Good point. The color photos especially of
Missouri show the same hue of ocean gray on monitors and in print, and match the Snyder and Short sample for late-WW2 ocean gray, so I conclude that this hue is accurate.
Chemistry offers a solution, and in more than one sense.
http://www.hnsa.org/doc/camo/index.htm has the USN WW2 instructions for blending ocean gray (5-O) and haze gray (5-H):
45 pints 5-O = 40 white + 5 tint --> 11.11% concentration of tint
42 pints 5-H = 40 white + 2 tint --> 4.76% concentration of tint
By those formulas:
3 ml 5-O contains 0.33 ml tint.
7 ml 5-H contains 0.33 ml tint.
7 ml 5-H = 3 ml 5-O + 4 ml white --> 4.76% concentration of tint
That is, blending of 3 ml ocean gray and 4 ml white yields 7 ml haze gray of the same hue.
The experiment will be to see whether the contrast between ocean gray and this blended haze gray matches the photographs of the actual destroyers. If the match is poor, then I'd conclude that these actual ships wore measure 32 colors, not measure 31.
Good point. The color photos especially of [i]Missouri[/i] show the same hue of ocean gray on monitors and in print, and match the Snyder and Short sample for late-WW2 ocean gray, so I conclude that this hue is accurate.
Chemistry offers a solution, and in more than one sense. [url]http://www.hnsa.org/doc/camo/index.htm[/url] has the USN WW2 instructions for blending ocean gray (5-O) and haze gray (5-H):
45 pints 5-O = 40 white + 5 tint --> 11.11% concentration of tint
42 pints 5-H = 40 white + 2 tint --> 4.76% concentration of tint
By those formulas:
3 ml 5-O contains 0.33 ml tint.
7 ml 5-H contains 0.33 ml tint.
7 ml 5-H = 3 ml 5-O + 4 ml white --> 4.76% concentration of tint
That is, blending of 3 ml ocean gray and 4 ml white yields 7 ml haze gray of the same hue.
The experiment will be to see whether the contrast between ocean gray and this blended haze gray matches the photographs of the actual destroyers. If the match is poor, then I'd conclude that these actual ships wore measure 32 colors, not measure 31.