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dimensional analysis

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• Units Of Measurement

Dimensional Analysis

We quantify the size and shape of things using Dimensional Analysis. It helps us study the nature of objects mathematically. It involves lengths and angles as well as geometrical properties such as flatness and straightness. The basic concept of dimension is that we can add and subtract only quantities with the same dimensions. Similarly, two physical quantities can be equal only if they have the same dimensions.

Dimensional Analysis Explained

The study of the relationship between physical quantities with the help of dimensions and units of measurement is termed dimensional analysis. Dimensional analysis is essential because it keeps the units the same, helping us perform mathematical calculations smoothly.

Unit Conversion and Dimensional Analysis

Dimensional analysis is also called Factor Label Method or Unit Factor Method because we use conversion factors to get the same units. To help you understand the stated better, let’s say you want to know how many metres make 3 km?

We know that 1000 metres make 1 km,

3 km = 3 × 1000 metres = 3000 metres

Here, the conversion factor is 1000 metres.

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 its fundamental units on each side of the equation, we get

• \(\begin{array}{l}[T]=\frac{[L][T]^{-1}}{L}=[T]^{-1}\,\,(Wrong) \end{array} \)
• \(\begin{array}{l}[T]=\frac{[L]}{[L][T]^{-1}}=[T]\,\,(Right) \end{array} \)

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

Read more like this:

Homogeneity Principle of Dimensional Analysis

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another. To better understand the principle, let us consider the following example:

Example 1: Check the correctness of physical equation s = ut + ½ 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 + 1/2at 2

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

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

=[L 1 M 0 T 0 ]………..(2)

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

Applications of Dimensional Analysis

Dimensional analysis is a fundamental aspect of measurement and is applied in real-life physics. We make use of dimensional analysis for three prominent reasons:

• To check the consistency of a dimensional equation
• To derive the relation between physical quantities in physical phenomena
• To change units from one system to another

Limitations of Dimensional Analysis

Some limitations of dimensional analysis are:

• It doesn’t give information about the dimensional constant.
• The formula containing trigonometric function, exponential functions, logarithmic function, etc. cannot be derived.
• It gives no information about whether a physical quantity is a scalar or vector.

Additional Solved Problems

1. Check the correctness of the physical equation v 2 = u 2 + 2as 2 . Solution:

The computations made on the L.H.S and R.H.S are as follows:

L.H.S: v 2 = [v 2 ] = [ L 1 M 0 T –1 ] 2 = [ L 1 M 0 T –2 ] ……………(1)

R.H.S: u 2 + 2as 2

Hence, [R.H.S] = [u] 2 + 2[a][s] 2

Hence, by the principle of homogeneity, the equation is not dimensionally correct.

2. Evaluate the homogeneity of the equation when the rate flow of a liquid has a coefficient of viscosity η through a capillary tube of length ‘l’ and radius ‘a’ under pressure head ‘p’ given as \(\begin{array}{l}\frac{dV}{dt}=\frac{\pi p a^4}{8l\eta}\end{array} \) Solution:   \(\begin{array}{l} \frac{ dV }{ dt } = \frac{ \pi p ^ { 4 }}{ 8 l \eta } \end{array} \)   \(\begin{array}{l} [\textup{L.H.S}] = \frac{ [dV] }{[dt]} = \frac{[M^{0} L^{3} T^ {0}]}{[M^{0} T^{0} T^{1}]} = [M^{0}L^{3}T^{-1}] \textup{ …..(1)} \end{array} \)   \(\begin{array}{l} [\textup{R.H.S}] = \frac{[p] [a] ^ {4}}{[l] [\eta]} \end{array} \)   \(\begin{array}{l} \therefore [\textup{R.H.S}] = \frac{ [M ^ { 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 } L ^ { 1 } T ^ { 0 }]^ { 4 }} { [M ^ { 0 } L ^ { 1 } T ^ { 0 }] [M ^ { 1 } L ^ { -1 } T ^{ -1 }] } \end{array} \)   \(\begin{array}{l} = \frac{ [M^{ 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 }L ^ { 4 } T ^ { 0 } ] }{ [M^{ 1 }L^ { 0 }T^ { -1 }] } \end{array} \)   \(\begin{array}{l} = \frac{ [M^ { 1 } L ^ { 3 } T^ { 2 }] }{ [M^ { 1 } L^ { 0 } T ^ { -1 }] } = [M^ { 0 }L ^ { 3 }T^{ -1 }] \textup{ …….(2)} \end{array} \)  

Hence, by the principle of homogeneity, the given equation is homogenous.

Frequently Asked Questions – FAQs

What is dimensional analysis, state true or false: dimensional analysis can not be used to find dimensionless constants., state the principle of homogeneity of dimensions, why do we use dimensional analysis.

• To change units from one system to another.

What are the limitations of dimensional analysis?

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Can the SI unit of a physical quantity be obtained knowing its dimensional formula? Explain with an example.

Yes! The unit of a physical quantity can be obtained by its dimensional formula.

As an example, let us consider force.

By Newton’s second of law of motion, we know that force is given by the following equation:

The above equation can be rewritten as

Rewriting the above equation using basic dimensions, we get

F=(m(d/t))/t

F=(md)/(t^2)=M^1L^1T^(-2)

Substituting the measuring unit for these basic dimensions, we get

(kg.m)/(s^2)

The above unit is nothing but the Newton, the SI unit of force. Newton is equal to the force that would give a mass of one kilogram an acceleration of one metre per second per second

This is how knowing the dimensional formula helps us find the unit of the physical quantity.

Thank you so much for providing me with all this DATA.

