The extraordinary genius of Albert Einstein (DOC) | ScienceDump

"Henry Wilson DSc." ..@.. wrote in message
news
On Thu, 31 May 2012 08:03:17 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
. ..
On Thu, 31 May 2012 07:15:36 +0100, "Androcles" wrote:

Read the Wiki article.

**** the wiki article, I'm reading the ignorant crap you write and
having a good laugh.

So is everyone who reads your posts.


Oh, you want to multiply by 2pi now?
You must have been listening to Tusseladd.

2pi, 4pi, 8pi....whateverpi...depending on the number of harmonics you
want.


Of course in practical situations, the string has mass and is
subject
to
damping and the vibration has to be maintained. The maths involve
very
complicated partial differential equations that I wont bother you
with.

Bwahahahahaha!
"For a string,
A[sin(wt-x/L) + sin(wt+x/L)] = 2[sin(wt).cos(x/L)] " -- Wilson.

Make it: A[sin(x/L+wt) + sin(x/L-wt)] = 2A[cos(wt).sin(x/L)].... if
you
wish.
It's all the same really.


I'll go with this instead:

http://publicliterature.org/tools/di...ver/index.html

You didn't even look, did you?

Actually, I did....but the equation here involves a trig identity and not
a
differential equation.

Yeah, right, that's why is it is called "differential_equation_solver".
Solve the differential equation d2y/dt2 = -y(t) for me, Wilson.
Hint: it's very easy if you know dsin(t)/dt = cos(t) and cos(t)
=sin(t+pi/2).
You can't bluff me, junior lab boy with the D.Sc.
You talk the talk just like Roberts, now walk the walk or go back to
selling VW camper vans with a 6 Volt battery.




All you have to do is add two traveling (sine) waves going in opposite
directions.

I thought that would be easy for any pommie engineer.

I would if you told me what speed the two mythical sine waves were
travelling at. I thought that would be easy for any ozzie sheep shagger.

The speed is the same for both and is included in the constants.

What is the speed, Wilson? I thought that would be easy for any
ozzie sheep shagger.

On Thu, 31 May 2012 23:32:03 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
news
On Thu, 31 May 2012 08:03:17 +0100, "Androcles" wrote:

http://publicliterature.org/tools/di...ver/index.html

You didn't even look, did you?

Actually, I did....but the equation here involves a trig identity and not
a
differential equation.

Yeah, right, that's why is it is called "differential_equation_solver".
Solve the differential equation d2y/dt2 = -y(t) for me, Wilson.

What are the boundary conditions?

Hint: it's very easy if you know dsin(t)/dt = cos(t) and cos(t)
=sin(t+pi/2).
You can't bluff me, junior lab boy with the D.Sc.
You talk the talk just like Roberts, now walk the walk or go back to
selling VW camper vans with a 6 Volt battery.




All you have to do is add two traveling (sine) waves going in opposite
directions.

I thought that would be easy for any pommie engineer.

I would if you told me what speed the two mythical sine waves were
travelling at. I thought that would be easy for any ozzie sheep shagger.

The speed is the same for both and is included in the constants.

What is the speed, Wilson?

L/t. It's the same in both directions.

"Henry Wilson DSc." ..@.. wrote in message
...
On Thu, 31 May 2012 23:32:03 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
news
On Thu, 31 May 2012 08:03:17 +0100, "Androcles" wrote:

http://publicliterature.org/tools/di...ver/index.html

You didn't even look, did you?

Actually, I did....but the equation here involves a trig identity and
not
a
differential equation.

Yeah, right, that's why is it is called "differential_equation_solver".
Solve the differential equation d2y/dt2 = -y(t) for me, Wilson.

What are the boundary conditions?

Bwahahahahaha!
Solve F = ma for me, Wilson, when the ball doesn't hit the ground!
"What are air resistance conditions?" -- Wilson.


Hint: it's very easy if you know dsin(t)/dt = cos(t) and cos(t)
=sin(t+pi/2).
You can't bluff me, junior lab boy with the D.Sc.
You talk the talk just like Roberts, now walk the walk or go back to
selling VW camper vans with a 6 Volt battery.




All you have to do is add two traveling (sine) waves going in opposite
directions.

I thought that would be easy for any pommie engineer.

