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Natural Units Environment

The Natural Units environment is an environment designed for computations with units. This environment is set up by using the command with(Units[Natural]), or alternatively by using the commands Units[UseMode](natural) and with(Units), in that order.

Various procedures are overloaded to handle units. For example, $\sin (3.0 degrees)$ evaluates the sine of 3 degrees and $\max (4 m, 10 ft)$ returns $4 ⟦m⟧$ .

Arithmetic operators are overloaded to extract units from their operands. For example, $3 m + 4 cm$ evaluates to $\frac{76}{25} ⟦m⟧$ and $\frac{53.32 km}{1.3 h}$ evaluates to $11.39316239 ⟦\frac{m}{s}⟧$ , which can be converted to $25.48577843 ⟦mph⟧$ .

Note: Although easier to use, it is slower, in general, to perform computations in the Natural Units environment than to use the conversion routines at the top-level.

If you do not want to interpret everything as a unit, you can either use the Units[Standard] package, where units must be specified with the Unit routine, or use the strict option to the Units[UseSystem] routine.

These computations can also be done in the Simple Units environment, the Standard Units environment, or at the top-level by using only conversion routines (see Default Units). Each example in this worksheet is also in the other worksheets to show how you can perform the computations in the other environments.

$restart$

>	$with (Units [Natural]) &colon;$

Simple Examples

Add 4 feet to 3 inches.

>	$4 ft + 3 inches$

$\frac{6477}{5000} ⟦m⟧$

(1.1)

>	$convert (, units, inches)$

$51 ⟦in⟧$

(1.2)

How many meters are equivalent to 4 yards?

$4 yd$

$4 ⟦yd⟧$

(1.3)

>	$convert (4 yd, units, m)$

$\frac{2286}{625} ⟦m⟧$

(1.4)

How many liters are equivalent to 5 UK gallons?

>	$convert (5 gal [UK], units, L)$

$\frac{454609}{20000} ⟦L⟧$

(1.5)

How many liters are equivalent to 5 US gallons?

>	$convert (5 gal [US_liquid], units, L)$

$\frac{473176473}{25000000} ⟦L⟧$

(1.6)

How many US liquid gallons are equivalent to a UK gallon?

>	$convert (gal [US_liquid], units, gal [UK])$

$\frac{473176473}{568261250} ⟦{gal}_{UK}⟧$

(1.7)

Unit Names and Symbols

Unit symbols and various spellings of unit names are recognized by the package.

>	$32000 m, 32000 meter, 32000 metres$

$32000 ⟦m⟧, 32000 ⟦m⟧, 32000 ⟦m⟧$

(2.1)

This feature is expandable so that, for example, you can specify metr (the Polish word for this unit) as an alternate spelling of meter. See the AddUnit help page for more information.

Some units can be recognized with SI or IEC prefixes.

>	$32 km, 32 kilometer, 32 kilometres$

$32 ⟦km⟧, 32 ⟦km⟧, 32 ⟦km⟧$

(2.2)

Sample Questions with Solutions

How many miles do you travel in 35 minutes moving at 55 miles per hour?

>	$distance ≔ 55 mph \cdot 35 minutes$

$distance := \frac{1290828}{25} ⟦m⟧$

(3.1.1)

>	$distance ≔ convert (distance, units, mi)$

$distance := \frac{385}{12} ⟦mi⟧$

(3.1.2)

>	$evalf (distance)$

$32.08333333 ⟦mi⟧$

(3.1.3)

How many seconds are equivalent to 3 weeks?

>	$convert (3 week, units, s)$

$1814400 ⟦s⟧$

(3.2.1)

How many inches are equivalent to 5 feet 4 inches?

>	$convert (5 ft + 4 inches, units, inches)$

$64 ⟦in⟧$

(3.3.1)

Convert 50 km/h to cm/s.

The cm/s must be quoted. Otherwise, $\frac{cm}{s}$ evaluates to $\frac{m}{100 s}$ .

