Anom Accel of Pioneer 10 for v>(GM/r)^1/2

In message , Aladar
writes
Jonathan Silverlight wrote in
message ...
In message , Lyndon Ashmore
writes
It shows that the shift in wavelengths (redshift) is not due to acceleration
but is due to an interaction of waves with matter. See 'ashmore's paradox'
www.lyndonashmore.com where I show that the Hubble constant at 64 km/s per
Mpc is nothing more than a combination of the parameters of the electron
multiplied by the planck constant. Therefore redshift cannot be due to
expansion, it must be due to an interaction with electrons.

An interaction that is completely independent of wavelength would be
quite a trick. That's why it's not considered as a cause of the Pioneer
effect (or of red shifts, come to that)
Aladar is the _only_ person who thinks the Pioneer affect is an excess
red shift - and the only person who thinks it's travelled 10 1/2 light
years, apparently!

Maybe I'm the _only_ person who freely voices that the Pioneer effect
is
an excess red shift - and the only person who thinks not that "it's
travelled 10 1/2 light years, apparently!" but that the residual
represents an
accumulated value, corresponding to the 10 1/2 years of light travel
time!

And it equals to the theoretical Hubble wavelength doubling time
constant
of 4.2 billion years or in linear approximation, for small distances
corresponds to a 'conventional' Hubble constant of 162 km/s per Mpc...

And please, Johnathan, allow me to formulate my position!

Sorry, but we've been here far too many times before. You can't equate
light travel time to the "arc" during which Anderson et al. collected
data.
And even if you could, your figure for the Hubble constant is
ridiculous. How do you explain the fact that it is more than twice the
various independent measurements? It doesn't even bear any relationship
to the anomalous acceleration.
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In message , Ned Wright
writes
Jonathan Silverlight
wrote in message ...
I haven't seen any evidence of an anomaly on Cassini.

Presumably the reason Cassini hasn't seen an acceleration is that it's
more than 20 x as massive.

Let's be clear he the Cassini spacecraft shows a LARGER
non-gravitational acceleration than the Pioneer's. But since it
was designed to radiate RTG waste heat primarily out the back, this
is not considered to be anomalous.

You're right, of course, and I knew that :-) People have been telling me
since my schooldays that my writing is too terse, and I omitted your
very important final word.
--
Rabbit arithmetic - 1 plus 1 equals 10
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Craig Markwardt replied to Jeff Root:

Yes, I can see that. It is why I estimate that only 20% of
the RTG surface area is visible to the back side of the antenna
dish and other parts of the spacecraft sunward of the RTGs,
rather than the approximately 45% that would be visible if they
were plain cylinders.

I think the fractional (i.e. edge-on) area that actually faces
the antenna is much smaller than the outward-facing area. How
do you get 20%?

Just from looking at photos and guessing. Nothing better.

Note that the 20% figure is for the area of the RTG which is
visible to the back of the dish, etc., and is not the most
important thing to measure here, but was a response to Bruce
Woodcock's assertion that:

every RTG is not a spherical black body, but rather has fins
that are "edge on" to the antenna, which means only 2.5% of
the surface area of the RTG is actually facing the antenna.

My 20% figure may be high, but Bruce's 2.5% figure has *got*
to be *way* too low. Or else he mis-stated what the figure
represented. As I said, if the RTGs were plain cylinders,
approximately 45% of their surfaces would be visible to the
back of the dish, etc. (Depending on how it is measured-- it
could be closer to 60%.) The fins reduce that greatly, but
not to 2.5%! Don't you agree?

The more important statement by Bruce was:

the RTGs are located at the ends of the booms, and they only
see the antenna "edge on", subtending an angle of about 1.5%
of 4 pi steradians.

This figure has also got to be way too low. Look at photos
or a model. No way can it be only 1.5%. My estimate was 5%.
Don't you agree that 5% looks more like it?

-- Jeff, in Minneapolis

..

(Jeff Root) writes:
Craig Markwardt replied to Jeff Root:

Yes, I can see that. It is why I estimate that only 20% of
the RTG surface area is visible to the back side of the antenna
dish and other parts of the spacecraft sunward of the RTGs,
rather than the approximately 45% that would be visible if they
were plain cylinders.

I think the fractional (i.e. edge-on) area that actually faces
the antenna is much smaller than the outward-facing area. How
do you get 20%?

