Discussion:
PHYSICS for Girls.
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Keith Stein
2021-02-23 17:28:39 UTC
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What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,

Newton's Three Laws of Motion.

1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.

2. Force = Mass * Acceleration.

3. To every action there is an equal and opposite reaction.

Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.

And even Anarchists obey Newton's Laws eh!

Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Python
2021-02-23 17:32:05 UTC
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Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Maciej Wozniak
2021-02-23 17:48:28 UTC
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Post by Python
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Said Python, a true master of barking.
Python
2021-02-23 18:32:08 UTC
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Post by Maciej Wozniak
Post by Python
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Said Python, a true master of barking.
At least I'm not rubbing my d*ck on the leg of everyone out here
as you usually are. You did notice that most people do not even
consider you any more but good samaritans of some kind?
Maciej Wozniak
2021-02-23 19:46:02 UTC
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Post by Python
Post by Python
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Said Python, a true master of barking.
At least I'm not rubbing my d*ck on the leg of everyone out here
as you usually are. You did notice that most people do not even
consider you any more but good samaritans of some kind?
Rave and spit, poor halfbrain (and, of course, bark).
You're too stupid for anything else, after all (though
I guess any dog could learn much from you).
Keith Stein
2021-02-23 18:38:50 UTC
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Post by Python
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Have you ever tried to teach a dog to bark ?
it's not as easy as you'd think eh!
And for sure i could teach even you a little physics, Missss Python.
Python
2021-02-23 18:42:07 UTC
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Post by Keith Stein
Post by Python
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Have you ever tried to teach a dog to bark ?
it's not as easy as you'd think eh!
And for sure i could teach even you a little physics, Missss Python.
Definitely not, idiot, neither "classical" nor "not classical".
Find another hobby and die in peace.
Paul Alsing
2021-02-24 02:31:53 UTC
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Post by Python
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Amen to that...
Maciej Wozniak
2021-02-24 06:53:13 UTC
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Post by Paul Alsing
Post by Python
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Amen to that...
Not as you, Al, at all. You could teach all dogs how to bark.
Together with Python.
A pity though that you both are able for nothing else.
Keith Stein
2021-02-24 07:09:54 UTC
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Post by Paul Alsing
Post by Python
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Keith, you are a senile idiot. You couldn't teach a dog how to bark.
Amen to that...
Any more sexist comments from you Mr.Alsing, and i'll set my dog on you.
Don't worry too much though Paul, I haven't taught her how to bite yet.

keith stein
Keith Stein
2021-02-23 20:09:52 UTC
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Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Is that true, do you think ? You might be forgiven for thinking that
it is NOT true, for surely in your experience bodies do NOT continue
in uniform motion in a straight line, but rather when anything is
set in motion barring some motor it tends to stop. So Newton was wrong,
you might think. But no, it took the genius of Newton to see that the
fact of stopping which we all witness was rather the exception bought
about by friction on this Earth, but the rule in the great cosmos out
there, was, despite appearance to the contrary, that things DO keep
going at constant velocity in a straight line, unless acted upon by
an external force.

Any questions on Newton's First Law of Motion ?
Odd Bodkin
2021-02-23 20:33:47 UTC
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Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Is that true, do you think ? You might be forgiven for thinking that
it is NOT true, for surely in your experience bodies do NOT continue
in uniform motion in a straight line, but rather when anything is
set in motion barring some motor it tends to stop. So Newton was wrong,
you might think. But no, it took the genius of Newton to see that the
fact of stopping which we all witness was rather the exception bought
about by friction on this Earth, but the rule in the great cosmos out
there, was, despite appearance to the contrary, that things DO keep
going at constant velocity in a straight line, unless acted upon by
an external force.
Any questions on Newton's First Law of Motion ?
Two comments:

1. The first law is due to Galileo and Newton called out that attribution
himself. The fact that you can’t get the attribution right and feel the
need to give Newton the credit he declined himself says something about
your knowledge of the subject.

