What is wrong with the 'Mainstream Scientific Establishment'?

On Tue, 22 Nov 2011 13:57:58 +0100, "Paul B. Andersen"
wrote:

On 21.11.2011 21:40, Henry Wilson DSc. wrote:
On Mon, 21 Nov 2011 12:42:23 +0100, "Paul B. Andersen"
wrote:



THis stuff is far too hard for you Wrmey, so please stay away.


Whatsa matter, Ralph, who is neither a Henry nor a Wilson and most
certainly not a DSc., are you afraid of discussing some physics?

You said, "the majority of perceived brightness variability is
consequence of the star's radial velocity changes as it orbits a
barycentre with one or more large satellites". I pointed out that
velocity changes result in Doppler shift, not brightness shift!

The VDoppler shifts of most variable stars are fairly small. Most of the
measurd shifts are due to ADoppler, which is phase with brightness curves
and likely to be in much larger than VDoppler, causing huge errors in
current theories.

The predicted Doppler shift caused by the acceleration is
not only is in phase with the brightness curve,
but is identical to the brightness curve.
The crazy spectra predicted by the Emission theory are
never observed.
Yet another falsification of the Emission theory.

This shows how little you understand BaTh.

Both ADoppler and VDoppler contribute to any observed spectral shifts.

The proportions can vary widely, depending on the source's acceleration at
emission time and the local extinction rate. My program allows the ratio to
be varied in order to match velocity curves.

That you can make your curve drawing program mimic
just about any periodic curve by fudging the parameters,
doesn't help.

Any 'fudging' would have to follow a very limited range of curve shapes.

Cepheid velocity curves are virtual mirror images of their brightness
curves, with blue shift corresponding with increased brighness. Clearly,
their observed spectral shifts are mainly ADoppler.

Assume that the intrinsic brightness of the source is constant.

That is another factor altogether. It might not be constant.,,,,but OK.

If the light that is emitted from the source during
the time dT is received by the observer during the time dt,
then the brightening of the source is dT/dt.
But the observed frequency is also changed by dT/dt.

The brightening and the Doppler shift caused by the motion
of the source are _always_ identical. This is independent
of theory, it follows from the definition of the entities.

So if the intrinsic brightness of the source is constant,
but the observed brightness is varying by a factor of two,
then the observed spectrum will also be shifted by a factor of two.

No it doesn't work like that as I've explained in my thesis.
Relative movement between individual photons can go on virtually
forever...but their 'ends' cannot do the same. So photons quickly settle
down to a stable length....but the change is still accelertion dependent.

see: http://www.scisite.info/fig11.jpg

So when your program produces a light curve, it has
also produced a curve showing how the spectrum should
be shifted.

The crazy spectra predicted by the Emission theory are
never observed.

Every observed spectrum is that resulting from a combination of ADoppler and
VDoppler. The task for astronomers is to try to xtract truth from the
willusion.

Yet another falsification of the Emission theory.

You failed again.

On Tue, 22 Nov 2011 08:01:06 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Tue, 22 Nov 2011 00:04:47 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Mon, 21 Nov 2011 23:09:19 -0000, "Androcles"
|
| |
| | That's a 1.02% error, almost equal to your IQ and certainly not
| beautiful,
| | pet chimp. Draw a line around a banana with a yellow crayon, did you?
| |
| | The exact figure is given in a different window.
| |
| That's far too wishy-washy.
| http://www.llamas.org/brawn/wishywashy.gif
| How about Wilson Acceleration, Non-Kepler Effect Rubbish, or ******?
|
| Oh, the old jealousy is appearing again.....
|
Try to suppress it, old chimp, being jealous of me won't do you any good.
Everybody else knows you can't figure out the piston frequency of the
Iron Duke at 60 mph. Test your IQ with this question instead: the length
of the connecting rod is the same as the radius of the crank, so the piston
is at the centre of the flywheel at bottom dead centre. Plot the curve of
the height of the piston. Feel free to use your wonderful BASIC.

No need. The piston is smashed into many pieces on the first rotation.

