The Dimensional Analysis Calculator is a free online tool that analyses the dimensions for two given physical quantities. STUDYQUERIES’S online dimensional calculator tool makes the calculation faster, and it analyses the two physical quantities in a fraction of seconds.

**How to Use the Dimensional Analysis Calculator?**

To use the Dimensional Analysis calculator, follow these steps:

**Step 1:**Enter two physical quantities in their respective input fields**Step 2:**To receive the analysis, click “Submit”**Step 3:**In the new window, you will see the dimensional analysis

Dimensional Analysis Calculator

**A Step by Step Guide to Dimensional Analysis**

The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers of symbols) representing the base quantities. The table lists the base quantities and the symbols used for their dimension.

For example, a measurement of length is said to have dimension $$[L]\ or\ [L^1]$$, a measurement of mass has dimension $$[M]\ or\ [M^1]$$, and a measurement of time has dimension $$[T]\ or\ [T^1]$$

Like units, dimensions obey the rules of the algebra. Thus, the area is the product of two lengths and so has dimension $$[L^2]$$, or length squared. Similarly, volume is the product of three lengths and has dimension $$[L^3]$$, or length cubed. Speed has dimension length over time, $$\left[\frac{L}{T} \right]\ or\ [L^1T^{-1}]$$ Volumetric mass density has dimension $$\left[\frac{M}{L^3} \right]\ or\ [M^1L^{–3}]$$, or mass over length cubed. In general, the dimension of any physical quantity can be written as

$$\left[{M^a}{L^b}{T^c}{I^d}{\Theta^e}{N^f}{J^g} \right]$$

for some powers a, b, c, d, e, f, and g. We can write the dimensions of a length in this form with a = 1 and the remaining six powers all set equal to zero:

$$\left[M^1\right] = \left[{M^1}{L^0}{T^0}{I^0}{\Theta^0}{N^0}{J^0}\right]$$

Any quantity with a dimension that can be written so that all seven powers are zero $$(that\ is\, its\ dimension\ is\ \left[{M^1}{L^0}{T^0}{I^0}{\Theta^0}{N^0}{J^0}\right] )$$ is called dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless quantities pure numbers.

$$Base\ Quantity\longrightarrow Symbol\ for\ Dimension$$

$$Length\longrightarrow [L]$$

$$Mass\longrightarrow [M]$$

$$Time\longrightarrow [T]$$

$$Electric\ Current\longrightarrow [I]$$

$$Thermodynamic\ Temperature\longrightarrow [\Theta]$$

$$Amount\ of\ Substance\longrightarrow [N]$$

$$Luminous\ Intensity\longrightarrow [J]$$

Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity.

For example, if r is the radius of a cylinder and h is its height, then we write [r] = L and [h] = L to indicate the dimensions of the radius and height are both those of length, or L. Similarly if we use the symbol A for the surface area of a cylinder and V for its volume, then [A] = L² and [V] = L³. If we use the symbol m for the mass of the cylinder and ρ for the density of the material from which the cylinder is made, then [m] = M and [ ρ ] = ML−3.

Dimension plays a vital role in mathematics since any equation relating to physical quantities must be dimensionally consistent, which means that it must obey the following rules:

- The terms of an expression must all have the same dimensions; one cannot add or subtract quantities of different dimensions (think of the old saying, “You can’t add apples and oranges”). The expressions on both sides of equality in an equation must have the same dimensions.
- The arguments of any standard mathematical functions such as trigonometric functions (such as sine and cosine), logarithms, or exponential functions must be dimensionless. The functions take pure numbers as inputs and return pure numbers as outputs.

**How to Write a Literary Analysis Essay**

An equation that violates either of these rules cannot possibly be a true statement of physical law because it cannot be dimensionally consistent. You can use this simple fact to check for typos or algebraic mistakes, to help remember the various laws of physics, and even to propose the form that new laws of physics might take. You will undoubtedly learn about this last use of dimensions later in your academic career since it is beyond the scope of this text.

**Using Dimensions to Remember an Equation**

Suppose we need the formula for the area of a circle for some computation. Like many people who learned geometry too long ago to recall with any certainty, two expressions may pop into our minds when we think of circles: $$\pi r^2\ and\ 2\pi r$$ One expression is the circumference of a circle of radius r and the other is its area. But which is which?

**Strategy**

Searching for information from a reputable source could take some time, but this would be a natural strategy. Furthermore, even if we trust the source, we shouldn’t believe everything we read. I like that I can double-check something just by thinking about it.

We might also be in a situation where we cannot look things up (such as during a test). By using the fact that dimensions follow algebraic rules, the strategy is to find the dimensions of both expressions. Either expression cannot possibly be the correct equation for the area of a circle if it does not have the same dimensions as the area.

**Solution**

We know the dimension of the area is $$[L^2]$$ Now, the dimension of the expression πr² is

$$[\pi r^2]=[π]⋅[r]^2=1⋅L^2=L^2$$

since the constant π is a pure number and the radius r is a length. Therefore, πr² has the dimension of the area. Similarly, the dimension of the expression 2πr is

$$[2\pi r]=[2]⋅[\pi]⋅[r]=1⋅1⋅L=L$$

since the constants 2 and π are both dimensionless and the radius r is a length. We see that 2πr has the dimension of length, which means it cannot possibly be an area.

