What is wrong with the 'Mainstream Scientific Establishment'?

On Fri, 25 Nov 2011 02:15:30 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Thu, 24 Nov 2011 20:28:44 -0000, "Androcles"
| wrote:


| 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.

You can't argue with reality.

Maybe the curves I producesd are not what you wanted.

What the **** are you talking about?

On Fri, 25 Nov 2011 12:04:14 +0100, "Paul B. Andersen"
wrote:

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:

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 much is slightly true...

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.

....except for the flat bit at the bottom.

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.

Oh dear! I feel embarrassed reading this.

No wonder you are silly enough to believe everything Einstein said.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Fri, 25 Nov 2011 02:15:30 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Thu, 24 Nov 2011 20:28:44 -0000, "Androcles"
| | wrote:
|
|
| | 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.
|
| You can't argue with reality.
|
| Maybe the curves I producesd are not what you wanted.
|
| What the **** are you talking about?
|
1) R/L = 1, we are finished with the family, ok? Just one curve.
There are no square roots or X coordinates in this, L & R are the
hypotenuse of two different triangles, one on top of the other.
|\
| \ L
| \
/|
/ | R
/ |

2) Height of crank = R.sin(t), so it's range is from -R to +R.
3) Height of piston = Height of crank +L.sin(t) for t=0 to pi,
so it's 0 to 2 to 0.
4) Height of piston = Height of crank -L.sin(t) for t=pi to 2pi,
so it's 0 to 0 to 0.
That's the red curve done, although you have to scale it to fit the
screen for display.

For the green curve:
5) GreenCurve[n]= RedCurve[n] - RedCurve[n-1]
That's the green curve done, although you have to scale (differently) it to
fit the screen for display.

Even if it looks strange at first, let the computer do its job.

On Fri, 25 Nov 2011 13:25:19 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Fri, 25 Nov 2011 02:15:30 -0000, "Androcles"
| wrote:

| |
| | 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.
|
| You can't argue with reality.
|
| Maybe the curves I producesd are not what you wanted.
|
| What the **** are you talking about?
|
1) R/L = 1, we are finished with the family, ok? Just one curve.
There are no square roots or X coordinates in this, L & R are the
hypotenuse of two different triangles, one on top of the other.
|\
| \ L
| \
/|
/ | R
/ |

2) Height of crank = R.sin(t), so it's range is from -R to +R.
3) Height of piston = Height of crank +L.sin(t) for t=0 to pi,
so it's 0 to 2 to 0.
4) Height of piston = Height of crank -L.sin(t) for t=pi to 2pi,
so it's 0 to 0 to 0.
That's the red curve done, although you have to scale it to fit the
screen for display.

No, the green and red curves are scaled the same, based on the value of R,
which is held constant. The piston moves the same distance 2*R irrespective
of rod length L.
Here is the code:

T1.Enabled = False
R = 100: L = CRat * R: LL = (1 + ((CRat - 1) / 3)) * R
n = 0: p = 20

End Sub
Public Sub T1_Timer()
If n 500 Then T1.Enabled = True Else T1.Enabled = False
A = (((L ^ 2) - ((R ^ 2) * (Sin(n / p) ^ 2))) ^ 0.5) + (R * Cos(n / p))
B = (((LL ^ 2) - ((R ^ 2) * (Sin(n / p) ^ 2))) ^ 0.5) + (R * Cos(n / p))
ADiff = A - B

PSet (n, 200 + A), 655244
PSet (n, 200 + B), 255
PSet (n, 100 + (500 * ADiff)), 255200
n = n + 1
End Sub


For the green curve:
5) GreenCurve[n]= RedCurve[n] - RedCurve[n-1]
That's the green curve done, although you have to scale (differently) it to
fit the screen for display.

The difference is actually the yellow curve....but it is rather similar to
the green. I had to change the scale for that one.

Even if it looks strange at first, let the computer do its job.

It doesn't look strange and the computer has done its job correctly.

