Derived Units and Physical Calculations

While the seven SI base units provide the foundational reference points for physical measurement, the vast majority of test and calibration instruments measure derived quantities. A derived unit is formed by algebraic multiplication, division, and exponentiation of base units according to the physical laws governing the system. When a derived unit contains no numerical multipliers other than the number 1, it is defined as a coherent derived unit.

To simplify communication and documentation, the General Conference on Weights and Measures (CGPM) has assigned special names and symbols to 22 derived units.

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Master Reference Table of SI Derived Units in Metrology

The following table details the derived units most frequently encountered in industrial calibration, their special names, equivalent expressions in other derived units, and their fundamental decomposition into the seven base SI units:

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Mechanical and Fluid Metrology Calculations

Force and Newton's Second Law

Force satisfies $F=ma$. A deadweight force machine uses calibrated masses and evaluated local gravity, with buoyancy and other relevant corrections. Proving rings and load cells instead respond elastically to applied force and are calibrated by comparison; they do not all generate their own force from suspended masses.

$$F = m \cdot g_{\text{local}}$$

Where:

• $m$ is the calibrated mass in kilograms ($\text{kg}$)
• $g_{\text{local}}$ is the true local acceleration due to gravity in meters per second squared ($\text{m/s}^2$)
• $F$ is the resulting force in newtons ($\text{N}$)

Pressure Standards and Conversions

Pressure is defined as force applied perpendicularly per unit area:

$$P = \frac{F}{A} = \frac{\text{N}}{\text{m}^2} = \text{Pa}$$

Because the pascal is a relatively small unit (a single dollar bill lying flat on a table exerts approximately $1\text{ Pa}$), calibration technicians frequently handle multiples and non-SI units. The technician must be adept at converting between these common units:

Step-by-Step Calibration Calculation: Deadweight Piston Gauge

A deadweight tester generates pressure from calibrated mass, local gravitational acceleration, and effective piston area. The following is a simplified model, with true masses, negligible pressure distortion and surface-tension effects. Use the instrument’s documented model for actual work; certificates may provide conventional masses that require conversion.

The Complete Physical Equation

In an accredited laboratory, calculating the true generated pressure requires correcting for local gravity, air buoyancy on the weights, and fluid head height differences:

$$P = \frac{\sum m \cdot g_{\text{local}}}{A_{\text{eff}} \left[1 + (\alpha_{\text{p}} + \alpha_{\text{c}})(T - T_0)\right]} \left(1 - \frac{\rho_{\text{air}}}{\rho_{\text{weights}}}\right) + \rho_{\text{fluid}} g_{\text{local}} \Delta h$$

Where:

• $\sum m$ is the total mass of the loaded weights plus the tare piston weight ($\text{kg}$)
• $g_{\text{local}}$ is the local acceleration of gravity ($\text{m/s}^2$)
• $A_{\text{eff}}$ is the effective area of the piston-cylinder assembly at reference temperature $T_0$ ($20^\circ\text{C}$)
• $\alpha_{\text{p}}, \alpha_{\text{c}}$ are the thermal expansion coefficients of the piston and cylinder
• $\rho_{\text{air}}$ is ambient laboratory air density (typically $\approx 1.2\text{ kg/m}^3$)
• $\rho_{\text{weights}}$ is the density of the stainless steel weights (typically $\approx 8,000\text{ kg/m}^3$)
• $\Delta h$ is the vertical difference defined positive when the UUT port is below the DWT reference level

Worked Example

A technician must calibrate a pressure transmitter at exactly $P = 400.00\text{ kPa}$.

• Effective piston area: $A_{\text{eff}} = 0.50000\text{ cm}^2 = 0.50000 \times 10^{-4}\text{ m}^2 = 5.0000 \times 10^{-5}\text{ m}^2$
• Local acceleration of gravity: $g_{\text{local}} = 9.80000\text{ m/s}^2$
• Ambient air density: $\rho_{\text{air}} = 1.20\text{ kg/m}^3$
• Weight alloy density: $\rho_{\text{weights}} = 8,000\text{ kg/m}^3$
• Neglecting fluid head difference and operating at reference temperature $20^\circ\text{C}$.

Step 1: Calculate the required upward force $F$:

$$F = P \times A_{\text{eff}} = (400,000\text{ Pa}) \times (5.0000 \times 10^{-5}\text{ m}^2) = 20.000\text{ N}$$

Step 2: Calculate the air buoyancy correction factor $B_{\text{cf}}$:

$$B_{\text{cf}} = 1 - \frac{\rho_{\text{air}}}{\rho_{\text{weights}}} = 1 - \frac{1.20\text{ kg/m}^3}{8,000\text{ kg/m}^3} = 1 - 0.000150 = 0.999850$$

Step 3: Calculate the required mass $m$:

$$F = m \cdot g_{\text{local}} \cdot B_{\text{cf}} \implies m = \frac{F}{g_{\text{local}} \cdot B_{\text{cf}}}$$ $$m = \frac{20.000\text{ N}}{(9.80000\text{ m/s}^2) \times (0.999850)} = \frac{20.000}{9.79853} = 2.04112\text{ kg}$$

If the technician had naively used standard gravity ($g_n = 9.80665\text{ m/s}^2$) and omitted air buoyancy, they would have calculated $m = 20.000 / 9.80665 = 2.03943\text{ kg}$—introducing a systematic calibration error of $-1.69\text{ grams}$, or $828\text{ ppm}$ ($0.083\%$), which could change a conformity decision near a limit.

