Dimensional Analysis: Reference Dimensions & Repeating Variables

In summary: You must write out the full equation before doing dimensional analysis.For example, your f(V) will in fact have a V2 term. Only with the full equation can you group or cancel similar dimensions.
  • #1
goonking
434
3
lets say for example, the air drag (ϑ) that wind exerts on a tile is a function of the tile's width (w), height (h), viscosity of air (μ), density of air (ρ) and velocity of the air (V)

then ϑ = f(w,h, μ, ρ, V)

ϑ : MLT-2
w: L
h: L
μ : ML-1T-1
ρ: ML-3
V: LT-1
I understand there are 3 basic dimensions : M, L, and T
but how do I decide which are 'reference dimensions', would choosing wisely simply the problem?also, w and h have the same dimensions (measure of length : L), so would they be considered 'repeating variables'?
 
Last edited:
Engineering news on Phys.org
  • #2
You must write out the full equation before doing dimensional analysis.
For example, your f(V) will in fact have a V2 term.
Only with the full equation can you group or cancel similar dimensions.
 
  • Like
Likes Simon Bridge
  • #3
please state the definitions for "reference dimension" and "repeating variable".
[edit: beaten to it...]
 
  • #4
Simon Bridge said:
please state the definitions for "reference dimension" and "repeating variable".
[edit: beaten to it...]

My book doesn't seem to clearly define "reference dimension", it states "Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T. However, in some instances perhaps only two dimensions, such as L and T, are required, or maybe just one, such as L"

A repeating variable must be dimensionally independent of the others (i.e., the dimensions of one repeating variable cannot be reproduced by some combination of products of powers of the remaining repeating variables). This means that the repeating variables cannot themselves be combined to form a dimensionless product.
 
  • #5
Baluncore said:
You must write out the full equation before doing dimensional analysis.
For example, your f(V) will in fact have a V2 term.
Only with the full equation can you group or cancel similar dimensions.
you mean the "pi" terms?
this book suggests using Buckingham pi theorem to find number of pi terms needed.
 
  • #6
goonking said:
My book doesn't seem to clearly define "reference dimension", it states "Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T. However, in some instances perhaps only two dimensions, such as L and T, are required, or maybe just one, such as L"
It's basically saying the basis dimensions are the members of the smallest set of dimensions that you can use to get the dimensions of every other variable in the problem. (Or more generally, in the system of units you want to use.)
You can finess it to the dimensions of the most commonly used variables.

Choosing wisely your basis dimensions is pretty much the same as choosing your system of units - it may simplify things but the process can be more trouble than it's worth. Look up "unified units".

[edit] One of the things that is really useful though, is to write out the whole equation you are concerned with.

A repeating variable must be dimensionally independent of the others (i.e., the dimensions of one repeating variable cannot be reproduced by some combination of products of powers of the remaining repeating variables). This means that the repeating variables cannot themselves be combined to form a dimensionless product.
... OK, so would two different variables that share dimensions fit that definition? Doesn't the definition you wrote say the dimensions have to be different?
 
Last edited:
  • #7
goonking said:
then ϑ = f(w,h, μ, ρ, V)
That does not help. You need the actual equation, not just a conceptual function of some variables.
 
  • Like
Likes Simon Bridge
  • #8
goonking said:
I understand there are 3 basic dimensions : M, L, and T
but how do I decide which are 'reference dimensions', would choosing wisely simply the problem?

Represent the variables as row vectors, using the values of the exponents of the fundamental dimensions as if they were scalar coefficients of some standard basis vectors.

##\vartheta: M^1L^1T^2:\ [ 1, 1, -2 ]##
##\ w: M^0L^1 T^0:\ [ 0,1, 0]##
##\ h: M^0L^1T^0:\ [0,1,0]##
##\mu:M L^{-1}T^{-1}:\ [1, -1, -1]##
##\rho:M^1L^{-3}T^0:\ [1, -3, 0]##
##\ V:M^0L^1T^{-1}:\ [0,1,-1]##

A repeating variable must be dimensionally independent of the others (i.e., the dimensions of one repeating variable cannot be reproduced by some combination of products of powers of the remaining repeating variables). This means that the repeating variables cannot themselves be combined to form a dimensionless product.

Then, for "repeating variables", you need to pick a set variables that are associated with a set of row vectors that can serve as a basis for the set of row vectors above. There may be different ways of doing this.

Scalar multiplication of a row vector by an integer represents finding the dimensions of a variable raised to that power.
For example, the dimensions of ##\vartheta^2## are ##2[ 1, 1, -2] = [2,2,-4] : M^2 L^2 T^{-4}##.

Addition of the row vectors corresponds to finding the dimensions of the multiplication of variables.
For example, ## \rho V^2:\ [1,-3,0] + 2[0,1,-1] = [ 1, -1, -2] :\ M L^{-1} T^{-2} ##

And, for example, from ##[1,-3,0] + 2[0,1,-1] + 2[0,1,0] = [1,1,-2]##, we see that the dimensions of ## \rho V^2 w^2 ## (or ##\rho V^2 wh##) match those of ##\vartheta##.

A "wise selection" of "repeating variables" allows you to express the other rows in terms of linear combinations of the rows of the repeating variables using small integer coefficients in the linear combinations.

You can use the usual procedures of linear algebra for finding a set of basis vectors for a given set of vectors, but these procedures will involve linear combinations with non-integer coefficients. The coefficients will be rational numbers, so you can transform such a linear combination to a linear combination with integer coefficients by multiplying every term by the common denominator of all the non-integer coefficients. However, this is somewhat of a nuisance. Picking a "wise choice" for the rows means to trying to avoid linear algebra that introduces a lot of fractions when you express the non-basis rows in terms of the basis rows.
 

What is dimensional analysis?

Dimensional analysis is a mathematical technique used in science to convert between different units of measurement and to ensure the consistency and accuracy of mathematical equations.

Why is dimensional analysis important?

Dimensional analysis is important because it allows scientists to compare and analyze physical quantities from different systems of units, ensuring that calculations and measurements are accurate and consistent.

What are reference dimensions in dimensional analysis?

Reference dimensions are the fundamental units of measurement that are used as a basis for all other measurements. They include length, mass, time, temperature, and electric current.

What are repeating variables in dimensional analysis?

Repeating variables are quantities that appear on both sides of an equation and have the same units. They are used to ensure that the equation is dimensionally consistent and mathematically accurate.

How is dimensional analysis used in real-world applications?

Dimensional analysis is commonly used in scientific research, engineering, and other fields to solve problems involving unit conversions, scaling relationships, and dimensional consistency in equations. It is also used in industry to optimize processes and design experiments.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
822
Replies
35
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
755
  • Advanced Physics Homework Help
Replies
2
Views
873
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
24
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top