Dimensional Analysis Calculator

Created by Purnima Singh, PhD
Reviewed by Steven Wooding
Last updated: Oct 03, 2022

Use the dimensional analysis calculator to convert between different units. You can also use this calculator to compare two different values of a physical quantity.

Have you ever wondered how many seconds are in 1 year or how the radius of a tennis ball compares to the radius of the earth? If you have, you have stumbled into the right place!

Even if you don't know what dimensional analysis is and what it is used for, worry not! We have got you covered too. Continue reading this article to know the definition of dimensional analysis and the examples of its application. You will also learn how to do dimensional analysis for converting from one system of units to another.

If you simply want to convert between imperial and metric units of length, our length converter might come in handy!

What is dimensional analysis? – Dimensional analysis definition

Dimensional analysis (or unit analysis) is a method in which we analyze physical quantities in terms of their units or dimensions. Using this process, we can convert between different units of a physical quantity. To do so, we need to find the conversion factor, which can be multiplied by the measurement in one system of units to convert it to other systems of units.

The dimensional analysis method is also known as the unit factor method or the factor label method.

Before going any further, let us first understand what units are.

Base and derived units

To express the value of physical quantities, a unit is necessary. How else would we tell the shopkeeper how much sugar, flour, oil, etc., we need? However, the number of physical quantities that we measure is vast! Imagine how difficult it would be for us to remember all of them if we were to define a separate unit for each.

Fortunately, we only need a small set of units to express most of the physical quantities. These units are called base units or fundamental units. We call the physical quantities that are represented by these base units as the seven dimensions of the physical world.

These seven physical quantities, their base units (in SI), and dimensions are:

• Length: $\text m$ (meter) [$\text L$];
• Mass: $\text{kg}$ (kilogram) [$\text M$];
• Time: $\text s$ (second) [$\text T$];
• Electric current: $\text A$ (Ampere) [$\text I$];
• Temperature: $\text K$ (Kelvin) [$\Theta$];
• Amount of substance: $\text {mol}$ (mole) [$\text N$]; and
• Luminous intensity: $\text {cd}$ (candela) [$\text J$].

Using these base units, we can express other quantities as well. For example, the area is calculated as a product of the length of sides, and hence, we can express it in units of $\text m^2$ and dimensions of [$\text L^2$]. The units that can be expressed in terms of base units are called derived units.

How to write the formula for dimensional analysis?

To use dimensional analysis, we need to write the dimensional formula for any physical quantity. As we can express most of the physical quantities in mechanics in terms of the dimensions [$\text L$], [$\text M$] and [$\text T$], it is a common practice to use these three dimensions while writing the dimensional formula.

For example, although we can express area as [$\text L^2$], we can also write its dimensional formula as [$\rm {M^0\ L^2\ T^0}$]. Since the area is independent of mass and time, we will say that it has zero dimensions in mass and time.

Some commonly used derived units and their dimensions are:

• Area: $\text m^2$ [$\rm {M^0\ L^2\ T^0}$];

• Volume: $\text m^3$ [$\rm {M^0\ L^3\ T^0}$];

• Velocity: $\text {m/s}$ [$\rm {M^0\ L^1\ T^{-1}}$];

• Acceleration: $\rm {m/s^2}$ [$\rm {M^0\ L^1\ T^{-2}}$];

• Density: $\rm {g/cm^3}$ [$\rm {M^1\ L^{-3}\ T^{0}}$];

• Force: $\rm {kg \cdot m/s^2}$ or $\rm N$ (newton)[$\rm {M^1\ L^{1}\ T^{-2}}$]; and

• Energy or work: $\rm {N \cdot m}$ or $\rm J$ (joule) [$\rm {M^1\ L^{2}\ T^{-2}}$].

How to do the dimensional analysis? – Understanding dimensional analysis equations

To do dimensional analysis, we should keep the following instructions in mind:

• Express a physical quantity in terms of the dimensions of the base quantities: For example, when we measure distance, we express it in the dimension of length, i.e., $\text L$ or $\rm {L^1}$. We can express the dimension of time as $\rm {T}$ or $\rm {T^1}$. Since speed is length over time, the dimensional formula of speed will be $\rm {L^1\ T^{-1}}$, and we can express the dimensional equation for speed as:

$\rm {[v]} = \rm {[L^1\ T^{-1}]}$

• Perform algebraic operations: Dimensions follow the rules of algebra, and we can add, subtract, multiply or divide them. For example, volume is the product of three lengths, and it has the dimensions $\rm L \cdot \rm L \cdot \rm L=\rm {L^3}$.

• Make sure equations are dimensionally consistent: We can add or subtract the magnitudes of physical quantities only when they have the same dimensions. Thus, we can't add density to mass or subtract length from time. This elementary principle is instrumental in checking if an equation is correct or not. If the dimensions of all the terms in an equation are not the same, it is wrong.

In the next section, we will use dimensional analysis for conversion from one system of units to another.

