# General Engineering Introduction/Units Dimensions/Dimensions

Dimensional analysis is routinely used to check the plausibility of derived equations. It is also used to create new equations.

## Base Units

A discussion of the dimensions of reality has to start with the Base Units: temperature (K for Kelvin), current (A for Ampere), light intensity (cd for candela), time (s for seconds), distance (m for meters), mass (kg for kilogram), and count (mol for mole).
Base Units are what all other units are built upon. In the graphic at the right are the 7 base units.
It is interesting that temperature (K) stands alone with no dependencies even though an argument can be made that it is the root mean square "average" vibration velocity or a form of potential kinetic energy or internal energy.
It is interesting that Ampere (A) and Candela (cd) are largely dependent upon other base units. Each captures a unique bit of the universe absent from the others. However, both are more descriptive of how the human race arbitrary defines the universe than the universe itself.
The arrow from time (s for seconds) to length (m for meters) is from the space-time relationship found in the study of general relativity and hinted at in special relativity.

## Dimensions

The dimensions of reality are the 7 base units. All other units can be expressed in terms of the base units. Here are some examples:
Some named units derived from SI base units :
| Name | Symbol | Quantity | Expression in terms of other units | Dimensions in SI base units |
| hertz | Hz | frequency | 1/s | s$^{−1}$ |
| radian | rad | angle | m/m | dimensionless |
| newton | N | force, weight | kg⋅m/s$^2$ | kg⋅m⋅s$^{−2}$ |
| pascal | Pa | pressure, stress | N/m$^2$ | kg⋅m$^{−1}$⋅s$^{−2}$ |
| joule | J | energy, work, heat | N⋅m = C⋅V = W⋅s | kg⋅m$^2$⋅s$^{−2}$ |
| watt | W | power, radiant flux | J/s = V⋅A | kg⋅m$^2$⋅s$^{−3}$ |
| coulomb | C | electric charge or quantity of electricity | s⋅A | s⋅A |
| volt | V | voltage, electrical potential difference, electromotive force | W/A = J/C | kg⋅m$^2$⋅s$^{−3}$⋅A$^{−1}$ |
| farad | F | electric capacitance | C/V | kg$^{−1}$⋅m$^{−2}$⋅s$^4$⋅A$^2$ |
| ohm | Ω | electric resistance, impedance, reactance | V/A | kg⋅m$^2$⋅s$^{−3}$⋅A$^{−2}$ |
| siemens | S | electrical conductance | 1/Ω = A/V | kg$^{−1}$⋅m$^{−2}$⋅s$^3$⋅A$^2$ |
| weber | Wb | magnetic flux | J/A | kg⋅m$^2$⋅s$^{−2}$⋅A$^{−1}$ |
| henry | H | inductance | V⋅s/A = Wb/A | kg⋅m$^2$⋅s$^{−2}$⋅A$^{−2}$ |
| degree Celsius | °C | temperature | K - 273.15 | K - 273.15 |

## Dimensionally Correct

Suppose we're given an equation (shown on the left) involving force F, radius r, length s, velocity v, distance d, and viscosity p and are asked to check if it is dimensionally correct.
       F = -2πrsvp/d
Is this formula dimensionally correct? We would have to know the dimensions of each of these symbols. We know most of them:
* F, force is ML/T$^2$
* r, radius is L
* s, length is L
* v, velocity is L/T
* d, distance is L
What is viscosity? The easiest way is to look them up at NIST: Viscosity (dynamic viscosity) has units of pascal second (Pa·s). Pascal is a unit of pressure or Force per unit area. Force is ML/T$^2$. Force per unit area is pressure M/LT$^2$. Viscosity is pressure multiplied by time: M/LT. Substituting back into the formula:
       ${\frac {ML}{T^{2}}}?=?L*L*{\frac {L}{T}}*{\frac {M}{LT}}*{\frac {1}{L}}$
Simplifying the right hand side, can see that the equation is dimensionally correct.
       ${\frac {ML}{T^{2}}}?=?{\cancel {L}}*{\cancel {L}}*{\frac {L}{T}}*{\frac {M}{{\cancel {L}}T}}*{\frac {1}{\cancel {L}}}$
Just because the dimensions are correct, doesn't mean that the formula is correct. However, most of the time this process works in undergraduate physic classes! Most physics tests throw numbers at you and expect you to combine them in dimensionally correct ways. Most students think physics is about memorizing formulas. In reality it is about knowing your units and being able to combine them in dimensionally correct ways. Do you understand why they are always taking points off if you don't write down your units?