Just from looking at photos and guessing. Nothing better.

Note that the 20% figure is for the area of the RTG which is
visible to the back of the dish, etc., and is not the most
important thing to measure here, but was a response to Bruce
Woodcock's assertion that:

every RTG is not a spherical black body, but rather has fins
that are "edge on" to the antenna, which means only 2.5% of
the surface area of the RTG is actually facing the antenna.

My 20% figure may be high, but Bruce's 2.5% figure has *got*
to be *way* too low. Or else he mis-stated what the figure

I wouldn't be too quick to dismiss the 2.5% amount without a more
rigorous factual basis. The majority portions of the RTGs which face
the dish are the endcaps, which are designed to be *non*-radiative.
This leaves the edges of the radiator fins, which are pretty thin
indeed, compared to the total surface area.

Don't you agree that 5% looks more like it?

I have no idea. It's a realm where eyeballing it won't help.

Craig


--
--------------------------------------------------------------------------
Craig B. Markwardt, Ph.D. EMAIL:
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
--------------------------------------------------------------------------

Jonathan Silverlight wrote in message ...
[...]

Maybe I'm the _only_ person who freely voices that the Pioneer effect
is
an excess red shift - and the only person who thinks [..] that the residual
represents an
accumulated value, corresponding to the 10 1/2 years of light travel
time!

And it equals to the theoretical Hubble wavelength doubling time
constant
of 4.2 billion years or in linear approximation, for small distances
corresponds to a 'conventional' Hubble constant of 162 km/s per Mpc...

And please, Johnathan, allow me to formulate my position!

Sorry, but we've been here far too many times before. You can't equate
light travel time to the "arc" during which Anderson et al. collected
data.

You pointed to the key of the problem! The bigbangology tends to
forget
the fact that the so called anomalous acceleration is the unexplained
portion of FREQUENCY SHIFT - (recalculated to Doppler velocity, hence
cm/s or m/s) - divided by the DISTANCE of photon travel - (measured in
light seconds)! We been here so many times, but somehow you tend to
forget - or intentionally misrepresent - the facts!

And even if you could, your figure for the Hubble constant is
ridiculous. How do you explain the fact that it is more than twice the
various independent measurements? It doesn't even bear any relationship
to the anomalous acceleration.

First of all, what you call 'measurements' - they are closer to
regilion
then to physics... Actually, there is only one factor which defines
the
alleged around H=70 km/s per Mpc value: the alleged similarity
of Supernovae. I would say it is ridiculous. Secondly: my Hd=4.1167
billion years (need this correction!) results in a very interesting
pictu the look-alike galaxies are the same sizes! I would say - it
is
an important proof of the correctness of my about H=160 km/s per Mpc -
equivalent theoretical value for the photon energy loss rate.

Indeed, the reported anomaly - frequency drift vs. distance - is the
Hubble redshift vs. distance law, and indeed it could be viewed as a
dedicated experiment to verify or falsify the tired light cause!
Sorry, but it put the last nails into the coffin of big bang hoax...

Cheers!
Aladar
http://www.stolmarphysics.com

Craig Markwardt replied to Jeff Root:

Note that the 20% figure is for the area of the RTG which is
visible to the back of the dish, etc., and is not the most
important thing to measure here, but was a response to Bruce
Woodcock's assertion that:

every RTG is not a spherical black body, but rather has fins
that are "edge on" to the antenna, which means only 2.5% of
the surface area of the RTG is actually facing the antenna.

My 20% figure may be high, but Bruce's 2.5% figure has *got*
to be *way* too low. Or else he mis-stated what the figure

I wouldn't be too quick to dismiss the 2.5% amount without a more
rigorous factual basis.

I'm not dismissing it-- I'm saying it looks wrong, so it should
be checked.

The majority portions of the RTGs which face the dish are the
endcaps, which are designed to be *non*-radiative.

That has very little effect on my analysis. I don't recall
taking into account that the ends were designed to not radiate,
but it was obvious that the area they expose to the dish and
body is much less than that of the sides and fins.

This leaves the edges of the radiator fins, which are pretty
thin indeed, compared to the total surface area.

Are you assuming that infrared from an RTG is radiated only
perpendicular to the surfaces? When I look at an RTG from near
the body of the spacecraft, I see a good deal of surface area
of the fins and of the cylinder itself. About 20% of the total
surface area, as I said.

