Chapter 4 Units and estimation

The central rules which we advise that you follow are:

Where do these rules come from?

Thus $d=7\mathrm{m}$ means $7\times$(one meter). The 7 and the ‘m’ are of equal status in any math you do.

Every line of every calculation should be dimensionally sensible. That is, the dimensions on the left of the equal sign should be consistent with the dimensions on the right, the same way numbers have to balance. Otherwise the equations are not equations.

For example, if two bicycles tied in a race you could say they were in some way equal. But even if you noticed that the weight difference between these equivalent bicycles was 10% more than 2 pounds you would not write

The equivalence between the two bikes in race times does not make eight kilograms equal to nine kilograms. In this same way it would be wrong to write

if you noticed that it takes a bug about a minute ($60$ seconds) to walk the length of your body (say about $60$ inches). The passing of a second corresponds to the passing of an inch, so for some purposes an inch is equivalent to a second. But that does not mean that an $\mathrm{inch}$ is a second. An inch has dimensions of length which cannot be equal to a second which has dimensions of time. Length can equal time no more than $8$ can equal $9$.

But it is correct to write that

Both centimeters and inches have dimensions of length and one inch is equivalent to 2.54 centimeters always (fig. 4.1). An equation where the units on both sides of the equation are the same physical quantities (length in the example above) is balanced with regard to units.

When you go from one line of a calculation to the next you should carry (keep written track of) the units with as much care as any other numerical or algebraic quantities.

When you do arithmetic and don’t forget any terms you have ‘carried’ the numbers from one line of calculation to the next. Similarly, carrying the units just means not forgetting them in your calculations

.

A bicycle goes $7$ meters in $2$ seconds so

Here we have divided 7 by 2 and also divided $\mathrm{m}$ by $\mathrm{s}$. But a meter ($\mathrm{m}$) is not a number, nor is a second ($\mathrm{s}$). So the ratio $\mathrm{m}/\mathrm{s}$ cannot be reduced more. In particular, the $\mathrm{m}/\mathrm{s}$ is not sitting next to the equation but is part of the equation: the velocity is not $v=3.5$ but rather $v=3.5\mathrm{m}/\mathrm{s}$.

The rest of this section is, more or less, a discussion of how and why to ‘carry your units.’

Distance has dimensions of length $[L]$ that can be measured with various units — centimeters ($\mathrm{cm}$), yards ( yd), or furlongs (an obsolete unit equal to $1/8$ mile). A meter is the standard unit of length in the SI system. In answer to the question ‘What is the length of a bicycle crank $\mathrm{\ell }$?’ we say ‘$\mathrm{\ell }$ is seven inches’ and write $\mathrm{\ell }=7\in$ or say ‘$\mathrm{\ell }$ is seventeen point seven centimeters’ and write $\mathrm{\ell }=17.7\mathrm{cm}$. In each case, a number multiplies a dimensional unit.

Force has dimension of mass times acceleration $[m\cdot a]$. Because acceleration itself has dimensions of length over time squared $[L/T^2]$, force also has dimensions of mass times length divided by time squared $[M\cdot L/T^2]$. Because force has such a central role in mechanics, it is often convenient to think of force as having its own units. Force then has dimensions of, simply, force $[F]$. The most common units for force are Newton ( N) and the pound force ($\mathrm{lbf}$). The ‘f’ in the notation for the pound $\mathrm{lbf}$ is to distinguish a pound force $\mathrm{lbf}$ from the pound mass $\mathrm{lbm}$, $1\mathrm{lbf}=\mathrm{lbm}\cdot g\approx 1\mathrm{lbm}\cdot 32.2\mathrm{ft}/{\mathrm{s}}^2\approx 32.2\mathrm{lbm}\cdot \mathrm{ft}/{\mathrm{s}}^2$. Some people use $\mathrm{lb}$ to mean pound force or pound mass, depending on context. To avoid confusion we use $\mathrm{lbm}$ for pound mass and $\mathrm{lbf}$ for pound force (see box 4.1 on page 4.1).

Units and dimensions are part of scientific notation just as spelling, punctuation, and grammar are parts of English composition. If used properly, they aid both thinking and the communication of these thoughts to others. If units and dimensions are used improperly they can impede communication, even with oneself, and convey the wrong meaning.

A gadget that breaks with a $300\text{N}$ $(300\mathrm{Newtons})$ load instead of a needed $300\mathrm{lbf}$ ($300\text{pounds}$ force) load is exactly as bad as one that breaks with a $67\mathrm{lbf}$ load instead of a needed $300\mathrm{lbf}$ load. An unsatisfied consumer will not be placated by learning that the engineer’s calculation was ‘numerically correct’.