>	$convert (50 \frac{km}{h}, units,' cm / s')$

$\frac{12500}{9} ⟦\frac{cm}{s}⟧$

(3.4.1)

>	$evalf ()$

$13.88888889 ⟦\frac{m}{s}⟧$

(3.4.2)

How many seconds does it take an object, released from rest, to fall 20 meters?

From the equation $d = \frac{g_{n} t^{2}}{2}$ , we derive:

>	$elapsed_time ≔ \sqrt{2 \frac{dist}{accel}}$

$elapsed_time := \sqrt{2} \sqrt{\frac{dist}{accel}}$

(3.5.1)

>	$eval (elapsed_time, [dist = 20 m, accel = gn])$

$\frac{1}{196133} \sqrt{2} \sqrt{20} \sqrt{20000} \sqrt{196133} ⟦s⟧$

(3.5.2)

>	$evalf ()$

$2.019619976 ⟦s⟧$

(3.5.3)

What is the rest energy of an electron?

Assume the mass of an electron is $9.11 10^{- 31}$ kilograms.

>	$mass ≔ 9.11 \cdot 10^{−31} kg$

$mass := 9.110000000 10^{-31} ⟦kg⟧$

(3.6.1)

Approximate the speed of light by $3.00 10^{8}$ meters per second.

>	$c ≔ 3.00 \cdot 10^{8} \frac{m}{s}$

$c := 3.000000000 10^{8} ⟦\frac{m}{s}⟧$

(3.6.2)

Using the formula $E = m c^{2}$ , you can approximate the rest energy of the electron.

>	$E ≔ mass \cdot c^{2}$

$E := 8.199000000 10^{-14} ⟦J⟧$

(3.6.3)

A more precise answer can be found by converting the known unit the electron mass (em) to joules using energy conversions.

>	$convert (1 em, units, J, energy)$

$8.187104141 10^{-14} ⟦J⟧$

(3.6.4)

Approximately what volume does 1,000,000,000 US dollars worth of gold occupy?

>	$cost ≔ 1210.00 \frac{USD}{oz [troy]} &semi; &num; November 18th, 2016$

$cost ≔ 38902.40335 ⟦\frac{USD}{kg}⟧$

(3.7.1)

>	$density ≔ 19.3 \frac{g}{{cm}^{3}}$

$density ≔ 19300.0 ⟦\frac{kg}{m^{3}}⟧$

(3.7.2)

>	$mass ≔ \frac{10^{9} USD}{cost}$

$mass ≔ 25705.35273 ⟦kg⟧$

(3.7.3)

>	$volume ≔ \frac{mass}{density}$

$volume ≔ 1.331883561 ⟦m^{3}⟧$

(3.7.4)

What is the length of one side of a cube with this volume?

>	$length_side ≔ \sqrt[3]{volume}$

$length_side ≔ 1.100243351 ⟦m⟧$

(3.7.5)

An Su-27 Flanker can travel at mach 1.1 at sea level. How fast is this in miles per hour, miles per second, and meters per second?

>	$convert (1.1 M, units, mph)$

$815.6003937 ⟦mph⟧$

(3.8.1)

>	$convert (1.1, units, M, \frac{mi}{s})$

$0.2265556649$

(3.8.2)

>	$convert (1.1 M, units, \frac{m}{s})$

$364.606 ⟦\frac{m}{s}⟧$

(3.8.3)

Approximately how many meters are there in 3.5 miles?

>	$convert (3.5 mi, units, m)$

$5632.704000 ⟦m⟧$

(3.9.1)

>	$round ()$

$5633 ⟦m⟧$

(3.9.2)

Given 50 US gallons of water, how many 750 mL bottles could you fill?

>	$\frac{50 gal [US_liquid]}{750 mL}$

$\frac{157725491}{625000}$

(3.10.1)

>	$floor ()$

$252$

(3.10.2)

Given nylon with a linear mass density of 20 deniers, what length of thread is used in an object weighing 12 grams?

