All the mathematical and logical support for this webpage is provided in: The Mathematical Laws of Nature.
Black holes have always been an enigma to the author of this webpage, even when no other explanation was offered for dark celestial objects. There was, and still is, no mathematical or logical argument for them.
To wit; you see a dark object in space. How would you explain it?
Dark simply means the object is not radiating electromagnetic energy (EME), which could mean one of two things;
a) It is not reflecting EME radiated onto it, or;
b) It is not generating EME within it; i.e. it is cold.
Hmmm!
'a)' is easily answered; the body is simply too far from any significant source of EME.
'b)' is also understandable; just because a body is cold, doesn't mean it doesn't exist.
Before blackholes were presented to the world, the apparent absence of a galactic forcecentre triggered the invention of dark matter, despite the fact that Newton's laws of orbital motion tell us that a gas cloud could never replace a forcecentre. These same scientists later invented an alternative theory that defied all known physics and logic (a hole in empty space!). This new theory did not comply with the 'conservation of energy' (which until then was accepted by every scientist since its discovery) and declared the existence of negative energy; something that remains unproven today. But even after replacing the need for universal dark matter  with the blackhole  most scientists today, still believe in the existence of dark matter, despite its original motive having been removed.
This approach to science has always been at the root of the author's scepticism.
Instead, those same scientists could simply have declared such bodies to be cold.
The apparently missing body at the centre of our galaxy, and in fact at the centre of all galaxies could be identified simply by using Isaac Newton's constant of proportionality;
K =t²/a³ = (2π)² / G.m₁
where:
t = the orbital period;
a = half the orbital major axis;
m₁ = the mass of the dark object.
Taking spin theory and core pressure into account, we could also estimate the body's density, diameter, and internal structure, despite it being invisible. All of it would have obeyed all the laws of known physics and the conservation of energy and required no invention.
In fact, any galactic body that has no satellites of its own to generate internal heat would appear dark to us. The back of Mercury, for example, would appear dark from outer space.
Our galactic orbital radius is approximately 1E+21m (NASA), whereas the bodyradius of our own galactic forcecentre is less than 3.5E+12m. The ratio of these two dimensions is less than 3.5E09. Therefore, searching for our galactic forcecentre would be like looking for a dark atom in the centre of a onemetre diameter black disk. That's why we can't see it!
The cold argument explains everything. So, one must ask the question; why is so much public money being squandered on searching for celestial objects that can't possibly exist (e.g. EHT)?
Before reading further, it is important for the reader to understand that whilst the maths provided below are correct, they are based upon a fundamental flaw; the photon
The original argument for black holes went something like this;
If a celestial object is dark, it must be absorbing light, therefore it must be sufficiently massive to prevent a photon from leaving its surface.
At no time did the inventor of black holes suggest that the body could be dark simply because it is cold; i.e. does not emit (radiate) EME!
Whilst the following mathematics are correct, they are provided here for information only. They cannot be regarded as justification for the existence of blackholes.
Schwarzschild identified the radius of a given mass (m₁) that would prevent a photon (m₂) from leaving its surface, so we must begin with his work:
A photon generates the following kinetic energy; KE = ½.m₂.c²
The potential energy in the mass (m₁) that would hold the photon's KE in check may be calculated by rearranging Newton's gravitational force formula 'F = G.m₁.m₂ / r²' to 'PE = G.m₁.m₂ / r';
First we must estimate a density for mass 'm₁' and then define its surface radius; r = ³√[3.m₁ / 4π.ρ]
assuming a density of ≈8000kg/m³ (≈iron)
PE = G.m₁.m₂ / ³√[3.m₁ / 4π.ρ] = KE
m₁ = √[3.KE³ / G³.4π.ρ.mₑ³] = 9.54527199116538E+37 kg
(where; m₂ = mₑ)
The surface radius of m₁ must be; r = ³√[3.m₁ / 4π.ρ] = 1.41754407340377E+11 m
Schwarzschild would have predicted:
Rs = 2.G.m/c² = 1.41754407340377E+11 m
So from here on, all the various definitions for dark stars and celestial bodies can be found using Schwarzschild's radius.
After a star has collected sufficient satellite mass (planets) to create the internal frictional heat through spin energy necessary to generate atomic fission it will have become active (see Stars & The Gas Planets). In this way, active stars create the universe's energybank together with the lighter elements and ultimately, hydrogen (protonelectron pairs).