Don't you agree that 5% looks more like it?

I have no idea. It's a realm where eyeballing it won't help.

I don't have any problem estimating the solid angle subtended
by the body of the spacecraft as seen from an RTG, based only
on what I can see.

-- Jeff, in Minneapolis

..

In message , Aladar
writes
Jonathan Silverlight
wrote in message

Sorry, but we've been here far too many times before. You can't equate
light travel time to the "arc" during which Anderson et al. collected
data.

You pointed to the key of the problem! The bigbangology tends to
forget
the fact that the so called anomalous acceleration is the unexplained
portion of FREQUENCY SHIFT - (recalculated to Doppler velocity, hence
cm/s or m/s) - divided by the DISTANCE of photon travel - (measured in
light seconds)!

OK, can you repeat why you think 10 1/2 light years is the distance of
photon travel? I think I know the answer but I want to knock it down
again.


And even if you could, your figure for the Hubble constant is
ridiculous. How do you explain the fact that it is more than twice the
various independent measurements? It doesn't even bear any relationship
to the anomalous acceleration.

First of all, what you call 'measurements' - they are closer to
regilion
then to physics... Actually, there is only one factor which defines
the
alleged around H=70 km/s per Mpc value: the alleged similarity
of Supernovae. I would say it is ridiculous. Secondly: my Hd=4.1167
billion years (need this correction!) results in a very interesting
pictu the look-alike galaxies are the same sizes! I would say - it
is
an important proof of the correctness of my about H=160 km/s per Mpc -
equivalent theoretical value for the photon energy loss rate.

Indeed, the reported anomaly - frequency drift vs. distance - is the
Hubble redshift vs. distance law, and indeed it could be viewed as a
dedicated experiment to verify or falsify the tired light cause!
Sorry, but it put the last nails into the coffin of big bang hoax...

As you are presumably aware, Anderson et al. write that several people
have noted that the anomalous acceleration (8 x 10^-8 cm s^2) is
approximately equal to c * H, if H =82km/s/Mpc (a bit high by modern
standards). It's fairly obvious that doesn't work for your figure.
And grabbing a web page at random
http://nedwww.ipac.caltech.edu/level5/Fukugita/Fukugita2_1.html I find
that the Tully-Fisher relationship gives H=80, type 1 supernovae give
H=50-55, and HST observations of Cepheids in the Virgo cluster give
h=65.
Three independent methods, and there are others.
--
Rabbit arithmetic - 1 plus 1 equals 10
Remove spam and invalid from address to reply.

(Jeff Root) writes:
Are you assuming that infrared from an RTG is radiated only
perpendicular to the surfaces? When I look at an RTG from near
the body of the spacecraft, I see a good deal of surface area
of the fins and of the cylinder itself. About 20% of the total
surface area, as I said.

Perhaps, but the surface emissivity does decrease as the cosine of the
viewing angle to the normal. Say the angle was 85 degrees, then
cos(85 deg) = 0.087, which when multiplied by your 20% figure is 1.7%.
I have no idea what the real angles are, but the radiation transfer is
more compilcated than just estimating the solid angle.

Craig

(Aladar) writes:

Craig Markwardt wrote in message ...
Now, what is this anomaly report? Where did they find the anomaly,
what is it?

In my understanding the summation of residual frequency drift shows
the photon energy loss up to a photon propagation time (accumulated)
of 10 1/2 years. And in my understanding the reported frequency
drift - of this anomaly - is an excess redshift or lower then
expected frequency, hence the negative 'velocity'.

I can speak from personal experience: your understanding is incorrect.

CM

its not anomalous except according to anticipation that a theory is
reality.

part of the rational of sending out space probes is to monitor the
actual reality and conditions of space verses theoretical space. I keep
arguing with orthodox believers that vacuum space is a lot fuller that
it initially appears. the two factors that lead to a higher density of
vacuum space are having a very large solar or planetary mass and / or
having a high relative velocity to the space per distance of a transit.

the soft vacuum radiometer experiments in my literature , and classic
texts, explains it perfectly. even the smallest spec of dust in the
solar system can behave as a comet if its out of sync with the solar
wind turn back orbital shells. the moons of jupiter and saturn offer a
bit of a contradiction to the solar system model thus until you look at
their thermodyamic stratifications as a matrix effect system with the
solar overlay. Its even tells you where you will expect to see rings
form.



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