Loaded weight

How much a full magazine should add to a gun's empty weight, and the test a record's two weights must pass.

The test

Equations 1 and 2 on the cartridge mass page give one round. A full magazine adds that many rounds and the magazine itself, each with its bounds, and a record's loaded weight minus its empty weight must fall between them.

m_{r}^{\pm} = m_{b} + k^{\pm} V, k^{-} = 1.8, k^{+} = 3.2 g {cm}^{- 3}

n m_{r}^{-} + M_{\min} \leq W_{L} - W_{E} \leq n m_{r}^{+} + M_{\max}

where

$m_{r}^{\pm}$: the round's lower and upper mass, from equation 2 with the lower and upper k
$n$: the magazine's capacity, in rounds
$M$: the empty magazine's mass: 15 to 450 g for a box, up to 2500 g for a drum or belt
$W_{L}, W_{E}$: the gun's loaded and empty weights as the record states them, in g

Table 1. How the difference between loaded and empty is judged. Rows are read in order; the first that applies decides.
#	When	Verdict	What it means
1	$W_{L} - W_{E} \leq 0$	Impossible	A loaded gun cannot weigh the same as, or less than, the same gun empty: one figure is the other measurement.
2	$n m_{r}^{-} + M_{\min} \leq W_{L} - W_{E} \leq n m_{r}^{+} + M_{\max}$	Holds	The rounds and the magazine account for the difference: the two weights agree.
3	$M_{\min} \leq W_{L} - W_{E} < n m_{r}^{-} + M_{\min}$	Review	A magazine's worth and no rounds: the loaded figure is really the weight with an empty magazine.
4	Anything else	Review	Neither figure fits the other: both weights go to a person to check against a source.

The band is wider than the fit needs (1.8 to 3.2 g per cm³ and 15 to 450 g for the magazine) because rimfire and the largest rounds have no stated mass to fit to. A difference below the rounds alone but inside a magazine's weight is reported as a loaded figure that is really the weight with an empty magazine: the commonest mistake by far.

Worked example: a Glock 17

From its record: a 7.45 g bullet in a 9x19mm case (9.93 mm base, 19.15 mm long), 17 rounds, 625 g empty and 863 g loaded.

Case envelope (1)	$V = \frac{π}{4} \times {0.993}^{2} \times 1.915 = 1.483 {cm}^{3}$
One round (2)	$m_{r} = 7.45 + 2.5 \times 1.483 = 11.16 g$
17 rounds	$n m_{r} = 17 \times 11.16 = 189.7 g$
A full magazine should add (3, 4)	$187 g \leq W_{L} - W_{E} \leq 657 g$
The record's difference	$W_{L} - W_{E} = 863 - 625 = 238 g$

A full magazine should add

187 g657 grecorded: 238 g

Answer: 238 g lies inside 187 to 657 g, so the loaded and empty weights agree.

The test

Worked example: a Glock 17

Checks that use this method