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• 1.4 Dimensional Analysis
• Introduction
• 1.1 The Scope and Scale of Physics
• 1.2 Units and Standards
• 1.3 Unit Conversion
• 1.5 Estimates and Fermi Calculations
• 1.6 Significant Figures
• 1.7 Solving Problems in Physics
• Key Equations
• Conceptual Questions
• Additional Problems
• Challenge Problems
• 2.1 Scalars and Vectors
• 2.2 Coordinate Systems and Components of a Vector
• 2.3 Algebra of Vectors
• 2.4 Products of Vectors
• 3.1 Position, Displacement, and Average Velocity
• 3.2 Instantaneous Velocity and Speed
• 3.3 Average and Instantaneous Acceleration
• 3.4 Motion with Constant Acceleration
• 3.5 Free Fall
• 3.6 Finding Velocity and Displacement from Acceleration
• 4.1 Displacement and Velocity Vectors
• 4.2 Acceleration Vector
• 4.3 Projectile Motion
• 4.4 Uniform Circular Motion
• 4.5 Relative Motion in One and Two Dimensions
• 5.2 Newton's First Law
• 5.3 Newton's Second Law
• 5.4 Mass and Weight
• 5.5 Newton’s Third Law
• 5.6 Common Forces
• 5.7 Drawing Free-Body Diagrams
• 6.1 Solving Problems with Newton’s Laws
• 6.2 Friction
• 6.3 Centripetal Force
• 6.4 Drag Force and Terminal Speed
• 7.2 Kinetic Energy
• 7.3 Work-Energy Theorem
• 8.1 Potential Energy of a System
• 8.2 Conservative and Non-Conservative Forces
• 8.3 Conservation of Energy
• 8.4 Potential Energy Diagrams and Stability
• 8.5 Sources of Energy
• 9.1 Linear Momentum
• 9.2 Impulse and Collisions
• 9.3 Conservation of Linear Momentum
• 9.4 Types of Collisions
• 9.5 Collisions in Multiple Dimensions
• 9.6 Center of Mass
• 9.7 Rocket Propulsion
• 10.1 Rotational Variables
• 10.2 Rotation with Constant Angular Acceleration
• 10.3 Relating Angular and Translational Quantities
• 10.4 Moment of Inertia and Rotational Kinetic Energy
• 10.5 Calculating Moments of Inertia
• 10.6 Torque
• 10.7 Newton’s Second Law for Rotation
• 10.8 Work and Power for Rotational Motion
• 11.1 Rolling Motion
• 11.2 Angular Momentum
• 11.3 Conservation of Angular Momentum
• 11.4 Precession of a Gyroscope
• 12.1 Conditions for Static Equilibrium
• 12.2 Examples of Static Equilibrium
• 12.3 Stress, Strain, and Elastic Modulus
• 12.4 Elasticity and Plasticity
• 13.1 Newton's Law of Universal Gravitation
• 13.2 Gravitation Near Earth's Surface
• 13.3 Gravitational Potential Energy and Total Energy
• 13.4 Satellite Orbits and Energy
• 13.5 Kepler's Laws of Planetary Motion
• 13.6 Tidal Forces
• 13.7 Einstein's Theory of Gravity
• 14.1 Fluids, Density, and Pressure
• 14.2 Measuring Pressure
• 14.3 Pascal's Principle and Hydraulics
• 14.4 Archimedes’ Principle and Buoyancy
• 14.5 Fluid Dynamics
• 14.6 Bernoulli’s Equation
• 14.7 Viscosity and Turbulence
• 15.1 Simple Harmonic Motion
• 15.2 Energy in Simple Harmonic Motion
• 15.3 Comparing Simple Harmonic Motion and Circular Motion
• 15.4 Pendulums
• 15.5 Damped Oscillations
• 15.6 Forced Oscillations
• 16.1 Traveling Waves
• 16.2 Mathematics of Waves
• 16.3 Wave Speed on a Stretched String
• 16.4 Energy and Power of a Wave
• 16.5 Interference of Waves
• 16.6 Standing Waves and Resonance
• 17.1 Sound Waves
• 17.2 Speed of Sound
• 17.3 Sound Intensity
• 17.4 Normal Modes of a Standing Sound Wave
• 17.5 Sources of Musical Sound
• 17.7 The Doppler Effect
• 17.8 Shock Waves
• B | Conversion Factors
• C | Fundamental Constants
• D | Astronomical Data
• E | Mathematical Formulas
• F | Chemistry
• G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

• Find the dimensions of a mathematical expression involving physical quantities.
• Determine whether an equation involving physical quantities is dimensionally consistent.

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. Table 1.3 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 algebra. Thus, 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, L/T or LT –1 . Volumetric mass density has dimension M/L 3 or ML –3 , or mass over length cubed. In general, the dimension of any physical quantity can be written as L a M b T c I d Θ e N f J g L a M b T c I d Θ e N f J g for some powers a , b , c , d , e , f , a , b , c , d , e , f , and g . We can write the dimensions of a length in this form with a = 1 a = 1 and the remaining six powers all set equal to zero: L 1 = L 1 M 0 T 0 I 0 Θ 0 N 0 J 0 . L 1 = L 1 M 0 T 0 I 0 Θ 0 N 0 J 0 . Any quantity with a dimension that can be written so that all seven powers are zero (that is, its dimension is L 0 M 0 T 0 I 0 Θ 0 N 0 J 0 L 0 M 0 T 0 I 0 Θ 0 N 0 J 0 ) is called dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless quantities pure numbers .

Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity. For example, if r r is the radius of a cylinder and h h is its height, then we write [ r ] = L [ r ] = L and [ h ] = L [ 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 A for the surface area of a cylinder and V V for its volume, then [ A ] = L 2 and [ V ] = L 3 . If we use the symbol m m for the mass of the cylinder and ρ ρ for the density of the material from which the cylinder is made, then [ m ] = M [ m ] = M and [ ρ ] = ML −3 . [ ρ ] = ML −3 .

The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities must be dimensionally consistent , which means the equation must obey the following rules:

• Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities of differing dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions on each side of the equality in an equation must have the same dimensions.
• The arguments of any of the standard mathematical functions such as trigonometric functions (such as sine and cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions require pure numbers as inputs and give pure numbers as outputs.

