Share
Share with:
Link:
Copy link
It's not clear to me how to construct this experiment in Japanese. Even in text with katakana words, there aren't usually enough characters in a word to jumble them up to any great extent. And if there are katakana words, well, we can tell from context what phrase looks like ã‚±â€¦ã‚¸ãƒ»ãƒ¦â€¦ã‚£. So please explain, preferably with some evidence that the experiment has been done.Takashi @ Apr 13 2015, 10:58 AM wrote:The fact that the effectiveness of pattern match does not change when we write Japanese that is an entirely different language with Kana character.
------Takashi @ Mar 31 2015, 09:58 AM wrote:2.2.2 Units with special names and symbols; units that incorporate special names and symbols (p.117)
Among these names and symbols the last four entries in Table 3 are of particular note
since they were adopted by the 15th CGPM (1975, Resolutions 8 and 9; CR, 105 and
Metrologia, 1975, 11, 180), the 16th CGPM (1979, Resolution 5; CR, 100 and
Metrologia, 1980, 16, 56) and the 21st CGPM (1999, Resolution 12; CR, 334-335
and Metrologia, 2000, 37, 95) specifically with a view to safeguarding human health.
You define your UUS units in terms of natural constants. So, technically speaking, the "definition" of your units never changes. But if we get revised values for those natural constants, based on more and more accurate measurements, then the actual values of your units need to change accordingly, if they are to keep with their definitions. But how are we to know exactly how "big" those values are, if we cannot measure them against some kind of stable reference standard?Takashi @ Nov 15 2016, 08:46 AM wrote: Updating the numerical value of CODATA is not a problem of the UUS, it is a matter of the SI unit system.
This is so garbled. \(d\mathbf{S}\) is a differential of a vector area. The integral of this over some surface \(S\) is not a volume and is not zero. It is simply the total area of the surface, i.e. \(\displaystyle\iint_S d\mathbf{S} = S\). Really now, this is pretty elementary calculus.wendy.krieger @ Jan 19 2017, 09:59 AM wrote: The vector area A does not change when the surface changes. It is a geometric trick, that relies on volume not changing with coordinate, and that volume is the moment of surface.
(1) \( V = \int x \cdot dS\quad \) volume = sum of coordinate dot surface normal.
(2) \( \int dS = 0 \quad \) volume does not depend on coordinate, has zero sum