What is wrong with the 'Mainstream Scientific Establishment'?

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

On Sat, 26 Nov 2011 03:39:53 -0000, "Androcles"
wrote:


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


| | 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.
|
That's not the right chaos.

I know all about the other aspects of chaos.

Don't you know any motion?
If you watch the video carefully you'll hear repeated references to
celestial mechanics -- and at 8:40 into the video he calls it "non-linear
dynamics", the subject I'm trying to get across.

I am aware of the chaotic nature of thre or more body systems.

The equation for the double pendulum is
y[t] = R.sin(wt) + L.sin(wt)
x[t] = R.cos(wt) + L.cos(wt)
but w isn't constant.

That's not sufficient. The two sections have entirely different 'w's and
neither is constant, although one might be pretty close if it relates to a
relatively long arm.

|
| 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.

I don't care if you drew a cow jumping over the moon for the little
dog to laugh at, I didn't ask for two elliptical orbits! I asked for ONE
red curve with R/L = 1, showing position of a piston, and ONE
difference curve for a piston with non-linear motion.

| 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.
|
Only if you could stay on target instead of rolling around the deck and
firing
off in all directions like the loose cannon you are.

ME? a loose cannon????? YOU have been rolling aimlessly around the deck for
years.

You don't even realise the importance of your chance discovery...and if I
hadn't mentioned it it would have gone completely unnoticed.

"Henry Wilson DSc." ..@.. wrote in message
news | On Sat, 26 Nov 2011 03:39:53 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
|
|
| | | 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.
| |
| That's not the right chaos.
|
| I know all about the other aspects of chaos.
|
| Don't you know any motion?
| If you watch the video carefully you'll hear repeated references to
| celestial mechanics -- and at 8:40 into the video he calls it "non-linear
| dynamics", the subject I'm trying to get across.
|
| I am aware of the chaotic nature of thre or more body systems.
|
| The equation for the double pendulum is
| y[t] = R.sin(wt) + L.sin(wt)
| x[t] = R.cos(wt) + L.cos(wt)
| but w isn't constant.
|
| That's not sufficient. The two sections have entirely different 'w's and
| neither is constant, although one might be pretty close if it relates to a
| relatively long arm.
|
| |
| | 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.
|
| I don't care if you drew a cow jumping over the moon for the little
| dog to laugh at, I didn't ask for two elliptical orbits! I asked for ONE
| red curve with R/L = 1, showing position of a piston, and ONE
| difference curve for a piston with non-linear motion.
|
| | 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.
| |
| Only if you could stay on target instead of rolling around the deck and
| firing
| off in all directions like the loose cannon you are.
|
| ME? a loose cannon????? YOU have been rolling aimlessly around the deck
for
| years.

Yes, you.

|
| You don't even realise the importance of your chance discovery...and if I
| hadn't mentioned it it would have gone completely unnoticed.
|
Are you going to do what I asked or not?

On Sat, 26 Nov 2011 10:04:50 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
news | On Sat, 26 Nov 2011 03:39:53 -0000, "Androcles"
| wrote:
|

| | I know all about Chaos. http://www.scisite.info/Henry.htm
| | ,note the background fractal.
| |
| That's not the right chaos.
|
| I know all about the other aspects of chaos.
|
| Don't you know any motion?
| If you watch the video carefully you'll hear repeated references to
| celestial mechanics -- and at 8:40 into the video he calls it "non-linear
| dynamics", the subject I'm trying to get across.
|
| I am aware of the chaotic nature of thre or more body systems.
|
| The equation for the double pendulum is
| y[t] = R.sin(wt) + L.sin(wt)
| x[t] = R.cos(wt) + L.cos(wt)
| but w isn't constant.
|
| That's not sufficient. The two sections have entirely different 'w's and
| neither is constant, although one might be pretty close if it relates to a
| relatively long arm.
|
| |
| | 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.
|
| I don't care if you drew a cow jumping over the moon for the little
| dog to laugh at, I didn't ask for two elliptical orbits! I asked for ONE
| red curve with R/L = 1, showing position of a piston, and ONE
| difference curve for a piston with non-linear motion.
|
| | 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.
| |
| Only if you could stay on target instead of rolling around the deck and
| firing
| off in all directions like the loose cannon you are.
|
| ME? a loose cannon????? YOU have been rolling aimlessly around the deck
for
| years.