If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope of this text, but is something you may learn later in your academic career.

Example 1.4

Using dimensions to remember an equation.

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

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

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

Significance

Check your understanding 1.5.

Suppose we want the formula for the volume of a sphere. The two expressions commonly mentioned in elementary discussions of spheres are 4 π r 2 4 π r 2 and 4 π r 3 / 3 . 4 π r 3 / 3 . One is the volume of a sphere of radius r and the other is its surface area. Which one is the volume?

Example 1.5

Checking equations for dimensional consistency.

• There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need only look at the dimensions of each term appearing in the equation. There are three terms, one in the left expression and two in the expression on the right, so we look at each in turn: [ s ] = L [ v t ] = [ v ] · [ t ] = LT −1 · T = LT 0 = L [ 0.5 a t 2 ] = [ a ] · [ t ] 2 = LT −2 · T 2 = LT 0 = L . [ s ] = L [ v t ] = [ v ] · [ t ] = LT −1 · T = LT 0 = L [ 0.5 a t 2 ] = [ a ] · [ t ] 2 = LT −2 · T 2 = LT 0 = L . All three terms have the same dimension, so this equation is dimensionally consistent.
• Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the dimensions of each of the three terms appearing in the equation: [ s ] = L [ v t 2 ] = [ v ] · [ t ] 2 = LT −1 · T 2 = LT [ a t ] = [ a ] · [ t ] = LT −2 · T = LT −1 . [ s ] = L [ v t 2 ] = [ v ] · [ t ] 2 = LT −1 · T 2 = LT [ a t ] = [ a ] · [ t ] = LT −2 · T = LT −1 . None of the three terms has the same dimension as any other, so this is about as far from being dimensionally consistent as you can get. The technical term for an equation like this is nonsense .
• This equation has a trigonometric function in it, so first we should check that the argument of the sine function is dimensionless: [ a t 2 s ] = [ a ] · [ t ] 2 [ s ] = LT −2 · T 2 L = L L = 1 . [ a t 2 s ] = [ a ] · [ t ] 2 [ s ] = LT −2 · T 2 L = L L = 1 . The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation: [ v ] = LT −1 [ sin ( a t 2 s ) ] = 1 . [ v ] = LT −1 [ sin ( a t 2 s ) ] = 1 .

The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.”

Check Your Understanding 1.6

Is the equation v = at dimensionally consistent?

One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t , we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t :

Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t :

By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.

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• Dimensional Analysis

What is Dimensional Analysis?

To solve the mathematical problems of physical quantities, it is important to have a brief knowledge of units and dimensions. The basic concept of dimensions is that only those quantities can be added or subtracted which have the same dimension. This concept helps us to derive relationships between physical quantities.

Dimensional analysis is the study of the relation between physical quantities based on their units and dimensions. It is used to convert a unit from one form to another. While solving mathematical problems, it is necessary to keep the units the same to solve the problem easily.

Do you know what is the significance of dimensional analysis? Well! In engineering and science, dimensional analysis describes the relationships between different physical quantities based on their fundamental qualities such as length, mass, time, and electric current, and units of measure like miles v/s kilometres, or pounds v/s kilograms.

In other words, in Physics, we study two types of physical quantities, i.e., fundamental and derived. The seven fundamental units include mass, length, amount of substance, time, luminous intensity, and electric current. However, if we combine two or more fundamental units, we get derived quantities.

For examples, we denote [M] for mass, [L] for length, and so on. Similarly, for speed, which is a derived quantity given by distance/time, we denote it with [M]/[L] or [ M L -1 ]. This is how we derive the dimensional formula of various quantities. 

The conversion factor used is based on the unit that we desire in the answer. Further, we will derive the dimensional formula of various quantities on this page.

How to Perform Dimensional Analysis?

(Image to be added soon)

Unit Conversion

Dimensional analysis is also called a Unit Factor Method or Factor Label Method because a conversion factor is used to evaluate the units.

For example, suppose we want to know how many meters there are in 4 km.

Normally, we calculate as-

1 km = 1000 meters

4 km = 1000 × 4 = 4000 meters

(Here the conversion factor used is 1000 meters)

Principle of Homogeneity of Dimensional Analysis

This principle depicts that, “the dimensions are the same for every equation that represents physical units. If two sides of an equation don’t have the same dimensions, it cannot represent a physical situation.”

For example, in the equation

[M a L b T c ] = MxLyTz

As per this principle, we have

Example of Dimensional Analysis

For using a conversion factor, it is necessary that the values must represent the same quantity. For example, 60 minutes is the same as 1 hour, 1000 meters is the same as 1 kilometre, or 12 months is the same as 1 year.

Let us try to understand it in this way. Imagine you have 15 pens and you multiply that by 1, now you still have the same number of 15 pens. If you want to find out how many packages of the pen are equal to 15 pens, you need the conversion factor.

Now, suppose you have a packaged set of ink pens in which each package contains 15 pens. Let's consider that you have 6 packages. To calculate the total pens, you have to multiply the number of packages by the number of pens in each package. This comes out to be:

15 × 6 = 90 pens

Some other examples of conversion factors that are used in day to day life are:

A simple example : the time taken by a harmonic oscillator.

A complex example: the energy of vibrating conduction or wire.

A third example : demand versus capacity for a disk that is rotating.

Applications of Dimensional Analysis

Dimensional analysis is an important aspect of measurement, and it has many applications in Physics. Dimensional analysis is used mainly because of five reasons, which are:

To check the correctness of an equation or any other physical relation based on the principle of homogeneity. There should be dimensions on two sides of the equation. The dimensional relation will be correct if the L.H.S and R.H.S of an equation have identical dimensions. If the dimensions on two sides are incorrect, then the relations will also be incorrect.

