New Kilogram, Volt and Ohm are ready for SI-2018

[Cerebus]

The SI-2018 Volt will change by 0.00844 ppm, and the Ohm will change by 0.00015ppm, compared to the current definition of the Josephson and the Klitzing constants.

Calibrated 3458A's will be going cheap on ebay soon 

In your dreams buddy! cf

[dr.diesel]

I frequently dream of cheap 3458As, 732s and Data Proof scanners.

[Echo88]

Dont you already have two 3458A?  But a 732B and a Scanner would be nice, yes. Hint: On ebay.de theres at the moment a pricey Keithley 7168.

[dr.diesel]

Dont you already have two 3458A?

Good things come in pairs! 

[texaspyro]

I frequently dream of cheap 3458As, 732s and Data Proof scanners.

I intercepted your dreams before you had them... 

I have bought (all working properly):
4 x 3458A's at $2000 - $2400 ea.  Two with the high-spec Vref.
1 x 732B at $600
2 x 732A at $300 ea (one missing battery assembly)
1 x SR104 for $200!

Good things come to those who wait.  It helps when sellers mis-label their auctions.

[Andreas]

The SI-2018 Volt will change by 0.00844 ppm,

Mhm,

so I will have to correct all my LTZ reference readings and my ADC reference voltage calibration values: but in which direction?
Assuming the LTZ now reads 7.200000000V will it be 61nV more or less after the change?

With best regards

Andreas

[HighVoltage]

I intercepted your dreams before you had them... 

I have bought (all working properly):
4 x 3458A's at $2000 - $2400 ea.  Two with the high-spec Vref.
1 x 732B at $600
2 x 732A at $300 ea (one missing battery assembly)
1 x SR104 for $200!

Good things come to those who wait.  It helps when sellers mis-label their auctions.

WOW!
I am also looking for miss - labeled items but you seem to have a skill 10 fold better!
Congratulations.
Are you willing to sell the 732B at a reasonable profit?

[saturnin]

The SI-2018 Volt will change by 0.00844 ppm,

Mhm,

so I will have to correct all my LTZ reference readings and my ADC reference voltage calibration values: but in which direction?
Assuming the LTZ now reads 7.200000000V will it be 61nV more or less after the change?

With best regards

Andreas

JVS generates a voltage that is given by following formula: U_J(n) = nf/K_J, where n, an integer, is the step number of the voltage and K_J = 2e/h is the Josephson constant (e - elementary charge and h - Planck's constant). If I am not mistaken, the current volt is defined by using Josephson constant called K_J-90 which is agreed to be 483 597.9 x 10⁹ Hz*V^-1 (https://physics.nist.gov/cgi-bin/cuu/Value?kj90).

Newly defined SI units will be based on exactly defined fundamental constants (values of e and h will shift slightly). Therefore, there will be also new value of exactly defined Josephson constant (you can calculate it based on the proposed values of e and h). Comparison of new K_J and actual K_J-90 will give you magnitude and direction of change of the Volt unit (and your LTZ reference too ).

Please, correct me if I made mistake anywhere...

[Dr. Frank]

The SI-2018 Volt will change by 0.00844 ppm,

Mhm,

so I will have to correct all my LTZ reference readings and my ADC reference voltage calibration values: but in which direction?
Assuming the LTZ now reads 7.200000000V will it be 61nV more or less after the change?

With best regards

Andreas

Hey, that's a really good question, which brings another very important feature to my mind.
The change of -8.4E-9 is from the latest CODATA-2014 definition of Kj to the upcoming one.

As saturnin argues, 'conventional units' for Volt and Ohm have been used, that is an 'adopted value' for each of K_j and R_k, called K_j90 and R_k90!
Therefore, these conventional Volt₉₀ and Ohm₉₀ deviate from the SI-Volt and the SI-Ohm

Definitely this definition of conventional units had been used in the calibration of instruments from 1.1.1990 onwards, also creating a jump of 7..8ppm with that date. (I have reported my experience with my then 2 months old hp3458A @ university).
Reference here: https://www.bipm.org/en/publications/mises-en-pratique/electrical-units.html

The deviation between the Volt₉₀ and the new SI-2018 definition is as big as -0.11ppm, and for the Ohm it's +0.018ppm.
Both deviations are very well noticeable on the calibration data of high-grade Volt and Ohms standards, and measurable (in difference mode) with 8.5 digits DMMs.. provided, if this old 1990 definition of conventional units is still in use as of today.