If anybody is ever to use your calculation, giving them the wrong units is just as bad as giving them the wrong numerical value.

Although using units properly often seems annoyingly tedious, it also often pays. If units are carried through honestly, not just tagged on to the end of an equation for appearance, you can check your work for dimensional consistency. If you are trying to find a speed and your answer comes out $13\mathrm{kg}\mathrm{\cdot }\mathrm{m}/\mathrm{s}$, you know you have made a mistake — $\mathrm{kg}\mathrm{\cdot }\mathrm{m}/\mathrm{s}$ just isn’t a speed. Such dimensional errors in a calculation often reveal corresponding algebraic or conceptual mistakes. Also, if a problem is based on data with mixed units, such as $\mathrm{cm}$ and  meters, or pound force and pound mass, you may often not know the units of your answer unless you properly ‘carry’ your units^††margin: ${(1\mathrm{lbf}/1\mathrm{pdl})}^2\approx 1000$. You can easily generate errors of approximately a factor of $1000$ with English units if you wishfully multiply or divide answers by $32.2$ (the value of $g$ in $\mathrm{ft}/{\mathrm{s}}^2$) at the end of a sloppy calculation. If you do it wrong you get an error of a factor of ${32.2}^2$ which is 3% greater than $1000$. Following sloppiness with unscrupulousness, some are tempted to then slide a decimal point three places to the right or left to ‘fix’ things. The decrepit insecurity that provokes such despicable crimes is avoided by going to a church, temple or mosque regularly, or by making a habit of carrying units. .

People are most pleased if you speak their language, speak correctly, and make sense. Similarly, scientists and engineers with whom you communicate will be most comfortable if you use the units they use and use them with correct notation. But most importantly, you should use units in a way that makes physical sense.

Just as the United Nations argues over which language to use for communication, educators, editors, and makers of standards have argued for decades over conventions for units: whether they should come in multiples of $10$, whether they should use the standard international scientific conventions, and whether they will be clear to someone who has worked in the stock room of a supplier of $\frac{1}{2}$-inch bolts for 35 years and thinks SI might be a friend of his cousin Amil.

Even if you are not fluent in someone’s favorite language, you can still say sensible things. Similarly, no matter what you or your work place’s choice of units (SI, English, or hodgepodge), no matter whether you use upper case and lower case correctly, you should make sense. Physically sensible units — that is, balanced units — should be used to make your equations dimensionally correct. Then you should work on refining your notation so as to be more professional.

So, in order of importance,

• 1.

  Use balanced units.

• 2.

  Use units of the type that are liked by your co-workers.

• 3.

  Spell and punctuate these units correctly.

If you are in a situation where your only problem is the third item on the list you are doing fine, unless you are really fussy, or work for someone who is really fussy. (e.g., the authors of this book only try for the first two items on this list.)

Calculators and computers generally do not keep track of units for you. In order for your numerical calculations to make sense you have the following choices.

For example, all lengths should be in the same units and the unit of force should equal the unit of mass times the unit of distance divided by the unit of time squared. Each row of the table below defines a consistent set of units for mechanics.

The radian is the unit of angle in all consistent unit systems. Whether or not a radian is a proper unit or not is an issue of some philosophical debate. Practically speaking, you can generally replace 1 radian with the number 1.

What, in the SI system, is the net braking force when a $2000\mathrm{lbm}$ car skids to a stop on level ground? For this units problem we skip the careful mechanics and just work with the formula

where $m$ is the mass of the car, $g$ is the local gravitational constant and $\mu$ is the coefficient of friction for sliding between the tire and the road. We won’t be off by more than a quarter of a percent using the standard rather than the local value of the gravitational constant, $g=32.2\mathrm{ft}/{\mathrm{s}}^2$. The coefficient of friction for rubber and dry road is about one, so we use $\mu =1$. We proceed by plugging in values into the formula and then multiplying by $1$ until things are in standard SI (Systémé Internationale) form. We use a table of units to make the various substitutions for 1.

A few of the detailed steps could be contracted. The approach below is only one, albeit an awkward one, of many routes to the answer.

The net braking force is $8.92\mathrm{kN}$. In each step of the calculation we accumulate what we had from the previous step and then multiply by 1, where 1 is the ratio of two quantities that have the same dimensions but different units.

Repeating, in engineering we do math not just with numbers, but with dimensional quantities. The bad habits of many of us not withstanding, there are good and useful standards for how to deal with units in calculations^††margin: An excellent description of good practice is the “Guide for the Use of the International System of Units (SI)” by Barry Taylor, 1998, NIST (National Institute of Standards and Technology) publication # 811. .