I would if you told me what speed the two mythical sine waves were
travelling at. I thought that would be easy for any ozzie sheep
shagger.

The speed is the same for both and is included in the constants.

What is the speed, Wilson?

L/t. It's the same in both directions.

1/t = f, so Lf?
You can't bluff me, junior lab boy with the D.Sc.

On Fri, 1 Jun 2012 00:17:49 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
On Thu, 31 May 2012 23:32:03 +0100, "Androcles" wrote:


Actually, I did....but the equation here involves a trig identity and
not
a
differential equation.

Yeah, right, that's why is it is called "differential_equation_solver".
Solve the differential equation d2y/dt2 = -y(t) for me, Wilson.

What are the boundary conditions?

Bwahahahahaha!
Solve F = ma for me, Wilson, when the ball doesn't hit the ground!
"What are air resistance conditions?" -- Wilson.

Come on Andro, that's an easy one.

something like: m.dv/dt = mg-kv

Hint: it's very easy if you know dsin(t)/dt = cos(t) and cos(t)
=sin(t+pi/2).
You can't bluff me, junior lab boy with the D.Sc.
You talk the talk just like Roberts, now walk the walk or go back to
selling VW camper vans with a 6 Volt battery.




All you have to do is add two traveling (sine) waves going in opposite
directions.

I thought that would be easy for any pommie engineer.

I would if you told me what speed the two mythical sine waves were
travelling at. I thought that would be easy for any ozzie sheep
shagger.

The speed is the same for both and is included in the constants.

What is the speed, Wilson?

L/t. It's the same in both directions.

1/t = f, so Lf?
You can't bluff me, junior lab boy with the D.Sc.

Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

"Henry Wilson DSc." ..@.. wrote in message
news
On Fri, 1 Jun 2012 00:17:49 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
. ..
On Thu, 31 May 2012 23:32:03 +0100, "Androcles" wrote:


Actually, I did....but the equation here involves a trig identity and
not
a
differential equation.

Yeah, right, that's why is it is called "differential_equation_solver".
Solve the differential equation d2y/dt2 = -y(t) for me, Wilson.

What are the boundary conditions?

Bwahahahahaha!
Solve F = ma for me, Wilson, when the ball doesn't hit the ground!
"What are air resistance conditions?" -- Wilson.

Come on Andro, that's an easy one.

something like: m.dv/dt = mg-kv

Let's put the boundary condition in, then.
Solve F = ma for me, Wilson, when the ball DOES hit the ground!


Come on Wilson, this is an easy one.
Solve the differential equation d2y/dt2 = -y(t) for me.
Something like http://mathworld.wolfram.com/EulerFormula.html



Hint: it's very easy if you know dsin(t)/dt = cos(t) and cos(t)
=sin(t+pi/2).
You can't bluff me, junior lab boy with the D.Sc.
You talk the talk just like Roberts, now walk the walk or go back to
selling VW camper vans with a 6 Volt battery.




All you have to do is add two traveling (sine) waves going in
opposite
directions.

I thought that would be easy for any pommie engineer.

I would if you told me what speed the two mythical sine waves were
travelling at. I thought that would be easy for any ozzie sheep
shagger.

The speed is the same for both and is included in the constants.

What is the speed, Wilson?

L/t. It's the same in both directions.

1/t = f, so Lf?
You can't bluff me, junior lab boy with the D.Sc.

Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

That's an interesting one. So v = 2v?
I stand by my conclusion... as barking mad as an aetherialist.

On Fri, 1 Jun 2012 09:02:44 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
news
On Fri, 1 Jun 2012 00:17:49 +0100, "Androcles" wrote:

Bwahahahahaha!
Solve F = ma for me, Wilson, when the ball doesn't hit the ground!
"What are air resistance conditions?" -- Wilson.

Come on Andro, that's an easy one.

something like: m.dv/dt = mg-kv

Let's put the boundary condition in, then.
Solve F = ma for me, Wilson, when the ball DOES hit the ground!


Come on Wilson, this is an easy one.
Solve the differential equation d2y/dt2 = -y(t) for me.

What is that supposed to represent?

Something like http://mathworld.wolfram.com/EulerFormula.html

Nothing like it.


What is the speed, Wilson?