Maybe you meant to say the length of the connecting rod is more than twice
the radius of the crank.

Anyway, the solution is trivial.

My Prius uses an Atkinson cycle engine, in which the crankshaft is offset at
time of explosion so that the resulting torque is increased.

....that's a good Japanese name, 'Atkinsono'....very clever engineers over
there....

"Henry Wilson DSc." ..@.. wrote in message
...
| On Tue, 22 Nov 2011 08:01:06 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Tue, 22 Nov 2011 00:04:47 -0000, "Androcles"
| | wrote:
| |
| |
| | "Henry Wilson DSc." ..@.. wrote in message
| | .. .
| | | On Mon, 21 Nov 2011 23:09:19 -0000, "Androcles"
| |
| | |
| | | That's a 1.02% error, almost equal to your IQ and certainly not
| | beautiful,
| | | pet chimp. Draw a line around a banana with a yellow crayon, did
you?
| | |
| | | The exact figure is given in a different window.
| | |
| | That's far too wishy-washy.
| | http://www.llamas.org/brawn/wishywashy.gif
| | How about Wilson Acceleration, Non-Kepler Effect Rubbish, or ******?
| |
| | Oh, the old jealousy is appearing again.....
| |
| Try to suppress it, old chimp, being jealous of me won't do you any good.
| Everybody else knows you can't figure out the piston frequency of the
| Iron Duke at 60 mph. Test your IQ with this question instead: the length
| of the connecting rod is the same as the radius of the crank, so the
piston
| is at the centre of the flywheel at bottom dead centre. Plot the curve of
| the height of the piston. Feel free to use your wonderful BASIC.
|
| No need. The piston is smashed into many pieces on the first rotation.
|
| Maybe you meant to say the length of the connecting rod is more than twice
| the radius of the crank.
|
| Anyway, the solution is trivial.

Far from trivial, the con rod small end reaches the centre at 90 degrees
rotation, remains there for 180 degrees, then, if you are lucky enough
to have opened the steam valve that pushes the piston out again, rises
to TDC once more. I meant to say exactly what I said. You failed again.

|
| My Prius uses an Atkinson cycle engine, in which the crankshaft is offset
at
| time of explosion so that the resulting torque is increased.
|
| ...that's a good Japanese name, 'Atkinsono'....very clever engineers over
| there....
|
Actually I was discussing the non-sinusoidal nature of light curves and
velocity curves in the context of the motion of a piston.
....that's a good Japanese name, 'Keplermoto San'....very clever engineers
over in Germany... none in Oz, though.
Still, your Atkinsono cycle will do. Assume the connecting rod is sqrt(2) *
crank radius and the piston path is tangential to the crank radius, plot the
curve of the height of the piston. Feel free to use your wonderful BASIC.

If you really want to draw curves that are a tad more complicated than
your erroneous 0.84 eccentricity that you tried to fake as 0.85
eccentricity,
take a look he
http://www.youtube.com/watch?v=H1IO2f1xvd0

On 11/22/11 3:38 PM, Henry Wilson DSc. wrote:
My Prius uses an Atkinson cycle engine, in which the crankshaft is offset at
time of explosion so that the resulting torque is increased.

What kind of mileage do you get?

On Tue, 22 Nov 2011 16:37:36 -0600, Sam Wormley wrote:

On 11/22/11 3:38 PM, Henry Wilson DSc. wrote:
My Prius uses an Atkinson cycle engine, in which the crankshaft is offset at
time of explosion so that the resulting torque is increased.

What kind of mileage do you get?

22-23 km/litre...~60 miles per gallon.

It's a hybrid of course. Infinitely variable gearing...What a car!

On Tue, 22 Nov 2011 22:31:39 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .

| |
| Try to suppress it, old chimp, being jealous of me won't do you any good.
| Everybody else knows you can't figure out the piston frequency of the
| Iron Duke at 60 mph. Test your IQ with this question instead: the length
| of the connecting rod is the same as the radius of the crank, so the
piston
| is at the centre of the flywheel at bottom dead centre. Plot the curve of
| the height of the piston. Feel free to use your wonderful BASIC.
|
| No need. The piston is smashed into many pieces on the first rotation.
|
| Maybe you meant to say the length of the connecting rod is more than twice
| the radius of the crank.
|
| Anyway, the solution is trivial.

