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.

Checks that use this method

Findings this method raised

Every task the check resting on this method put on the review queue, open and closed, with where each one stands.

Reading the findings…

Task	Record	Check	Status	Opened	Closed