Dimensions500.101

Dimensional Reasoning

Dimensions500.101 Dimensions and Measurements

• “Dimension” is characteristic of the object, condition, or event and is described quantitatively in terms of defined “units”.

• A physical quantity is equal to the product of two elements:– A quality or dimension– A quantity expressed in terms of “units”

• Dimensions– Physical things are measurable in terms of three primitive qualities

(Maxwell 1871)• Mass (M)• Length (L)• Time (T)

Note: (Temperature, electrical charge, chemical quantity, and luminosity were added as “primitives” some years later.)

Dimensions500.101

– Examples• Length (L)• Velocity (L/T)• Force (ML/T2)

• Units– Measurements systems--cgs, MKS, SI--define units– SI units are now the international standard (although many

engineers continue to use Imperial or U.S.)

Dimensions and Measurements (cont.)

Dimensions500.101 SI Primitives

DIMENSION UNIT SYMBOL for UNIT

Length meter m

Mass kilogram kg

Time second s

Elec. Current ampere A

luminous intensity candela cd

amount of substance mole mol

Dimensions500.101 SI Derived units

DESCRIPTION DERIVED UNIT SYMBOL DIMENSION

Force newton N mkg/s2

Energy joule J m2kg/s2

Pressure pascal Pa kg/(ms2)

Power watt W m2kg/s3

Dimensions500.101 Dimensional analysis

• Fundamental rules:– All terms in an equation must reduce to identical primitive dimensions

– Dimensions can be algebraically manipulated, e.g.

– Example:

• Uses:– Check consistency of equations– Deduce expression for physical phenomenon

TLTL

2

21 ats LT

TL

22

Dimensions500.101 Dimensional analysis

distance s = s0 +vt2 + 0.5at3

constant = p + ρgh +ρv2/2

volume of a torus = 2π2(Rr)2

222

)sin()sin()sin(cba

Dimensions500.101 Deduce expressions

• Example: What is the period of oscillation for a pendulum?

Possible variables: length l [L], mass m [M], gravity g i.e. period P = f(l , m, g) Period = [ T ], so combinations of variables must be equivalent to [ T

].

LT 2

m

lg

Dimensions500.101 Dimensional analysis (cont.)

• Example: What is the period of oscillation for a pendulum?

Possible variables: length l [L], mass m [M], gravity g i.e. period P = f(l , m, g) Period = [ T ], so combinations of variables must be

equivalent to [ T ].

Only possible combination is

Note: mass is not involved

LT 2

~lg

m

lg

Dimensions500.101 Quantitative considerations

• Each measurement carries a unit of measurement– Example: it is meaningless to say that a board is “3” long. “3”

what? Perhaps “3 meters” long.• Units can be algebraically manipulated (like dimensions)• Conversions between measurement systems can be

accommodated, e.g., 1 m = 100 cm,

or or

Example:

cmm

100mcm

100

1

cmmcmm

mcmmm 300100310033

Dimensions500.101 Quantitative considerations (cont.)

• Arithmetic manipulations between terms can take place only with identical units.

Example:

but,

?23 cmm

cmcmmcmm 30221003

Dimensions500.101

Buckingham Pi Theorem (1915)

• Pi theorem tells how many dimensionless groups define a problem.

• Theorem: If n variables are involved in a problem and these are expressed using k primitive dimensions, then (n-k) dimensionless groups are required to characterize the problem.

Example: in the pendulum, the variables were time [T],

gravity [L/T2], length [L], mass [M] . So, n = 4 k = 3. So, only one dimensionless group describes the system. 2/1/ lgt

Dimensions500.101 Buckingham Pi Theorem (cont.)

• How to find the dimensional groups:– Pendulum example:

where a,b,c,d are coefficients to be determined.

In terms of dimensions:

),,,( 321 knf

dcba mglt1

000)()2(2 )( TLMMLTMLTLT dcbcadcba

Dimensions500.101 Buckingham Pi Theorem (cont.)

Therefore:a - 2c = 0b + c = 0 d = 0

Arbitrarily choose a = 1. Then b = -1/2, c = 1/2, d = 0.