Dimensional analysis is used to convert the value of a physical quantity from one system of units to another system of units.

It is used to represent the nature of physical quantity.

The expressions of dimensions can be manipulated as algebraic quantities.

Dimensional analysis is used to derive formulas.

Limitations of Dimensional Analysis

Some major limitations of dimensional analysis are:

Dimensional analysis doesn't provide information about the dimensional constant.

Dimensional analysis cannot derive trigonometric, exponential, and logarithmic functions.

It doesn't give information about the scalar or vector identity of a physical quantity.

Example of Dimensional Formula: Derivation for Kinetic Energy

The dimensional formula of any physical entity is the mathematical expression representing the powers to which the fundamental units (mass M, length L, time T) are to be raised to obtain one unit of a derived quantity. 

Let us now understand the dimensional formula with an example. Now, we know that kinetic energy is one of the fundamental parts of Physics, hence its formula plays a vital role in many fields of Physics. So, let us derive the dimensional formula of kinetic energy. 

The kinetic energy has a dimensional formula of,

[M L 2 T -2 ]

M = Mass of the object

L = Length of the object

T = Time taken

Kinetic energy (K.E) is given by = \[\frac {1} {2}\]    [Mass x Velocity 2 ] ---- (I)

The dimensional formula of Mass is = [ M 1 L 0 T 0 ] --- (ii)

We know that,

Velocity = Distance × Time -1

= L x T -1 (dimensional formula)

Velocity has a dimensional formula [ M 0 L 1 T -1 ] ----- (iii)

On substituting equation (ii) and iii) in the above equation (i) we get,

Kinetic energy (K.E) = \[\frac {1} {2}\]    [Mass x Velocit y 2 ]

Or, K.E = [ M 1 L 0 T 0 ] [ M 0 L 1 T -1 ] 2 = [ M (0 +1) L (1 + 1) T (-1 + -1) ]

Therefore, on solving, we get the dimensional formula for kinetic as [ M 0 L 2 T -2 ].

From this context, we understand that in dimensional analysis a set of units helps us establish the form of an equation and to check that the answer is free of even minute errors. 

Solved Example

1. Find out how many feet are there in 300 centimeters (cm).

Ans. We need to convert cm into feet.

Firstly, we have to convert cm into inches, and then inches into feet, as we can't directly convert cm into feet.

The calculation of two conversion factors is required here:

Then, 300 cm = 300 x \[\frac {1} {30.48}\]      feet

= 9.84 feet

FAQs on Dimensional Analysis

1. Find out how Many Feet are in 300 Centimeters (cm).

We need to convert cm into feet.

Then, 300 cm = 300×  \[\frac {1} {30.48}\]       feet

 2. Check the Consistency of the Dimensional Equation of Speed.

Dimensional analysis is used to check the consistency of an equation.

Speed = \[\frac {Distance} {Time}\]  

[LT -1 ] = \[\frac {L} {T}\]  

[LT -1 ] = [LT -1 ]

This equation is correct dimensionally because it has the same dimensions of speed on both sides. This basic test of dimensional analysis is used to check the consistency of equations, but it doesn't check the correctness of an equation.

By this method, the constants of some physical quantities cannot be determined.

Sine of angle =  \[\frac {Length} {Length}\]   , and hence it is unit-less.

So, it is a dimensionless quantity.

3. How to Check For Dimensional Consistency

Let us consider one of the equations, let say, constant acceleration,

The equation is given by

s = ut +  \[\frac {1} {2}\]       at 2 

s: displacement = it is a unit of length, 

Hence its dimension is L

ut: velocity x time, its dimension is LT -1  x T = L

\[\frac {1} {2}\]   at 2   = acceleration x time, its dimension is LT -2  x T 2   = L

All these three terms must have the same dimensions in order to be correct.

As these terms have units of length, the equation is dimensionally correct.

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Dimensional Analysis: Definition, Examples, And Practice

If you’ve heard the term “dimensional analysis,” you might find it a bit overwhelming. While there’s a lot to “unpack” when learning about dimensional analysis, it’s a lot easier than you might think. Learn more about the basics and a few examples of how to utilize the unique method of conversion.

Dimensional Analysis: Definition, Examples, and Practice

As we mentioned, you may hear dimensional analysis referred to as unit analysis; it is often also known as factor-label method or the unit factor method. A formal definition of dimensional analysis refers to a method of analysis “in which physical quantities are expressed in terms of their fundamental dimensions that is often used.”

Most people might agree that this definition needs to be broken down a bit and simplified. It might be easier to understand this method of analysis if we look at it as a method of solving problems by looking converting one thing to another.

While dimensional analysis may seem like just another equation, one of the unique (and important) parts of the equation is that the unit of measurement always plays a role in the equation (not just the numbers).

We use conversions in everyday life (such as when following a recipe) and in math class or in a biology course. When we think about dimensional analysis, we’re looking at units of measurement, and this could be anything from miles per gallon or pieces of pie per person.

Many people may “freeze up” when they see a dimensional analysis worksheet or hear about it in class, but if you’re struggling with some of the concepts, just remember that it’s about units of measurements and conversion. Dimensional analysis is used in a variety of applications and is frequently used by chemists and other scientists.

The Conversion Factor in Dimensional Analysis

One important thing to consider when using dimensional analysis is the conversion factor. A conversion factor , which is always equal to 1, is a fraction or numerical ratio that can help you express the measurement from one unit to the next.

When using a conversion factor, the values must represent the same quantity. For example, one yard is the same as three feet or seven days is the same as one week. Let’s do a quick example of a conversion factor.

Imagine you have 20 ink pens and you multiply that by 1; you still have the same amount of pens. You might want to find out how many packages of pens that 20 pens equal and to figure this out, you need your conversion factor.