We rule out 2πr because it is not dimensionally consistent with being an area. We see that πr² is dimensionally consistent with being an area, so if we have to choose between these two expressions, πr² is the one to choose.

**Significance**

It may seem like a silly example, but these ideas are very general. When we know the dimensions of the physical quantities in an equation, we can check whether it is dimensionally consistent. Alternatively, since true equations are dimensionally consistent, we can match expressions from imperfect memories to quantities for which they might be expressions.

While this will not help us remember dimensionless factors that appear in the equations (for example, if you accidentally conflated the two expressions from the example into 2πr²), it will help us remember the correct basic form of the equation.

**Conversion Ratios**

It is possible to make simple unit conversions mentally or with a simple one-step multiplication or division. There are, however, more complex conversions that require several conversions between numerous units.

A conversion ratio (or unit factor) is a ratio equal to one. This ratio carries the names of the units to be used in the conversion. It can be used for conversions within the English and Metric Systems, as well as for conversions between the systems. The conversion ratio is based upon the concept of equivalent values. In the example below, one foot is substituted for its equivalent measure of 12 inches.

$$\frac{12\ Inches}{12\ Inches}=\frac{1\ Foot}{12\ Inches}$$

The main idea in Dimensional Analysis is to create a conversion ratio (unit factor) that has the units you want in the numerator and the units you already have in the denominator. It may be necessary to multiply by more than one conversion ratio in more difficult problems. Remember that you are setting up for one, or more, of the units to cancel until only the desired units remain. ALL conversion ratios (unit factors) must equal one.

$$Units\ You\ Have\times \frac{Units\ You\ Want}{Units\ You\ Have}=Units\ You\ Want$$

**Unit Conversion**

We use conversion factors to get the same units when performing dimensional analysis, which is also called Factor Label Method or Unit Factor Method. How many meters does it take to make 3 kilometers? Let’s use an example to help you better understand the statement.

We know that 1000 meters make 1 km, Therefore, 3 km = 3 × 1000 meters = 3000 meters

Here, the conversion factor is 1000 meters.

**Using Dimensional Analysis to Check the Correctness of Physical Equation**

Let’s say that you don’t remember whether

**time = speed/distance, or**

**time = distance/speed**

We can check this by making sure the dimensions on each side of the equations match.

Reducing both the equations to their fundamental units on each side of the equation, we get

$$[T]=\left[\frac{LT^{-1}}{L}\right]=[T^{-1}]\ (Wrong)$$

$$[T]=\left[\frac{L}{LT^{-1}}\right]=[T^{1}]\ (Right)$$

However, it should be kept in mind that dimensional analysis cannot help you determine any dimensionless constants in the equation.

**Homogeneity Principle of Dimensional Analysis**

According to Homogeneity, each term in a dimensional equation should have the same dimensions on both sides. Our ability to convert units from one form to another is made possible by this principle. Let’s see an example to better understand the concept:

Check the correctness of the physical equation $$s=ut+\frac{1}{2}at^2$$ In the equation, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time in which change occurs.

We know that $$L.H.S = s\ and\ R.H.S = ut+\frac{1}{2}at^2$$

The dimensional formula for the L.H.S can be written as $$s = [L^1M^0T^0]\longrightarrow(1)$$

We know that R.H.S is ut + ½ at² , simplifying we can write R.H.S as [u][t] + [a] [t]²

$$R.H.S=[L^0 M^0 T^{-1}][L^0 M^0 T^1] +[L^1 M^0 T^{-2}][L^0 M^0 T^2]$$

$$=[L^1M^0T^0]\longrightarrow(2)$$

From (1) and (2), we have [L.H.S] = [R.H.S]

Hence, by the principle of homogeneity, the given equation is dimensionally correct.

**Applications of Dimensional Analysis**

Real-life physics relies on dimension analysis as a fundamental aspect of measurement. There are three major reasons for using dimensional analysis:

- Checking the consistency of a dimensional equation
- To derive the relationship between physical quantities in physical phenomena
- Changing units from one system to another

**Limitations of Dimensional Analysis**

Dimensional analysis has the following limitations:

- It does not provide information about the dimensional constant.
- Formulas containing trigonometric functions, exponential functions, logarithmic functions, etc. cannot be derived.
- The function doesn’t indicate whether a physical quantity is a scalar or a vector.

**FAQs**

**What is dimensional analysis?**

A method of analysis in which physical quantities are expressed in terms of their fundamental dimensions is often used when there is not enough information to set up precise equations.

**What is the principle of Homogeneity of dimension?**

The principle of homogeneity states that the dimensions of each of the terms of a dimensional equation on both sides are the same.

**How many kilograms go into a gram?**

1 gram (g) is equal to 0.001 kilograms (kg).

**What is the purpose of dimensional analysis?**

dimensional analysis, a technique used in the physical sciences and engineering to reduce physical properties, such as acceleration, viscosity, energy, and others, to their fundamental dimensions of length (L), mass (M), and time (T).