All the heavy elements are created in the ultimatebody; the only place with sufficient gravitational (magnetic charge) energy to generate the fusion required to create them.
During its life, an active star continually emits (radiates) the energy generated by its nuclear fission in the form of alpha and beta particles, and electromagnetic radiation.
Apart from a little core heat, the only energy a star will possess at the end of its life will be gravitational, but it will have retained all the mass it had when originally ejected from the ultimatebody.
A star's mass, and therefore its gravitational energy, does not [significantly] alter throughout its life. However, as it grows older, having converted most of its matter to hydrogen, gravitational attraction at a star's surface decreases as it grows in size.
i.e. the greater a star's density, the greater its gravitational attraction at its surface
and the greater a star's mass, the greater its gravitational energy
A dead star is one that has converted all its matter to lighter elements and hydrogen atoms, and therefore generates insufficient internal frictional heat for fission to occur.
There are no such things as BlackHoles
Blackholes were predicted by astronomists because they couldn't see [detect] the large 'massive' bodies that were attracting other celestial bodies and deflecting light more than expected. So the blackhole was invented by scientists to explain something they couldn't understand, exactly as had mankind has done since time immemorial.
The reason they couldn't be seen, however, is because these bodies are cold, they emit almost no electromagnetic energy.
And the reason they are cold is because they have no forcecentres of their own, so they cannot generate mantle heat through spin.
The bodies at the centre of spiral galaxies are simply large cold masses of the same matter as all other matter in the universe, and the same as that comprising the UltimateBody; galactic forcecentres.
See Hades for the properties of the galactic forcecentre at the centre of our Milky Way.
As a theoretical exercise (that has no place in reality) it is possible to predict today's misconception of a blackhole via its Schwarzschild radius:
It is currently believed that a blackhole is body with sufficient density to prevent photons from escaping its surface.
For example; VY Canis Majoris has a radius (9.879082E+11 m) almost 20 times larger than that of the minimum iron star that could be expected to collapse to an instant blackhole (5.3218365E+10 m) according to the Calculations below⁽¹⁾
Whilst CalQlata has adopted 'U' to symbolise energy, with due respect to Henri Poincaré we have adopted his symbol 'E' for this page
From Isaac Newton's gravitational force: F = G.m₁.m₂ / R² → m.g = G.m₁.m₂ / R² → g = G.m / R²
The following Table contains the principle formulas used in the Calculations (below) to determine the properties of collapsed and active stars and blackholes.
Formula  Variables & Constants  Comments 

V = 4/3 π.R³  V = volume of a sphere R = radius of a sphere 

ρ = m/V  V = volume of a sphere ρ = density m = mass 

R = ³√[3.m / 4π.ρ]  R = radius of a sphere ρ = density m = mass 
Calculating a star's radius based upon its known density and mass 
F = G.m₁.m₂ / R²  F = centripetal force between a force centre and an orbiting body G = Newton's gravitational constant m₁ = mass of the force centre m₂ = mass of the orbiting body R = the straightline distance between the centres of the force centre and the orbiting body 
The attracting (gravitational) force between a force centre (a sun) and an orbiting body (a planet) 
F = m.a = m.g  F = force m = mass a = acceleration g = gravitational acceleration 
As described by Isaac Newton 
g = G.m / R²  g = gravitational acceleration at 'R' G = Newton's gravitational constant m = mass of force centre (star) R = radius of the force centre (star) 
Used to determine the gravitational acceleration of a mass at radial distance 'R' from its centre 
p = ρ.R  p = core masspressure of the hydrogen gas ρ = average planet density R = planet radius 
Masspressure at planet core 