Now, imagine that you found the packaging for a set of ink pens and the label says that there are 10 pens to each package. Your conversion factor ends up being your conversion factor. The equation might look something like this:

20 ink pens x 1 package of pens/10 pens = 2 packages of ink pens. We’ve canceled out the pens (as a unit) and ended up with the package of pens.

​ While this is a basic scenario, and you probably wouldn’t need to use a conversion factor to figure out how many pens you have, it gives you an idea of what it does and how it works. As you can see, conversion factors work a lot like fractions (working with numerators and denominators)

Even though you’re more likely to work with more complex units of measurement while in chemistry, physics, or other science and math courses, you should have a better understanding of using the conversion factor in relation to the units of measurement.

Steps For Working Through A Problem Using Dimensional Analysis

Like many things, practice makes perfect and dimensional analysis is no exception. Before you tackle a dimensional analysis that your instructor hands to you, here are some tips to consider before you get started.

• Read the problem carefully and take your time
• Find out what unit should be your answer
• Write down your problem in a way that you can understand
• Consider a simple math equation and don’t forget the conversion factors
• Remember, some of the units should cancel out, resulting in the unit you want
• Double-check and retry if you have to
• The answer you come up with should make sense to you

To help you understand the basic steps we are using an easy problem that you could probably figure out fairly quickly. The question is: How many seconds are in a day?

First, you need to read the question and determine the unit you want to end up with; in this case, you want to figure out “seconds in a day.” To turn this word problem into a math equation, you might decide to put seconds/day or sec/day.

The next step is to figure out what you already know. You know that there are 60 seconds to one minute and you also know that there are 24 hours in one day; all of these units work together, and you should be able to come up with your final unit of measurement. Again, it’s best to write down everything you know into an equation.

After you’ve done a little math, your starting factor might end up being 60 seconds/1 minute. Next, you will need to work your way into figuring out how many seconds per hour. This equation will be 60 seconds/1 minute x 60 minutes/1 hour. The minutes cancel themselves out, and you have seconds per hour.

Remember, you want to find out seconds per day so you’ll need to add another factor that will cancel out the hours. The equation should be 60 seconds/1 minute x 60 minutes/1 hour x 24 hours/1 day. All units but seconds per day should cancel out and if you’ve done your math correctly 86,400 seconds/1 day.

When doing a dimensional analysis problem, it’s more important to pay attention to the units and make sure you are canceling out the right ones to get the final product. Doing your math correctly important, but it’s easier to double-check than trying to backtrack and figure out how you ended up with the wrong unit.

Our example is relatively simple, and you probably had no problem getting the right answer or using the right units. As you work through your science courses, you will be faced with more difficult units to understand. While dimensional analysis will undoubtedly be more challenging, just keep your eye on the units, and you should be able to get through a problem just fine.

Why Use Dimensional Analysis?

As we’ve demonstrated, dimensional analysis can help you figure out problems that you may encounter in your everyday. While you’re likely to explore dimensional analysis a bit more as you take science courses, it can be particularly helpful for Biology students to learn more.

Some believe that dimensional analysis can help students in Biology have a “better feel for numbers” and help them transition more easily into courses like Organic Chemistry or even Physics (if you haven’t taken those courses yet).

Can you figure out a math equation or a word problem without dimensional analysis? Of course, and many people have their own ways of working through a problem. If you do it correctly, dimensional analysis can actually help you answer a problem more efficiently and accurately.

Ready To Test Your Dimensional Analysis Skills?

If you want to practice dimensional analysis, there are dozens of online dimensional analysis worksheets. While many of them are pretty basic or geared towards specific fields of study like Chemistry, we found a worksheet that has an interesting variety. Test out what we’ve talked about and check your answers when you’re done.

• How many minutes are in 1 year?
• Traveling at 65 miles/hour, how many minutes will it take to drive 125 miles to San Diego?
• Convert 4.65 km to meters
• Convert 9,474 mm to centimeters
• Traveling at 65 miles/hour, how many feet can you travel in 22 minutes? (1 mile = 5280 feet)

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Dimensional analysis

Dimensional analysis is a method for converting one unit to another using the relationships between various physical quantities. Dimensional analysis is a skill that is used widely in science and engineering. It can help with understanding how to convert between different units of measurement.

Converting between different units of measurement typically involves the use of a conversion factor. When you convert between different units of measurement, you are using dimensional analysis, even if you are not doing so explicitly.

1. In the United States, weight is most commonly referenced in terms of pounds . In most other countries, kilograms are used instead. In order to convert from pounds to kilograms, or vice versa, we use a conversion factor that relates pounds and kilograms, along with dimensional analysis.

1 kilogram ≈ 2.2 pounds

We can rearrange this relationship and say that there are 2.2 pounds per kilogram, where the "per" indicates division:

This is the conversion factor we can use to convert betweeen these two measurements of weight. Convert 135 pounds to kilograms using dimensional analysis:

The unit of pounds cancels out, leaving us with just kilograms.

2. Time is another quantity that we convert frequently. There are 60 seconds in one minute, 60 minutes in 1 hour, and 24 hours in 1 day. Convert 1.5 days to minutes:

As can be seen from the examples above, dimensional analysis, though sometimes tedious, can be very helpful for helping us see the relationships between various quantities, as well as for converting quantities from one to another using basic algebra .

Did you know?

In science and engineering, it is common to perform dimensional analysis to confirm that a derived equation or computation has the expected dimensions. If it does, then it makes the result more plausible. If however, the dimensions are not the expected dimensions, it is likely a mistake was made along the way.

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Deducing the Process of Arriving at a Solution

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Dimensional analysis is a method of using the known units in a problem to help deduce the process of arriving at a solution. These tips will help you apply dimensional analysis to a problem.

How Dimensional Analysis Can Help

In science , units such as meter, second, and degree Celsius represent quantified physical properties of space, time, and/or matter. The International System of Measurement (SI) units that we use in science consist of seven base units, from which all other units are derived.

