[QUOTE=VictordeHolland;472864]

You would think it would be possible to define the kg or gram using a similar technique. A xxxxxx amount of y atoms would come to mind (with for instance Carbon-12 with unified atomic mass (12u) and Avogadro's number for the amount of atoms). As Avogadro's number is defined as the amount of molecules/atoms in 12 gram of Carbon-12. As I understand it, this what they are trying to achieve with a "Silicon sphere kg".

Another approach is using E=mc^2 , as we already have defined c (the speed of light), but we would need something for E (energy). They are now proposing to use E=hv (Planck constant x frequency) or similar technique to define mass in terms of the Planck constant. But then we would need to define/measure the Planck constant first. As I understand it this is the approach they are now pursuing. The new measurement of the Planck constant to 13 parts per billion would meet the requirement for defining a 'new' kg and tying it to a universal constant.
[/QUOTE]

These are all theoretically "easy" methods of course; the limiting factor is not coming up with ideas for the precise redefinition, but rather choosing the new definition whose engineering requirements to reproduce the measurement are the least burden. It's not so hard to acquire some caseisum-133 and measure its state-transition properties, but e.g. things like the silicon sphere require extraordinarily difficult and expensive setups. The new definition must be practically usable.

An amusing alternative approach to a "standard kilogram" is mentioned in [url=https://arxiv.org/pdf/physics/0612087.pdf]A Proposed Exact Integer Value for Avogadro’s Number[/url]

[QUOTE=rogue;471889][URL="https://home.cern/about/updates/2017/11/how-much-does-kilogram-weigh"]How much does a kilogram weigh?[/URL][/QUOTE]Where?

[QUOTE=Dr Sardonicus;472871]An amusing alternative approach to a "standard kilogram" is mentioned in [url=https://arxiv.org/pdf/physics/0612087.pdf]A Proposed Exact Integer Value for Avogadro’s Number[/url][/QUOTE]

the problem then is you also can't define a kg becasue if this equates toa new 12 grams then the kg would have to be 83.333333 times it so the value must be divisible by 3 ( okay technically it is in this case).

[QUOTE=science_man_88;472876]the problem then is you also can't define a kg becasue if this equates toa new 12 grams then the kg would have to be 83.333333 times it so the value must be divisible by 3 ( okay technically it is in this case).[/QUOTE]
I doubt we'll ever experimentally find the value of the avogrado constant down to high enough precision that that would be an issue. Note that the constant is 6,xxx 10^23, so if you round off to the nearest whole atom, that is not going to dramatically significantly change the outcome.

[QUOTE=VictordeHolland;472864]Interesting!

The short answer would be 1000 gram :razz:.
[/QUOTE]
Shorter answer: 1kg. :razz:

This reminds me when I was in grade 2 or so, a joker uncle of mine came to visit and my parents told him ho good I was at math (I could multiply two and three digits numbers in my head, with some difficulty, but correct, from an early age, before I even went to school) and he told me "Really? Let's see... there are some geese on the street and they walk in line. How many geese there are if the one in front has two in the back and the one in the back has two in front?" and I replied very fast "Five!" and everybody laugh and I was very proud that he didn't trick me into saying "six", hehe... It didn't last long...

[QUOTE=Dr Sardonicus;472871]An amusing alternative approach to a "standard kilogram" is mentioned in [URL="https://arxiv.org/pdf/physics/0612087.pdf"]A Proposed Exact Integer Value for Avogadro’s Number[/URL][/QUOTE]
Beautiful! I love that! 84446886^3 ! We already memorized it!

[QUOTE=VictordeHolland;472908]I doubt we'll ever experimentally find the value of the avogrado constant down to high enough precision that that would be an issue. Note that the constant is 6,xxx 10^23, so if you round off to the nearest whole atom, that is not going to dramatically significantly change the outcome.[/QUOTE]

the new units actually fix the constants and base the units off them. so we would need to fix the constant and work from that. also rounding by .5 atoms would be off by 41.6666666666 atoms at the kg scale. we'd also need to drop some prefixes to get whole numbers for all of them.

[QUOTE=VictordeHolland;472908]I doubt we'll ever experimentally find the value of the avogrado constant down to high enough precision that that would be an issue. Note that the constant is 6,xxx 10^23, so if you round off to the nearest whole atom, that is not going to dramatically significantly change the outcome.[/QUOTE][QUOTE=science_man_88;472916]the new units actually fix the constants and base the units off them.
...[/QUOTE]The [url=https://www.bipm.org]Bureau International des poids et des mesures[/url] is indeed planning to set the Avogadro constant to an a defined number (an integer equal to 6,02214129 10^23) in their [url=https://www.bipm.org/en/measurement-units/rev-si/#communication]9th draft[/url] :"The number of entities in one mole is equal to the numerical value of the Avogadro constant which is known as the Avogadro number. As a consequence of the definition of the mole, the Avogadro constant, and the Avogadro number have no uncertainty."

The new proposed definitions and the history of the definitions of the basic units are an interesting reading.

Jacob

[QUOTE=S485122;472946]The [url=https://www.bipm.org]Bureau International des poids et des mesures[/url] is indeed planning to set the Avogadro constant to an a defined number (an integer equal to 6,02214129 10^23) in their [url=https://www.bipm.org/en/measurement-units/rev-si/#communication]9th draft[/url] :"The number of entities in one mole is equal to the numerical value of the Avogadro constant which is known as the Avogadro number. As a consequence of the definition of the mole, the Avogadro constant, and the Avogadro number have no uncertainty."

The new proposed definitions and the history of the definitions of the basic units are an interesting reading.

Jacob[/QUOTE]

luckily that divides by 3 as 602214129 divides by 3, which means the kg would be: 5.018451075 * 10^25 molecules exactly when this defines 12 grams. which means down to picograms are an exact whole number of molecules each.

[QUOTE=LaurV;472909]Shorter answer: 1kg. :razz:

This reminds me when I was in grade 2 or so, a joker uncle of mine came to visit and my parents told him ho good I was at math (I could multiply two and three digits numbers in my head, with some difficulty, but correct, from an early age, before I even went to school) and he told me "Really? Let's see... there are some geese on the street and they walk in line. How many geese there are if the one in front has two in the back and the one in the back has two in front?" and[U] I replied very fast "Five!" and everybody laugh and I was very proud that he didn't trick me into saying "six",[/U] hehe... It didn't last long...[/QUOTE]
I am going out on a limb, here. Why isn't the answer "three"?
[CODE]G1 G2 G3[/CODE]Goose 1 has 2 behind. Goose 3 has 2 in front.