What is wrong with the 'Mainstream Scientific Establishment'?

On 11/23/11 3:06 PM, Henry Wilson DSc. wrote:
On Wed, 23 Nov 2011 09:41:53 -0600, Sam wrote:

On 11/23/11 12:11 AM, Henry Wilson DSc. wrote:
On Tue, 22 Nov 2011 22:50:10 -0600, Sam wrote:

On 11/22/11 10:47 PM, Henry Wilson DSc. wrote:
I corrected for the fact that the speedo reads 2% high, which is apparently
compulsory for all cars.

That has nothing to do with the determination of km/litre, Henry!



I assumed the odometer reading is also 2% high...but I could be wrong. It
also depends on the type pressure.



Odometer readings are affected by tire wear, tire pressure,
tire temperature and so one. One's mileage is based on distance
traveled divided by fuel consumes. If you "measure" distance
via your odometer, then you must take those variables, I mentioned,
into the error calculation.

I did. I also checked the odometer over a five km test section of a highway
and found that it was indeed reading about 2% high. I since learned that
speedos have to be at least 2% high by law in Australia. I don't know about
other countries.


Thanks -- You get way better averages than my 2005 Prius.

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"
| wrote:

| | 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.

Here's the answer.

www.scisite.info/AtheA.exe

The L/R ratio must be greater than 1 or the program tries to get a sqrt of a
negative number. It can't be less than 1 for practical reasons.

"Henry Wilson DSc." ..@.. wrote in message
...
| 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"
| | wrote:
|
| | | 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.
|
| Here's the answer.
|
| www.scisite.info/AtheA.exe
|
| The L/R ratio must be greater than 1 or the program tries to get a sqrt of
a
| negative number. It can't be less than 1 for practical reasons.

I won't download unsigned executables, it upsets my security.
Just plot the graph on a jpeg. Better yet, show a family of curves with R/L
set to
1:1
1:1.1
1:1.2
1:1.3
1:1.5
1: 1.8
1: 2
1: 3
1:5
1:10
1:20
1:50
1:100

On Thu, 24 Nov 2011 08:37:11 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Wed, 23 Nov 2011 04:11:52 -0000, "Androcles"
| wrote:
|

|
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.
|
| Here's the answer.
|
| www.scisite.info/AtheA.exe
|
| The L/R ratio must be greater than 1 or the program tries to get a sqrt of
a
| negative number. It can't be less than 1 for practical reasons.

I won't download unsigned executables, it upsets my security.

Tuen it off.

Just plot the graph on a jpeg. Better yet, show a family of curves with R/L
set to
1:1
1:1.1
1:1.2
1:1.3
1:1.5
1: 1.8
1: 2
1: 3
1:5
1:10
1:20
1:50
1:100

http://www.scisite.info/AtheA1.jpg

As L/R is increased the curve tends toward a sine wave.
As it aproaches 1, the top becomes flat and the width of hte top equals the
width of the dip.

The equation is A = sqrt(L^2 -(R ^ 2 * (Sin(x) ^ 2))) + (R * cos (x))

Why couldn't you do that?
35 minutes of my time = 1 box of red.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Thu, 24 Nov 2011 08:37:11 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Wed, 23 Nov 2011 04:11:52 -0000, "Androcles"
| | wrote:
| |
|
| |
|
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.
| |
| | Here's the answer.
| |
| | www.scisite.info/AtheA.exe
| |
| | The L/R ratio must be greater than 1 or the program tries to get a sqrt
of
| a
| | negative number. It can't be less than 1 for practical reasons.
|
| I won't download unsigned executables, it upsets my security.
|
| Tuen it off.
|
| Just plot the graph on a jpeg. Better yet, show a family of curves with
R/L
| set to
| 1:1
| 1:1.1
| 1:1.2
| 1:1.3
| 1:1.5
| 1: 1.8
| 1: 2
| 1: 3
| 1:5
| 1:10
| 1:20
| 1:50
| 1:100
|
| http://www.scisite.info/AtheA1.jpg
|
| As L/R is increased the curve tends toward a sine wave.
| As it aproaches 1, the top becomes flat and the width of hte top equals
the
| width of the dip.
|
| The equation is A = sqrt(L^2 -(R ^ 2 * (Sin(x) ^ 2))) + (R * cos (x))
|
| Why couldn't you do that?
| 35 minutes of my time = 1 box of red.