>	$\frac{12 g}{20 deniers}$

$5400 ⟦m⟧$

(3.11.1)

What is the volume in cubic inches of a 2 liter engine.

>	$convert (2 L, units, {inches}^{3})$

$\frac{250000000}{2048383} ⟦{in}^{3}⟧$

(3.12.1)

>	$evalf ()$

$122.0474882 ⟦{in}^{3}⟧$

(3.12.2)

For a given phenomena with a frequency of 1.420,405,761 GHz, find the:

1. period,

2. number of cycles per year, and

3. number of cycles since the beginning of the earth.

First you must convert the frequency from GHz to Hz (cycles per second).

>	$frequency ≔ 1.420405761 GHz$

$frequency := 1.420405761 ⟦GHz⟧$

(3.13.1)

The period is:

>	$\frac{1}{frequency}$

$7.040241792 10^{-10} ⟦s⟧$

(3.13.2)

>	$convert (, units, nanoseconds)$

$0.7040241792 ⟦ns⟧$

(3.13.3)

The number of cycles per year is:

>	$convert (frequency, units,' 1 / yr')$

$4.482363838 10^{16} ⟦\frac{1}{yr}⟧$

(3.13.4)

The number of cycles since the beginning of the earth is:

>	$frequency \cdot 10000000000 yr$

$4.482363838 10^{26}$

(3.13.5)

Given inductance and capacitance, find the resistance in microohms.

The following formula relates the resistance to the inductance and capacitance.

>	$resistance ≔ \sqrt{\frac{inductance}{capacitance}}$

$resistance ≔ \sqrt{\frac{inductance}{capacitance}}$

(3.14.1)

Use an inductance of 124 nanohenries and a capacitance of 3.52 microfarads.

>	$eval (resistance, [inductance = 124. nH, capacitance = 3.52 uF]) &semi;$

$0.005935256754 \sqrt{1000} ⟦Ω⟧$

(3.14.2)

>	$evalf ()$

$0.1876892984 ⟦Ω⟧$

(3.14.3)

>	$convert (, units, uOmega)$

$187689.2984 ⟦μΩ⟧$

(3.14.4)

Given a molar energy, find the mass energy in Btu's per pound.

>	$molar_energy ≔ 523.432 \frac{Btu}{mol (carbon)}$

$molar_energy := 5.518806676 10^{5} ⟦\frac{J}{mol (carbon)}⟧$

(3.15.1)

>	$carbon_mass ≔ 12 \frac{kg}{mol (carbon)}$

$carbon_mass := 12 ⟦\frac{kg}{mol (carbon)}⟧$

(3.15.2)

>	$mass_energy ≔ \frac{molar_energy}{carbon_mass}$

$mass_energy := 45990.05563 ⟦\frac{m^{2}}{s^{2}}⟧$

(3.15.3)

>	$convert (, units,' Btu / lb')$

$19.78539678 ⟦\frac{Btu}{lb}⟧$

(3.15.4)

Given a distance function, find the speed by differentiation, speed at 2.5 seconds and at 5 blinks by evaluation, and distance traveled between 1 and 2.5 seconds by integration and by subtraction of the distance function values.

Note: t is a unit of mass, the tonne. Therefore, t1 is used to represent the time variable.

>	$distance ≔ \frac{5}{1 + 4 \cdot {&ExponentialE;}^{- 3 \cdot t1}} m$

$distance ≔ \frac{5}{1 + 4 {&ExponentialE;}^{- 3 t1}} ⟦m⟧$

(3.16.1)

To find the speed function, differentiate the distance function.

>	$speed ≔ diff (distance, t1 \cdot s)$

$speed ≔ \frac{60 {&ExponentialE;}^{- 3 t1}}{{(1 + 4 {&ExponentialE;}^{- 3 t1})}^{2}} ⟦\frac{m}{s}⟧$

(3.16.2)

To find the speed at 2.5 seconds, evaluate the speed function at t1=2.5.