Yes, you.

|
| You don't even realise the importance of your chance discovery...and if I
| hadn't mentioned it it would have gone completely unnoticed.
|
Are you going to do what I asked or not?

PLease state precisely what you are asking. I believe I have already
answered it twice.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Sat, 26 Nov 2011 10:04:50 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| news | | On Sat, 26 Nov 2011 03:39:53 -0000, "Androcles"
| | wrote:
| |
|
| | | I know all about Chaos. http://www.scisite.info/Henry.htm
| | | ,note the background fractal.
| | |
| | That's not the right chaos.
| |
| | I know all about the other aspects of chaos.
| |
| | Don't you know any motion?
| | If you watch the video carefully you'll hear repeated references to
| | celestial mechanics -- and at 8:40 into the video he calls it
"non-linear
| | dynamics", the subject I'm trying to get across.
| |
| | I am aware of the chaotic nature of thre or more body systems.
| |
| | The equation for the double pendulum is
| | y[t] = R.sin(wt) + L.sin(wt)
| | x[t] = R.cos(wt) + L.cos(wt)
| | but w isn't constant.
| |
| | That's not sufficient. The two sections have entirely different 'w's
and
| | neither is constant, although one might be pretty close if it relates
to a
| | relatively long arm.
| |
| | |
| | | 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.
| |
| | I don't care if you drew a cow jumping over the moon for the little
| | dog to laugh at, I didn't ask for two elliptical orbits! I asked for
ONE
| | red curve with R/L = 1, showing position of a piston, and ONE
| | difference curve for a piston with non-linear motion.
| |
| | | 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.
| | |
| | Only if you could stay on target instead of rolling around the deck
and
| | firing
| | off in all directions like the loose cannon you are.
| |
| | ME? a loose cannon????? YOU have been rolling aimlessly around the deck
| for
| | years.
|
| Yes, you.
|
| |
| | You don't even realise the importance of your chance discovery...and if
I
| | hadn't mentioned it it would have gone completely unnoticed.
| |
| Are you going to do what I asked or not?
|
| PLease state precisely what you are asking. I believe I have already
| answered it twice.

Draw a curve
red[t] = R.sin(wt) + L.sin(wt) for t=0 to pi, R = 1, L = 1, w = 1
red[t] = R.sin(wt) - L.sin(wt) for t=pi to 2pi

Draw a second curve
green[t] = red[t]-red[t-1]

Normalise both curves to the same amplitude to fit the screen.
Do not change the phase.

On Sat, 26 Nov 2011 22:12:16 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Sat, 26 Nov 2011 10:04:50 -0000, "Androcles"
| wrote:
|

|
| |
| | You don't even realise the importance of your chance discovery...and if
I
| | hadn't mentioned it it would have gone completely unnoticed.
| |
| Are you going to do what I asked or not?
|
| PLease state precisely what you are asking. I believe I have already
| answered it twice.

Draw a curve
red[t] = R.sin(wt) + L.sin(wt) for t=0 to pi, R = 1, L = 1, w = 1
red[t] = R.sin(wt) - L.sin(wt) for t=pi to 2pi

Why not simply: red[t] = R.sin(wt) + L.sin(wt) for t=0 to 2pi,
R = 1, L = 1, w = 1

Draw a second curve
green[t] = red[t]-red[t-1]

HAHAHAHHHAHAHA! I don't think that's what you want.



Normalise both curves to the same amplitude to fit the screen.
Do not change the phase.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Sat, 26 Nov 2011 22:12:16 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Sat, 26 Nov 2011 10:04:50 -0000, "Androcles"
| | wrote:
| |
|
| |
| | |
| | | You don't even realise the importance of your chance discovery...and
if
| I
| | | hadn't mentioned it it would have gone completely unnoticed.
| | |
| | Are you going to do what I asked or not?
| |
| | PLease state precisely what you are asking. I believe I have already
| | answered it twice.
|
| Draw a curve
| red[t] = R.sin(wt) + L.sin(wt) for t=0 to pi, R = 1, L = 1, w = 1
| red[t] = R.sin(wt) - L.sin(wt) for t=pi to 2pi
|
| Why not simply: red[t] = R.sin(wt) + L.sin(wt) for t=0 to 2pi,
| R = 1, L = 1, w = 1
|
Because the piston is constrained by the cylinder.


| Draw a second curve
| green[t] = red[t]-red[t-1]
|
| HAHAHAHHHAHAHA! I don't think

Everybody knows that already, there is no need to advertise it.
Leave the thinking to engineers; just do as you are told, lab boy.