Far from trivial, the con rod small end reaches the centre at 90 degrees
rotation, remains there for 180 degrees, then, if you are lucky enough
to have opened the steam valve that pushes the piston out again, rises
to TDC once more. I meant to say exactly what I said. You failed again.

I must be missing something. How can the connecting rod be only as long as
the radius of the crank?

|
| My Prius uses an Atkinson cycle engine, in which the crankshaft is offset
at
| time of explosion so that the resulting torque is increased.
|
| ...that's a good Japanese name, 'Atkinsono'....very clever engineers over
| there....
|
Actually I was discussing the non-sinusoidal nature of light curves and
velocity curves in the context of the motion of a piston.
...that's a good Japanese name, 'Keplermoto San'....very clever engineers
over in Germany... none in Oz, though.
Still, your Atkinsono cycle will do. Assume the connecting rod is sqrt(2) *
crank radius and the piston path is tangential to the crank radius, plot the
curve of the height of the piston. Feel free to use your wonderful BASIC.

I could do it in my head...but I don't have time...

If you really want to draw curves that are a tad more complicated than
your erroneous 0.84 eccentricity that you tried to fake as 0.85
eccentricity,
take a look he
http://www.youtube.com/watch?v=H1IO2f1xvd0

My curves are far more accurate than any brightness curve.

On 11/22/11 7:14 PM, Henry Wilson DSc. wrote:
On Tue, 22 Nov 2011 16:37:36 -0600, Sam wrote:

On 11/22/11 3:38 PM, Henry Wilson DSc. wrote:
My Prius uses an Atkinson cycle engine, in which the crankshaft is offset at
time of explosion so that the resulting torque is increased.

What kind of mileage do you get?

22-23 km/litre...~60 miles per gallon.

It's a hybrid of course. Infinitely variable gearing...What a car!


I get 17.8 km/litre in the winter
21.3 km/litre in the summer

19.6 km/litre average

So I find your numbers hard to believe. But good for you if you
really do achieve 22-23 km/litre based on distance traveled
divided by litres purchased.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Tue, 22 Nov 2011 22:31:39 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
|
| | |
| | Try to suppress it, old chimp, being jealous of me won't do you any
good.
| | Everybody else knows you can't figure out the piston frequency of the
| | Iron Duke at 60 mph. Test your IQ with this question instead: the
length
| | of the connecting rod is the same as the radius of the crank, so the
| piston
| | is at the centre of the flywheel at bottom dead centre. Plot the curve
of
| | the height of the piston. Feel free to use your wonderful BASIC.
| |
| | No need. The piston is smashed into many pieces on the first rotation.
| |
| | Maybe you meant to say the length of the connecting rod is more than
twice
| | the radius of the crank.
| |
| | Anyway, the solution is trivial.
|
| Far from trivial, the con rod small end reaches the centre at 90 degrees
| rotation, remains there for 180 degrees, then, if you are lucky enough
| to have opened the steam valve that pushes the piston out again, rises
| to TDC once more. I meant to say exactly what I said. You failed again.
|
| I must be missing something. How can the connecting rod be only as long as
| the radius of the crank?

With the crank at bottom dead centre the con rod reaches the wheel centre.
In a steam locomotive a piston rod is attached between the con rod and the
piston, held in place by slide guides.
http://upload.wikimedia.org/wikipedi..._reversing.gif
Ignore the valve hookup, shorten the con rod and extend the piston rod
and the slide guides to the wheel centre.
What I'm try to get you to do is grasp the mathematics.



