Dimensional analysis: Difference between revisions

As an example of the usefulness of the first approach, suppose we wish to calculate the [[trajectory#Range and height|distance a cannonball travels]] when fired with a vertical velocity component <math>V_v_\mathrmtext{y}</math> and a horizontal velocity component {{tmath|V_v_\mathrmtext{x} }}, assuming it is fired on a flat surface. Assuming no use of directed lengths, the quantities of interest are then {{mvar|R}}, the distance travelled, with dimension L, {{tmath|V_v_\mathrmtext{x} }}, {{tmath|V_v_\mathrmtext{y} }}, both dimensioned as T⁻¹L, and {{mvar|g}} the downward acceleration of gravity, with dimension T⁻²L.

: <math>R \propto V_v_\text{x}^a\,V_v_\text{y}^b\,g^c .\,</math>

: <math>\mathsf{L} = \left(\frac{\mathsf{LT}^{-1}{\mathsf{T}L}\right)^{a+b} \left(\frac{\mathsf{LT}^{-2}{\mathsf{T}^2L}\right)^c\,</math>

from which we may deduce that <math>a + b + c = 1</math> and {{tmath|1=a + b + 2c = 0}}, which leaves one exponent undetermined. This is to be expected since we have two fundamental dimensions T and L, and four parameters, with one equation.

However, if we use directed length dimensions, then <math>V_v_\mathrm{x}</math> will be dimensioned as T⁻¹L_{{math|x}}, <math>V_v_\mathrm{y}</math> as T⁻¹L_{{math|y}}, {{mvar|R}} as L_{{math|x}} and {{mvar|g}} as T⁻²L_{{math|y}}. The dimensional equation becomes:

\left(\frac{\mathsf{LT}_\mathrm^{x-1}}{\mathsf{TL}_\mathrm{x}}\right)^a

\left(\frac{\mathsf{LT}_\mathrm^{y-1}}{\mathsf{TL}_\mathrm{y}}\right)^b

\left(\frac{\mathsf{LT}_\mathrm^{y-2}}{\mathsf{TL}_\mathrm{y}^2}\right)^c