This means that a good knowledge of the units you're using for a problem can help you figure out how to approach a science problem, especially early on when the equations are simple and the biggest hurdle is memorization. If you look at the units provided within the problem, you can figure out some ways that those units relate to each other and, in turn, this might give you a hint as to what you need to do to solve the problem. This process is known as dimensional analysis.

A Basic Example

Consider a basic problem that a student might get right after starting physics. You're given a distance and a time and you have to find the average velocity, but you're completely blanking on the equation you need to do it.

Don't panic.

If you know your units, you can figure out what the problem should generally look like. Velocity is measured in SI units of m/s. This means that there is a length divided by a time. You have a length and you have a time, so you're good to go.

A Not-So-Basic Example

That was an incredibly simple example of a concept that students are introduced to very early in science, well before they actually begin a course in physics . Consider a bit later, however, when you've been introduced to all kinds of complex issues, such as Newton's Laws of Motion and Gravitation. You're still relatively new to physics, and the equations are still giving you some trouble.

You get a problem where you have to calculate the gravitational potential energy of an object. You can remember the equations for force, but the equation for potential energy is slipping away. You know it's kind of like force, but slightly different. What are you going to do?

Again, a knowledge of units can help. You remember that the equation for gravitational force on an object in Earth's gravity and the following terms and units:

F g = G * m * m E / r 2

• F g is the force of gravity - newtons (N) or kg * m / s 2
• G is the gravitational constant and your teacher kindly provided you with the value of G , which is measured in N * m 2 / kg 2
• m & m E are mass of the object and Earth, respectively - kg
• r is the distance between the center of gravity of the objects - m 
• We want to know U , the potential energy, and we know that energy is measured in Joules (J) or newtons * meter 
• We also remember that the potential energy equation looks a lot like the force equation, using the same variables in a slightly different way

In this case, we actually know a lot more than we need to figure it out. We want the energy, U , which is in J or N * m. The entire force equation is in units of newtons, so to get it in terms of N * m you will need to multiply the entire equation a length measurement. Well, only one length measurement is involved - r - so that's easy. And multiplying the equation by r would just negate an r from the denominator, so the formula we end up with would be:

F g = G * m * m E / r

We know the units we get will be in terms of N*m, or Joules. And, fortunately, we did study, so it jogs our memory and we bang ourselves on the head and say, "Duh," because we should have remembered that.

But we didn't. It happens. Fortunately, because we had a good grasp on the units we were able to figure out the relationship between them to get to the formula that we needed.

A Tool, Not a Solution

As part of your pre-test studying, you should include a bit of time to make sure you're familiar with the units relevant to the section you're working on, especially those that were introduced in that section. It is one other tool to help provide physical intuition about how the concepts you're studying are related. This added level of intuition can be helpful, but it shouldn't be a replacement for studying the rest of the material. Obviously, learning the difference between gravitational force and gravitational energy equations is far better than having to re-derive it haphazardly in the middle of a test.

The gravity example was chosen because the force and potential energy equations are so closely related, but that isn't always the case and just multiplying numbers to get the right units, without understanding the underlying equations and relationships, will lead to more errors than solutions.

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• Published: 12 February 2014

How dimensional analysis can explain

• Mark Pexton 1 , 2  

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Dimensional analysis can offer us explanations by allowing us to answer What-if–things-had-been-different? questions rather than in virtue of, say, unifying diverse phenomena, important as that is. Additionally, it is argued that dimensional analysis is a form of modelling as it involves several of the aspects crucial in modelling, such as misrepresenting aspects of a target system. By highlighting the continuities dimensional analysis has with forms of modelling we are able to describe more precisely what makes dimensional analysis explanatory and understand otherwise puzzling aspects of dimensional reasoning, such as introducing fictitious dimensions and excluding dimensionally relevant information to characterise some systems. Finally, thinking of dimensional arguments as a form of modelling allows an explication of the role abstraction and multiple realisability; not as compatibility with other possible worlds but as compatibility with different fictional descriptions of our own world.

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Stellar structure is the internal structure of stars: the composition, size, pressure gradients, and energy transport mechanisms, and how these interact to produce the properties of the star.

Essentially we assume that \(\rho (\hbox {r}){/}\!<\rho >\) the density as a function of radius as a fraction of average density is independent of total mass.

I am grateful to Marc Lange (private communication) for clarification of this point.

Noether’s theorem is a classic example of finding independent theoretical justification for a symmetry constraint on laws. Consider a symmetry assumption that physics should be spatially translationally invariant. In other words, all things being equal, the laws of physics shouldn’t alter because we are in Leeds rather than London. Noether was able to show that conservation of linear momentum follows from translational symmetry, likewise temporal symmetry leads to the principle of conservation of energy and rotational symmetry leads to conservation of angular momentum. So when we say both Newton’s and Einstein’s laws adhere to conservation of energy and momentum, we are constraining them via a very basic symmetry assumption. Similarly dimensional explanations offer explanations by locating relational facts about the dimensional constraints inherent within a system.

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Acknowledgments

I would like to thank Juha Saatsi without who’s supervision this work would have not been possible, and Marc Lange for helpful comments, as well as the comments of the editors and referees. This work was supported by the Arts and Humanities Research Council and Templeton Foundation.

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Dimensional Analysis and Its Applications

Dimensional analysis answers some very interesting questions. Which is more, one meter or one second? Are they even comparable? Can you add a kilogram and a dozen eggs to each other? No, right? After studying this section, you will be able to understand how dimensional analysis answers such questions. Let’s begin!

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Dimensional analysis.

Dimensional analysis is the use of dimensions and the dimensional formula of physical quantities to find interrelations between them. It is based on the following facts:

Browse more Topics under Units And Measurement

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The Physical laws

The physical laws are independent of the units in which a quantity is measured.If   \(n_1 a_1\) is the measured value of a physical quantity in one system of units and \(n_2 a_2\) is the value in another system of units then, from the above reasoning, these two must be equal.

