Talk:Grave (unit)

message: This page reads more like a petition to the CGPM than an encyclopedia article. Grouse 15:56, 6 September 2006 (UTC)
replies: text: I think the article as it stands now is fairly good from a NPOV perspective. I have deleted an uncited POV section about what "many purists" believe about the name, and how it is superior to other potential names for the kilogram. Using weasel words does not make something less POV. Nor has the notability of this practice been established. Grouse 09:54, 9 September 2006 (UTC)
thread_title: POV

message: I reverted a dubious edit made by an anonymous user on 11 december 2010. To the existing sentence 'the decision of the Republican government to name this new unit the “kilogramme” had been mainly politically motivated, because the name “grave” was at that time considered politically incorrect' he added 'as it made for too easy puns about Robespierre's activities.' Which easy puns? In English someone might associate the word grave with death by guillotine and Robespierre, but not in French. Ceinturion (talk) 10:28, 4 August 2011 (UTC)
thread_title:  Easy puns about Robespierre's activities? 

message: Lavoisier had the equipment for the high precision measurement of the weight of 1 dm3 of water, and he was waiting for the completion of the new meridian measurement that would define the true meter and the true volume of 1 dm3. However, in 1793 he was asked to quickly provide a provisional value, less precise but good enough for the new currency system with a 1 franc coin that by law had to weigh 5.00 gram within 0.5%. He determined that the weight of 1 provisional cubic decimeter (1 dm3prov) of water at 0 °C, was 18841 grains (equivalent to a density of 999.67 g/dm3, in modern units;  0.09 g/dm3 less than the modern value at that temperature). After Lavoisier's death by guillotine, Lefevre-Gineau was asked to complete the high precision measurement. In 1798, when the meridian measurement was completed, he used Lavoisier's equipment and determined that the weight of 1 dm3 of water at 4.0 °C was 18827 grains (density 1000 g/dm3; 0.03 g/dm3 more than the modern value at that temperature).   
Lavoisier used a hollow brass cylinder and a precision balance to determine the density of water. The cylinder of known volume V was weighed first in air, and then submerged in water. Its apparent mass above water was M1, and under water M2. These masses were expressed in a traditional unit of mass, the grain. The mass of the displaced water was ΔM=M1-M2, so the density of water was ρwater=ΔM/V, in grain/dm3prov. Hence the grave, defined as the weight of 1 dm3prov water, was ρwater·(1 dm3prov). 
There is a remarkable discrepancy between the conditions specified in the definition and those in the actual measurements. The grave was defined as the weight of 1 dm3 of water at 0 °C in vacuum. However, Lavoisier was in a hurry and had no ice bath, so he measured the weight at 6 °C, and compensated by subtracting 0.01%. Water would boil and evaporate in vacuum, so he measured in atmospheric air, and compensated by adding 0.11%. 
Lefevre-Gineau defined the kilogram as 1 dm3 of water at 4 °C in vacuum. However, the temperature had to be steady for a fairly long time because the density measurement was repeated many times (~50) to improve the accuracy, so the central water bath was embedded in an ice water bath at a steady temperature of 0 °C. The water bath remained slightly warmer, 0,3 °C, to prevent unwanted ice formation on the brass cylinder. Lefevre-Gineau then calculated the density at 4 °C by adding 0.013%. Like Lavoisier he measured in ordinary atmospheric air instead of vacuum, and applied the same correction.
Ceinturion (talk) 16:26, 16 May 2016 (UTC)
thread_title:  Experimental detail 

message: This section is a copy-paste from the Kilogram article. See that article for details of the edit history. -Arch dude (talk) 19:30, 23 May 2019 (UTC)
thread_title:  Provenance for ''Kilogram des Archives'' section 

message: This wikipedia article  says that Lefevre's new "true" kilogram was 99.9265% of the old provisional kilogram. However, another wikipedia article (The history of the metric system ) says Lefevre's kilogram was 99.92072% of the provisional kilogram. These two conflicting values required some digging in the original report of Lefevre's work in 1799. Apparently some readers were confused by Lefevre's "arbitrary unit". 
According to the report, Lefevre wanted to weigh the hollow cylinder, in air and submerged in water, using a precision balance with a special set of coherent counterbalance weights that he made for this experiment. First he made an "arbitrary unit" which was for some reason about 0.01% heavier than the provisional kilogram, and he then carefully made a series of smaller weights, the decimal submultiples down to 1 milligram (-0.01%). 
The arbitrary unit weighed 18842.088 grains; the provisional kilogram weighed 18841 grains. 
The hollow cylinder displaced 11.2796203 dm3 of water at 4 °C, weighing 11.2706787 arbitray units. Therefore 1 dm3 of water at 4 °C weighs 18827.15 grains. This was Lefevre's new kilogram. 
So Lefevre's kilogram is 18827.15 / 18841 = 0.999265 of the provisional unit, and it is 18841.088 / 18841 = 0.9992072 of the arbitrary unit. So the statement in the other wikipedia article (The history of the metric system) is wrong.
I suppose the confusion is due to a particular sentence in the original report: "one single cubic decimetre of water at its maximum of density weighs 0.9992072 of the 'unity', which is the true kilogramme of the new metrical system." A sloppy reader might have missed that the previous paragraph explained that the 'unity' is the arbitrary unit: "The weights were arbitrary, though nearly the intended standard; it being of more consequence that they should be adjusted to the most precise equality, than that they should agree with any definite weight." Ceinturion (talk) 10:44, 10 July 2020 (UTC)
replies: text: This wikipedia article  provides Zupko (Revolution in Measurement ) as the source for the statement that Lefevre's kilogram was 99.9265% of the provisional kilogram. However, Zupko is not a source for that statement. Zupko does not contain the value "99.9265%". It only contains the weight of the provisional kilogram, which could be used in a calculation of this percentage, but unfortunately Zupko made a typo in that weight: 18848.25 instead of 18841. Ceinturion (talk) 11:27, 10 July 2020 (UTC)
thread_title:  True kilogram 99.9265% or 99.92072% of the provisional kilogram?