p = g/G  p = masspressure g = gravitational acceleration G = Newton's gravitational constant 
special case: unknown 
m = p.R²  m = mass of body p = masspressure at interface of iron core and hydrogen gas R = radius of star 
Used to determine the mass of a star of known radius from its core pressure⁽³⁾ 
p₁/R₁ = p₂/R₂  R₁ = radius of star₁ R₂ = radius of star₂ p₁ = masspressure at core of R₁ p₂ = masspressure at core of R₂ 
A relationship between any two masses of the same average density⁽³⁾ 
R₁/R₂ = g₁/g₂  R₁ = radius of star₁ R₂ = radius of star₂ g₁ = gravitational acceleration₁ g₂ = gravitational acceleration₂ 
A relationship between any two masses of the same average density⁽³⁾ 
R₁/ρ₁ = R₂/ρ₂  R₁ = radius of star₁ R₂ = radius of star₂ ρ₁ = density of star₁ ρ₂ = density of star₂ 
A relationship between any two masses of the same material (e.g. hydrogen)⁽³⁾ 
R₂ = (3.m₂.R₁ / 4.π.ρ₁)⁰˙²⁵  R₁ = radius of star₁ R₂ = radius of star₂ ρ₁ = density of star₁ m₂ = mass of star₂ 
Used to find the radius of a star of known mass of the same material (e.g. hydrogen)⁽³⁾ 
E = m.c²  E = relativistic momentum (energy) in 'm' at 'c' m = mass of star c = speed of light in a vacuum 
Limiting momentum (energy) in a body of mass (m) when moving at the speed of light (c) 
E = m.g.R  E = gravitational energy in 'm' m = mass of body R = radius of 'm' g = gravitational acceleration at 'R' 
Gravitational energy in a body of mass (m) at radius 'R' 
E = G.m²/R  G = Newton's gravitational constant E = gravitational energy of star m = mass of star R = radius of star 
Gravitational energy 
v² = 2.a.R  c = speed of light g = gravitational acceleration at 'R' R = radius of body (star) 
Relationship between velocity and acceleration 
c² = 2.g.R  c = speed of light g = gravitational acceleration at 'R' R = radius of body (star) 
Used to determine the equivalent limiting radius 'R' below which light cannot escape a given mass 
v² = 2.G.m/R  G = Newton's gravitational constant m = mass of star R = radius of star v = velocity 
Newton's escape velocity for a planetary body 
m = ¾(E/G)⁰˙⁶ / ρ⁰˙²  m = mass of body (star) E = gravitational energy G = Newton's gravitational constant ρ = density of body 
Alternative formula for the mass of a body 
m = (3.c⁶ / 32.π.ρ.G³)⁰˙⁵ 
m = mass of body (star) c = speed of light G = Newton's gravitational constant ρ = density of body 
Minimum (limiting) mass of a star that can be expected to collapse directly to form a black hole 
The following comprises a series of calculations that predict the expected ultimate destiny of three stars according to the above formulas and data developed in CalQlata's 'Laws of Motion' web page.
Known Properties (active):
mass (m) = 1.98850E+30 kg
radius (R) = 6.95710E+08 m
Calculated Properties (active):
Volume (V) = 4/3 . π.R³ = 1.4105E+27 m³
average density (ρ) = m/V = 1409.7829 kg/m³
gravitational acceleration (g) = G.m/R² = 274.1755834 m/s²
gravitational energy (E) = m.g.R = 3.793E+41 J
Known Properties (dead) assuming the density of iron:
average density (ρ) = 7870 kg/m³ ⁽⁴⁾
mass (m) ≈1.98850E+30 kg
Calculated Properties (dead):
volume (V) ≈ 4/3 . π.R³ ≈ 2.52668E+26 m³
radius (R) ≈ ³√[3.m / 4π.ρ] ≈ 3.921818E+08 m
gravitational acceleration (g) ≈ G.m/R² ≈ 862.8 m/s²
2.g.R ⁽⁵⁾ ≈ 6.767493E+11 m²/s² (c² = 8.987552E+16)
Given that 2.g.R is so much less than c², it is unlikely that the earth's sun will become a blackhole immediately after collapse
Known Properties (active):
Radius (R) = 9.87908E+11 m
mass (m) = 5.967E+31 kg
Calculated Properties (active):
volume (V) = 4/3 . π.R³ = 4.03867E+36 m³
average density (ρ) = m/V = 1.477467E05 kg/m³
gravitational acceleration (g) = G.m/R² = 4.0802E03 m/s²
gravitational energy (E) = m.g.R = 2.4052224E+41 J
Known Properties (dead): (assuming the density of iron)
mass (m) = 5.967E+31 kg
average density (ρ) = 7870 kg/m³ ⁽⁴⁾
Calculated Properties (dead):
volume (V) ≈ 4/3 . π.R³ ≈ 7.58196E+27 m³
radius (R) ≈ ³√[3.m / 4π.ρ] ≈ 1.21870E+09 m
gravitational acceleration (g) ≈ G.m/R² ≈ 2681.156 m/s²
2.g.R ⁽⁵⁾ ≈ 6.53505E+12 m²/s² (c² = 8.9875518E+16)
Whilst VY Canis Majoris is very large, it has a mass little larger than our sun and therefore is unlikely to collapse into an instant black hole