Good, even if upside down, but that's to be expected from an ozzie.
Turn the white diagram on the right over and it'll be fine.

The next step is to plot the difference between each pair of points
(you can use a different colour). You can use arrays to do that.
For n = 1 to k
B[n] = scalefactor *( [A[n]-A[n-1] )
Next n
B[0] = scalefactor *(A[0]-A[k])

Make sure the difference plot has the same phase as the red plot.

On Thu, 24 Nov 2011 20:28:44 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Thu, 24 Nov 2011 08:37:11 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Wed, 23 Nov 2011 04:11:52 -0000, "Androcles"
| | wrote:
| |
|
| |
|
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.
| |
| | Here's the answer.
| |
| | www.scisite.info/AtheA.exe
| |
| | The L/R ratio must be greater than 1 or the program tries to get a sqrt
of
| a
| | negative number. It can't be less than 1 for practical reasons.
|
| I won't download unsigned executables, it upsets my security.
|
| Tuen it off.
|
| Just plot the graph on a jpeg. Better yet, show a family of curves with
R/L
| set to
| 1:1
| 1:1.1
| 1:1.2
| 1:1.3
| 1:1.5
| 1: 1.8
| 1: 2
| 1: 3
| 1:5
| 1:10
| 1:20
| 1:50
| 1:100
|
| http://www.scisite.info/AtheA1.jpg
|
| As L/R is increased the curve tends toward a sine wave.
| As it aproaches 1, the top becomes flat and the width of hte top equals
the
| width of the dip.
|
| The equation is A = sqrt(L^2 -(R ^ 2 * (Sin(x) ^ 2))) + (R * cos (x))
|
| Why couldn't you do that?
| 35 minutes of my time = 1 box of red.

Good, even if upside down, but that's to be expected from an ozzie.
Turn the white diagram on the right over and it'll be fine.

The next step is to plot the difference between each pair of points
(you can use a different colour). You can use arrays to do that.
For n = 1 to k
B[n] = scalefactor *( [A[n]-A[n-1] )
Next n
B[0] = scalefactor *(A[0]-A[k])

Wait a minute.
I have to use some kind of fixed ratio for L/R.

For instance I can compare the curve for L/R = 4 with that for L/R = 2 ,
then maybe L/R = 1.4 with L/R = 1.2.

What are you actually looking for?

Make sure the difference plot has the same phase as the red plot.

I don't need arrays because the time scale is the same on both. I can
subtract directly.

.....(two boxes)

On Thu, 24 Nov 2011 20:28:44 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Thu, 24 Nov 2011 08:37:11 -0000, "Androcles"
| wrote:

|
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.
| |
| | Here's the answer.
| |
| | www.scisite.info/AtheA.exe
| |
| | The L/R ratio must be greater than 1 or the program tries to get a sqrt
of
| a
| | negative number. It can't be less than 1 for practical reasons.
|
| I won't download unsigned executables, it upsets my security.
|
| Tuen it off.
|
| Just plot the graph on a jpeg. Better yet, show a family of curves with
R/L
| set to
| 1:1
| 1:1.1
| 1:1.2
| 1:1.3
| 1:1.5
| 1: 1.8
| 1: 2
| 1: 3
| 1:5
| 1:10
| 1:20
| 1:50
| 1:100
|
| http://www.scisite.info/AtheA1.jpg
|
| As L/R is increased the curve tends toward a sine wave.
| As it aproaches 1, the top becomes flat and the width of hte top equals
the
| width of the dip.
|
| The equation is A = sqrt(L^2 -(R ^ 2 * (Sin(x) ^ 2))) + (R * cos (x))
|
| Why couldn't you do that?
| 35 minutes of my time = 1 box of red.