\(n_1 a_1\) = \(n_2 a_2\)    …… (1)

The principle of Homogeneity

The equations depicting physical situations must have the same dimensions throughout. If two sides of an equation have different dimensions, that equation can’t represent any physical situation. This is known as the Principle of Homogeneity. For example, if

\( [M]^{a} \) \( [L]^{b} \) \( [T]^{c} \) = \( [M]^{x} \) \( [L]^{y} \) \( [T]^{z} \)

then from the principle of Homogeneity, we have:

a = x; b = y; c = z

Applications of the Dimensional Analysis

Conversion of units

The dimensions of a physical quantity are independent of the system of units used to measure the quantity in. Let us suppose that  \(M_1\) , \(L_1\) and \(T_1\) and \(M_2\) , \(L_2\) and \(T_2\) are the fundamental quantities in two different systems of units. We will measure a quantity Q (say) in both these systems of units. Suppose, a, b, c be the dimensions of the quantity respectively.

In the first system of units, Q = \(n_1\) \(u_1\) = \(n_1\) [ \(M_1^a\) \(L_1^b\) \(T_1^c\) ]      …. (2)

In the second system of units, Q = \(n_2\) \(u_2\) = \(n_1\) [ \(M_2^a\) \(L_2^b\) \(T_2^c\) ] …. (3)

\(n_1\) [\(M_1^a\) \(L_1^b\) \(T_1^c\)] = \(n_2\) [ \(M_2^a\) \(L_2^b\) \(T_2^c\) ]   … (4)  [using (1)]

Substitution of the respective values will give the value of \(n_1\) or \(n_2\)

Checking the consistency of an equation

All the physical equations must be consistent. The reverse may not be true. For example, the following equations are not consistent because of the dimensions of the L.H.S. ≠ Dimensions of the R.H.S.

F = \( m^2 \)× a

F = m × \( a^2\)

Finding relations between physical quantities in a physical phenomenon

The principle of  Homogeneity can be used to derive the relations between various physical quantities in a physical phenomenon. Let us see it with the help of an example.

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Solved Example for You

Example 1: The Time period (T) of a simple pendulum is observed to depend on the  following factors:

• length of the pendulum (L),
• mass of the Bob (m)
• acceleration due to gravity (g)

Sol. Let T ∼  L α m β g γ or …… (5)

[T] = [L] α [M] β [L] γ [T] -2γ or

Solving the above for α, β and γ we have:

β = 0;  α + γ = 0;

-2 γ = 1 or γ= -1/2

Using these in equation (5), we have:

T ∼ \( \sqrt[]{L/g} \)

Example 2: Convert 1 J to erg.

Sol. Joule is the S.I. unit of work. Let this be the first system if units. Also erg is the unit of work in the cgs system of units. This will be the second system of units. Also \(n_1\) = 1J and we have to find the value of \(n_2\)

From equation (4), we have:

[\(M_1^a\) \(L_1^b\) \(T_1^c\)] = \(n_2\) [ \(M_2^a\) \(L_2^b\) \(T_2^c\) ]

The dimensional formula of work is  [\(M^1\) \(L^2\) \(T^{-2}\)]. As a result a = 1, b = 2 and c = -2.

[\(M_1^1\) \(L_1^2\) \(T_1^-2\)] = \(n_2\) [ \(M_2^1\) \(L_2^2\) \(T_2^{-2}\) ]

Also \(M_1\) = 1 kg , \(L_1\) = 1 m,  \(T_1\)=1 s

and \(M_2\) = 1 g , \(L_2\) = 1 cm,  \(T_2\)=1 s

\(n_2\) = [\(M_1^1\) \(L_1^2\) \(T_1^{-2}\)] / [ \(M_2^1\) \(L_2^2\) \(T_2^{-2}\) ]

= [\(1kg^1\) \(1g^2\) \(1s^{-2}\)] /  [ \(1g^1\) \(1cm^2\) \(1s^{-2}\) ]

\(n_2\) = \( 10^7 \)

Hence, 1J = \( 10^7 \) erg

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Dimensional Analysis

Core Concepts

In this tutorial, you will learn what dimensional analysis is in the field of chemistry, how to use it, see examples, and learn how it can be applied to chemistry. 

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What is Dimensional Analysis?

What is the definition of dimensional analysis? Dimensional analysis is an essential skill used widely in the field of chemistry. Using this technique can answer questions like: “How much of this chemical do I need in my reaction?” and “What is the concentration of my solution?” At its simplest form, dimensional analysis is the methodical canceling-out of units. Take the example below:

In more real-world applications, dimensional analysis is used to convert between different units of measurement, and find unknown characteristics from those that we do know. 

Unit Conversions

It is often necessary to convert between units of measurement. Units of measurement are used to define the qualities of something. 

• A gold block weighs 12 kilograms . 
• There are 150 milliliters of water in the container. 

You will encounter Imperial and SI ( International System ) Units. Imperial units are measurements like feet, inches, and pounds. SI units are measurements like meters, centimeters, and kilograms. SI units are most common in chemistry; furthermore, one of the most important units of measurement is the mole (mol).

Below are some of the relationships between these units you might see throughout various sources, such as textbooks, or online:

Additionally, it is also possible to convert between the two systems of measurement. 

If the gold block weighs 12 kilograms, how many pounds does it weigh?

1.0 pound is about 0.45 kilograms, so we find …

We are left with the solution: For every 12 kilograms, there are 26 pounds. 

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Finding Unknowns

Dimensional analysis is not only useful for converting between one unit to another, but can help in solving for a number of different properties. It becomes important to be aware of how quantities like mass, volume, and density are related:

It is also increasingly important to pay attention to units of measurement. 

If there are of water in a container, and the water in the container weighs , what is the density of the water?

In this case, mass is given in grams, and volume is given in milliliters. We define density as mass/volume, so for this case density is grams/milliliter.