At the moment, I can't find any hint at NIST, to which definition Volt calibrations are currently performed, I'll have to ask them.

For the future, from 2019 onwards, there already exists a draft for a new mise-en-pratique of the electrical units, which definitely foresees usage of K_j and R_k. That's logical, because from then on, the kilogram will be mise-en-pratique by the Kibble balance, and therefore the correct SI-definition must be used.
The final definition of e, h is not yet incorporated, see paragraphs 5. and 6.:
https://www.bipm.org/cc/CCEM/Allowed/26/CCEM-09-05.pdf

Therefore, I have to cancel my statement, maybe in 2019 there will be a considerable step in calibration of Volt and Ohm.

Frank

PS: on the PTB site, there's a new hint: The new SI will formally go effective on the World Metrology Day, that will be 20th May 2019.

[texaspyro]

Are you willing to sell the 732B at a reasonable profit?

Not even for an unreasonable profit!   Unless the price is very close to a new one.  That one came from France.   Reminds me, time to change the battery.

Oh, and I forgot to mention the $200 Fluke 720 KVD.  Listed properly.  For who knows what reason, I was the only bidder.   Seller was not too pleased with the closing price but accepted the deal.

[tggzzz]

Are you willing to sell the 732B at a reasonable profit?

Not even for an unreasonable profit!   Unless the price is very close to a new one.  That one came from France.   Reminds me, time to change the battery.

Oh, and I forgot to mention the $200 Fluke 720 KVD.  Listed properly.  For who knows what reason, I was the only bidder.   Seller was not too pleased with the closing price but accepted the deal.

I got my 7 decade KVD for £180. Listed as "Vintage Julie Research Laboratories New York VDR 106/7 Voltage Divider Mod A" for collection only, which is correct but doesn't mention it is a KVD.

I got it from an antiques dealer that had got it from an acquaintance suffering from dementia. When I entered the house it reeked of fags, and I mentally went "oh no". Fortunately the KVD had been briefly stored in an outdoor garage.

I'm still playing around with it using my Solartron 7081 in ration mode. So far the quick experiments show errors of the order of 0.5ppm, which is very acceptable.

No, it won't be for sale for a while, and I'm not entirely sure how to ship it since it weighs ~43lbs/20kg.

I could have picked this https://www.ebay.co.uk/itm/Vintage-Julie-Research-Laboratories-New-York-Null-Detector-ND106/282626942722 up at the same time, but I don't have the space.

[saturnin]

At the moment, I can't find any hint at NIST, to which definition Volt calibrations are currently performed, I'll have to ask them.

This document (ftp://ftp.nist.gov/pub/physics/neilz/papers/15_8_Measure_on_impacts.pdf) suggests the "conventional" volt, V₉₀, is used for calibrations (since 1990):

"...Thus, given the likely relative step change (see attached table) in voltage of 1×10^-7, it is likely that there will be a moderate step change for a number of calibrations..."

I didn't know there is such huge discrepancy between the volt in use (V₉₀) and SI volt. I thought it would be only a cosmetic (formal) update, but obviously not...

[Echo88]

My 720A came from ebay and was described as working, but the function switch was broken. So i bargained with the seller and paid only about 150€ in the end for it. I was able to repair the function switch with epoxy and love it. Extremely accurate, only the 6. and 7. decade have some slight dial contact corrosion, which produce errors at some settings. Fingers crossed that the function switch doesnt break again. In the future i gotta document the factory-changes it has built in, some reed-relays and connectors on the backside, for calibration purposes i assume.

[VintageNut]

I have two KVDs per my signature. The oldest one was $50 and looks horrible with scratches and missing binding post plastics. The other is a newer D model and looks very clean. I paid $300 for the nice newer one.

Both agree with one another very close. I can advance every digit of my calibrated KE 2000 DMM. Its really sweet to see every digit of the 10V and 1V range marching up and down.

Both KVDs are aged for decades and they appear to be very stable.

At this point I am waiting for a few calibration cycles of my 731Bs before I start to really use the KVDs. I need a good stable well-known voltage source to start adjusting some of the un-calibrated equipment.