Good, even if upside down, but that's to be expected from an ozzie.
Turn the white diagram on the right over and it'll be fine.

The next step is to plot the difference between each pair of points
(you can use a different colour). You can use arrays to do that.
For n = 1 to k
B[n] = scalefactor *( [A[n]-A[n-1] )
Next n
B[0] = scalefactor *(A[0]-A[k])

Make sure the difference plot has the same phase as the red plot.

OK. You have discovered something. I expect half share of your fame....(and
three boxes of red).

http://www.scisite.info/AtheA2.jpg

The yellow curve is the difference.

The interesting aspect is that the frequency of the difference is double
that of the others UNLESS both curves are pure sinewaves...in other words
when L is infinite.

If you can translate this into some kind of stellar model, it could explain
the existence of what appear to be 1st harmonics in many brightness curves.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Thu, 24 Nov 2011 20:28:44 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Thu, 24 Nov 2011 08:37:11 -0000, "Androcles"
| | wrote:
| |
| |
| | "Henry Wilson DSc." ..@.. wrote in message
| | .. .
| | | On Wed, 23 Nov 2011 04:11:52 -0000, "Androcles"
| | | wrote:
| | |
| |
| | |
| |
|
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.
| | |
| | | Here's the answer.
| | |
| | | www.scisite.info/AtheA.exe
| | |
| | | The L/R ratio must be greater than 1 or the program tries to get a
sqrt
| of
| | a
| | | negative number. It can't be less than 1 for practical reasons.
| |
| | I won't download unsigned executables, it upsets my security.
| |
| | Tuen it off.
| |
| | Just plot the graph on a jpeg. Better yet, show a family of curves
with
| R/L
| | set to
| | 1:1
| | 1:1.1
| | 1:1.2
| | 1:1.3
| | 1:1.5
| | 1: 1.8
| | 1: 2
| | 1: 3
| | 1:5
| | 1:10
| | 1:20
| | 1:50
| | 1:100
| |
| | http://www.scisite.info/AtheA1.jpg
| |
| | As L/R is increased the curve tends toward a sine wave.
| | As it aproaches 1, the top becomes flat and the width of hte top equals
| the
| | width of the dip.
| |
| | The equation is A = sqrt(L^2 -(R ^ 2 * (Sin(x) ^ 2))) + (R * cos (x))
| |
| | Why couldn't you do that?
| | 35 minutes of my time = 1 box of red.
|
| Good, even if upside down, but that's to be expected from an ozzie.
| Turn the white diagram on the right over and it'll be fine.
|
| The next step is to plot the difference between each pair of points
| (you can use a different colour). You can use arrays to do that.
| For n = 1 to k
| B[n] = scalefactor *( [A[n]-A[n-1] )
| Next n
| B[0] = scalefactor *(A[0]-A[k])
|
| Wait a minute.
| I have to use some kind of fixed ratio for L/R.
|
| For instance I can compare the curve for L/R = 4 with that for L/R = 2 ,
| then maybe L/R = 1.4 with L/R = 1.2.
|
| What are you actually looking for?

L/R = 1, and normalize the plots to 1.


|
| Make sure the difference plot has the same phase as the red plot.
|
| I don't need arrays because the time scale is the same on both. I can
| subtract directly.
|
| ....(two boxes)

You won't know the value of scalefactor without an array or a second
pass of A[n]-A[n-1], but do it any way you want.
You do know what "normalise" means, I trust?
the application of a constant amount of gain in order to bring the average
or peak amplitude to a target level (the norm).

