.. _pitheorem:
Buckingham Pi Theorem
=====================
`Buckingham π theorem`_ states that an equation involving *n* number of
physical variables which are expressible in terms of *k* independent fundamental
physical quantities can be expressed in terms of *p = n - k* dimensionless
parameters.
.. testsetup:: *
from pint import UnitRegistry
ureg = UnitRegistry()
Q_ = ureg.Quantity
To start with a very simple case, consider that you want to find a dimensionless
quantity involving the magnitudes `V`, `T` and `L` with dimensions `[length]/[time]`,
`[time]` and `[length]` respectively.
.. doctest::
>>> from pint import pi_theorem
>>> pi_theorem({'V': '[length]/[time]', 'T': '[time]', 'L': '[length]'})
[{'V': 1.0, 'T': 1.0, 'L': -1.0}]
The result indicates that a dimensionless quantity can be obtained by
multiplying `V` by `T` and the inverse of `L`.
Which can be pretty printed using the `Pint` formatter:
.. doctest::
>>> from pint import formatter
>>> result = pi_theorem({'V': '[length]/[time]', 'T': '[time]', 'L': '[length]'})
>>> print(formatter(result[0].items()))
T * V / L
You can also apply the Buckingham π theorem associated to a Registry. In this case,
you can use derived dimensions such as speed:
.. doctest::
>>> from pint import UnitRegistry
>>> ureg = UnitRegistry()
>>> ureg.pi_theorem({'V': '[speed]', 'T': '[time]', 'L': '[length]'})
[{'V': 1.0, 'T': 1.0, 'L': -1.0}]
or unit names:
.. doctest::
>>> ureg.pi_theorem({'V': 'meter/second', 'T': 'second', 'L': 'meter'})
[{'V': 1.0, 'T': 1.0, 'L': -1.0}]
or quantities:
>>> Q_ = ureg.Quantity
>>> ureg.pi_theorem({'V': Q_(1, 'meter/second'),
... 'T': Q_(1, 'second'),
... 'L': Q_(1, 'meter')})
[{'V': 1.0, 'T': 1.0, 'L': -1.0}]
Application to the pendulum
---------------------------
There are 3 fundamental physical units in this equation: time, mass, and length, and 4 dimensional variables, T (oscillation period), M (mass), L (the length of the string), and g (earth gravity). Thus we need only 4 − 3 = 1 dimensionless parameter.
.. doctest::
>>> ureg.pi_theorem({'T': '[time]',
... 'M': '[mass]',
... 'L': '[length]',
... 'g': '[acceleration]'})
[{'T': 2.0, 'g': 1.0, 'L': -1.0}]
which means that the dimensionless quantity is:
.. math::
\Pi = \frac{g T^2}{L}
and therefore:
.. math::
T = constant \sqrt{\frac{L}{g}}
(In case you wonder, the constant is equal to 2 π, but this is outside the scope of this help)
Pressure loss in a pipe
-----------------------
What is the pressure loss `p` in a pipe with length `L` and diameter `D` for a fluid with density `d`, and viscosity `m` travelling with speed `v`? As pressure, mass, volume, viscosity and speed are defined as derived dimensions in the registry, we only need to explicitly write the density dimensions.
.. doctest::
>>> ureg.pi_theorem({'p': '[pressure]',
... 'L': '[length]',
... 'D': '[length]',
... 'd': '[mass]/[volume]',
... 'm': '[viscosity]',
... 'v': '[speed]'
... }) # doctest: +SKIP
[{'p': 1.0, 'm': -2.0, 'd': 1.0, 'L': 2.0}, {'v': 1.0, 'm': -1.0, 'd': 1.0, 'L': 1.0}, {'L': -1.0, 'D': 1.0}]
The second dimensionless quantity is the `Reynolds Number`_
.. _`Buckingham π theorem`: http://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem
.. _`Reynolds Number`: http://en.wikipedia.org/wiki/Reynolds_number