Dimensional analysis example

As an example, let us convert $\text{1 N}$ into $\text{dyne}$. We know that newton is the S.I. (mks – meter-kilogram-second) unit of force and dyne is the cgs (centimeter-gram-second) unit of force.

1. We will express the dimensional formula for force as:

$\rm {F} = \rm {[M^1\ L^1\ T^{-2}]}$

2. For conversion between mks system to cgs system, we can write the dimensional analysis equation:

$\rm n_1 \cdot \rm {[M_1^1\ L_1^1\ T_1^{-2}]} = n_2 \cdot \rm {[M_2^1\ L_2^1\ T_2^{-2}]}$

where:

• $\rm n_1 = 1\ N$, $\rm M_1 = 1\ kg$, $\rm L_1 = 1\ m$, and $\rm T_1 = 1\ s$, i.e., the units in S.I. system or the mks system.
• $\rm n_2 = ?\ dyn$, $\rm M_2 = 1\ g$, $\rm L_2 = 1\ cm$, and $\rm T_2 = 1\ s$, i.e., the units in the cgs system.
3. To determine the value of $\rm{n_2}$, we will take the ratio:

$\rm n_2 = n_1 \cdot [\frac{M_1}{M_2}]^1\ [\frac{L_1}{L_2}]^1\ [\frac{T_1}{T_2}]^{-2}$

4. Now, we will substitute the given values in the above equation:

$\rm n_2 = 1 \cdot [\frac{1\ kg}{1\ g}]^1\ [\frac{1\ m}{1\ cm}]^1\ [\frac{1\ s}{1\ s}]^{-2}$

$\rm n_2 = 1 \cdot [\frac{1000\ g}{1\ g}]^1\ [\frac{100\ cm}{1\ cm}]^1\ [\frac{1\ s}{1\ s}]^{-2}$

$\rm n_2 = 10^5$

Hence, $\rm 1\ N = 10^5\ dyn$

How to use the dimensional analysis calculator to solve dimensional analysis problems?

Let us see how we can use the dimensional analysis calculator (or unit analysis calculator) to convert the value of $g$ (acceleration due to gravity) from $\rm {m/s^2}$ to $\rm {km/h^2}$.

1. Choose the option convert units.

2. Using the drop-down menu, choose between the base and derived units. For this example, select "Derived units".

3. Select acceleration under the physical quantity section.

4. Enter the value and unit of the physical quantity, i.e., 9.8 and $\rm {m/s^2}$.

5. In the unit conversion section, choose your desired unit, for example, $\rm {km/h^2}$.

6. The unit analysis calculator will convert the value of g from $9.8\ \rm {m/s^2}$ to $127,008\ \rm {km/h^2}$.

7. You can also compare the dimensions of two measurements by using the dimensional analysis calculator. We recommend you cross-check that the Earth's radius ($6378\ \rm km$) is $1.8759 \times 10^8$ times larger than the radius of a tennis ball ($3.4\ \rm cm$)!

Applications of dimensional analysis

We can apply dimensional analysis in physics, engineering, mathematics, medicine, and chemistry problems. Some common applications of dimensional analysis are:

• Conversion of one system of units into another – You can use our dimensional analysis calculator for this purpose.

• Checking the formulae for correctness – By checking the equations for dimensional consistency, we can verify if a given formula is correct or not. For example, let us consider the equation for the uniform motion:

$s = vt$

Here $s$ is the displacement, i.e., it has dimensions of length [$\rm {L}$], $v$ is the velocity [$\rm {L\ T^{-1}}$] and $t$ is the time [$\rm {T}$]. Using the dimensional formula of these physical quantities in the R.H.S of the above equation, we get:

$\rm {L\ T^{-1}} \cdot T= L$

$\implies \rm L.H.S = R.H.S$

Hence, the equation is dimensionally consistent and correct.

• Derivation of formulae – We can also derive the formula of a physical quantity by checking for dimensional consistency.

FAQ

What is dimensional analysis used for?

Dimensional analysis is mostly used for converting a physical quantity from one unit to another. But we can also use it to verify various formulae and equations.

Use dimensional analysis to find how many minutes are in 180 days?

We know that there are 60 minutes in 1 hour and 24 hours in 1 day. Hence, to find out how many minutes are in 180 days using dimensional analysis, proceed as follows:

1. Multiply 60 by 24 to determine the conversion factor (the number of minutes in 1 day), i.e., 1440.
2. Multiply the conversion factor by 180 to find the number of minutes in 180 days.
3. Congrats! You have determined the number of minutes in 180 days is 259,200 minutes.

What are the limitations of dimensional analysis?

The limitations of dimensional analysis are:

• It doesn't give us any information about dimensionless constants like π.
• We cannot use it for deriving formulae with trigonometric functions, log functions, etc.
• It does not give us the exact form of relation.
• It does not tell us whether a physical quantity is a scalar or vector.

What is the dimensional formula of power?

[M¹ L² T⁻³], as power is work done per unit time, i.e., [M¹ L² T⁻²]/[T]. The dimensional formula for work, [M¹ L² T⁻²], comes from the newton, which has units kg•m/s².