"Henry Wilson DSc." ..@.. wrote in message
...
| On Thu, 24 Nov 2011 20:28:44 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Thu, 24 Nov 2011 08:37:11 -0000, "Androcles"
| | wrote:
|
| |
|
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.
| | |
| | | Here's the answer.
| | |
| | | www.scisite.info/AtheA.exe
| | |
| | | The L/R ratio must be greater than 1 or the program tries to get a
sqrt
| of
| | a
| | | negative number. It can't be less than 1 for practical reasons.
| |
| | I won't download unsigned executables, it upsets my security.
| |
| | Tuen it off.
| |
| | Just plot the graph on a jpeg. Better yet, show a family of curves
with
| R/L
| | set to
| | 1:1
| | 1:1.1
| | 1:1.2
| | 1:1.3
| | 1:1.5
| | 1: 1.8
| | 1: 2
| | 1: 3
| | 1:5
| | 1:10
| | 1:20
| | 1:50
| | 1:100
| |
| | http://www.scisite.info/AtheA1.jpg
| |
| | As L/R is increased the curve tends toward a sine wave.
| | As it aproaches 1, the top becomes flat and the width of hte top equals
| the
| | width of the dip.
| |
| | The equation is A = sqrt(L^2 -(R ^ 2 * (Sin(x) ^ 2))) + (R * cos (x))
| |
| | Why couldn't you do that?
| | 35 minutes of my time = 1 box of red.
|
| Good, even if upside down, but that's to be expected from an ozzie.
| Turn the white diagram on the right over and it'll be fine.
|
| The next step is to plot the difference between each pair of points
| (you can use a different colour). You can use arrays to do that.
| For n = 1 to k
| B[n] = scalefactor *( [A[n]-A[n-1] )
| Next n
| B[0] = scalefactor *(A[0]-A[k])
|
| Make sure the difference plot has the same phase as the red plot.
|
| OK. You have discovered something. I expect half share of your
fame....(and
| three boxes of red).
|
| http://www.scisite.info/AtheA2.jpg
|
| The yellow curve is the difference.
|
| The interesting aspect is that the frequency of the difference is double
| that of the others UNLESS both curves are pure sinewaves...in other words
| when L is infinite.
|
| If you can translate this into some kind of stellar model, it could
explain
| the existence of what appear to be 1st harmonics in many brightness
curves.


Oh dear... can you not see that where two red dots are furthest apart
is the greatest difference, and where they are closest is the least
difference?
The height of the yellow curve should be at a maximum where the slope
of the red curve is greatest. That is, the yellow curve should be phase
shifted 90 degrees from the red curve.

On 23.11.2011 16:41, Sam Wormley wrote:
On 11/23/11 12:11 AM, Henry Wilson DSc. wrote:
On Tue, 22 Nov 2011 22:50:10 -0600, Sam wrote:

On 11/22/11 10:47 PM, Henry Wilson DSc. wrote:
I corrected for the fact that the speedo reads 2% high, which is apparently
compulsory for all cars.

That has nothing to do with the determination of km/litre, Henry!



I assumed the odometer reading is also 2% high...but I could be wrong. It
also depends on the type pressure.



Odometer readings are affected by tire wear, tire pressure,
tire temperature and so one. One's mileage is based on distance
traveled divided by fuel consumes. If you "measure" distance
via your odometer, then you must take those variables, I mentioned,
into the error calculation.

While we are off topic, just a little even more off topic
comment because it is an interesting peculiarity:

The odometer/speedometer readings are much less dependent
on the tire pressure than one would be inclined to think.
That's because modern tires usually are steel-belted
radial tires, where the circumference of the tire is very
stable and little dependent on the pressure.
So when the wheel has made one revolution, it will have
advanced one circumference.

The peculiarity is that if r is the distance from the hub
to the ground, the wheel will advance more than 2 pi r
per revolution.
How is that possible?
The sides of the tire are flexing.


--
Paul

http://www.gethome.no/paulba/