Formula Parser & Unit Checker
Check physical formulas for unit consistency and dimensional homogeneity. Ensure your equations make physical sense.
Formula
[LHS] = [RHS]\ne.g., [L/T] = [L] / [T]
Dimensional Homogeneity states that an equation is only physically valid if every term has the same physical dimensions (e.g., you cannot add meters to seconds).
Worked Examples
Example 1: Velocity - The Basic Consistency Check
Problem:Check that v = d / t is dimensionally sound, where v is in metres per second, d in metres and t in seconds.
Solution:Assign base dimensions - L for length, T for time, M for mass: v has dimensions [L T^-1] d has dimensions [L] t has dimensions [T] Evaluate the right-hand side: [d] / [t] = [L] / [T] = [L T^-1] Compare with the left-hand side: [L T^-1] = [L T^-1] - consistent. Why this survives a change of units: the same check holds whether you work in metres per second or miles per hour, because dimensions describe the kind of quantity while units describe the scale. That is exactly Fourier's principle - a physical law cannot depend on the yardstick you chose.
Result:Consistent. [L T^-1] on both sides, in any unit system.
Example 2: Catching a Real Error - Energy Versus Momentum
Problem:A draft page states kinetic energy as E = 1/2 m v instead of E = 1/2 m v^2. Both look plausible. Does dimensional analysis catch it?
Solution:Correct form, E = 1/2 m v^2: [M] x [L T^-1]^2 = [M] x [L^2 T^-2] = [M L^2 T^-2] That is the dimension of the joule - consistent with energy. The erroneous form, E = 1/2 m v: [M] x [L T^-1] = [M L T^-1] That is the dimension of momentum, the kilogram metre per second - a different physical quantity entirely. The check fails, and it fails loudly: the two sides differ by a factor of [L T^-1]. This is the single most valuable use of dimensional analysis, because a dropped exponent produces an answer that is numerically plausible and silently wrong. Note what the check cannot do: the leading 1/2 is dimensionless, so dimensional analysis would equally accept E = m v^2 or E = 7 m v^2. It validates structure, never coefficients.
Result:Caught. [M L T^-1] is momentum, not the [M L^2 T^-2] of energy - though the 1/2 coefficient is invisible to the method.
Example 3: A Dimensionless Group - The Reynolds Number
Problem:Verify that Re = rho x v x L / mu is dimensionless, where rho is density, v velocity, L a characteristic length and mu dynamic viscosity.
Solution:Dimensions of each term: rho (density, kg/m^3) = [M L^-3] v (velocity, m/s) = [L T^-1] L (length, m) = [L] mu (dynamic viscosity, Pa s) = [M L^-1 T^-1] Numerator: [M L^-3] x [L T^-1] x [L] = [M L^(-3+1+1) T^-1] = [M L^-1 T^-1] Divide by the denominator: [M L^-1 T^-1] / [M L^-1 T^-1] = [1] Re is dimensionless - every exponent cancels. That is what allows a scale model in a wind tunnel to predict full-size behaviour: match the Reynolds number and the flow regime transfers regardless of physical size. The method's blind spot: torque and energy both carry [M L^2 T^-2], yet a newton metre of torque and a joule of work are not interchangeable. Dimensional homogeneity is a necessary condition for a correct equation, never a sufficient one.
Result:Dimensionless, [1]. All exponents cancel - which is why Reynolds scaling works across model sizes.
Frequently Asked Questions
What is Dimensional Analysis?
A method to understand relationships between physical quantities by identifying their base dimensions (mass, length, time) and tracking these dimensions as calculations or comparisons are performed.
Can this prove a formula is correct?
No. It can only prove a formula is <em>incorrect</em> (if units mismatch). A consistent formula might still be wrong (e.g., missing a constant like 1/2).
What are the 7 base dimensions?
Mass (M), Length (L), Time (T), Current (I), Temperature (Θ), Amount of Substance (N), Luminous Intensity (J).
What is a dimensionless quantity?
A quantity with no physical dimension, usually a ratio (like angle in radians, or Mach number). Its dimension is 1.
Why did the Mars Climate Orbiter crash?
Lockheed Martin used Imperial units (lb-s) while NASA used Metric (N-s) for impulse. The numbers looked 'right' but the dimensions were wrong.
How do I denote dimensions?
Square brackets are common, e.g., [v] = L T⁻¹.
Can I take the log of a dimensioned number?
Strictly speaking, no. Transcendental functions (sin, cos, log, exp) arguments must be dimensionless.
What is F = ma in dimensions?
[F] = [m][a] -> Force = M * (L/T²).
Does this work for Imperial units?
Yes, dimensional analysis is unit-agnostic. L is L, whether it's measured in feet or meters.
How do significant figures affect unit conversions?
Your converted result should have the same number of significant figures as your original measurement. If you measure 5.2 inches (2 significant figures), converting to centimeters gives 13 cm, not 13.208 cm. Using excessive decimal places implies false precision.
Background & Theory
Base Dimensions
In the SI system, there are 7 base dimensions. Most mechanics problems use just 3:
- M: Mass (kg)
- L: Length (m)
- T: Time (s)
The Principle of Homogeneity
You can only add or subtract quantities with the same dimensions. (5 meters + 2 seconds is meaningless). You can multiply or divide different dimensions to create new ones (meters / seconds = velocity).
Common Dimensions
- Velocity: L/T (m/s)
- Acceleration: L/T² (m/s²)
- Force: M·L/T² (Newtons)
- Energy: M·L²/T² (Joules)
- Pressure: M/(L·T²) (Pascals)
Why check consistency?
If your derivation leads to an equation where the units don't match, you made an algebra error. It is a powerful "sanity check" for students and engineers.
History
Fourier's Contribution
The concept of Dimensional Analysis was formalized by Joseph Fourier in 1822 in his work Théorie analytique de la chaleur. He established that physical laws should be independent of the units used to measure them.
Buckingham π Theorem
In 1914, Edgar Buckingham formalized the "Pi Theorem", which states that any physically meaningful equation involving n variables can be rewritten as an equation of n-k dimensionless parameters. This is the foundation of fluid dynamics scaling (e.g., wind tunnel testing).
Modern Engineering
Today, automated dimensional analysis is built into tools like Mathematica and MATLAB to prevent "Mars Climate Orbiter" style disasters, where unit mismatches (Imperial vs Metric) caused the loss of a $327 million spacecraft.