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Engineering formulas & units

SI units, conversions and dimensions for engineers

The SI base and derived units, prefixes, the conversions engineers use every day, and how dimensional analysis catches mistakes in a formula.

On this page
  1. SI base units
  2. Derived units
  3. Prefixes
  4. Conversion table
  5. Worked examples
  6. Dimensional analysis
  7. Good habits

Most mistakes in engineering calculations are unit mistakes: a diameter in millimetres put into a formula that expects metres, or a pressure in psi read as bar. Working consistently in SI units, and checking the dimensions of every equation, removes most of them. Keep this page next to your calculations.

SI base units

Seven base units define everything else. Mechanical engineering uses the first five most.

QuantityUnitSymbolDimension
MasskilogramkgM
LengthmetremL
TimesecondsT
TemperaturekelvinKΘ
Electric currentampereAI
Amount of substancemolemolN
Luminous intensitycandelacdJ

Derived units

kgmassmlengthstimeNewton, Nforce = kg·m/s²Pascal, Papressure, stress = N/m²Joule, Jenergy, work = N·mWatt, Wpower = J/s÷ s
Every mechanical unit is built from a few base units. The newton comes from kilograms, metres and seconds; the pascal, joule and watt come from the newton.
QuantityUnitSymbolIn base unitsDimensions
ForcenewtonNkg·m/s²MLT⁻²
Pressure, stresspascalPaN/m²ML⁻¹T⁻²
Energy, work, heatjouleJN·mML²T⁻²
PowerwattWJ/sML²T⁻³
Torque, momentnewton metreN·mkg·m²/s²ML²T⁻²
FrequencyhertzHz1/sT⁻¹
Density—kg/m³kg/m³ML⁻³
Dynamic viscositypascal secondPa·skg/(m·s)ML⁻¹T⁻¹
Kinematic viscosity—m²/sm²/sL²T⁻¹
Angular velocity—rad/s1/sT⁻¹

Torque and energy share the same base units, but they are different quantities. Write torque as N·m and energy as J to keep them apart.

Prefixes

PrefixSymbolFactorExample
gigaG10⁹Young’s modulus of steel, 200 GPa
megaM10⁶Yield strength, 250 MPa
kilok10³Load, 30 kN
millim10⁻³Dimension, 25 mm
microµ10⁻⁶Tolerance or surface finish, 0.8 µm

A useful identity for stress: 1 MPa = 1 N/mm². If forces are in newtons and lengths in millimetres, stresses come out directly in MPa, and Young’s modulus must be in N/mm² too (steel: 200 000 N/mm²).

Conversion table

FromTo SINotes
1 inch (in)25.4 mmExact
1 foot (ft)0.3048 mExact
1 pound mass (lb)0.4536 kg
1 pound force (lbf)4.448 N
1 kilogram force (kgf)9.807 NStill found on older drawings and gauges
1 psi (lbf/in²)6.895 kPa1 ksi = 6.895 MPa
1 bar100 kPa1 bar ≈ 14.5 psi
1 standard atmosphere101.325 kPa
1 horsepower (hp)745.7 WMetric horsepower (PS) = 735.5 W
1 lbf·ft1.356 N·mTorque
1 kWh3.6 MJ
1 BTU1055 J
1 calorie4.184 J
1 rev/min2π/60 = 0.1047 rad/s
1 litre0.001 m³
1 US gallon3.785 L
1 km/h0.2778 m/sDivide by 3.6
1 centipoise (cP)1 mPa·sWater at 20 °C is about 1 cP
°F to °CT(°C) = (T(°F) − 32) / 1.8
°C to KT(K) = T(°C) + 273.15Temperature differences are equal in K and °C

Worked examples

Worked example 1: hydraulic pressure

A hydraulic pump is rated at 3000 psi. What is that in MPa and bar?

3000×6.895 kPa=20 685 kPa3000 \times 6.895\ \text{kPa} = 20\,685\ \text{kPa}
Answer20.7 MPa, or 207 bar

Worked example 2: power from torque and speed

A gearbox output shaft carries 200 N·m at 1500 rev/min. Find the power.

P=T ω=2πNT60=2π×1500×20060=31 400 WP = T\,\omega = \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 200}{60} = 31\,400\ \text{W}
AnswerP = 31.4 kW (42 hp)

Worked example 3: mixed units

A drawing gives a 1.5 in diameter steel pin. Find its area in mm² and the shear stress under a 2000 lbf single-shear load.

d=1.5×25.4=38.1 mm,A=π×38.124=1140 mm2d = 1.5 \times 25.4 = 38.1\ \text{mm}, \qquad A = \frac{\pi \times 38.1^2}{4} = 1140\ \text{mm}^2F=2000×4.448=8896 N,τ=88961140=7.8 N/mm2F = 2000 \times 4.448 = 8896\ \text{N}, \qquad \tau = \frac{8896}{1140} = 7.8\ \text{N/mm}^2
Answerτ = 7.8 MPa

Dimensional analysis

Every term in a correct equation has the same dimensions. Checking this takes a minute and catches missing factors, wrong powers and unit slips.

Worked example 4: is Bernoulli's equation consistent?

Check the three terms of p+12ρV2+ρgzp + \tfrac{1}{2}\rho V^2 + \rho g z.

  • Pressure pp: ML⁻¹T⁻²
  • ρV2\rho V^2: (ML⁻³)(L²T⁻²) = ML⁻¹T⁻²
  • ρgz\rho g z: (ML⁻³)(LT⁻²)(L) = ML⁻¹T⁻²
AnswerAll three terms are pressures, so the equation is dimensionally consistent

Dimensional analysis also builds dimensionless groups such as the Reynolds number, ρVD/μ\rho V D / \mu. Its dimensions are (ML⁻³)(LT⁻¹)(L) / (ML⁻¹T⁻¹), which cancel completely. Groups like this let a test on a small model predict the behaviour of a full-size part.

Good habits

  • Convert everything to base SI units (or to a consistent N–mm–MPa set) before you start.
  • Carry units through every line of a calculation, not only the answer.
  • Write the answer with sensible precision: three significant figures is usually enough.
  • On drawings, state the unit system once in the title block (for example “All dimensions in mm”) and stay with it.

Key points

  • Seven base units define all others; mechanical work uses kg, m, s and K most.
  • 1 MPa = 1 N/mm², so the N–mm system gives stresses in MPa directly.
  • Use absolute temperature (K) in gas laws and efficiency formulas.
  • Check that every term in an equation has the same dimensions.

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