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Home Test for Planck's Constant Calculation

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  Home Test for Planck's Constant Calculation Alan Folmsbee <omnilobe@gmail.com> - 2016-10-25 18:25 -0700
    Re: Home Test for Planck's Constant Calculation noTthaTguY <abu.kuanysh05@gmail.com> - 2016-10-26 12:16 -0700
      Re: Home Test for Planck's Constant Calculation Alan Folmsbee <omnilobe@gmail.com> - 2016-10-28 16:31 -0700
    Re: Home Test for Planck's Constant Calculation Serigo <invalid@invalid.com> - 2016-10-26 15:39 -0500
      Re: Home Test for Planck's Constant Calculation Alan Folmsbee <omnilobe@gmail.com> - 2016-10-27 12:46 -0700
        Re: Home Test for Planck's Constant Calculation poraty350@gmail.com - 2016-10-30 01:05 -0700
    Re: Home Test for Planck's Constant Calculation Poutnik <poutnik4nntp@gmail.com> - 2016-10-30 19:22 +0100

#602503 — Home Test for Planck's Constant Calculation

Here is the proposed formula for h, based on a home gravity test:

h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)

where 
z = height fallen in 1 second gravity test at radius R from planet center
a = Bohr Radius = .5 Angstrom
k = Coulomb Constant
m = proton mass
R = radius from planet center to test object that falls under gravity
N = number of baryons in planet
q = proton charge
G = Newton's Constant
Conclusion: Planck's Constant can be calculated from dropping a rock for 1 second on any star or planet. The test drops a rock a distance of z in a 1 second test. Calculate h for any planet, same result. z is 4.9033 meters on Earth with R = 6378000 meters, N = 3.569*10^51 baryons. The derivation is available. Do the test on your next planet or moon.

h = (zakm) * ((4 pi R)^2) / (NqG * 1 second) 

verified magnitude within 2% on Earth.

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#602579

the secondpower of pi-times-half of the diameter,
all muiltiplexed with mkaz by one GqN-th and by one second-th, but
not so sure of variable assignment

> h = (zakm) * ((4 pi R)^2) / (NqG * 1 second) 
> 
> verified magnitude within 2% on Earth.

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#602841

On Wednesday, October 26, 2016 at 9:16:27 AM UTC-10, noTthaTguY wrote:
> the secondpower of pi-times-half of the diameter,
> all muiltiplexed with mkaz by one GqN-th and by one second-th, but
> not so sure of variable assignment
> 
> > h = (zakm) * ((4 pi R)^2) / (NqG * 1 second) 
> > 
> > verified magnitude within 2% on Earth.

The 4 pi is squared for an area of a sphere around a baryon times 4 pi for the area of a sphere around a star. It is below:

Conclusion for Home Test of h: Gravity and Planck's Constant have been related in a formula. h has 4 factors that are logically grouping universal facts:

h = A*B*C*D

h = (za/N seconds) * (m/q) * (k/G) * (4 pi R)^2 

A = za / (N * 1.000000000 seconds) 

B = m/q is a ratio like in the Bohr Magneton formula

C = k/G is a ratio from force formulas with similar math

D = (4 pi R)^2  = 4 pi * 4 pi * R^2

D is using 4 pi as a factor for a sphere around a baryon, multiplied by a second factor of 4 pi for the area of a sphere around a star.

BC is called The Traction Ratio : it is mk / qG which is a force divided by a force.

A = za / (N * 1.000 seconds) = height fallen times Bohr radius per N per 1 second
The variables for the one second home test are z, N, and R, times a constant. Any time can be used other than 1 sec if the calculus version is used.

h = (zaR^2/(N*(1 second))= mk/Gq * 16 pi^2 

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#602596

On 10/25/2016 8:25 PM, Alan Folmsbee wrote:
> Here is the proposed formula for h, based on a home gravity test:
>
> h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)

> where z = height fallen in 1 second gravity test at radius R from
> planet center a = Bohr Radius = .5 Angstrom k = Coulomb Constant m =
> proton mass R = radius from planet center to test object that falls
> under gravity N = number of baryons in planet q = proton charge G =
> Newton's Constant Conclusion: Planck's Constant can be calculated
> from dropping a rock for 1 second on any star or planet. The test
> drops a rock a distance of z in a 1 second test. Calculate h for any
> planet, same result. z is 4.9033 meters on Earth with R = 6378000
> meters, N = 3.569*10^51 baryons. The derivation is available. Do the
> test on your next planet or moon.
>
> h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)
>
> verified magnitude within 2% on Earth.
>

so, how do you know earths radius to 2% ?
how do you know the number of baryons in planet to 2%
what air pressure ?  Shape of rock ?

and you would have to know the accuracy of these z,a,k,m,R,N,q,G all 
within 0.02% to get to your 2% error, right ?

