Re: [Boost-users] boost::units primer explained / conversion factors required

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.

So, by your description:

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::type funky_factor_dimension

typedef unit funky_factor;

quantity factor(quantitysi::pressure p, quantitysi::length l, quantitysi::mass_density d) { return (p / l) / d; }

Alternatively:

template < typename System > quantity> factor( quantity> p , quantity> l , quantity> 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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