2. Why do you think posting about high school physics on a relativity group
is going to be helpful or useful to anyone?
--
Odd Bodkin -- maker of fine toys, tools, tables
Keith Stein
2021-02-23 21:25:07 UTC
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Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Is that true, do you think ? You might be forgiven for thinking that
it is NOT true, for surely in your experience bodies do NOT continue
in uniform motion in a straight line, but rather when anything is
set in motion barring some motor it tends to stop. So Newton was wrong,
you might think. But no, it took the genius of Newton to see that the
fact of stopping which we all witness was rather the exception bought
about by friction on this Earth, but the rule in the great cosmos out
there, was, despite appearance to the contrary, that things DO keep
going at constant velocity in a straight line, unless acted upon by
an external force.
Any questions on Newton's First Law of Motion ?
1. The first law is due to Galileo and Newton called out that attribution
himself. The fact that you can’t get the attribution right and feel the
need to give Newton the credit he declined himself says something about
your knowledge of the subject.
Look up "Newton's First Law" in ANY PHYSICS TEXT BOOK, Mr.Bodkin and it
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Not everything i am going to teach these girls will be in agreement with
what they would find in a physics text book, but Newton's First Law of
Motion is something on which text books and i do agree. The important
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
not who wrote it eh!
Post by Odd Bodkin
2. Why do you think posting about high school physics on a relativity group
is going to be helpful or useful to anyone?
You may be surprised to see how fast my girls learn physics Mr.Bodkin ?
We'll be tackling problem's beyond your own competence before this
course is over Mr.Bodkin, and i don't think that would take very many
weeks in fact. But then you're only a chippy really eh! :)

keith stein
Python
2021-02-23 21:29:33 UTC
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Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Is that true, do you think ? You might be forgiven for thinking that
it is NOT true, for surely in your experience bodies do NOT continue
in uniform motion in a straight line, but rather when anything is
set in motion barring some motor it tends to stop. So Newton was wrong,
you might think. But no, it took the genius of Newton to see that the
fact of stopping which we all witness was rather the exception bought
about by friction on this Earth, but the rule in the great cosmos out
there, was, despite appearance to the contrary, that things DO keep
going at constant velocity in a straight line, unless acted upon by
an external force.
Any questions on Newton's First Law of Motion ?
1. The first law is due to Galileo and Newton called out that attribution
himself. The fact that you can’t get the attribution right and feel the
need to give Newton the credit he declined himself says something about
your knowledge of the subject.
Look up "Newton's First Law" in ANY PHYSICS TEXT BOOK, Mr.Bodkin and it
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Not everything i am going to teach these girls will be in agreement with
what they would find in a physics text book, but Newton's First Law of
Motion is something on which text books and i do agree.  The important
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
not who wrote it eh!
Post by Odd Bodkin
2. Why do you think posting about high school physics on a relativity group
is going to be helpful or useful to anyone?
You may be surprised to see how fast my girls learn physics Mr.Bodkin ?
We'll  be tackling problem's beyond your own competence before this
course is over Mr.Bodkin, and i don't think that would take very many
weeks in fact.   But then you're only a chippy really eh! :)
keith stein
You're a pathetic senile kook Mr Stein. Everyone and his dog know
more than you about "classical physics" than you.