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#602689

On Wednesday, October 26, 2016 at 10:39:14 AM UTC-10, Serigo wrote:
> On 10/25/2016 8:25 PM, Alan Folmsbee wrote:
> > Here is the proposed formula for h, based on a home gravity test:
> >
> > h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)
> 
> > where z = height fallen in 1 second gravity test at radius R from
> > planet center a = Bohr Radius = .5 Angstrom k = Coulomb Constant m =
> > proton mass R = radius from planet center to test object that falls
> > under gravity N = number of baryons in planet q = proton charge G =
> > Newton's Constant Conclusion: Planck's Constant can be calculated
> > from dropping a rock for 1 second on any star or planet. The test
> > drops a rock a distance of z in a 1 second test. Calculate h for any
> > planet, same result. z is 4.9033 meters on Earth with R = 6378000
> > meters, N = 3.569*10^51 baryons. The derivation is available. Do the
> > test on your next planet or moon.
> >
> > h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)
> >
> > verified magnitude within 2% on Earth.
> >
> 
> so, how do you know earths radius to 2% ?
> how do you know the number of baryons in planet to 2%
> what air pressure ?  Shape of rock ?
> 
> and you would have to know the accuracy of these z,a,k,m,R,N,q,G all 
> within 0.02% to get to your 2% error, right ?

Hello Nothatguy and Sergio, the derivation is inspired by The Bohr 
Magneton (mu), which can 
be calculated using Error Bars for Sergio:

mu = q*h / (4 pi m) = Bohr Magneton for proton
so
h = 4 pi * mu * (m/q)

That formula for Planck's Constant is used as a model for my new formula. 
But instead of a "magneton" I use the baryon's momentum of free space (p) 
times the Bohr Radius (a).
N*p = z*A/(1 second)

h = 4 pi (zA/(N*1 second)) * a * (m/q)*(k/G)

A = 4 pi R^2 for area of star or planet in rock drop test of h

h = 4 pi (z*4*pi*R^2/(N*1 second)) * a * (m/q)*(k/G)
where m/q is dimensionless at abstraction level 2, and k/G is a dimensionless ratio using the mass-area theorem. The first 4 pi is for a proton spherical area, the second 4 pi is for the star's spherical area.

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
h = z*(4*pi*R)^2/(N*1 second)) * a * (m/q)*(k/G)
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

ERROR BARS INCLUDED using 10 digits
z is measured when you drop a rock for 1.000000000 seconds using an atomic clock
t = 1.000000000 second
z = 1/2 g t^2
g = 9.806650000 meter per second^2
z = 4.903320000 meter
R = 6,378,140.000 meters
N = M/m baryons in the star or planet
m = (1.674928600 + 1.6726231 )*10^-27 kg / 2 average of neutron and proton
m = 1.67377585 * 10^-27 kg
M = 5.972300000 * 10^24 kg
N = M/m = 3.56820000 * 10^51 baryons
a = 5.291772109 * 10^-11 meter
q = 1.602176621 * 10^-19 Coulombs
k = 8.987551787 * 10^9 (meter/second^2 when charge=area in E-continuum)
G = 6.674080000 *10^-11 9 (meter/second^2 when mass=area in G-continuum)
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

Error bar summary: The least precise variable is N, with 5 digits of precision. 
A loss of a digit would cause an error bar of 1/3568 = 0.028 %. 
Also,  G has only 6 digits of precision, so increase the error bar total to 0.04%.
Multiply the latest numbers' mantissas:

h = z*(4*pi*R)^2/(N*1 second)) * a * (m/q)*(k/G)
h = (774.301232 R^2/N) * 7.444542916
R^2 / N = 1.1400893 mantissa

h = 6.571838130 * 10^-34 Js

difference from standard h : 
6.626070040 - 6.571838130  = 0.05423191

Ratio
0.05423191 / 6.626070040  = 0.82% error
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

Conclusion, This suggests the mass of the Earth is 0.8% more than the current estimate.