Now, the interesting aspect of this is that it can be applied to elliptical
orbits.
The difference between the radial velocity curves of two elliptial orbit
with the same periods but different eccentricities is a curve with TWICE the
frequency.
That situation is impossible in real planetary systems but if one planet's
orbit period was slightly different from the other, the resulting star orbit
would produce a brightness curve which features an apparent 'harmonic' that
continually changes phase.
That kind of brightness curve is observed quite frequently.

I will add to my program this kind of three body system, assuming it is
stable for a fairly long time.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Fri, 25 Nov 2011 13:25:19 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Fri, 25 Nov 2011 02:15:30 -0000, "Androcles"
| | wrote:
|
| | |
| | | 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.
| |
| | You can't argue with reality.
| |
| | Maybe the curves I producesd are not what you wanted.
| |
| | What the **** are you talking about?
| |
| 1) R/L = 1, we are finished with the family, ok? Just one curve.
| There are no square roots or X coordinates in this, L & R are the
| hypotenuse of two different triangles, one on top of the other.
| |\
| | \ L
| | \
| /|
| / | R
| / |
|
| 2) Height of crank = R.sin(t), so it's range is from -R to +R.
| 3) Height of piston = Height of crank +L.sin(t) for t=0 to pi,
| so it's 0 to 2 to 0.
| 4) Height of piston = Height of crank -L.sin(t) for t=pi to 2pi,
| so it's 0 to 0 to 0.
| That's the red curve done, although you have to scale it to fit the
| screen for display.
|
| No,

There is no "no" about it, sin(t) ALWAYS lies between -1 and 1 for all t.
Why you have to argue over the obvious is just stupidity.


the green and red curves are scaled the same, based on the value of R,
| which is held constant.

Jesus wept, I ****ing said that above.
2) Height of crank = R.sin(t), so it's range is from -R to +R.

**** the value of R, set it to 1 and ignore it. When Excel plots a graph
it allows the user to stretch the display window, it doesn't change the
numbers on the axes. Keep your displays separate from the data and
you can scale to any size window or screen.



| The piston moves the same distance 2*R irrespective
| of rod length L.

Of course it does, but not linearly. The sinusoidal curve
is only a true sine wave when L = infinity. When L = R
the curve is a straight line for half the period. L can never
be less than R. What we are exploring is the non-linearity
which I'll point out when you draw what I ask.


| Here is the code:
|
| T1.Enabled = False
| R = 100: L = CRat * R: LL = (1 + ((CRat - 1) / 3)) * R
| n = 0: p = 20
|

| End Sub
|

Gawd knows what all those numbers are for.
In one degree steps:
For t = 0 to pi step pi/180
y[t] = R*sin(t) + L*sin(t) /*y in range 0 to 2 for L = R = 1 */
(call plot routine with parameters t, y[t], screen height, screen width,
colour)
Next t
For t = pi to 2*pi step pi/180
y[t] = R*sin(t) - L*sin(t) /*y in range 0 to 0 for L = R = 1 */
(call plot routine)
Next t
Red curve done.

On 25.11.2011 12:29, Henry Wilson DSc. wrote:
On Fri, 25 Nov 2011 12:04:14 +0100, "Paul B. Andersen"
wrote:

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:

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 much is slightly true...

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.

...except for the flat bit at the bottom.

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.

Oh dear! I feel embarrassed reading this.

If you didn't understand it, you should indeed
be embarrassed.

Whenever you are told the simplest fact, you refuse to believe it.
Why is that?

But I am sure you will be able to understand this,
if you give yourself a chance.

Consider this. And think!

You have a thin, flat, flexible steel band.
You bend it in a circle.
You place a hub in the centre, and attach
(rather strong) rubber bands as spokes.
Now you have something like a bicycle wheel.

(A steel-belted radial tire is something like this.
The sidewalls of the tire are flexible, like
the spokes in my analogy. See URL below.)

Now you push the hub towards the ground so that
the bottom part of the wheel is flattened.
What happens to the length of the steel band?
What happens to the overall shape of the steel band?
What happens to the rubber bands?