Moreover pretending to teach to "girls" especially, while there are
hundreds of thousand of them out there that can kick your ass and
know far more about physics than you (this is not a big deal) is
ridiculous and sexist. So you're not only an idiot, you are a
despicable person. Quite a usual scheme for cranks around here.
Odd Bodkin
2021-02-23 22:37:51 UTC
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Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Is that true, do you think ? You might be forgiven for thinking that
it is NOT true, for surely in your experience bodies do NOT continue
in uniform motion in a straight line, but rather when anything is
set in motion barring some motor it tends to stop. So Newton was wrong,
you might think. But no, it took the genius of Newton to see that the
fact of stopping which we all witness was rather the exception bought
about by friction on this Earth, but the rule in the great cosmos out
there, was, despite appearance to the contrary, that things DO keep
going at constant velocity in a straight line, unless acted upon by
an external force.
Any questions on Newton's First Law of Motion ?
1. The first law is due to Galileo and Newton called out that attribution
himself. The fact that you can’t get the attribution right and feel the
need to give Newton the credit he declined himself says something about
your knowledge of the subject.
Look up "Newton's First Law" in ANY PHYSICS TEXT BOOK, Mr.Bodkin and it
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Yes, though Newton in Principia was careful to say this was Galileo’s, and
YOU said it was Newton’s genius that produced it. Some good textbooks also
make note of this attribution, and even when they don’t, they don’t credit
Newton incorrectly for the insight.

My comment about the shallowness of your training in the subject, even at
the high school level, stands.
Post by Keith Stein
Not everything i am going to teach these girls will be in agreement with
what they would find in a physics text book, but Newton's First Law of
Motion is something on which text books and i do agree. The important
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
not who wrote it eh!
If who wrote it doesn’t matter to you, don’t credit Newton’s genius for
producing it eh!
Post by Keith Stein
Post by Odd Bodkin
2. Why do you think posting about high school physics on a relativity group
is going to be helpful or useful to anyone?
You may be surprised to see how fast my girls learn physics Mr.Bodkin ?
We'll be tackling problem's beyond your own competence before this
course is over Mr.Bodkin, and i don't think that would take very many
weeks in fact. But then you're only a chippy really eh! :)
A woodworker, yes, who has read books. What you’re talking about is high
school physics.
Post by Keith Stein
keith stein
--
Odd Bodkin -- maker of fine toys, tools, tables
Maciej Wozniak
2021-02-24 06:50:45 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
Is that true, do you think ? You might be forgiven for thinking that
it is NOT true, for surely in your experience bodies do NOT continue
in uniform motion in a straight line, but rather when anything is
set in motion barring some motor it tends to stop. So Newton was wrong,
you might think. But no, it took the genius of Newton to see that the
fact of stopping which we all witness was rather the exception bought
about by friction on this Earth, but the rule in the great cosmos out
there, was, despite appearance to the contrary, that things DO keep
going at constant velocity in a straight line, unless acted upon by
an external force.
Any questions on Newton's First Law of Motion ?
1. The first law is due to Galileo and Newton called out that attribution
himself. The fact that you can’t get the attribution right
Only the poor idiot woodworker knows how to get it
right.
Keith Stein
2021-02-23 20:54:32 UTC
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Permalink
Newton's 2nd Law of Motion

Force = Mass * Acceleration.

That is the very definition of "Force", and it is measured in "Newtons".
GRAVITY is a FORCE DOWN due to THE ACCELERATION OF GRAVITY of 9.8 m/s/s

In general ACCELERATION is RATE OF CHANGE OF VELOCITY.

So by Newton's First Law of Motion Velocity (of anything) is
Constant,unless a Force acts on the body.

Newton's Second Law tells us how quickly the velocity of the body
differs from constant, when a force does act on the body. We can see
from Newton's Second Law that the rate of change of velocity of
"Acceleration" is inversely proportional to the mass of the body,
which is to say that big things change slow, while little things
can change their velocity very quickly indeed. So unlike the first
law, this second law accords with common experience.

