of the Post 1460, as it doesn’t have A/B scales. Let’s say we have R = 12 \(\Omega\) and XL = 22 \(\Omega\). We can set the left index to 12 (R1 scale)2, move HL to 22 (R1 scale), read 3.362 (C scale), add 13 and move HL to 4.362 (C scale), then read R1 scale for Z = 25.06 \(\Omega\).
Let’s try another example where R = 50 and XL = 100. This time we use the right index and HL over 5 (50 \(\Omega\)) on R2 scale, then move the HL over 1 (100 \(\Omega\)) on R1 scale, reading 4 on C scale, adding 1, moving HL to 5, then reading 1.12 (112 \(\Omega\)) on R1 scale. The phase angle \(\theta = \arctan{\frac{X}{R}}\). As we’re looking for the inverse tangent, we use the CI scale, where we set the HL to 2 (100/50), flip the rule to read the red T scale for 63.4o. So, the complex impedance in polar form is \(112 \Omega \angle 63.4^{o}\).
Polar to Rectangular Conversion
Now, let’s transform the above value to its rectangular form by \(Z(\cos{\theta} + j\sin{\theta})\). Set index over Z on D scale, HL over cosine of angle (Cos scale, R to L). Read resistive Z on D scale. Move HL over sine of angle (S scale, L to R), read reactive Z on D scale. If angle is negative, write as \(r-jx\).
So, for \(112 \Omega \angle 63.4^{o}\), we set the index at 112 (D scale), move the HL over 63.4 (Cos scale, which reads from right to left), then note the value (50) on the D scale. Now move the HL to 63.4 (S scale, which reads from left to right), and note 100 on the D scale. This is written as \(50 \Omega + j 100 \Omega\).
Just for fun, let’s do a negative angle for contrast. For \(Z = 12 \angle -41^{o}\), set the index on 12 (D scale), move the HL over cosine of angle (S scale, R to L) and read resistive Z = 9.06 \(\Omega\). Slide the HL over sine of 41 (S scale, L to R) and read reactive Z = 7.88 \(\Omega\). As the angle is negative, we write \(9.06 \Omega - j 7.88 \Omega\).
And that’s enough for this post. Have a great day in the Lord Jesus Christ, The Word. Happy Holidays, and God Bless you and yours!