You mark the point at the bottom, and move
the wheel along the ground, so that the bottom
part still is flattened.
When the mark again is at the bottom, what
is then the distance the wheel has advanced?
How many revolutions has the hub made?

You cannot fail to understand that the wheel
necessarily must have advanced a distance equal
to the constant circumference of the wheel,
even if the shape of the wheel is distorted.
And of course the hub has made exactly one revolution.
(The spokes are not winding up.)

The distance the wheel advances per revolution
is independent of how much the bottom part is flattened!

So if r is the distance from the hub to the ground,
the wheel will advance more than (2 pi r) per revolution.

I will strongly advice you to think carefully before
you again make a fool of yourself by revealing
that you still don't understand this.

http://www.idiomorf.nl/infographic%20tire.html
the blue "thing" is the steel-belt.

--
Paul

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

On Fri, 25 Nov 2011 21:13:47 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Fri, 25 Nov 2011 13:25:19 -0000, "Androcles"
| wrote:
|

| | You can't argue with reality.
| |
| | Maybe the curves I producesd are not what you wanted.
| |
| | What the **** are you talking about?
| |
| 1) R/L = 1, we are finished with the family, ok? Just one curve.
| There are no square roots or X coordinates in this, L & R are the
| hypotenuse of two different triangles, one on top of the other.
| |\
| | \ L
| | \
| /|
| / | R
| / |
|
| 2) Height of crank = R.sin(t), so it's range is from -R to +R.
| 3) Height of piston = Height of crank +L.sin(t) for t=0 to pi,
| so it's 0 to 2 to 0.
| 4) Height of piston = Height of crank -L.sin(t) for t=pi to 2pi,
| so it's 0 to 0 to 0.
| That's the red curve done, although you have to scale it to fit the
| screen for display.
|
| No,

There is no "no" about it, sin(t) ALWAYS lies between -1 and 1 for all t.
Why you have to argue over the obvious is just stupidity.


the green and red curves are scaled the same, based on the value of R,
| which is held constant.

Jesus wept, I ****ing said that above.
2) Height of crank = R.sin(t), so it's range is from -R to +R.

**** the value of R, set it to 1 and ignore it. When Excel plots a graph
it allows the user to stretch the display window, it doesn't change the
numbers on the axes. Keep your displays separate from the data and
you can scale to any size window or screen.



| The piston moves the same distance 2*R irrespective
| of rod length L.

Of course it does, but not linearly. The sinusoidal curve
is only a true sine wave when L = infinity. When L = R
the curve is a straight line for half the period. L can never
be less than R. What we are exploring is the non-linearity
which I'll point out when you draw what I ask.


| Here is the code:
|
| T1.Enabled = False
| R = 100: L = CRat * R: LL = (1 + ((CRat - 1) / 3)) * R
| n = 0: p = 20
|

| End Sub
|

Gawd knows what all those numbers are for.
In one degree steps:
For t = 0 to pi step pi/180
y[t] = R*sin(t) + L*sin(t) /*y in range 0 to 2 for L = R = 1 */
(call plot routine with parameters t, y[t], screen height, screen width,
colour)
Next t
For t = pi to 2*pi step pi/180
y[t] = R*sin(t) - L*sin(t) /*y in range 0 to 0 for L = R = 1 */
(call plot routine)
Next t
Red curve done.

That's not the right equation.

Don't you know any trigonometry?

On Fri, 25 Nov 2011 22:49:47 +0100, "Paul B. Andersen"
wrote:

On 25.11.2011 12:29, Henry Wilson DSc. wrote:
On Fri, 25 Nov 2011 12:04:14 +0100, "Paul B. Andersen"
wrote:

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 much is slightly true...

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.

...except for the flat bit at the bottom.

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.

Oh dear! I feel embarrassed reading this.

If you didn't understand it, you should indeed
be embarrassed.

Whenever you are told the simplest fact, you refuse to believe it.
Why is that?

You did not make clear whether you took into account the flattening of the
tyre when spcified the radius.

Any normal reader would take it to mean the uncompressed radius of the tyre.