>	$\binom{speed}{} \| \binom{}{t1 = 2.5}$

$0.03303871496 ⟦\frac{m}{s}⟧$

(3.16.3)

To find the speed at 5 blinks, evaluate the speed function at t1=5. $⟦blink⟧$ , converted to seconds.

>	$dt ≔ convert (5., units, blink, s)$

$dt ≔ 4.320000000$

(3.16.4)

>	$\binom{speed}{} \| \binom{}{t1 = dt}$

$0.0001411518555 ⟦\frac{m}{s}⟧$

(3.16.5)

By using a definite integral, you can determine the distance traveled from the speed function.

>	$\int_{1}^{2.5} speed &DifferentialD; t1$

$0.8193365798 ⟦\frac{m}{s}⟧$

(3.16.6)

The distance traveled can also be calculated directly from the distance function.

>	$\binom{distance}{} \|\binom{}{t1 = 2.5} - \binom{distance}{}\| \binom{}{t1 = 1}$

$(4.988962733 - \frac{5}{1 + 4 {&ExponentialE;}^{- 3}}) ⟦m⟧$

(3.16.7)

>	$evalf ()$

$0.819336584 ⟦m⟧$

(3.16.8)

Find the minimum and maximum length of 1.2 yards, 1 meter, 3.2 feet, and 0.6 fathoms.

>	$\min (1.2 yd, 1 m, 3.2 ft, 0.6 fathom)$

$0.9753600000 ⟦m⟧$

(3.17.1)

>	$\max (1.2 yd, 1 m, 3.2 ft, 0.6 fathom)$

$1.097280000 ⟦m⟧$

(3.17.2)

Given a torque of 3 newton meters, how much energy is required to move a lever through 10 degrees?

The energy required is the product of the torque and the angle in radians.

>	$energy ≔ torque \cdot angle$

$energy := torque angle$

(3.18.1)

>	$eval (energy, [torque = 3 N \cdot m (radius), angle = 10 \deg])$

$\frac{1}{6} π ⟦J⟧$

(3.18.2)

>	$evalf ()$

$0.5235987758 ⟦J⟧$

(3.18.3)

The Hyper-X can travel at speeds up to 7200 miles per hour. How long would it take to circle the earth at maximum speed (assuming it could carry sufficient fuel)? How far does it travel in a 10 second flight at maximum speed?

To find the time to circle the earth, divide the distance by the speed.

The meter was originally defined as 1/10,000,000 th the distance from the North Pole to the Equator on the meridian passing through Paris. Therefore, 40,000 kilometers is a good approximation of the circumference of the earth.

>	$\frac{40000 km}{7200 mph}$

$\frac{156250000}{12573} ⟦s⟧$

(3.19.1)

>	$convert (, units, h)$

$\frac{390625}{113157} ⟦h⟧$

(3.19.2)

>	$evalf ()$

$3.452062179 ⟦h⟧$

(3.19.3)

To find the distance traveled, multiply the speed by the time.

>	$7200 mph \cdot 10 s$

$\frac{804672}{25} ⟦m⟧$

(3.19.4)

>	$evalf ()$

$32186.88000 ⟦m⟧$

(3.19.5)

>	$convert (, units, km)$

$32.18688000 ⟦km⟧$

(3.19.6)

Given 1032 UK gallons of oil, how many cylindrical cans with a height of 1.2 feet and diameter of 0.9 feet could you fill?

The volume of a cylinder is $h π {(\frac{d}{2})}^{2}$ .

>	$vol ≔ convert (1.2 ft \cdot π \cdot {(\frac{0.9 ft}{2})}^{2}, units, gal [UK])$

$vol := 4.755136685 ⟦{gal}_{UK}⟧$

(3.20.1)

>	$\frac{1032 gal [UK]}{vol}$

$217.0284617$

(3.20.2)

Given an power gain from 332 microwatts to 23 milliwatts, what is the gain in decibels? What would the decibel gain be if the increase were a voltage increase?