L/t. It's the same in both directions.

1/t = f, so Lf?
You can't bluff me, junior lab boy with the D.Sc.

Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

That's an interesting one. So v = 2v?
I stand by my conclusion... as barking mad as an aetherialist.

L is a half wave.

"Henry Wilson DSc." ..@.. wrote in message
...
On Fri, 1 Jun 2012 09:02:44 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
news
On Fri, 1 Jun 2012 00:17:49 +0100, "Androcles" wrote:

Bwahahahahaha!
Solve F = ma for me, Wilson, when the ball doesn't hit the ground!
"What are air resistance conditions?" -- Wilson.

Come on Andro, that's an easy one.

something like: m.dv/dt = mg-kv

Let's put the boundary condition in, then.
Solve F = ma for me, Wilson, when the ball DOES hit the ground!


Come on Wilson, this is an easy one.
Solve the differential equation d2y/dt2 = -y(t) for me.

What is that supposed to represent?

Something like http://mathworld.wolfram.com/EulerFormula.html

Nothing like it.

I'll tell you since you know nothing about differential equations.
The solution to a differential equation is a function, Wilson.
In this case y(t) = sin(t) or cos(t) or sin(t+theta).
The first derivative of sin(t) is cos(t),
the second derivative is -sin(t),
the third derivative is -cos(t)
and the fourth derivative is sin(t),
which is itself.
Hence y''(t) = -y(t)
( y'(t) is just a different notation for dy/dt ).




What is the speed, Wilson?

L/t. It's the same in both directions.

1/t = f, so Lf?
You can't bluff me, junior lab boy with the D.Sc.

Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

That's an interesting one. So v = 2v?
I stand by my conclusion... as barking mad as an aetherialist.

L is a half wave.

So x/L is 2x because x is the length of half an organ string or
a violin pipe?

On Fri, 1 Jun 2012 12:33:57 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
On Fri, 1 Jun 2012 09:02:44 +0100, "Androcles" wrote:

Come on Andro, that's an easy one.

something like: m.dv/dt = mg-kv

Let's put the boundary condition in, then.
Solve F = ma for me, Wilson, when the ball DOES hit the ground!


Come on Wilson, this is an easy one.
Solve the differential equation d2y/dt2 = -y(t) for me.

What is that supposed to represent?

Something like http://mathworld.wolfram.com/EulerFormula.html

Nothing like it.

I'll tell you since you know nothing about differential equations.
The solution to a differential equation is a function, Wilson.
In this case y(t) = sin(t) or cos(t) or sin(t+theta).
The first derivative of sin(t) is cos(t),
the second derivative is -sin(t),
the third derivative is -cos(t)
and the fourth derivative is sin(t),
which is itself.
Hence y''(t) = -y(t)
( y'(t) is just a different notation for dy/dt ).

That's pretty simple then...but is it the only solution?


Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

That's an interesting one. So v = 2v?
I stand by my conclusion... as barking mad as an aetherialist.

L is a half wave.

So x/L is 2x because x is the length of half an organ string or
a violin pipe?

The string is only half the fundamental's wavelength. It is the full length
of the first harmoonic.

"Henry Wilson DSc." ..@.. wrote in message
...
On Fri, 1 Jun 2012 12:33:57 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
. ..
On Fri, 1 Jun 2012 09:02:44 +0100, "Androcles" wrote:

Come on Andro, that's an easy one.

something like: m.dv/dt = mg-kv

Let's put the boundary condition in, then.
Solve F = ma for me, Wilson, when the ball DOES hit the ground!


Come on Wilson, this is an easy one.
Solve the differential equation d2y/dt2 = -y(t) for me.

What is that supposed to represent?

Something like http://mathworld.wolfram.com/EulerFormula.html

Nothing like it.

I'll tell you since you know nothing about differential equations.
The solution to a differential equation is a function, Wilson.
In this case y(t) = sin(t) or cos(t) or sin(t+theta).
The first derivative of sin(t) is cos(t),
the second derivative is -sin(t),
the third derivative is -cos(t)
and the fourth derivative is sin(t),
which is itself.
Hence y''(t) = -y(t)
( y'(t) is just a different notation for dy/dt ).

That's pretty simple then...but is it the only solution?