You did not consider compression of the tyre in the flattened region, which
must be considerable even in steel belted tyres.

But I am sure you will be able to understand this,
if you give yourself a chance.

Consider this. And think!

You have a thin, flat, flexible steel band.
You bend it in a circle.
You place a hub in the centre, and attach
(rather strong) rubber bands as spokes.
Now you have something like a bicycle wheel.

(A steel-belted radial tire is something like this.
The sidewalls of the tire are flexible, like
the spokes in my analogy. See URL below.)

Now you push the hub towards the ground so that
the bottom part of the wheel is flattened.
What happens to the length of the steel band?
What happens to the overall shape of the steel band?
What happens to the rubber bands?

You mark the point at the bottom, and move
the wheel along the ground, so that the bottom
part still is flattened.
When the mark again is at the bottom, what
is then the distance the wheel has advanced?
How many revolutions has the hub made?

You cannot fail to understand that the wheel
necessarily must have advanced a distance equal
to the constant circumference of the wheel,
even if the shape of the wheel is distorted.
And of course the hub has made exactly one revolution.
(The spokes are not winding up.)

The distance the wheel advances per revolution
is independent of how much the bottom part is flattened!

If it was made with an incompressible steel band that would be approximately
true. The band is covered with a thick layer of rubber which IS
compressible.

So if r is the distance from the hub to the ground,
the wheel will advance more than (2 pi r) per revolution.

What hte hell is 'r'? Why don't you define it properly?

I will strongly advice you to think carefully before
you again make a fool of yourself by revealing
that you still don't understand this.

http://www.idiomorf.nl/infographic%20tire.html
the blue "thing" is the steel-belt.

Not many tyres are steel belted.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Fri, 25 Nov 2011 21:13:47 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Fri, 25 Nov 2011 13:25:19 -0000, "Androcles"
| | wrote:
| |
|
| | | You can't argue with reality.
| | |
| | | Maybe the curves I producesd are not what you wanted.
| | |
| | | What the **** are you talking about?
| | |
| | 1) R/L = 1, we are finished with the family, ok? Just one curve.
| | There are no square roots or X coordinates in this, L & R are the
| | hypotenuse of two different triangles, one on top of the other.
| | |\
| | | \ L
| | | \
| | /|
| | / | R
| | / |
| |
| | 2) Height of crank = R.sin(t), so it's range is from -R to +R.
| | 3) Height of piston = Height of crank +L.sin(t) for t=0 to pi,
| | so it's 0 to 2 to 0.
| | 4) Height of piston = Height of crank -L.sin(t) for t=pi to 2pi,
| | so it's 0 to 0 to 0.
| | That's the red curve done, although you have to scale it to fit the
| | screen for display.
| |
| | No,
|
| There is no "no" about it, sin(t) ALWAYS lies between -1 and 1 for all t.
| Why you have to argue over the obvious is just stupidity.
|
|
| the green and red curves are scaled the same, based on the value of R,
| | which is held constant.
|
| Jesus wept, I ****ing said that above.
| 2) Height of crank = R.sin(t), so it's range is from -R to +R.
|
| **** the value of R, set it to 1 and ignore it. When Excel plots a graph
| it allows the user to stretch the display window, it doesn't change the
| numbers on the axes. Keep your displays separate from the data and
| you can scale to any size window or screen.
|
|
|
| | The piston moves the same distance 2*R irrespective
| | of rod length L.
|
| Of course it does, but not linearly. The sinusoidal curve
| is only a true sine wave when L = infinity. When L = R
| the curve is a straight line for half the period. L can never
| be less than R. What we are exploring is the non-linearity
| which I'll point out when you draw what I ask.
|
|
| | Here is the code:
| |
| | T1.Enabled = False
| | R = 100: L = CRat * R: LL = (1 + ((CRat - 1) / 3)) * R
| | n = 0: p = 20
| |
|
| | End Sub
| |
|
| Gawd knows what all those numbers are for.
| In one degree steps:
| For t = 0 to pi step pi/180
| y[t] = R*sin(t) + L*sin(t) /*y in range 0 to 2 for L = R = 1 */
| (call plot routine with parameters t, y[t], screen height, screen
width,
| colour)
| Next t
| For t = pi to 2*pi step pi/180
| y[t] = R*sin(t) - L*sin(t) /*y in range 0 to 0 for L = R = 1 */
| (call plot routine)
| Next t
| Red curve done.
|
| That's not the right equation.
|
| Don't you know any trigonometry?
|
http://www.youtube.com/watch?v=fPbExSYcQgY