## Dimensionless

Radians, degrees are dimensionless. Trig functions work with dimensionless numbers. Logrithmic functions work with dimensionless numbers. Dimensionless numbers may have names such as dB's (decibels).

## Equation Creation

Engineers create equations all the time that are not in physics books. They are called empirical equations because they have never been traced back to first principles. But they have been experimentally verified.
The purpose of equations is to predict behavior. Being able to describe behavior through an equation communicates more detail than a photo or video. So engineers create and use equations that science never sees. Ever seen the orifice equation in a physics book? Do you know it only works when the pipe is full of water and that there is another equation for when there is air in the pipe?
       ${\dot {m}}=\rho \;Q=C\;A_{2}\;{\sqrt {2\;\rho \;(P_{1}-P_{2})}}$
The base unit of meter is given the symbol L for length. The base unit of seconds is given the symbol T for time. The base unit of kg is given the symbol M for mass.

### Harmonic Oscillator

We are trying to create a formula that predicts the period T of a spring attached to a ceiling with a mass M at the other end that is bouncing up and down due to a spring constant K in the presence of gravity g.
       $T\propto M^{a}\,g^{c}\,K^{d}$
Time has the dimension T, Mass has the dimension M, Length has the dimension L and g has the dimension L/T$^2$. The spring constant has dimensions M/T$^2$. When writing the dimension equation, we are not concerned with the exact values of numbers, only the exponents. The unknown exponents a, b and c are what we are trying to determine. The dimensional analysis equation associated with the above is thus:
       $T=M^{a}\,\left({\frac {L}{T^{2}}}\right)^{c}\,\left({\frac {M}{T^{2}}}\right)^{d}$
Comparing the left with the right, we can write an equation in each dimension:
   * looking at time: 1 = -2c -2d
   * looking at mass: 0 = a + d
   * looking at length: 0 = c
This means that gravity has nothing to do with it, (c=0), d = -1/2 and a = 1/2. So the formula looks like this:
       $T\propto {\sqrt {m/K}}$

### Cannon Ball Range

We are trying to create a formula that estimates the range R of a cannon ball. R has the dimension L (length). Suppose V$_x$ and V$_y$ refer to horizontal and verticle velocities (L/T dimensions) and g gravity (L/T$^2$ dimensions) are related by this equation:
       $R\propto {\frac {V_{x}\,V_{y}}{g}}\,$
We don't know what the exponents, a, b and c are. Our goal is to use dimensional analysis to compute them or check them.
The proportional sign can be replaced with an equal sign when comparing the dimensions:
    $L_{x}=(L_{x}/T)^{a}\,(L_{y}/T)^{b}(L_{y}/T^{2})^{c}\,$
Comparing the left with the right, we can write an equation in each dimension:
   * In the horizontal (x) direction : 1=a
   * In the verticle (y) direction: 0 = b+c
   * For time: 0 = -a-b-2c
There are only three equations possible because only only three dimensions in the appear in the dimensional analysis. Doing some algebra, the only solution can be found:
   * a = 1
   * b = 1
   * c = -1
so the original equation can be written:
       $R\propto {\frac {V_{x}\,V_{y}}{g}}\,$
The proportional sign can be replaced by an equal sign and a multiplicative constant k. The constant could be determined through experimentation and the choice of units.
       $R=k*{\frac {V_{x}\,V_{y}}{g}}\,$

### Vibrating Wire

We are trying to write a formula for the energy E (ML$^2$/T$^2$) of a vibrating wire of length h (L) vibrating with an amplitude s (L). The wire has a linear density p (M/L) and is under tension F (ML/T$^2$).
       $E\propto h^{a}\,s^{b}\,p^{c}\,F^{d}\,$
We don't know what the exponents, a, b, c and d. Our goal is to use dimensional analysis to compute them or check them.
The proportional sign can be replaced with an equal sign when comparing the dimensions:
       ${\frac {ML^{2}}{T^{2}}}=L^{a}\,L^{b}\,\left({\frac {M}{L}}\right)^{c}\,\left({\frac {ML}{T^{2}}}\right)^{d}\,$
Comparing the left with the right, we can write an equation in each dimension:
* Looking at Mass, 1 = c + d
* Looking at Length, 2 = a + b - c + d
* Looking at Time, -2 = d
Three equations, four unknowns means that there is not one unique answer. But certain values are known:
* d=-2
* c=3
* a+b = 7
An experiment is going to have to be done that holds everything constant but varies the amplitude (s) or length (h). This way either a or b can be found. Say we expect to do an experiment to vary s to find b. Then a = 7-b and we can finish of this formula the best we can at the moment by saying:
       $E\propto {\frac {h^{7-b}s^{b}p^{3}}{F^{2}}}$