Many standard empirical formulas, formulas based on experience and not theory, are presented in an ‘undimensional’ or numerical form. The units are not part of the equations. We present the approach here, not because we want to promote it, we don’t. But we don’t want the more formal approach to units we advocate here to stop you from reading and using empirical sources.

For example, Mark’s Handbook for Mechanical Engineers (8th edition, page 8-138) presents the following useful formula to describe the working life of commercially manufactured ball bearings:

where
$L_{10}$ $=$ the number of hours that pass before 10% of the bearings fail, $N$ $=$ the rotational speed in revolutions per minute $C$ $=$ the rated load capacity of the bearing in $\mathrm{lbf}$, $P$ $=$ the actual load on the bearing in $\mathrm{lbf}$, and $K$ $=$ $3$ for ball bearings, $10/3$ for roller bearings.

In this approach the idea of dimensional consistency has been disguised for the sake of brevity. $L_{10}$, $N$, $C$, and $P$ are just numbers. Such an equation is sometimes called a ‘numerical equation’. It is a relation between numerical quantities. If you happen to know the rotation speed of the shaft in radians per second instead of revolutions per minute you will have to first convert before plugging in the formula. Unlike a dimensional formula, the formula does not help you to convert these units. An alternative to this ‘numerical formula’ approach for empirical formulas is in box 4.1 on page 4.1.

How can we write tables of values without the clutter of units in every entry?

The simplest way to use dimensionless variables, though not necessarily the best, is to do something that involves notational compromise. For example, let $x$ represent dimensionless distance rather than distance. That is, $x$ represents distance divided by $1\mathrm{mi}$. Similarly, $t$ is time divided by $1\mathrm{hr}$. And $dx/dt$ is dimensionless distance differentiated with respect to dimensionless time, which is, evidently, dimensionless speed. In this example, recovering the dimensional speed is common sense: speed is in $\mathrm{mi}/\mathrm{hr}$. The notational compromise is that $v$ is being used to represent both dimensional and dimensionless speed, with the precise meaning depending on context

Using notational compromise we can use the formula $x=vt$ with $v=50\mathrm{mi}/\mathrm{hr}$ to do a set of calculations. Say we want to know the distance $x$ every quarter of an hour for two hours. So we multiply $50$ by $.25,.5,.75,\mathrm{\dots }$ and thus make a table with two columns labeled $t(\mathrm{hr})$ and $x(\mathrm{mi})$.

This approach has some ambiguity to some eyes. Here is a more clear way to make the same table.

The exact meaning of the columns in the above example are a little ambiguous. We can make it more precise by labeling the columns as follows

That is, the columns of numbers are dimensionless. The first column, is the time divided by one hour the second is distance divided by one mile.

Finally, the way things are most often done in science, and sometimes in engineering practice, is to only use clearly defined and distinct dimensionless variables (i.e., not to use $v$ both for the speed and for the speed as measured in $\mathrm{m}/\mathrm{s}$. This approach is more precise, if cumbersome, than using $v$ to be both dimensional and dimensionless depending on context.

If we take $x$ to be dimensional distance, $t$ to be dimensional time, and $v$ to be dimensional speed, we can define new dimensionless variables. $t^{\ast }=t/(1\mathrm{hr})$, $x^{\ast }=x/(1\mathrm{mi})$, and $v^{\ast }=v/(1\mathrm{mi}/\mathrm{hr})$. Now there is no ambiguity: $x$ is dimensional and $x^{\ast }$ is dimensionless. Dividing the equation $x=vt$ on both sides by one mile, and multiplying the right side by $1$, in the form of $1=(1\mathrm{hr}/1\mathrm{hr})$ we get:

which is, using the dimensionless variables,

Because $v$ is $50\mathrm{mi}/\mathrm{hr}$, $v^{\ast }=50$ as we can show formally as follows:

The dimensionless speed $v^{\ast }$ is just the dimensionless number $50$. Now we can make a table by multiplying $50$ by $.25,.5,.75,\mathrm{\dots }$. The columns of the table can be labeled $t^{\ast }$ and $x^{\ast }$ and all variables are clearly defined.

In practice most non-theoretiticians will not go to the trouble of defining a whole set of dimensionless variables. But it can be helpful if you have got confused with the difference between a pound force and a pound mass, or from some variables being measured in meters and others in feet, etc.

A horse has a kinetic energy of 10,000 slug (furlongs/fortnight)², what is its kinetic energy in MKS units?