You don't know any physics!
Come on, think, Daisy. Draw the correct difference curve and I'll show
you something.

On Sat, 26 Nov 2011 01:39:55 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Fri, 25 Nov 2011 21:13:47 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Fri, 25 Nov 2011 13:25:19 -0000, "Androcles"
| | wrote:
| |
|
| | | You can't argue with reality.
| | |
| | | Maybe the curves I producesd are not what you wanted.
| | |
| | | What the **** are you talking about?
| | |
| | 1) R/L = 1, we are finished with the family, ok? Just one curve.
| | There are no square roots or X coordinates in this, L & R are the
| | hypotenuse of two different triangles, one on top of the other.
| | |\
| | | \ L
| | | \
| | /|
| | / | R
| | / |
| |
| | 2) Height of crank = R.sin(t), so it's range is from -R to +R.
| | 3) Height of piston = Height of crank +L.sin(t) for t=0 to pi,
| | so it's 0 to 2 to 0.
| | 4) Height of piston = Height of crank -L.sin(t) for t=pi to 2pi,
| | so it's 0 to 0 to 0.
| | That's the red curve done, although you have to scale it to fit the
| | screen for display.
| |
| | No,
|
| There is no "no" about it, sin(t) ALWAYS lies between -1 and 1 for all t.
| Why you have to argue over the obvious is just stupidity.
|
|
| the green and red curves are scaled the same, based on the value of R,
| | which is held constant.
|
| Jesus wept, I ****ing said that above.
| 2) Height of crank = R.sin(t), so it's range is from -R to +R.
|
| **** the value of R, set it to 1 and ignore it. When Excel plots a graph
| it allows the user to stretch the display window, it doesn't change the
| numbers on the axes. Keep your displays separate from the data and
| you can scale to any size window or screen.
|
|
|
| | The piston moves the same distance 2*R irrespective
| | of rod length L.
|
| Of course it does, but not linearly. The sinusoidal curve
| is only a true sine wave when L = infinity. When L = R
| the curve is a straight line for half the period. L can never
| be less than R. What we are exploring is the non-linearity
| which I'll point out when you draw what I ask.
|
|
| | Here is the code:
| |
| | T1.Enabled = False
| | R = 100: L = CRat * R: LL = (1 + ((CRat - 1) / 3)) * R
| | n = 0: p = 20
| |
|
| | End Sub
| |
|
| Gawd knows what all those numbers are for.
| In one degree steps:
| For t = 0 to pi step pi/180
| y[t] = R*sin(t) + L*sin(t) /*y in range 0 to 2 for L = R = 1 */
| (call plot routine with parameters t, y[t], screen height, screen
width,
| colour)
| Next t
| For t = pi to 2*pi step pi/180
| y[t] = R*sin(t) - L*sin(t) /*y in range 0 to 0 for L = R = 1 */
| (call plot routine)
| Next t
| Red curve done.
|
| That's not the right equation.
|
| Don't you know any trigonometry?
|
http://www.youtube.com/watch?v=fPbExSYcQgY

I know all about Chaos. http://www.scisite.info/Henry.htm
,note the background fractal.

You don't know any physics!

You don't know any engineering.

Come on, think, Daisy. Draw the correct difference curve and I'll show
you something.

I've drawn the bloody difference curve. It has double the frequency.
I have now drawn the difference curve between radial velocities of two
elliptical orbits and the frequency is again doubled.
This provides a missing link in BaTh that you aren't even aware of.


With your knack of fluking discoveries and my brains, we could go a long way
together.