Known Properties (dead):
c² = 2.g.R = 8.9875518E+16 m²/s²
average density (ρ) = 7870 kg/m³ (assuming the density of iron)
Calculated Properties (dead):
mass (m) = √(3.c⁶ / 32π.ρ.G³) = 9.6237854E+37 kg
volume (V) = m/ρ = 1.22284441E+34 m³
radius (R) = ³√(3.V / 4π) = 1.42920392E+11 m ⁽⁵⁾
gravitational acceleration (g) = G.m/R² = 314425.10331 m/s²
Known Properties (active):
mass (m) = 9.62378552E+37 kg
Calculated Properties (active):
radius (R) = ⁴√(R₁⁴ . m₁/m₂) = 5.802742927E+10 m
volume (V) = 4/3π.R³ = 8.1844330782E+32 m³
average density (ρ) = m/V = 117586.46475 kg/m³ ⁽⁴⁾
gravitational acceleration (g) = G.m/R² = 1907387.5622 m/s²
gravitational energy (E) = m.g.R = 1.0651682497E+55 J
2.g.R = 8.9875517874E+16 m²/s² (c² = 8.9875517874E+16)
A blackhole at the time of collapse (see Neutron Star below)
The above results are summarised in the following Table (input data):
Units  The Earth's Sun  VY Canis Majoris  Minimum Potential BlackHole Star  

Active  
mass  kg  1.98850E+30  5.967E+31  9.6237855E+37 
radius  m  6.95710E+08  9.879082E+11  5.802743E+10 
volume  m³  1.41050E+27  4.03867E+36  8.1844331E+32 
average density  kg/m³  1409.78293  1.477467E05  117586.46475 
gravitational acceleration  m/s²  274.175583  4.08021E03  1907387.5622 
gravitational energy  J  3.79299803E+41  2.405222E+41  1.06516825E+55 
Dead  
mass  kg  1.98850E+30  5.967E+31  9.6237855E+37 
radius  m  392181784  1.218702E+09  1.429204E+11 
volume  m³  2.52668E+26  7.58196E+27  1.2228444E+34 
average density ⁽⁴⁾  kg/m³  7870  7870  7870 
gravitational acceleration  m/s²  862.800588  2681.14672  314425.10331 
2.g.R ⁽⁵⁾  m²/s²  6.76749348E+11  6.5350384E+12  8.9875518E+16 
c²  m²/s²  8.9875518E+16  8.9875518E+16  8.9875518E+16 
From our formulas above:
v² = 2.a.R
Special [limiting] case to trap photons: c² = 2.g.R → g = c² / 2.R
Gravitational (potential) energy: E = m.g.R → E = m.R.c² / 2.R
Limiting gravitational energy that will trap photons: E = m.c² / 2
i.e.: Kinetic energy of the mass at light speed: E = ½.m.c²
If we apply these calculations to the above (iron) limiting blackhole:
E = m.g.R = 9.6287855E+37 x 314425.10331 x 1.4292039174E+11 = 4.32696038E+54 J
KE = ½.m.c² = 9.6287855E+37 x 2997924582² ÷ 2 = 4.32696038E+54 J
CalQlata has therefore shown that the gravitational energy in a planet that will trap photons is equal to its kinetic energy at light speed
Assuming that the limiting mass for a star that can be expected to collapse to form an instant black hole (trap highenergy photons) is governed by the relationship 2.g.R = c², its mass can be calculated thus:
m = √(3.c⁶ / 32.π.ρ.G³) = 9.62378551609E+37
and its radius; R = 5.3218365E+10 m⁽¹⁾
However, it is claimed that all black holes are neutron stars, i.e. they collapse to the density of a neutron. If this is so its properties may be calculated as follows:
The properties of a neutron:
m = 1.6749272928E27 kg
R = 1.11328405737E15 m
V = 5.77971706488E45 m³
ρ = 2.8979399407E+17 kg/m³ ⁽⁶⁾
The radius of this star is related to its mass thus:
R³ = 3.m / 4.π.ρ
R = ³√(3.m / [4 x π x 2.8979399407E+17])
R = ³√m . ³√(3 / [4 x π x 2.8979399407E+17])
R = 9.37433933214E07 x ³√m
According to Newton: v² = 2.G.m/R
Limiting condition: c² = 2.G.m/R
R = 2.G.m / c²
2.G.m / c² = 9.37433933214E07 x ³√m
m/³√m = ¹˙⁵√m = 9.37433933214E07 x c²/2.G
m = (9.37433933214E07 x c² ÷ 2.G)¹˙⁵ = 1.58677073182E+31 kg
{check: m = √(3.c⁶ / 32.π.ρ.G³) = 1.58677073182E+31 kg}
which represents the minimum mass that could create a neutron star on collapse
Its radius: R = 2.G.m / c² = 23556.57050854 m
The gravitational acceleration at its surface is:
g = c² / 2.R = 1.90765285297E+12 m/s²
{check: g = G.m/R² = 1.90765285297E+12}
KE = ½.m.c² = 7.13059206345E+47 J
Eg = m.g.R = 7.13059206345E+47 J
confirming CalQlata's claim that the gravitational energy of a blackhole is equal to its kinetic energy at light speed
The problem is, however, whether or not iron can collapse to the density of a neutron with the above mass(es), because in order to do so, there can be no protons or electrons (or photons) within the body.