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#602943

On Thursday, October 27, 2016 at 10:46:58 PM UTC+3, Alan Folmsbee wrote:
> On Wednesday, October 26, 2016 at 10:39:14 AM UTC-10, Serigo wrote:
> > On 10/25/2016 8:25 PM, Alan Folmsbee wrote:
> > > Here is the proposed formula for h, based on a home gravity test:
> > >
> > > h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)
> > 
> > > where z = height fallen in 1 second gravity test at radius R from
> > > planet center a = Bohr Radius = .5 Angstrom k = Coulomb Constant m =
> > > proton mass R = radius from planet center to test object that falls
> > > under gravity N = number of baryons in planet q = proton charge G =
> > > Newton's Constant Conclusion: Planck's Constant can be calculated
> > > from dropping a rock for 1 second on any star or planet. The test
> > > drops a rock a distance of z in a 1 second test. Calculate h for any
> > > planet, same result. z is 4.9033 meters on Earth with R = 6378000
> > > meters, N = 3.569*10^51 baryons. The derivation is available. Do the
> > > test on your next planet or moon.
> > >
> > > h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)
> > >
> > > verified magnitude within 2% on Earth.
> > >
> > 
> > so, how do you know earths radius to 2% ?
> > how do you know the number of baryons in planet to 2%
> > what air pressure ?  Shape of rock ?
> > 
> > and you would have to know the accuracy of these z,a,k,m,R,N,q,G all 
> > within 0.02% to get to your 2% error, right ?
> 
> Hello Nothatguy and Sergio, the derivation is inspired by The Bohr 
> Magneton (mu), which can 
> be calculated using Error Bars for Sergio:
> 
> mu = q*h / (4 pi m) = Bohr Magneton for proton
> so
> h = 4 pi * mu * (m/q)
> 
> That formula for Planck's Constant is used as a model for my new formula. 
> But instead of a "magneton" I use the baryon's momentum of free space (p) 
> times the Bohr Radius (a).
> N*p = z*A/(1 second)
> 
> h = 4 pi (zA/(N*1 second)) * a * (m/q)*(k/G)
> 
> A = 4 pi R^2 for area of star or planet in rock drop test of h
> 
> h = 4 pi (z*4*pi*R^2/(N*1 second)) * a * (m/q)*(k/G)
> where m/q is dimensionless at abstraction level 2, and k/G is a dimensionless ratio using the mass-area theorem. The first 4 pi is for a proton spherical area, the second 4 pi is for the star's spherical area.
> 
> $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
> h = z*(4*pi*R)^2/(N*1 second)) * a * (m/q)*(k/G)
> $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
> 
> ERROR BARS INCLUDED using 10 digits
> z is measured when you drop a rock for 1.000000000 seconds using an atomic clock
> t = 1.000000000 second
> z = 1/2 g t^2
> g = 9.806650000 meter per second^2
> z = 4.903320000 meter
> R = 6,378,140.000 meters
> N = M/m baryons in the star or planet
> m = (1.674928600 + 1.6726231 )*10^-27 kg / 2 average of neutron and proton
> m = 1.67377585 * 10^-27 kg
> M = 5.972300000 * 10^24 kg
> N = M/m = 3.56820000 * 10^51 baryons
> a = 5.291772109 * 10^-11 meter
> q = 1.602176621 * 10^-19 Coulombs
> k = 8.987551787 * 10^9 (meter/second^2 when charge=area in E-continuum)
> G = 6.674080000 *10^-11 9 (meter/second^2 when mass=area in G-continuum)
> $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
> 
> Error bar summary: The least precise variable is N, with 5 digits of precision. 
> A loss of a digit would cause an error bar of 1/3568 = 0.028 %. 
> Also,  G has only 6 digits of precision, so increase the error bar total to 0.04%.
> Multiply the latest numbers' mantissas:
> 
> h = z*(4*pi*R)^2/(N*1 second)) * a * (m/q)*(k/G)
> h = (774.301232 R^2/N) * 7.444542916
> R^2 / N = 1.1400893 mantissa
> 
> h = 6.571838130 * 10^-34 Js
> 
> difference from standard h : 
> 6.626070040 - 6.571838130  = 0.05423191
> 
> Ratio
> 0.05423191 / 6.626070040  = 0.82% error
> $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
> 
> Conclusion, This suggests the mass of the Earth is 0.8% more than the current estimate.
=====================
E photon = h f
is not good enough for you ???
----------
 OLD Catto said 
THE SIMPLER - THE BETTER !!
(:-)
Y.Porat
===========================================

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#602958

Dne 26/10/2016 v 03:25 Alan Folmsbee napsal(a):
> Here is the proposed formula for h, based on a home gravity test:
> 
> h = (zakm) * ((4 pi R)^2) / (NqG * 1 second)
> 
> where 
> z = height fallen in 1 second gravity test at radius R from planet center
> a = Bohr Radius = .5 Angstrom
> k = Coulomb Constant
> m = proton mass
> R = radius from planet center to test object that falls under gravity
> N = number of baryons in planet
> q = proton charge
> G = Newton's Constant
> Conclusion: Planck's Constant can be calculated from dropping a rock for 1 second on any star or planet. The test drops a rock a distance of z in a 1 second test. Calculate h for any planet, same result. z is 4.9033 meters on Earth with R = 6378000 meters, N = 3.569*10^51 baryons. The derivation is available. Do the test on your next planet or moon.
> 
> h = (zakm) * ((4 pi R)^2) / (NqG * 1 second) 
> 
> verified magnitude within 2% on Earth.
> 

Any calculation of h is useless,
if multiplicative/dividing terms
have bigger relative confidential interval then of the h.

It is the direct consequence of applying of error propagation rules

h = 6.626070040(81)E−34 	J⋅s 	
G = 6.67408(31)E−11   m3 kg− 1  s−2

https://en.wikipedia.org/wiki/Planck_constant
https://en.wikipedia.org/wiki/Gravitational_constant


-- 
Poutnik ( The Pilgrim, Der Wanderer )
Knowledge makes great men humble, but small men arrogant.

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