No. Just as x^3 = 1 has three solutions for x, namely
[1],
[cos(2pi/3), i.sin(2pi/3)],
[cos(4pi/3), i.sin(4pi/3)]
and x^4 = 1 has four solutions,
1,
i,
-1,
-i
there are many solutions for a differential equation
(all of which are functions, not values), but the
ones that are interesting and relevant to physics are
exp(), sin() and cos(), which are a family as Euler
showed.
When you open a tap at the bottom of a U-tube,
one side of which is water filled, the water flows
from one side to the other until the level in each
side of U-tube is the same. You can control the
flow by how much you open the tap, but no
matter what you do the flow will reduce as the
height of water in the tubes approaches equality.
If the water has inertia it will overshoot and
oscillate, but ignore that. Concentrate on the
function that describes the flow, it is exp(-t)
and approaches zero as t reaches infinity.
The flow stops as t increases.
In electronics, the current stops when the
capacitor is charged, and it is charged when
the voltage across the capacitor reaches the
voltage across the battery. Charging a capacitor
through a resistor is just like opening the tap
a little in the U-tube. A partly open tap resists the flow.






Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

That's an interesting one. So v = 2v?
I stand by my conclusion... as barking mad as an aetherialist.

L is a half wave.

So x/L is 2x because x is the length of half an organ string or
a violin pipe?

The string is only half the fundamental's wavelength. It is the full
length
of the first harmoonic.

I've heard Loona is another name for the Moon, and loonatics
have harmoonics when hooling at it like doogs and woolves...
and of course werefools in sheep's clothing (wool). Are you a
wearwool, Wilson?

On Sat, 2 Jun 2012 00:23:23 +0100, "Androcles" wrote:


"Henry Wilson DSc." ..@.. wrote in message
.. .
On Fri, 1 Jun 2012 12:33:57 +0100, "Androcles" wrote:


What is that supposed to represent?

Something like http://mathworld.wolfram.com/EulerFormula.html

Nothing like it.

I'll tell you since you know nothing about differential equations.
The solution to a differential equation is a function, Wilson.
In this case y(t) = sin(t) or cos(t) or sin(t+theta).
The first derivative of sin(t) is cos(t),
the second derivative is -sin(t),
the third derivative is -cos(t)
and the fourth derivative is sin(t),
which is itself.
Hence y''(t) = -y(t)
( y'(t) is just a different notation for dy/dt ).

That's pretty simple then...but is it the only solution?

No. Just as x^3 = 1 has three solutions for x, namely
[1],
[cos(2pi/3), i.sin(2pi/3)],
[cos(4pi/3), i.sin(4pi/3)]
and x^4 = 1 has four solutions,
1,
i,
-1,
-i

e^-it is another solution.


there are many solutions for a differential equation
(all of which are functions, not values), but the
ones that are interesting and relevant to physics are
exp(), sin() and cos(), which are a family as Euler
showed.
When you open a tap at the bottom of a U-tube,
one side of which is water filled, the water flows
from one side to the other until the level in each
side of U-tube is the same. You can control the
flow by how much you open the tap, but no
matter what you do the flow will reduce as the
height of water in the tubes approaches equality.
If the water has inertia it will overshoot and
oscillate, but ignore that. Concentrate on the
function that describes the flow, it is exp(-t)
and approaches zero as t reaches infinity.
The flow stops as t increases.
In electronics, the current stops when the
capacitor is charged, and it is charged when
the voltage across the capacitor reaches the
voltage across the battery. Charging a capacitor
through a resistor is just like opening the tap
a little in the U-tube. A partly open tap resists the flow.

Yes we know all that.


Here's some more bluffing then:

v = 2L/T, where T is the period and L the length of the string.

That's an interesting one. So v = 2v?
I stand by my conclusion... as barking mad as an aetherialist.

L is a half wave.

So x/L is 2x because x is the length of half an organ string or
a violin pipe?

The string is only half the fundamental's wavelength. It is the full
length
of the first harmoonic.

I've heard Loona is another name for the Moon, and loonatics
have harmoonics when hooling at it like doogs and woolves...
and of course werefools in sheep's clothing (wool). Are you a
wearwool, Wilson?

......(in other words, Androcles doesn't understand what I'm talking about.)