On Sun, 27 Nov 2011 00:11:50 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Sat, 26 Nov 2011 22:12:16 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Sat, 26 Nov 2011 10:04:50 -0000, "Androcles"
| | wrote:
| |
|
| |
| | |
| | | You don't even realise the importance of your chance discovery...and
if
| I
| | | hadn't mentioned it it would have gone completely unnoticed.
| | |
| | Are you going to do what I asked or not?
| |
| | PLease state precisely what you are asking. I believe I have already
| | answered it twice.
|
| Draw a curve
| red[t] = R.sin(wt) + L.sin(wt) for t=0 to pi, R = 1, L = 1, w = 1
| red[t] = R.sin(wt) - L.sin(wt) for t=pi to 2pi
|
| Why not simply: red[t] = R.sin(wt) + L.sin(wt) for t=0 to 2pi,
| R = 1, L = 1, w = 1
|
Because the piston is constrained by the cylinder.

If L = R, the piston ain't goin' nowhere boy....except into little pieces.

| Draw a second curve
| green[t] = red[t]-red[t-1]
|
| HAHAHAHHHAHAHA! I don't think

Everybody knows that already, there is no need to advertise it.
Leave the thinking to engineers; just do as you are told, lab boy.

No wonder the Concorde crashed..

"Henry Wilson DSc." ..@.. wrote in message
...
| On Sun, 27 Nov 2011 00:11:50 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Sat, 26 Nov 2011 22:12:16 -0000, "Androcles"
| | wrote:
| |
| |
| | "Henry Wilson DSc." ..@.. wrote in message
| | .. .
| | | On Sat, 26 Nov 2011 10:04:50 -0000, "Androcles"
| | | wrote:
| | |
| |
| | |
| | | |
| | | | You don't even realise the importance of your chance
discovery...and
| if
| | I
| | | | hadn't mentioned it it would have gone completely unnoticed.
| | | |
| | | Are you going to do what I asked or not?
| | |
| | | PLease state precisely what you are asking. I believe I have already
| | | answered it twice.
| |
| | Draw a curve
| | red[t] = R.sin(wt) + L.sin(wt) for t=0 to pi, R = 1, L = 1, w = 1
| | red[t] = R.sin(wt) - L.sin(wt) for t=pi to 2pi
| |
| | Why not simply: red[t] = R.sin(wt) + L.sin(wt) for t=0 to 2pi,
| | R = 1, L = 1, w = 1
| |
| Because the piston is constrained by the cylinder.
|
| If L = R, the piston ain't goin' nowhere boy....except into little pieces.

As I said up front a week ago and repeated last Wednesday, but you
took no notice of, being the useless snipping ignorant failure that you a
"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."


| | Draw a second curve
| | green[t] = red[t]-red[t-1]
| |
| | HAHAHAHHHAHAHA! I don't think
|
| Everybody knows that already, there is no need to advertise it.
| Leave the thinking to engineers; just do as you are told, lab boy.
|
| No wonder the Concorde crashed..
|
One more time: are you going to do what I asked or not, ya ****ing failure?

On Sun, 27 Nov 2011 07:33:31 -0000, "Androcles"
wrote:


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

| | Draw a curve
| | red[t] = R.sin(wt) + L.sin(wt) for t=0 to pi, R = 1, L = 1, w = 1
| | red[t] = R.sin(wt) - L.sin(wt) for t=pi to 2pi
| |
| | Why not simply: red[t] = R.sin(wt) + L.sin(wt) for t=0 to 2pi,
| | R = 1, L = 1, w = 1
| |
| Because the piston is constrained by the cylinder.
|
| If L = R, the piston ain't goin' nowhere boy....except into little pieces.

As I said up front a week ago and repeated last Wednesday, but you
took no notice of, being the useless snipping ignorant failure that you a
"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."

L has to be considerably longer than R.....probably at least three times.

| | Draw a second curve
| | green[t] = red[t]-red[t-1]
| |
| | HAHAHAHHHAHAHA! I don't think
|
| Everybody knows that already, there is no need to advertise it.
| Leave the thinking to engineers; just do as you are told, lab boy.
|
| No wonder the Concorde crashed..
|
One more time: are you going to do what I asked or not, ya ****ing failure?

I'm already doing far more than anything you asked.