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The US fluid ounce is based on the US gallon, which in turn is based on the wine gallon of 231 cubic inches that was used in the United Kingdom prior to 1824. With the adoption of the international inch , the US fluid ounce became 1 ⁄ 128 gal × 231 in 3 /gal × (2.54 cm/in) 3 = 29.5735295625 mL exactly, or about 4% larger than the imperial unit.
The tun (Old English: tunne, Latin: tunellus, Medieval Latin: tunna) is an English unit of liquid volume (not weight), used for measuring wine, [1] oil or honey. Typically a large vat or vessel, most often holding 252 wine gallons, but occasionally other sizes (e.g. 256, 240 and 208 gallons) were also used. [2] The modern tun is about 954 litres.
[nb 2] Note that a 252-gallon tun of wine has a mass of approximately 2060 pounds, [4] between a short ton (2000 pounds) and a long ton (2240 pounds). The tun is approximately the volume of a cylinder with both diameter and height of 42 inches, as the gallon was originally a cylinder with diameter of 7 inches and height of 6.
A pipe is a tubular section or hollow cylinder, usually but not necessarily of circular cross-section, used mainly to convey substances which can flow — liquids and gases , slurries, powders and masses of small solids. It can also be used for structural applications; a hollow pipe is far stiffer per unit weight than the solid members.
Earlier, another size of whiskey barrel was the most common size; this was the 40 US gallons (33.3 imp gal; 151.4 L) barrel for proof spirits, which was of the same volume as five US bushels. However, by 1866, the oil barrel was standardized at 42 US gallons.
In US customary units, most units of volume exist both in a dry and a liquid version, with the same name, but different values: the dry hogshead, dry barrel, dry gallon, dry quart, dry pint, etc. The bushel and the peck are only used for dry goods. Imperial units of volume are the same for both dry and liquid goods. They have a different value ...
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Much more work is needed to find the volume if we use disc integration.First, we would need to solve = () for x.Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow.