2.2 Squares, Cubes, and Roots

2.2 Squares, Cubes, and Roots

In aviation maintenance mathematics, you will frequently encounter quadratic and cubic relationships. Physical properties such as electrical resistance, kinetic energy, and alternating current parameters do not scale linearly. This section covers the calculations of squares, cubes, square roots, and cube roots, and details their essential applications in aircraft systems.

Squares and Cubes

• Squaring a number means multiplying the number by itself. It is represented as $x^2$.
• Cubing a number means multiplying the number by itself twice (three factors of the number). It is represented as $x^3$.

In EASA exams, you will often need to calculate these values without the aid of a calculator. Let's review key arithmetic properties, particularly involving fractions and decimals:

1. Decimals: When squaring a decimal, the number of decimal places in the result is twice the number of decimal places in the original number. For example, $(0.05)^2 = 0.05 \times 0.05 = 0.0025$ (two decimal places in the base become four in the result). For cubing, it is three times: $(0.2)^3 = 0.2 \times 0.2 \times 0.2 = 0.008$ (one decimal place becomes three).
2. Fractions: To square or cube a fraction, square or cube the numerator and denominator independently: $$\left(\frac{3}{4}\right)^2 = \frac{3^2}{4^2} = \frac{9}{16}$$ $$\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$$

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Square Roots and Cube Roots

• The square root of a number $x$, denoted as $\sqrt{x}$ or $x^{1/2}$, is a value that, when multiplied by itself, equals $x$.
• The cube root of a number $x$, denoted as $\sqrt[3]{x}$ or $x^{1/3}$, is a value that, when multiplied by itself twice, equals $x$.

Manual Estimation Technique for Roots

When calculators are not permitted, finding roots of non-perfect squares requires estimation or prime factorization.

• Estimation Method: To find $\sqrt{55}$, identify the nearest perfect squares: $7^2 = 49$ and $8^2 = 64$. Since $55$ is between $49$ and $64$, its square root is between $7$ and $8$. A linear interpolation suggests $\sqrt{55} \approx 7.4$.
• Fractions and Decimals: $$\sqrt{0.09} = 0.3 \quad (\text{since } 0.3 \times 0.3 = 0.09)$$ $$\sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} = 0.8$$ $$\sqrt[3]{0.008} = 0.2 \quad (\text{since } 0.2 \times 0.2 \times 0.2 = 0.008)$$

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Aviation Application: Circular Mil Area (CMA)

In aircraft electrical maintenance, wire sizes are defined by their cross-sectional area. Standard commercial aircraft contain miles of wiring, and selecting the correct gauge is critical to prevent overheating and voltage drop.

Instead of measuring wire area in square inches or square millimeters (which requires multiplying by $\pi$), the aviation industry utilizes the circular mil (cmil).

• Mil: A unit of length equal to $0.001\text{ inches}$ ($10^{-3}\text{ in}$).
• Circular Mil: The area of a circle with a diameter of $1\text{ mil}$.

The formula to calculate the Circular Mil Area (CMA) of a conductor is:

$$\text{CMA} = d^2$$

Where $d$ is the diameter of the conductor expressed in mils.

Worked Example: Calculating Wire CMA

An aircraft technician is replacing a generator power feeder cable. The manual specifies a conductor diameter of $0.080\text{ inches}$. What is its area in circular mils, and what would the diameter be if the CMA was $10,000\text{ circular mils}$?

1. Convert Diameter to Mils: $$d = 0.080\text{ in} \times 1000\text{ mils/in} = 80\text{ mils}$$
2. Calculate CMA: $$\text{CMA} = d^2 = 80^2 = 6400\text{ circular mils}$$
3. Find Diameter from a Given CMA of 10,000: $$d = \sqrt{\text{CMA}} = \sqrt{10000} = 100\text{ mils}$$ $$\text{Diameter in inches} = 100\text{ mils} \times 0.001\text{ in/mil} = 0.100\text{ inches}$$

Using circular mils simplifies calculations for engineers, as resistance per unit length is directly inversely proportional to the CMA. Additionally, circular mils and square mils are related as follows: $1\text{ circular mil} = \frac{\pi}{4}\text{ square mils} \approx 0.7854\text{ square mils}$.

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Aviation Application: Kinetic Energy ($E_k = \frac{1}{2}m v^2$)

During aircraft landing operations or a rejected takeoff (RTO), the aircraft's kinetic energy must be completely absorbed and dissipated as heat by the wheel brakes. Kinetic energy ($E_k$) is governed by the formula:

$$E_k = \frac{1}{2}m v^2$$

Where:

• $m$ is the mass of the aircraft (in kg).
• $v$ is the velocity of the aircraft (in m/s).

Because velocity is squared, the kinetic energy scales quadratically with speed. This means a minor increase in speed results in a disproportionately large increase in kinetic energy, which can exceed the thermal limits of the brakes and cause brake fires or tire blowouts.

Worked Example: Brake Energy Scaling

An aircraft with a landing mass of $12,000\text{ kg}$ is landing at a speed of $45\text{ m/s}$ (approx. $87\text{ knots}$).

1. Calculate the baseline Kinetic Energy:

   $$E_k = 0.5 \times 12000\text{ kg} \times (45\text{ m/s})^2 = 6000 \times 2025 = 12,150,000\text{ Joules} = 12.15\text{ MJ}$$

2. Calculate Energy if landing speed increases to $60\text{ m/s}$ (due to tailwind or flap failure):

   $$E_k = 0.5 \times 12000\text{ kg} \times (60\text{ m/s})^2 = 6000 \times 3600 = 21,600,000\text{ Joules} = 21.60\text{ MJ}$$

Although the speed increased by only $33.3\%$ (from $45\text{ to } 60\text{ m/s}$), the kinetic energy that the brakes must dissipate increased by $77.8\%$ (from $12.15\text{ to } 21.60\text{ MJ}$). This highlights the mathematical danger of high-speed landings.

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Aviation Application: Root-Mean-Square (RMS) Voltage

Commercial aircraft electrical systems typically use $115\text{ V}$ Alternating Current (AC) at a frequency of $400\text{ Hz}$. The voltage of an AC system varies as a sine wave over time. The nominal value of $115\text{ V}$ is not the peak voltage, but rather the Root-Mean-Square (RMS) value.

RMS represents the equivalent Direct Current (DC) voltage that would deliver the same electrical power (heating effect) to a resistive load. To calculate RMS mathematically:

1. Square the values of the sine wave.
2. Find the Mean of these squared values.
3. Take the Square Root of the mean.

For a sinusoidal AC voltage, the relationships between RMS voltage ($V_{rms}$) and Peak voltage ($V_{peak}$) are:

$$V_{rms} = \frac{V_{peak}}{\sqrt{2}} \approx 0.707 \times V_{peak}$$ $$V_{peak} = V_{rms} \times \sqrt{2} \approx 1.414 \times V_{rms}$$

Worked Example: Insulation Voltage Rating

An avionics technician needs to select wire insulation for a $115\text{ V}$ AC bus. The insulation must withstand the peak voltage of the system without breaking down. What is the peak voltage of the $115\text{ V}$ AC system?

$$V_{peak} = 115\text{ V} \times \sqrt{2} \approx 115 \times 1.4142 \approx 162.63\text{ V}$$

The wire insulation must be rated to handle at least $163\text{ V}$ of electrical potential, even though the nominal system voltage is only $115\text{ V}$.