Hmm, let's see. I think it's more like this, (ML^-1T^-2 * L^-1) *   ML^-3 * (ML^-1T^-2 * L^-1)^-1 = L^-1T^-2, Pressure Gradient divided by Fluid (or Mass) Density.

I think I follow; doing my home work, one bar being a kilopascal (kPa), or (10e3 * NL^-2), or (10e3 * MLT^-2), so we have (10e3 * MLT^-2 * L^-1).

Okay, then yes we divide by mass density of ML^-3, or multiply by (ML^-3)^-1 if you prefer.

So we have (10e3 * MLT^-2 * L^-1) * (ML^-3)^-1.

Hope my rusty dimensional analysis skills are showing... :-)

Okay, so we can do some reductions I think, (10e3 * L^3 * T^-2). Am I reading this correctly? Is this the rate at which a volume transfers? Something along these lines. Really not up on my dimensional analysis like I should be; but I WILL be.

However it reduced, please verify I am reducing correctly, I don't think the units are supposed to make sense; we're arriving at an intermediate conversion factor I believe. At least that's how it is explained to me.

On Thu, Jul 21, 2011 at 5:23 PM, Noah Roberts wrote:
On 7/21/2011 3:16 PM, Michael Powell wrote:
Okay, here's what we need to get at, for starters. And maybe an
illustration or three and a little exchange will go a long way towards
helping my better comprehend units.

I'm starting with a set of SI calculations for oil and gas constants
calculations. Eventually we will need to accommodate US units as well.
But not quite yet.

We need to get at a calculation involving Pressure Gradient, which ends
up being metric::bar/si::meter (bars over meters) in specific units, or
I suppose si::pressure/si::length might also work.

Then we need to get after Fluid Density, which ends up being
si::kilogram/si:meter^3 (kilograms over cubic meters) in specific units,
or I suppose si::mass/si::meter^3 (I don't know what this looks like in
terms of boost::units, maybe one of the volumes?), or perhaps make use
of mass_density?

Yes, it's mass_density.

We take all that and divide Pressure Gradient by Fluid Density to arrive
at what we hope will be the the conversion factor: 0.0000981. Which we
could specify that as a constant, but I like proving it through the
software first (plausibly once) when we ask for it.

ML^-3 * (ML^-1T^-2 * L^-1)^-1 = L^-1T^-2.

This latter part is the dimension your factor is in.

using namespace boost::units;

typedef derived_dimension<length_base_dimension, -1, time_base_dimension, -2>::type funky_factor_dimension

typedef unit<funky_factor_dimension, si::system> funky_factor;

quantity<funky_factor> factor(quantity<si::pressure> p, quantity<si::length> l, quantity<si::mass_density> d)
{
return (p / l) / d;
}

Alternatively:

template < typename System >
quantity<unit<funky_factor_dimension, System>>
factor( quantity<unit<pressure_dimension,System>> p
, quantity<unit<length_dimension,System>> l
, quantity<unit<mass_density_dimension,System>> d )
{
return (p / l) / d;

}

Similar type calculations would follow for US units involving gallons,
cubic inches, inches, and inches per foot, along these lines.

As long as you use a coherent system, the template version above should work (assuming I got all the types correct).

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