What is Dimensional Analysis? Know More About it Principle, Example, Applications and Limitations

When we conduct and try to solve experiments, we come across numerical of different units. In order to solve numerical values which have different S.I. unit values, we need to convert them from one form to another to come to a solution. This conversion method is called dimensional analysis. 

In this article, we will discuss dimensional analysis example.

Definition of Dimensional Analysis:

The conversion of one set of units to another form presenting the same physical quantity is dimensional analysis. For example, if we have to measure meters and require to show it in centimetres, we must be aware of the dimensional factor, which is also called as unit conversion factor or level factor method. 

Meter and centimetre represent one physical quantity called length. One more example can be the conversion of units from gram to kilogram. Both represent the same physical quantity called mass. Dimensional analysis helps measure three broad categories like mass (M), length (L) and time (T).

Dimensional Analysis Example:

Q.1 Let us convert 25.6 meters into centimetres. 

Ans: 1 meter = 100 centimeters

Unit factor = 1 = 100cms/1meter

25.6 meters = 25.6 meters *1 meter

25.6 metes = 25.6 meter  * 100 centimetre/1meter

25.6 meters = 25.6 meters * 100 centimeters.

25.6 meters = 2560 meters.

The most important factor in solving dimensional analysis is that they must represent the same physical quantity. Hence it is rightly called the conversion method.

Q.2. Convert 1.50 cubic kilometres to litres. 

Here, we’ll need to use the compound conversion method. It is important to convert a set of units to which it is directly linked. Hence double conversion is required. We know that kilometre is directly converted to meter, we’ll also find another unit connected to litres.

Ans: 1 liter = 1 cubic decimeter.

1 meter = 10 decimeters

1 kilo meter = 1000 meters

1.50 cubic kilometer = 1.50 cubic kilometer * 1 (1= 1000 meteres/1km)

Since we have a cubic value or any exponential value or power to the given unit, we can take power on the other side of the conversion factor.

1.50 cubic kilometer = 1.50 km3 * (103m/1km)3 unit factor 1.50 cubic kilometer = 1.50 km3 * (109m3/ 1 km3)

1.50 cubic kilometer = 1.50 * 109m3

We are still to convert 1.50 * 109m3 to cubic decimetre

1 meter = 10 decimeter

The unit factor is 1 = 10 dm/1m

1 = 1.50 * 109m3*(10dm/1m)3

1 = 1.50 * 109m3*(103dm3/1m3)

1 =  1.50 * 109103*dm3

Since the base is same, the powers can be added

1 = 1.50 * 1012 dm3

1 dm = 1 Liter = 1.50*1012Liters

1.50 km3 = 1.50*1012L

Q.3. A piece of metal is 3 inches long. What is its length in centimetres?

Ans: 1 inch = 2.54 cm

Unit factor 1 = 2.5cm/1 inch

3 inches = 3*inches*2.54cm/1inch

3 inches = 7.62 cm

There are seven fundamental dimensions:

Name	Symbol	SI unit
Mass	M	Kg
Length	L	Meter
Time	T	Seconds
Temperature	O	Kelvin
Amount of substance	N	Moles
Light intensity	J	Candela
Electric current	I	Amper

Derived dimensions:

Secondary dimensions are a combination of a few-dimensional quality put together. Also called a derived quantity. 

When we look at the formula for Force:

F = m*a, Where ‘F’ is force, ‘m’ is mass and ‘a’ is acceleration. From the above table, it is clear that mass is a fundamental dimension. Still, acceleration is a derived quantity since we arrive at an answer after using a combination of fundamental dimensions.  

When we look at formulas like these, the power of fundamental dimensions (M, L, T etc.) should be the same of both the right-hand and left-hand sides of the equation for it to be a dimensionally homogeneous equation. The dimensional formula of the physical quantity does not depend on the system units.

F = m*a

Acceleration ‘a’ = m/s = L/T2 = L*T-2

Therefore, F = m * L*t-2

F = MLT-2

Application of dimensional analysis:

1. To convert a physical quantity from one system of the unit to another.
2. To check the dimensional correctness of various formulae.
3. To derive a relationship between different physical quantities.
4. Checking the dimensional consistency of the equation.
5. Deducing relation among the physical quantities.

Limitations of dimensional analysis:

1. It does not give information about the dimensionless constant in the formula.
2. If any quantity depends on three or more factors, having dimension, the formula cannot be derived. We can obtain three equations from which only three unknown dimensions can be calculated by equating the power of M, L and T on both sides of the dimensional equation.
3. The formula cannot be derived which contain trigonometric functions, exponential functions, log functions etc., which have no dimensions.
4. The dimension method cannot derive the exact form of relationship when it includes more than one part of any side.
5. It does not give any information on whether it is a scalar or a vector quantity.

Conclusion 

In engineering and science, the dimensional analysis serves a major purpose in identifying base quantities. This is how two different quantities can be measured.