[nixxon]

Follow-up video of current event from Veritasium:

https://youtu.be/c_e1wITe_ig

[metrologist]

Did they vote?

[schmitt trigger]

This is one tattoo that I would wear:

[tomato]

I whipped up a quick spreadsheet to estimate the rate of change of local g due to the apparent movement of the Sun and Moon.

https://docs.google.com/spreadsheets/d/1KM6urh6FdOvXzF-nFcXSEUT2DH5a1Cda5evWhQFkZoo

The Sun adds an acceleration of +/- 6 mm/s^2 with a period of one day. A little less at Paris, but I'll ignore that. The Moon's effect is around 180 times less.

Adding the vectors together, the magnitude of the total changes the fastest when the body is at 90 degrees to the local gravity field i.e. rising or setting.

For the Sun, the relative change in the total local gravity is 4.4e-08 per second. For the Moon, 2.4e-10 per second.

If you're trying to determine local G -- and then 1 kg -- to 1e-8 then you're going to have to measure g and then use g within a quarter of a second. Or at the very least, include compensation in your calculations. Or work only at noon or midnight, at new moon or full moon, when things will change more slowly.

You need to check your calculations.  Tidal effects on local gravity (g) are less than 1 mgal, which is about 10e-6 relative.  (1 gal = 1 cm/s2,  g ~ 980 gal).  Tide models are good to about 1 µgal and commercial instruments can measure g with an absolute accuracy of 1 µgal in minutes. 

[texaspyro]

Lady Heather can calculate and plot the vertical gravity offset in micro-gals.  Also the solid earth tides in mm.

[LapTop006]

Did they vote?

It starts about two hours and 10 minutes into this:

[brucehoult]

I whipped up a quick spreadsheet to estimate the rate of change of local g due to the apparent movement of the Sun and Moon.

https://docs.google.com/spreadsheets/d/1KM6urh6FdOvXzF-nFcXSEUT2DH5a1Cda5evWhQFkZoo

The Sun adds an acceleration of +/- 6 mm/s^2 with a period of one day. A little less at Paris, but I'll ignore that. The Moon's effect is around 180 times less.

Adding the vectors together, the magnitude of the total changes the fastest when the body is at 90 degrees to the local gravity field i.e. rising or setting.

For the Sun, the relative change in the total local gravity is 4.4e-08 per second. For the Moon, 2.4e-10 per second.

If you're trying to determine local G -- and then 1 kg -- to 1e-8 then you're going to have to measure g and then use g within a quarter of a second. Or at the very least, include compensation in your calculations. Or work only at noon or midnight, at new moon or full moon, when things will change more slowly.

You need to check your calculations.  Tidal effects on local gravity (g) are less than 1 mgal, which is about 10e-6 relative.  (1 gal = 1 cm/s2,  g ~ 980 gal).  Tide models are good to about 1 µgal and commercial instruments can measure g with an absolute accuracy of 1 µgal in minutes.

Where is the error? It's a very simple calculation.

- mass of Earth, Moon, Sun are well established, as are the distances from the surface of the Earth to the centres of those bodies
- force due to gravity = G*m1*m2/r^2, acceleration = F/m1, so reduces to k*m/r^2
- the force from a spherically symmetrical body is the same as if the entire mass is concentrated at its centre
 

[tomato]

I whipped up a quick spreadsheet to estimate the rate of change of local g due to the apparent movement of the Sun and Moon.

https://docs.google.com/spreadsheets/d/1KM6urh6FdOvXzF-nFcXSEUT2DH5a1Cda5evWhQFkZoo

The Sun adds an acceleration of +/- 6 mm/s^2 with a period of one day. A little less at Paris, but I'll ignore that. The Moon's effect is around 180 times less.

Adding the vectors together, the magnitude of the total changes the fastest when the body is at 90 degrees to the local gravity field i.e. rising or setting.

For the Sun, the relative change in the total local gravity is 4.4e-08 per second. For the Moon, 2.4e-10 per second.

If you're trying to determine local G -- and then 1 kg -- to 1e-8 then you're going to have to measure g and then use g within a quarter of a second. Or at the very least, include compensation in your calculations. Or work only at noon or midnight, at new moon or full moon, when things will change more slowly.