This yields

.1 constlgt

Dimensions500.101

A star undergoes some mode of oscillation. How does the frequency of oscillation ω depend upon the properties of the star? Certainly the density ρ and the radius R are important; we'll also need the gravitational constant G which appears in Newton's law of universal gravitation. We could add the mass m to the list, but if we assume that the density is constant, then m = ρ(4πR3/3) and the mass is redundant. Therefore, ω is the governed parameter, with dimensions [ω] = T-1, and (ρ; R; G) are the governing parameters, with dimensions [ρ] = ML-3, [R] = L, and [G] = M-1L3T-2 (check the last one). You can easily check that (ρ; R;G) have independent dimensions; therefore, n = 3; k = 3, so the function Φ is simply a constant in this case. Next, determine the exponents:

[ω] = T-1 = [ρ]a[R]b[G]c = Ma-cL-3a+b+3cT-2c

Equating exponents on both sides, we have a - c = 0; -3a + b + 3c = 0; -2c = -1Solving, we find a = c = 1/2, b = 0, so that ω = C(Gσ)1/2, with C a constant. We see that the frequency of oscillation is proportional to the square root of the density, and independent of the radius.

Buckingham Pi Theorem (cont.) Oscillations of a star

Dimensions500.101 “Dimensionless” Quantities

• Dimensional quantities can be made “dimensionless” by “normalizing” them with respect to another dimensional quantity of the same dimensionality.

Example: speed V (m/s) can be made "dimensionless“ by dividing by the velocity of sound c (m/s) to obtain M = V/c, a dimensionless speed known as the Mach number. M>1 is faster than the speed of sound; M<1 is slower than the speed of sound.

Other examples: percent, relative humidity, efficiency

• Equations and variables can be made dimensionless, e.g., Cd = 2D/(ρv2A)

• Useful properties:– Dimensionless equations and variable are independent of units.– Relative importance of terms can be easily estimated.– Scale (battleship or model ship) is automatically built into the dimensionless

expression.

Dimensions500.101 Dimensionless quantities (cont.)

– Reduces many problems to a single problem through normalization.

Example: Convert a dimensional stochastic variable x to anon-dimensional variable

to represent its position with respect to a Gaussian curve--N(0,1),e.g., grades on an exam

xx

Dimensions500.101

The area of any triangle depends on its size and shape, which can be unambiguously identified by the length of one of its edges (for example, the largest) and by any two of its angles (the third being determined by the fact that the sum of all three is π). Thus, recalling that an area has the dimensions of a length squared, we can write:

area = largest edge2 • f (angle1, angle2),where f is an nondimensional function of the angles.Now, referring to the figure at right, if we divide a right triangle in two smaller ones by tracing the segment perpendicular to its hypotenuse and passing by the opposite vertex, and express the obvious fact that the total area is the sum of the two smaller ones, by applying the previous equation we have:

c2 • f (α, π/2) = a2 • f (α, π/2) + b2 • f (α, π/2).And, eliminating f:

c2 = a2 + b2, Q.E.D.

Proof of the Pythagorean Theorem

Dimensions500.101 Scaling, modeling, similarity

• Types of “similarity” between two objects/processes.

– Geometric similarity – linear dimensions are proportional; angles are the same.

– Kinematic similarity – includes proportional time scales, i.e., velocity, which are similar.

– Dynamic similarity – includes force scale similarity, i.e., equality of Reynolds number (inertial/viscous), Froud number (inertial/buoyancy), Rossby number (inertial/Coriolis), Euler number (inertial/surface tension).

Dimensions500.101

• Distorted models– Sometimes it’s necessary to

violate geometric similarity: A 1/1000 scale model of the Chesapeake Bay is ten times as deep as it should be, because the real Bay is so shallow that, with proportional depths, the average model depth would be 6mm, too shallow to exhibit stratified flow.

Scaling, modeling, similarity

Dimensions500.101

Dimensions500.101 Scaling, modeling, similarity

• Scaling– What’s the biggest elephant? If one tries to keep similar

geometric proportions, weight L3, where L is a characteristic length, say height.

– However, an elephant’s ability to support his weight is proportional to the cross-sectional area of his bones, say R2.

– Therefore, if his height doubles, his bones would have to increase in radius as 22 R, not 2R.

– [Note: A cross-section of 8 R2 = (22 R)2]. So, with increasing size, an elephant will eventually have legs whose cross-sectional area will extend beyond its body

Dimensions500.101

Biological scaling