Tools / Dimension solver
Dimension-exponent solver
Enter the dimensions of each variable. The solver reduces the dimension matrix exactly and returns the independent dimensionless groups (the Buckingham π theorem), or the powers of your variables that give a dimension you choose.
Start from an example
Base dimensions
Dimension matrix
Each column is one variable, each row one base dimension. Type its exponent: for a force, M = 1, L = 1, T = −2. Use whole numbers, decimals or fractions such as 1/2. Tick Repeat under the variables you want as repeating variables (usually a density, a speed and a length); with none ticked, the variables listed first are used. Add a subscript with an underscore, for example c_p.
| Dimension | |||||
|---|---|---|---|---|---|
| Repeat | |||||
| M Mass | |||||
| L Length | |||||
| T Time |
What to find
Dimensionless groups
- Π1 = F / (ρ U2 L2)exponents (−1, −2, −2, 1, 0) · A·x = 0 checked exactly
- Π2 = μ / (ρ U L)exponents (−1, −1, −1, 0, 1) · A·x = 0 checked exactly
Repeating variables: ρ, U, L.
Exponents are listed in the column order: ρ, U, L, F, μ.
These are one valid basis. Reciprocals, products and powers of the groups are equally valid. For example, μ/(ρUL) is the reciprocal of the Reynolds number ρUL/μ.
How it works
- Set up the matrix A. Each column holds the exponents of M, L, T (and any other base dimensions) for one variable.
- Solve Ax = 0. Exact row reduction with fractions gives the rank r. Each free column gives one vector x, and the product of the variables raised to the powers in x has no dimensions.
- Count the groups. There are n − r independent groups. Any product or power of them is also a valid group, so reciprocals and different bases are equally correct.
- Check. Multiplying A by each vector gives zero in every row, which the solver confirms exactly.
For a target dimension b the solver solves Ax = b instead. It reports a unique solution, a family of solutions, or that the dimensions are inconsistent, meaning no combination of these variables has that dimension.