Dimensional Analysis
Derived quantities versus fundamental quantities
To measure physical quantities we use two types of units: fundamental units and derived units. There are innumerous derived units, but the number of fundamental units is kept to minimum. In Mechanics there are three fundamental units: unit of length [L], unit of time [T], and unit of mass [M]. To measure the fundamental units we use scales called standards. Standards may be different: meter or foot, second or year, kilogram or slag. Expression of a derived unit versus fundamental units is called dimensionality.
The scales can be different but the relationships between the quantities must remain the same: the ratio of two numerical values of the same quantity must not depend on the scale used in their measurement. For instance, the ratio of two areas, e.g. room floor and parking lot, does not depend on whether we measure them in square meters or squatter feet. Hence, dimension of a derived physical quantity must be a power term of the fundamental units:
[D]=[L]^m [T]^n [M]^k
Indeed, let’s consider a quantity D=Lm, e.g. area D of a square with a side L. Then, the ratio of two squares is D1/D2=(L1/L2)m. Now let us increase the scale of L by  times (going from meters to feet 3); then D1()/D2()=(L1/L2)m =D1/D2. That would not be the case if, e.g. D=e^L. For instance, force is a derived unit; its dimensionality is determined as: [F]=[M][L][T]2.
Equation verification
All equations must be consistent with respect to dimensions. That is, the dimensions of the physical quantities on the left hand side of an equation must be equal to the dimensions of the physical quantities on the right hand side. For instance, the dimension of physical work is [W]=[M][L]2[T]2 . Let’s verify if the laymen’s definition of physical work as “force times distance” is proper.
[F][L]=[M][L][T]2[L] =[M][L]2[T]2= [W] 
Relationship between physical quantities
Dimensional analysis can also be used to obtain relationships between physical quantities without a thorough analysis of the phenomenon. Let us consider as an example period of a simple pendulum. A simple pendulum has its mass concentrated at a point (called the center of mass). Although the definition of a point is a position without dimension, the length of the string supporting the mass (the bob) must be large in comparison with the diameter of the bob. The question before us is: How does its period P depend on the physical parameters of the pendulum? We may speculate that the period of a simple pendulum depends upon the length of the pendulum L, the mass of the pendulum bob m, and the magnitude of the bob’s displacement from the position of equilibrium h.

Let us try the formula:
P=L^m m^n h^k
There are three problems with this formula:
[P]=[L]/[T] but there is no [M] in the left hand side
There is no [T] in the right hand side
[h]=[L], then how do we combine h with L?
Problem 1 means that P is independent of m; this is an important conclusion of the Dimensional Analysis. Problem 2 means that we have to be looking for another quantity that affects motion of a pendulum. Such quantity is gravity and we should include acceleration due to gravity g into the list. Now the formula looks like that:
P=L^m g^n h^k
Actually, Problem 3 is the hardest. It is not possible to resolve the issue of h. This requires experimental study, which shows that the bob’s displacement from the equilibrium does not affect the period unless the angular displacement is too large. Finally we try the formula:
P=L^m g^n
Taking into account that [g]=[L] [T]2 the only possibility to satisfy this is the formula
This is a great formula, which tells us a lot about motion of a pendulum. The exact formula has a coefficient 2, which can be found experimentally.
Thus, a combination of Dimensional Analysis and Experiment is a powerful scientific tool.