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Electrical Derived Units and Calculations

Coherent derived electrical units are linked by Ohm's Law and Joule's Law of electric power:

$$V = I \cdot R, \quad P = V \cdot I = I^2 R = \frac{V^2}{R}, \quad E = P \cdot t$$

Decomposing the ohm into base units reveals the fundamental dimensional relationships:

$$\Omega = \frac{\text{V}}{\text{A}} = \frac{\text{W/A}}{\text{A}} = \frac{\text{J}/(\text{s}\cdot\text{A})}{\text{A}} = \frac{\text{N}\cdot\text{m}}{\text{s}\cdot\text{A}^2} = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}}{\text{s}\cdot\text{A}^2} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}$$

Worked Example: Precision Current Shunt Resistor Dissipation

A calibration technician is verifying a $100\text{ A}$ precision current shunt used in DC power calibration. The shunt has a nominal resistance of $R = 0.0010000\ \Omega$ ($1.0000\text{ m}\Omega$). During a calibration run at full-scale current ($I = 100.00\text{ A}$):

1. Calculated Voltage Output:

$$V = I \cdot R = (100.00\text{ A}) \times (0.0010000\ \Omega) = 0.10000\text{ V} = 100.00\text{ mV}$$

1. Calculated Power Dissipation:

$$P = I^2 \cdot R = (100.00\text{ A})^2 \times (0.0010000\ \Omega) = (10,000\text{ A}^2) \times (0.0010000\ \Omega) = 10.00\text{ W}$$

1. Heat Dissipation Over Time: If the test runs continuously for $t = 15.0\text{ minutes}$ ($900\text{ seconds}$), calculate energy dissipated in joules and kilowatt-hours:

$$E = P \cdot t = (10.00\text{ W}) \times (900\text{ s}) = 9,000\text{ J} = 9.00\text{ kJ}$$ $$E = \frac{10.00\text{ W} \times (15 / 60\text{ h})}{1,000\text{ W/kW}} = 0.0025\text{ kWh}$$

Metrological Insight: Continuous dissipation of $10\text{ W}$ causes self-heating in the manganin resistance element. In precision shunt calibration, the technician must document the ambient temperature and allow thermal equilibrium to stabilize, applying the manufacturer's power coefficient of resistance (PCR) to correct for self-heating resistance shift.

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Thermal Metrology & Temperature Conversions

Calibration technicians must navigate four temperature scales across international and domestic test protocols:

Point Formulas (Absolute State Points)

When converting an individual temperature reading from one scale to another:

$$T_{^\circ\text{C}} = T_{\text{K}} - 273.15$$ $$T_{\text{K}} = T_{^\circ\text{C}} + 273.15$$ $$T_{^\circ\text{F}} = \frac{9}{5} T_{^\circ\text{C}} + 32 = 1.8 T_{^\circ\text{C}} + 32$$ $$T_{^\circ\text{C}} = \frac{5}{9} (T_{^\circ\text{F}} - 32)$$ $$T_{^\circ\text{R}} = T_{^\circ\text{F}} + 459.67 = \frac{9}{5} T_{\text{K}}$$

The Critical Metrological Distinction: Temperature Intervals vs. Points

When a specification states that an environmental chamber must maintain a stability of $\pm 1.00^\circ\text{C}$, this is an interval (difference between two readings), not a state point:

$$\Delta T_{\text{K}} = \Delta T_{^\circ\text{C}}$$ $$\Delta T_{^\circ\text{F}} = \frac{9}{5} \Delta T_{^\circ\text{C}} = 1.8 \times \Delta T_{^\circ\text{C}}$$ $$\Delta T_{^\circ\text{C}} = \frac{5}{9} \Delta T_{^\circ\text{F}} = \frac{\Delta T_{^\circ\text{F}}}{1.8}$$

• A tolerance of $\pm 0.50^\circ\text{C}$ equals $\pm 0.50\text{ K}$ and $\pm 0.90^\circ\text{F}$.
• If a technician incorrectly adds $32$ to the interval, they would compute an absurd tolerance of $\pm 32.9^\circ\text{F}$!

Plane angles and degrees

The coherent SI unit of plane angle is the radian. The degree is a non-SI unit accepted for use with the SI: $1^\circ=\pi/180\text{ rad}$. Thus $90^\circ=\pi/2\text{ rad}$. A degree contains sixty arcminutes, and an arcminute contains sixty arcseconds. Use radians in small-angle formulas such as displacement approximately equal to offset times angle; inserting a number in degrees without conversion produces a large error. The blueprint names degree among the unit skills to apply, but that does not make degree a coherent SI unit.