What is wrong with the 'Mainstream Scientific Establishment'?

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

We engineers test our designs to destruction. Fortunately we don't
destroy computers with code.
Are you going to do as I asked or not?


| | | 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.
|

Are you going to do as I asked or not?

On Sun, 27 Nov 2011 10:23:47 -0000, "Androcles"
wrote:


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

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

We engineers test our designs to destruction. Fortunately we don't
destroy computers with code.
Are you going to do as I asked or not?

The plain fact is I HAVE ALREADY DONE IT.

WWW.SCISITE.info/athea2.jpg

I have not used L = R because that is impossible.

I have used L = 1.01 R and L = 1.003 R which is close enough to 1.

| | | 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.
|

Are you going to do as I asked or not?

It has been done.

The results are sensational...even if you will never be able to understand
why.

"Henry Wilson DSc." ..@.. wrote in message
...
| On Sun, 27 Nov 2011 10:23:47 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
|
| |
| | L has to be considerably longer than R.....probably at least three
times.
|
| We engineers test our designs to destruction. Fortunately we don't
| destroy computers with code.
| Are you going to do as I asked or not?
|
| The plain fact is I HAVE ALREADY DONE IT.
|
| WWW.SCISITE.info/athea2.jpg
|
| I have not used L = R because that is impossible.
|
| I have used L = 1.01 R and L = 1.003 R which is close enough to 1.
|
What you should have is this:
http://www.androcles01.pwp.blueyonde...pistonY&dY.JPG
Microsoft Excel can draw the impossible, stupid Wilson can't follow a
simple instruction. **** off, you are a moron.

On 26.11.2011 00:16, Henry Wilson DSc. wrote:
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.

A normal reader would understand what 'the distance
from the hub to the ground' means.

This is but a stupid excuse for not understanding
whatI wrote.


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

Because it is of no importance.

The part of the tire that has contact with the ground is flat!
The length of the flat part is in the order of 15 cm.

It is the length of the circumference that matters,
and this length doesn't get shorter because the rubber
between the steel-belt and the ground is a bit compressed
where the tire is flat.


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.

It's the normal in Europa.

--
Paul

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

"Paul B. Andersen" wrote in
:
[...]

Not many tyres are steel belted.

It's the normal in Europa.


I'm unaware of any normal cases where you wouldn't use them in the US.

Why is Ralph arguing about this?

On Nov 25, 5:16*pm, ..@..(Henry Wilson DSc.) wrote:
On Fri, 25 Nov 2011 22:49:47 +0100, "Paul B. Andersen"

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.

Try reading the original problem statement again:
"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."

I understood the answer within a few seconds of reading the
problem statement, yet I hesitated because I thought the solution
to be too simple and obvious to constitute any sort of real
challenge. I wasted a couple of minutes trying to find a deeper
meaning before concluding that there wasn't any.

What is -your- excuse?

Jerry

On Mon, 28 Nov 2011 05:31:03 -0000, "Androcles"
wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
| On Sun, 27 Nov 2011 10:23:47 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
|
| |
| | L has to be considerably longer than R.....probably at least three
times.
|
| We engineers test our designs to destruction. Fortunately we don't
| destroy computers with code.
| Are you going to do as I asked or not?
|
| The plain fact is I HAVE ALREADY DONE IT.
|
| WWW.SCISITE.info/athea2.jpg
|
| I have not used L = R because that is impossible.
|
| I have used L = 1.01 R and L = 1.003 R which is close enough to 1.
|
What you should have is this:
http://www.androcles01.pwp.blueyonde...pistonY&dY.JPG
Microsoft Excel can draw the impossible, stupid Wilson can't follow a
simple instruction. **** off, you are a moron.

I have produced your blue curve for different values of L/R as the ratio
approaches 1. Mine just happens to be upside down, as is normal on
computers.

Gawd knows what your pink curve is.

On Mon, 28 Nov 2011 13:39:03 +0100, "Paul B. Andersen"
wrote:

On 26.11.2011 00:16, Henry Wilson DSc. wrote:
On Fri, 25 Nov 2011 22:49:47 +0100, "Paul B. Andersen"
wrote:

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.

A normal reader would understand what 'the distance
from the hub to the ground' means.

This is but a stupid excuse for not understanding
what I wrote.

A normal reader would assume that the tyre radius was the measurement when
it was off the ground.

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

Because it is of no importance.

The part of the tire that has contact with the ground is flat!
The length of the flat part is in the order of 15 cm.

It is the length of the circumference that matters,
and this length doesn't get shorter because the rubber
between the steel-belt and the ground is a bit compressed
where the tire is flat.

The wheel doesn't roll on the steel belts. It has outside layers of rubber
that compress. There is still an error athough the steel no doubt reduces
it considerably.

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


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.

It's the normal in Europa.

Discussion terminated.

On Mon, 28 Nov 2011 11:16:55 -0800 (PST), Jerry
wrote:

On Nov 25, 5:16*pm, ..@..(Henry Wilson DSc.) wrote:
On Fri, 25 Nov 2011 22:49:47 +0100, "Paul B. Andersen"

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.

Try reading the original problem statement again:
"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."

There is no mention of a car....just a wheel.

I understood the answer within a few seconds of reading the
problem statement, yet I hesitated because I thought the solution
to be too simple and obvious to constitute any sort of real
challenge. I wasted a couple of minutes trying to find a deeper
meaning before concluding that there wasn't any.

What is -your- excuse?

Jerry

"Henry Wilson DSc." ..@.. wrote in message
...
| On Mon, 28 Nov 2011 05:31:03 -0000, "Androcles"
| wrote:
|
|
| "Henry Wilson DSc." ..@.. wrote in message
| .. .
| | On Sun, 27 Nov 2011 10:23:47 -0000, "Androcles"
| | wrote:
| |
| |
| | "Henry Wilson DSc." ..@.. wrote in message
| | .. .
| |
| | |
| | | L has to be considerably longer than R.....probably at least three
| times.
| |
| | We engineers test our designs to destruction. Fortunately we don't
| | destroy computers with code.
| | Are you going to do as I asked or not?
| |
| | The plain fact is I HAVE ALREADY DONE IT.
| |
| | WWW.SCISITE.info/athea2.jpg
| |
| | I have not used L = R because that is impossible.
| |
| | I have used L = 1.01 R and L = 1.003 R which is close enough to 1.
| |
| What you should have is this:
| http://www.androcles01.pwp.blueyonde...pistonY&dY.JPG
| Microsoft Excel can draw the impossible, stupid Wilson can't follow a
| simple instruction. **** off, you are a moron.
|
| I have produced your blue curve for different values of L/R as the ratio
| approaches 1. Mine just happens to be upside down, as is normal on
| computers.
|
| Gawd knows what your pink curve is.
|
|
Don't worry about it, a ****ing moron that can't do as asked can never
understand what it is. I have used L = R because the ****ing idiot Wilson
says it is impossible.