A gain is a quotient of the final value divided by the initial value.

>	$gain ≔ \frac{23 mW}{332 uW (base)}$

$gain := \frac{5750}{83} ⟦\frac{W}{W (base)}⟧$

(3.21.1)

To determine the decibel gain, first take the ln of the gain.

>	$\ln (gain)$

$\frac{1}{2} \ln (\frac{5750}{83}) ⟦Np⟧$

(3.21.2)

>	$convert (, units, dB)$

$\frac{10 \ln (\frac{5750}{83}) ⟦dB⟧}{\ln (10)}$

(3.21.3)

>	$evalf ()$

$18.40589752 ⟦dB⟧$

(3.21.4)

Power is proportional to the square of the voltage. Therefore, the decibel increase corresponding to the voltage gain should be a factor of 2 times that of the power gain.

>	$voltage_gain ≔ \frac{23 mV}{332 uV (base)}$

$voltage_gain := \frac{5750}{83} ⟦\frac{V}{V (base)}⟧$

(3.21.5)

>	$\ln (voltage_gain)$

$\ln (\frac{5750}{83}) ⟦Np⟧$

(3.21.6)

>	$convert (, units, dB)$

$\frac{20 \ln (\frac{5750}{83}) ⟦dB⟧}{\ln (10)}$

(3.21.7)

>	$evalf ()$

$36.81179504 ⟦dB⟧$

(3.21.8)

Given an initial velocity of 2.4 meters per second and an otherwise unspecified uniform acceleration, what is the generic formula for position? For starting at the origin and acceleration being gravity in the opposite direction to the initial velocity? After 0.4 seconds?

The general formula for velocity in a uniform acceleration is $v = v0 + a \cdot t$ . In order to use this formula in the Natural Units environment, we need to express this in terms of unit-free quantities:

>	$v0 ≔ v0U \cdot \frac{m}{s}$

$v0 ≔ v0U ⟦\frac{m}{s}⟧$

(3.22.1)

>	$a ≔ aU \cdot \frac{m}{s^{2}}$

$a ≔ aU ⟦\frac{m}{s^{2}}⟧$

(3.22.2)

>	$t ≔ tU \cdot s$

$t ≔ tU ⟦s⟧$

(3.22.3)

Now the unassigned variables, $v0U$ , $aU$ , and $tU$ , are unit-free quantities, in terms of which we have defined the quantities $v0$ , $a$ , and $t$ as quantities with a unit.

>	$v ≔ v0 + a \cdot t$

$v ≔ (aU tU + v0U) ⟦\frac{m}{s}⟧$

(3.22.4)

Substitute the given starting velocity.

>	$vs ≔ \binom{v}{} \| \binom{}{v0U = 2.4}$

$vs ≔ (aU tU + 2.4) ⟦\frac{m}{s}⟧$

(3.22.5)

Integrating this over time answers the first question. This requires defining the initial location in terms of the unit-free quantity $x0U$ .

>	$x0 ≔ x0U \cdot m$

$x0 ≔ x0U ⟦m⟧$

(3.22.6)

>	$x ≔ x0 + int (vs, t = 0 .. t0U)$

$x ≔ (x0U + 0.5000000000 aU {t0U}^{2} + 2.400000000 t0U) ⟦m⟧$

(3.22.7)

Starting at the origin and with the specified gravity:

>	$gravity ≔ evalf (ScientificConstants :- Constant (g)) &semi;$

$gravity ≔ 9.80665$

(3.22.8)

>	$\binom{x}{} \| \binom{}{[x0U = 0, aU = - gravity]}$

$(- 4.903325000 {t0U}^{2} + 2.400000000 t0U) ⟦m⟧$

(3.22.9)

After 0.4 seconds, this is the location:

>	$\binom{}{} \| \binom{}{t0U = 0.4}$

$0.1754680000 ⟦m⟧$

(3.22.10)

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