You need to check your calculations.  Tidal effects on local gravity (g) are less than 1 mgal, which is about 10e-6 relative.  (1 gal = 1 cm/s2,  g ~ 980 gal).  Tide models are good to about 1 µgal and commercial instruments can measure g with an absolute accuracy of 1 µgal in minutes.

Where is the error? It's a very simple calculation.

- mass of Earth, Moon, Sun are well established, as are the distances from the surface of the Earth to the centres of those bodies
- force due to gravity = G*m1*m2/r^2, acceleration = F/m1, so reduces to k*m/r^2
- the force from a spherically symmetrical body is the same as if the entire mass is concentrated at its centre

You calculated the forces at the mean earth-sun and earth-moon distances, and then assumed that is how much they vary over the course of a day.  The forces actually vary by a small fraction of that as the earth rotates (and revolves around the sun), because the earth-sun and earth-moon distances vary by a small fraction.

[brucehoult]

I whipped up a quick spreadsheet to estimate the rate of change of local g due to the apparent movement of the Sun and Moon.

https://docs.google.com/spreadsheets/d/1KM6urh6FdOvXzF-nFcXSEUT2DH5a1Cda5evWhQFkZoo

The Sun adds an acceleration of +/- 6 mm/s^2 with a period of one day. A little less at Paris, but I'll ignore that. The Moon's effect is around 180 times less.

Adding the vectors together, the magnitude of the total changes the fastest when the body is at 90 degrees to the local gravity field i.e. rising or setting.

For the Sun, the relative change in the total local gravity is 4.4e-08 per second. For the Moon, 2.4e-10 per second.

If you're trying to determine local G -- and then 1 kg -- to 1e-8 then you're going to have to measure g and then use g within a quarter of a second. Or at the very least, include compensation in your calculations. Or work only at noon or midnight, at new moon or full moon, when things will change more slowly.

You need to check your calculations.  Tidal effects on local gravity (g) are less than 1 mgal, which is about 10e-6 relative.  (1 gal = 1 cm/s2,  g ~ 980 gal).  Tide models are good to about 1 µgal and commercial instruments can measure g with an absolute accuracy of 1 µgal in minutes.

Where is the error? It's a very simple calculation.

- mass of Earth, Moon, Sun are well established, as are the distances from the surface of the Earth to the centres of those bodies
- force due to gravity = G*m1*m2/r^2, acceleration = F/m1, so reduces to k*m/r^2
- the force from a spherically symmetrical body is the same as if the entire mass is concentrated at its centre

You calculated the forces at the mean earth-sun and earth-moon distances, and then assumed that is how much they vary over the course of a day.  The forces actually vary by a small fraction of that as the earth rotates (and revolves around the sun), because the earth-sun and earth-moon distances vary by a small fraction.

I assumed that the magnitude of the acceleration from, for example, the Sun is near enough to constant over the course of a day, but varies in direction relative to the local vertical for someone standing on the surface of the Earth in the tropics.

[texaspyro]

Here is a plot showing the vertical gravity offset and solid earth tides over a 24 hour period in Dallas, Texas.

[tomato]

I assumed that the magnitude of the acceleration from, for example, the Sun is near enough to constant over the course of a day, but varies in direction relative to the local vertical for someone standing on the surface of the Earth in the tropics.

Let's look at a simple case.  You calculated a change of ± 6mm/s2 due to the sun, which is roughly 0.1% relative change. At noon everything should weigh about 0.1% less (because the sun would pull "up" on you) and at midnight everything should weigh about 0.1% (because the sun would pull "down" on you.)  That's would be huge change -- one that would disrupt all kinds of activities. 

The problem is that your calculation basically assumes that the earth and sun are separated by a distance (r), but they are motionless in space. Of course, they are not motionless, they are orbiting around their center of mass.  From our perspective, we are in a constant state of free fall (acceleration) towards the sun, and this free fall is just the amount to cancel the force you have calculated. 

Tidal forces arise because the gravitational force from the sun does vary slightly on the surface of the earth, because the distance from the sun varies slightly as the earth spins.  This force is much smaller, however, than the 6 mm/s2 you calculated. (Basically, you need to scale your result by (r/R), where r= radius of earth and R = earth-sun distance.)