The Nomorule in Electronic Calculations
April 1967 Radio-Electronics

April 1967 Radio-Electronics [Table of Contents] Wax nostalgic about and learn from the history of early electronics. See articles from Radio-Electronics, published 1930-1988. All copyrights hereby acknowledged.

 In the March 1967 issue of Radio-Electronics magazine, John Fasal introduced his "Nomorule" (a nomograph) as a method of simplifying routine electrical design calculations. In the days prior to smartphones, desktop computers, and electronic calculators, the choices for such figuring were paper and pencil, slide rule, or nomograph - or any combination thereof. The nice thing about a nomograph is the visual presentation of how varying one or more values affects the result. When the equation at hand is linear, the mental work of extrapolating numbers is not so hard, but when exponents, logarithms, or trigonometry functions are involved, things can get tough. Mr. Fasal instructs you on how to generate a custom nomograph for calculating just about any equation. The only restriction is that in general only three scales can be in play at a time - akin to a slide rule - unless other scales happen to fall within the realm of the primary three. Otherwise, if more than three numbers are involved, equations must be constructed that combine two or more quantities, with selected ones fixed at a certain value as a multiplier of the chosen variable. Then, it would be necessary to construct separate scales for each case. For instance, if you needed to calculate the voltage across a resistor that was in parallel with a capacitor and an inductor at various frequencies, you would need to reduce the equation to just three variables and hold one as fixed. That might be the case for a resonant circuit across a resistive load, so if the inductor is fixed in value but the capacitor is adjustable for tuning, an equation would be generated with the L value as a coefficient and f, C, and R as the nomograph scales. See Nomorule Part 1 in the March edition.

The Nomorule in Electronic Calculations

Part  2- More uses for last month's Nomorule: How to multiply and divide by using logarithms; trig functions and some practical examples

By John H. Fasal*

The basic Nomorule described in Part 1 is repeated here as Fig. 1. We showed that you can multiply two numbers by aligning a straightedge across one on scale A and the other on scale C, and reading off the product on scale B. If you have more than two factors, you must transfer the first product back to either scale A or scale C, to use it as a factor in the next product.

If any number at B is aligned with 1 on scale C, the line will intersect the same number at A. This is true because B = AC, and in this case C = 1, so B = A. Thus we can transfer the number from B to A without changing it.

We can also transfer a number from B to C. Let a line intersect the number at B with the top index (0.1) at D. At C we will find 0.1 B. The equation used is C = B/A = BD, and D = 0.1 in this case. We must remember to multiply the transferred number (A) by 10 to make it equal the original number at B.

Another transfer is possible using 10 on scale C as pole (or pivot point). Align it with the number at B, and find the answer at A. We are using the equation A = B/C with C = 10. Therefore A will be 0.1 B and it must be multiplied by 10 to equal B.

The Nomorule is ideally suited for calculations with trigonometric. functions. To find the sine of an angle (such as 30°) look for the angle on the "sin" scale and align it with either point R or point S on the fourth scale (Fig. 2). The answer is found on B, the center scale. Values of sine for angles between 6° and 90° lie between 0.1 and 1. For better accuracy, align with point R if the angle is smaller than 18° (midway up the "sin" scale). Use S for angles larger than 18°.

Sine or tangent values for smaller angles, between 36' and 6°, are based on the "sin-tan" scale and are calculated in the same way. All readings found on the B scale, lie between 0.01 and 0.1.

To find the tangent of an angle larger than 6° use the "tan" scale. Align the given angle (for example, 20° as shown) with either point P or point Q of the second scale. The answer is tan 20° = 0.364, as shown in Fig. 3. An angle smaller than 45° has a tangent smaller than 1.

When the angle is greater than 45°, use the equation tan B = cot (90 - B) = 1/tan (90 - B). For example, tan 70° = cot 20° = 1/tan 20°. To find tan 70°, determine tan 20°. Transfer the reading from B to the inverse scale D as shown by the phantom lines of Fig. 3. In other words, we have transferred tan 20° to scale A, then found its reciprocal at D. This indicates 1/ tan 20° or tan 70°, which is 2.74.

For cotangents of angles smaller than 45°, use the formula cot B = 1/tan B. Determine tan B on scale B, then transfer reading to inverse scale D. For angles larger than 45°, use cot B = tan (90 - B). The reading appears on scale B directly.

You can find common logarithms by using the "log" scale with point P or point Q on the second scale, and the B scale. Align the number (for example, 6) at scale B with P or Q and read off the logarithm on the "log" scale, as 0.78. To find log 6,000, write it as 6 X 10³, and add the corresponding logs. The answer is 3.78. Log 2 is also shown in Fig. 4: 0.301.

Now how about a couple of practical electronic examples worked out on the Nomorule? Let's say we want to find the resonant frequency of a circuit having L = 2.5 X 10^-1 henries and C = 1.6 x 10^-9 farads. The frequency is f = 

Align 2.5 on scale A with 1.6 on scale C, as in Fig. 5, and find the product 4 on scale B. The square root is read off directly as 2 on scale E. A point on scale B may be transferred to either scale A or C, as we've said before. But you can't do it that way from E. Instead, use a "cross alignment" onto scale C, like this: Connect 2 on scale E with (1) on scale C, intersecting the "tan" scale. Through this intersection draw a line through 1 on scale E, giving a point on C which is identical to the original point on E.

The next step is to multiply by 2π on A. This gives us 12.56 (scale B). Transfer this value to A, using 10 (scale C) as pole. We are using the relation A = B/C where C is 10. Therefore, the value at A has been reduced by a factor of 10. It is restored to its correct value by multiplying by 10. Instead of 1.256 (scale A) we actually have 12.56. The last step is to find the reciprocal of 12.56, which is 0.0795 (scale D). Multiplying by 10⁵ (see formula above), we get the frequency: 7.95 kHz.

Several problems are worked out in Fig. 6. The inductive reactance of a coil (X_L) is given by ωL where ω is 2πf. Let f = 6 x 10, and L = 5.6 x 10^-1.

First find 2πf = 37.7 x 10 as shown. Now transfer this result from B to A, using 10 on C as pole. We know that this transfer divides the number by 10, so we must multiply by 10 to compensate. We have, therefore, 3.77 x 10² at A. Now multiply this by the value of L (at C) and obtain the reactance as 21.1 x 10 at scale B.

Other examples in Fig. 6 show how to convert frequency to wavelength. The formula is f 300,000/λ, where f is in kHz, λ in meters, and 300,000 is the velocity of light in km/sec. The Nomorule shows that 416 meters corresponds to 720 kHz; 857 meters corresponds to 350 kHz and 250 meters to 1.2 MHz.

As a last example, let us calculate the Q at 5 kHz of a coil with 0.4 henry inductance and 628 ohms resistance, in accordance with the formula Q = 2πfL/R = 2πf(5x10^{^3})(4x10^-1)/628.

Fig. 7 shows how. The first product is read off as 31.4 x 10³ on B. Transfer this result to C by using the top index (0.1) on D as pole. Line TR shows this. Align this result with 4 x 10^-1 at A, and we find 12.56 x 10³ at B. Transfer this to C (line TR'), then divide by 0.628 x 10³ at D. The result is 2 x 10.

* Assistant chief engineer, Alarms Engineering, Walter Kidde Co., Belleville, N.J.

Posted April 8, 2024