|
| |
| | My Prius uses an Atkinson cycle engine, in which the crankshaft is
offset
| at
| | time of explosion so that the resulting torque is increased.
| |
| | ...that's a good Japanese name, 'Atkinsono'....very clever engineers
over
| | there....
| |
| Actually I was discussing the non-sinusoidal nature of light curves and
| velocity curves in the context of the motion of a piston.
| ...that's a good Japanese name, 'Keplermoto San'....very clever engineers
| over in Germany... none in Oz, though.
| Still, your Atkinsono cycle will do. Assume the connecting rod is sqrt(2)
*
| crank radius and the piston path is tangential to the crank radius, plot
the
| curve of the height of the piston. Feel free to use your wonderful BASIC.
|
| I could do it in my head...but I don't have time...
|
You sound just like Dork Van der Belgian Waffle the fumble mumbler, he
fumbles and then says he'll wait and see if anyone notices it.

| If you really want to draw curves that are a tad more complicated than
| your erroneous 0.84 eccentricity that you tried to fake as 0.85
| eccentricity,
| take a look he
| http://www.youtube.com/watch?v=H1IO2f1xvd0
|
| My curves are far more accurate than any brightnss curve.
|
You can't draw an ellipse with 0.85 eccentricity and don't check
your own work. Your bragging doesn't fool anyone but you.

On Wed, 23 Nov 2011 04:11:52 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Tue, 22 Nov 2011 22:31:39 -0000, "Androcles"

| | Maybe you meant to say the length of the connecting rod is more than
twice
| | the radius of the crank.
| |
| | Anyway, the solution is trivial.
|
| Far from trivial, the con rod small end reaches the centre at 90 degrees
| rotation, remains there for 180 degrees, then, if you are lucky enough
| to have opened the steam valve that pushes the piston out again, rises
| to TDC once more. I meant to say exactly what I said. You failed again.
|
| I must be missing something. How can the connecting rod be only as long as
| the radius of the crank?

With the crank at bottom dead centre the con rod reaches the wheel centre.
In a steam locomotive a piston rod is attached between the con rod and the
piston, held in place by slide guides.
http://upload.wikimedia.org/wikipedi..._reversing.gif
Ignore the valve hookup, shorten the con rod and extend the piston rod
and the slide guides to the wheel centre.
What I'm try to get you to do is grasp the mathematics.

They aren't difficult.

| |

| If you really want to draw curves that are a tad more complicated than
| your erroneous 0.84 eccentricity that you tried to fake as 0.85
| eccentricity,
| take a look he
| http://www.youtube.com/watch?v=H1IO2f1xvd0
|
| My curves are far more accurate than any brightnss curve.
|
You can't draw an ellipse with 0.85 eccentricity and don't check
your own work. Your bragging doesn't fool anyone but you.

I don't need to.

I don't care if it is 0.855 or 0.845. The exact figure is provided in my
program as a double precision number if anyone wants it.

For instance, my 0.75 eccentricities are really 0.746849573384083

Smaller eccentricity values are much closer to the selected one.
0.2 is actually 0.1990828.........etc

These are calculated using the major and minor axes and the focal point in
double precision.

On Tue, 22 Nov 2011 21:11:34 -0600, Sam Wormley wrote:

On 11/22/11 7:14 PM, Henry Wilson DSc. wrote:
On Tue, 22 Nov 2011 16:37:36 -0600, Sam wrote:

On 11/22/11 3:38 PM, Henry Wilson DSc. wrote:
My Prius uses an Atkinson cycle engine, in which the crankshaft is offset at
time of explosion so that the resulting torque is increased.

What kind of mileage do you get?

22-23 km/litre...~60 miles per gallon.

It's a hybrid of course. Infinitely variable gearing...What a car!


I get 17.8 km/litre in the winter
21.3 km/litre in the summer

19.6 km/litre average

Very good.
What car? It must be a diesel...

So I find your numbers hard to believe. But good for you if you
really do achieve 22-23 km/litre based on distance traveled
divided by litres purchased.

I get 21 km/litre on short runs.
I did a 3000 km trip recently mainly on country roads and got 23 km/l
according to both the car's own calculation and my own based on what I put
into it.
I corrected for the fact that the speedo reads 2% high, which is apparently
compulsory for all cars.