Moreover, its ultimate density is governed by the compressibility or iron.
For all stars (and planets): 2.π.R.G.ρ / g = 1.5 {radians}
Proof: 2.π.R.G.ρ / 1.5 g = 1
4/3.π.R.G.(m/(4/3.π.R²)) / g
G.m / g.R² = 1
g = G.m / R²
We can check the above calculations as follows:
The Earth's Sun
active: 2.π.G.ρ.R / g = 2.π.G x 1409.782932 x 6.95710E+08 ÷ 274.0806631 = 1.5 {radians}
dead: 2.π.G.ρ.R / g = 2.π.G x 7870 x 3.921817841E+08 ÷ 862.5018838 = 1.5 {radians}
VY Canis Majoris
active: 2.π.G.ρ.R / g = 2.π.G x 1.477467E05 x 9.87908E+11 ÷ 4.07880E03 = 1.5 {radians}
dead: 2.π.G.ρ.R / g = 2.π.G x 7870 x 1.21870E+09 ÷ 2680 = 1.5 {radians}
Minimum Potential BlackHole Star
active: 2.π.G.ρ.R / g = 2.π.G x 117601 x 5.80350E+10 ÷ 1907222 = 1.5 {radians}
dead: 2.π.G.ρ.R / g = 2.π.G x 7870 x 1.42945138E+11 ÷ 314370.67118 = 1.5 {radians}
This constant also applies to planets (e.g. the earth):
2.π.G.ρ.R / g = 2.π.G x 5508.25830105342 x 6371000.685 ÷ 9.80663139027614 = 1.5 {radians}
A useful constant (K), based upon Newton's theories, for all matter is:
K = g / ρ.R = 2.7954278140935E10 m³/s²/kg
and
G = K / 4/3π = 6.6735923200433E11 m³/s²/kg
Looking at this recently released photograph, we (at CalQlata) are left with a number of concerns:
1) There is no such thing as a singularity. A black hole is simply a large body of matter too cold to emit electromagnetic energy strong enough for us to detect.
2) And even if there were, by definition you would not be able to see it. Therefore, this cannot be a photograph of a blackhole. It is simply a photograph of a black space.
3) Singularities do not exist because they defy the laws of thermodynamics
4) Even if the dark area in the centre of the photograph is a singularity, why is it so large? The Schwarzschild's radius for a singularity would be much smaller than depicted in this photograph
5) Einstein's theories of relativity have already been proven incorrect; so there is no such thing as an 'eventhorizon'
6) The photograph was taken using radiotelescopes, which cannot define colour, so why is the gasring coloured orange?
7) The image looks exactly like a large star orbiting a large galactic forcecentre; the bottomfront, 'bright' part of the image is our side of the orbit and the upper dark area looks like the back of the orbit
8) The variable nature of the light could be scattered matter left behind in the orbital trails or simply timelapse photography. At such an orbital radius, the star's velocity would be very high indeed
9) The scientists presenting this discovery used the terms "I think" and "we think" far too many times to convince us of their interpretation
10) Standard procedure in science should be; to explain an event mathematically, and subsequently search for physical proof (of the mathematical explanation). The scientists here saw something and then paid computer analysts to create a mathematical model that bestmatched their argument. This is not proof, it is manipulation.
It is our (CalQlata's) opinion that the European money would have been better spent investigating our own galactic forcecentre'
All of the above, whilst accurate, is fiction!.
You will find further reading on this subject in reference publications^{(55, 60, 61, 62, 63 & 64)}