Re: [Boost-users] boost::units - converting from one derived_dimension to another across systems (imperial to metric)

Subject: Re: [Boost-users] boost::units - converting from one derived_dimension to another across systems (imperial to metric)
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2018-11-20 05:06:17

AMDG

On 11/19/2018 09:09 PM, Matt Vinson via Boost-users wrote:
>  Eisuke Kawashima: Thank You.  I'll have a look.  When I saw Boost.Units did not have imperial and we had to create it ourselves, I thought: the US is s 5 trillion dollar market, most of those markets need programming, all are on imperial.  Boost.Units needs imperial and programmers would find and use it.
> 
> Steven Watanabe:  Thank You.  
> This solution doesn't seem correct as it does not consider "length"....

How is it wrong?  length cancels out.  The divides version
will give exactly the same type.

> boost::units::derived_dimension<
> Â  boost::units::mass_base_dimension, 1,
> Â  boost::units::time_base_dimension, -2>::typeI thought I'd at it (L M T^-2 L-1) but this didn't do well.:Â Â  Â Â Â Â  typedef boost::units::derived_dimension<
> Â Â  Â Â Â  Â Â Â  Â boost::units::length_base_dimension, 1,
> Â Â  Â Â Â  Â Â Â  Â boost::units::mass_base_dimension, 1,
> Â Â  Â Â Â  Â Â Â  Â boost::units::time_base_dimension, -2,
> Â Â  Â Â Â  Â Â Â  Â boost::units::length_base_dimension, -1
> Â Â  Â Â Â  Â >::type
> Â Â  Â Â Â  Â Â Â  Â lineal_force_dimension;
> 

derived_dimension can't handle multiple uses of the
same base dimension.  Even if it did, it would just
cancel them out, so this serves to purpose.

> So, I tried the " typedef boost::mpl::divides<...>" solution as it has force and length.  I am closer, I hope.
> *.h------------------------------------------------------------------------------------------------------------------------------------------
> namespace dimensional_analysis {
> Â Â  Â Â  typedef boost::units::length_base_dimension::dimension_type length_dimension;
> Â Â  Â Â  typedef boost::units::mass_base_dimension::dimension_type mass_dimension;
> 
> Â Â  Â Â  typedef boost::units::make_system<
> Â Â  Â Â Â  Â Â Â  Â boost::units::us::inch_base_unit,
> Â Â  Â Â Â  Â Â Â  Â boost::units::us::pound_base_unit
> Â Â  Â Â Â  Â >::type ip_system;
> 
> Â Â Â  namespace lineal_force {
> Â Â  Â Â Â Â  typedef boost::mpl::divides<
> Â Â  Â Â Â  Â Â Â  Â boost::units::force_dimension,
> Â Â  Â Â Â  Â Â Â  Â boost::units::length_dimension>::type
> Â Â  Â Â Â  Â Â Â  Â lineal_force_dimension;
> Â Â  Â Â Â Â  namespace imperial {
> Â Â  Â Â Â  Â Â Â Â  typedef boost::units::unit<
> Â Â  Â Â Â  Â Â Â  Â Â Â  Â lineal_force_dimension,
> Â Â  Â Â Â  Â Â Â  Â Â Â Â  dimensional_analysis ::ip_system
> Â Â  Â Â Â  Â Â Â  Â > lineal_force_unit;
> Â Â  Â Â Â  Â Â Â  Â typedef boost::units::quantity<lineal_force_unit, double> pound_per_in_quantity;
> Â Â  Â Â Â  Â Â Â  Â BOOST_UNITS_STATIC_CONSTANT(pound_per_inch, pound_per_in_quantity);
> Â Â  Â Â Â Â  }
> Â Â  Â Â Â  Â namespace si {
> Â Â  Â Â Â  Â Â Â Â  typedef boost::units::unit<
> Â Â  Â Â Â  Â Â Â  Â Â Â  Â lineal_force_dimension,
> Â Â  Â Â Â  Â Â Â  Â Â Â  Â boost::units::si::system
> Â Â  Â Â Â  Â Â Â  Â > lineal_force_unit;

This is not the unit you want.  The SI
unit of force is N, not kg.  kilogram-force
is not currently provided by Boost.Units
and you'll need to define it yourself.
(As in your previous post.)

> Â Â  Â Â Â  Â Â Â  Â typedef boost::units::quantity<lineal_force_dimension, double> kg_per_meter_quantity;
> Â Â  Â Â Â  Â Â Â  Â BOOST_UNITS_STATIC_CONSTANT(kg_per_meter, kg_per_meter_quantity);
> Â Â  Â Â Â  Â }
> Â Â  Â }//lineal_force
> }//dimensional_analysis
>  <snip>

In Christ,
Steven Watanabe