Dimensional Analysis Examples in Chemistry

In chemistry, dimensional analysis can be used in the ways described above, but it can also be used to relate the quantities of chemicals in a reaction. Ratios can be created using stoichiometric relationships.

Since there are two mols of Iodide (I – ) for every one mol of Calcium Iodide (CaI 2 ), the following ratio can be written: 

How many moles of iodide ions are produced when 4 mols of CaI 2 are fully dissolved?

For a more in depth tutorial, see Solving Stoichiometry Problems .

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MgCl 2 and AgNO 3 react according to the following equation:

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Lakers vs. Cavs takeaways: Max Christie and Anthony Davis shine during win

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CLEVELAND — Here are the key takeaways from the Lakers’ 121-115 win over the Cavaliers on Saturday night.

In a lot of ways, a game like this from second-year guard Max Christie was the expectation, the response to the promise he showed as a rookie and this past summer as an exciting prospect for the Lakers ready to grab a rotation spot.

But those minutes weren’t really there, with the Lakers and coach Darvin Ham looking instead to Gabe Vincent and eventually Cam Reddish , before Christie got the nod.

While he waited, though, Christie cemented his reputation as a true professional, the kind of compliment rarely paid to 20-year-olds in an occupation where patience doesn’t always pay.

It did against the Cavaliers and Donovan Mitchell on Saturday, Christie delivering one of the best games of his young career in his second start for the injured Reddish.

“I think when you’re trusted by your coaches, one, that helps. So knowing that he’s going to get his number called, he’s been ready. And he stepped up,” LeBron James said of Christie. “And he was big-time tonight. Big-time, one of the tough matchups in this league is Donovan Mitchell, obviously. His ability to score on all three levels — from the three line, from the midrange, get into the paint — I just think he did a good job of just trying to keep his body on him, make him take tough shots and not foul him. And he made some key shots for us, too. Key plays for us. He was big-time.”

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Nov. 22, 2023

Christie finished with 12 points, his most this year. He scored twice on dunks off cuts to the basket and late in the fourth, he bailed the team out on a tough jumper to end a broken possession.

But it was his defense, primarily on Mitchell, that had the Lakers so excited Saturday.

“Max just wants to win. He plays hard and is going to take that challenge against a guy who is an elite scorer from all three levels,” Anthony Davis said. “He made sure that he knows his tendencies and took on that challenge, especially late game when they know — the whole arena knows — they’re going to him, especially with Darius Garland out and stuff like that. So he was the guy and he had two big stops for us. There’s nothing much more you can ask for. For him to only be in his second year, to take on that challenge, it shows a lot about him.”

Passing fancy

The Lakers spent a lot of time during training camp hoping their continuity from last season would help kick-start their offensive flow and allow the team to play with more ball movement.

Through the first quarter of the season, the results have been mixed, the Lakers’ passing sometimes looking a tick slow as the ball stuck for a blink too long on the perimeter.

Saturday, though, the Lakers attacked the Cavs in different ways, using dribble penetration to initiate a lot of their passing, keep Cleveland moving and unable to key on one particular area. The result was tremendous balance, seven players scoring at least 10 points and an eighth, Taurean Prince, adding seven. Four different Lakers, including Prince, had at least five assists, with Austin Reaves dishing out 10.

For Christian Wood, Lakers’ role is easy to define but hard to attain

Lakers reserve Christian Wood did not have a big game in a loss to his former team, the Dallas Mavericks. Metrics, though, show he can contribute.

Nov. 23, 2023

The Lakers’ 34 assists as a team was the second most this season. Last week against Houston, the Lakers had 35 assists.

“Just sharing the basketball — if you don’t have a shot, move it on to the next guy and they either have a shot or move it on to the next guy. Just playing stress free and letting the ball dictate the type of shot that we get instead of guys being selfish or trying to find their own look,” Davis said. “Obviously, you’re going to have that throughout the course of the game if a guy is hot or has an advantage, but for the most part, ball and body movement is something we preach.”

Reaves said the assists were proof the Lakers were playing his preferred style — “the right way.”

“Any time you can get that assist number up it means you’re playing the game the right way. You’re making the extra pass, playing unselfish and that’s the goal,” Reaves said. “When you do that, everybody feels good and everybody has touched the ball and when it gets swung their way, they feel confident with it in their hands. That’s the main thing for us as a unit: Play the game the right way and get those assist numbers up.”

After a quiet second half in a tight loss to Dallas on Wednesday , Davis scored 15 points in the third and eight in the fourth on Saturday, dominating against Cleveland’s super-sized frontcourt.

“Everybody’s quick to murder AD when he has an off night. I think if you go back and look at his off nights, in some way he’s affecting the game more than what the average human is picking up on,” Reaves said. “They go up these [box score], look and see how many points he had and go straight to their phones and start tweeting stuff and bashing him for not having 25 or 30. But he affects the game in so many ways that on his bad nights, he’s still very productive for us. Any time he plays like this, I think we’ll be pretty tough to beat. He’s super efficient.”

Davis gets another big matchup on Monday against the reigning MVP Joel Embiid when the Lakers will try and keep him involved as much as he was on Saturday.

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“He sets 95% of our pick-and-rolls and when he has an opportunity to catch it in the pocket, or if the bigs are up and it allows us to find him in transition and he gets an opportunity to seal or whatever, we have to find him,” James said. “I mean, it’s that simple. And there’s also times where we have certain play calls that we can call on the fly so we can get him the ball as well. But even if he’s not shooting, it’s just him touching the ball. And I think from the beginning of the game, I think right after my and-one, when we got an opportunity to call our first set, I think he touched the ball and throughout the whole game he continued to touch the ball.

“And that’s big-time for our ball club.”

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Dan Woike is the Lakers beat writer for the Los Angeles Times after spending two seasons covering the league as a whole, with an emphasis on Los Angeles’ teams.

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