symunit

Before specifying units, load units by using symunit. Then, specify a unit by using dot notation.

Specify a length of 3 meters. You can also use aliases u.meter or u.metre.

u = symunit;
length = 3*u.m

length = $$3\hspace{0.17em}\mathrm{m}\mathrm{"meter\; -\; a\; physical\; unit\; of\; length."}$$

Specify the acceleration due to gravity of 9.81 meters per second squared. Because units are symbolic expressions, numeric inputs are converted to exact symbolic values. Here, 9.81 is converted to 981/100.

g = 9.81*u.m/u.s^2

g = 
$$\frac{981}{100}\hspace{0.17em}\frac{\mathrm{m}\mathrm{"meter\; -\; a\; physical\; unit\; of\; length."}}{{\mathrm{s}\mathrm{"second\; -\; a\; physical\; unit\; of\; time."}}^2}$$

For more information about the differences between symbolic and numeric arithmetic, see Choose Numeric or Symbolic Arithmetic.

Units behave like symbolic expressions when you perform standard operations on them. For numeric operations, separate the value from the units, substitute for any symbolic parameters, and convert the result to double.

Find the speed required to travel 5 km in 2 hours.

u = symunit;
d = 5*u.km;
t = 2*u.hr;
s = d/t

s = 
$$\frac{5}{2}\hspace{0.17em}\frac{\mathrm{km}\mathrm{"kilometer\; -\; a\; physical\; unit\; of\; length."}}{\mathrm{h}\mathrm{"hour\; -\; a\; physical\; unit\; of\; time."}}$$

The value 5/2 is symbolic. You may prefer double output, or require double output for a MATLAB® function that does not accept symbolic values. Convert to double by separating the numeric value using separateUnits and then using double.

[sNum,sUnits] = separateUnits(s)

sNum = 
$$\frac{5}{2}$$

sUnits = 
$$\frac{\mathrm{km}\mathrm{"kilometer\; -\; a\; physical\; unit\; of\; length."}}{\mathrm{h}\mathrm{"hour\; -\; a\; physical\; unit\; of\; time."}}$$

sNum = double(sNum)

sNum = 
2.5000

For the complete units workflow, see Units of Measurement Tutorial.

Use your preferred unit by rewriting units using unitConvert. Also, instead of specifying specific units, you can specify that the output should be in terms of SI units.

Calculate the force required to accelerate 2 kg by 5 m/s². The expression is not automatically rewritten in terms of Newtons.

u = symunit;
m = 2*u.kg;
a = 5*u.m/u.s^2;
F = m*a

F = 
$$10\hspace{0.17em}\frac{\mathrm{kg}\mathrm{"kilogram\; -\; a\; physical\; unit\; of\; mass."}\hspace{0.17em}\mathrm{m}\mathrm{"meter\; -\; a\; physical\; unit\; of\; length."}}{{\mathrm{s}\mathrm{"second\; -\; a\; physical\; unit\; of\; time."}}^2}$$

Convert the expression to newtons by using unitConvert.

F = unitConvert(F,u.N)

F = $$10\hspace{0.17em}\mathrm{N}\mathrm{"newton\; -\; a\; physical\; unit\; of\; force."}$$

Convert 5 cm to inches.

length = 5*u.cm;
length = unitConvert(length,u.in)

length = 
$$\frac{250}{127}\hspace{0.17em}\mathrm{in}\mathrm{"inch\; -\; a\; physical\; unit\; of\; length."}$$

Convert length to SI units. The result is in meters.

length = unitConvert(length,'SI')

length = 
$$\frac{1}{20}\hspace{0.17em}\mathrm{m}\mathrm{"meter\; -\; a\; physical\; unit\; of\; length."}$$

Simplify expressions containing units of the same dimension by using simplify. Units are not automatically simplified or checked for consistency unless you call simplify.

u = symunit;
expr = 300*u.cm + 40*u.inch + 2*u.m

expr = $$300\hspace{0.17em}\mathrm{cm}\mathrm{"centimeter\; -\; a\; physical\; unit\; of\; length."}+40\hspace{0.17em}\mathrm{in}\mathrm{"inch\; -\; a\; physical\; unit\; of\; length."}+2\hspace{0.17em}\mathrm{m}\mathrm{"meter\; -\; a\; physical\; unit\; of\; length."}$$

expr = simplify(expr)

expr = 
$$\frac{752}{125}\hspace{0.17em}\mathrm{m}\mathrm{"meter\; -\; a\; physical\; unit\; of\; length."}$$

simplify automatically chooses the unit to return. To convert to a specific unit, use unitConvert.

expr = unitConvert(expr,u.ft)

expr = 
$$\frac{7520}{381}\hspace{0.17em}\mathrm{ft}\mathrm{"foot\; -\; a\; physical\; unit\; of\; length."}$$

By default, temperatures are assumed to represent temperature differences. For example, 5*u.Celsius represents a temperature difference of 5 degrees Celsius. This assumption allows arithmetical operations on temperature values and conversion between temperature scales.

To represent absolute temperatures, use degrees kelvin so that you do not have to distinguish an absolute temperature from a temperature difference.

Convert 23 degrees Celsius to K, treating the temperature first as a temperature difference and then as an absolute temperature.

u = symunit;
T = 23*u.Celsius;
diffK = unitConvert(T,u.K)

diffK = $$23\hspace{0.17em}\mathrm{K}\mathrm{"kelvin\; -\; a\; physical\; unit\; of\; temperature."}$$

absK = unitConvert(T,u.K,'Temperature','absolute')

absK = 
$$\frac{5923}{20}\hspace{0.17em}\mathrm{K}\mathrm{"kelvin\; -\; a\; physical\; unit\; of\; temperature."}$$