Any questions on Newton's Second Law of Motion ?
Odd Bodkin
2021-02-23 21:36:50 UTC
Reply
Permalink
Post by Keith Stein
Newton's 2nd Law of Motion
Force = Mass * Acceleration.
That is the very definition of "Force", and it is measured in "Newtons".
GRAVITY is a FORCE DOWN due to THE ACCELERATION OF GRAVITY of 9.8 m/s/s
In general ACCELERATION is RATE OF CHANGE OF VELOCITY.
So by Newton's First Law of Motion Velocity (of anything) is
Constant,unless a Force acts on the body.
Newton's Second Law tells us how quickly the velocity of the body
differs from constant, when a force does act on the body. We can see
from Newton's Second Law that the rate of change of velocity of
"Acceleration" is inversely proportional to the mass of the body,
which is to say that big things change slow, while little things
can change their velocity very quickly indeed. So unlike the first
law, this second law accords with common experience.
Any questions on Newton's Second Law of Motion ?
How about you cast it the way Newton actually described it. What you said
is not accurate.
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-24 09:42:10 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Newton's 2nd Law of Motion
Force = Mass * Acceleration.
That is the very definition of "Force", and it is measured in "Newtons".
GRAVITY is a FORCE DOWN due to THE ACCELERATION OF GRAVITY of 9.8 m/s/s
In general ACCELERATION is RATE OF CHANGE OF VELOCITY.
So by Newton's First Law of Motion Velocity (of anything) is
Constant,unless a Force acts on the body.
Newton's Second Law tells us how quickly the velocity of the body
differs from constant, when a force does act on the body. We can see
from Newton's Second Law that the rate of change of velocity of
"Acceleration" is inversely proportional to the mass of the body,
which is to say that big things change slow, while little things
can change their velocity very quickly indeed. So unlike the first
law, this second law accords with common experience.
Any questions on Newton's Second Law of Motion ?
How about you cast it the way Newton actually described it. What you said
is not accurate.
I'm sure all the girls here would love to hear anything further you can
tell us about Newton's Second Law of Motion, Mr Bodkin. Pay attention
girls. Mr.Bodkin has deep knowledge of physics (he has read a lot of
books on the subject), and his skill in the subject is about as good
as a would be that of a cabinet maker who had read a lot of books
on "How to Make Cabinets" eh! :)

The floor is yours >>>>>>>>> Mr. Bodkin:
Odd Bodkin
2021-02-24 15:29:43 UTC
Reply
Permalink
Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Newton's 2nd Law of Motion
Force = Mass * Acceleration.
That is the very definition of "Force", and it is measured in "Newtons".
GRAVITY is a FORCE DOWN due to THE ACCELERATION OF GRAVITY of 9.8 m/s/s
In general ACCELERATION is RATE OF CHANGE OF VELOCITY.
So by Newton's First Law of Motion Velocity (of anything) is
Constant,unless a Force acts on the body.
Newton's Second Law tells us how quickly the velocity of the body
differs from constant, when a force does act on the body. We can see
from Newton's Second Law that the rate of change of velocity of
"Acceleration" is inversely proportional to the mass of the body,
which is to say that big things change slow, while little things
can change their velocity very quickly indeed. So unlike the first
law, this second law accords with common experience.
Any questions on Newton's Second Law of Motion ?
How about you cast it the way Newton actually described it. What you said
is not accurate.
I'm sure all the girls here would love to hear anything further you can
tell us about Newton's Second Law of Motion, Mr Bodkin. Pay attention
girls. Mr.Bodkin has deep knowledge of physics (he has read a lot of
books on the subject), and his skill in the subject is about as good
as a would be that of a cabinet maker who had read a lot of books
on "How to Make Cabinets" eh! :)
This is not new stuff, Mr Stein. Though he wrote Principia in Latin, the
English translation “motion” that Newton used corresponds to what is now
called momentum.

The net force is the time derivative of momentum.

As you might well understand (or maybe not), the actual statement is
broader than the one you stated, and moreover it is STILL correct
relativistically.
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-24 17:14:21 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
Newton's 2nd Law of Motion
Force = Mass * Acceleration.
That is the very definition of "Force", and it is measured in "Newtons".
GRAVITY is a FORCE DOWN due to THE ACCELERATION OF GRAVITY of 9.8 m/s/s
In general ACCELERATION is RATE OF CHANGE OF VELOCITY.
So by Newton's First Law of Motion Velocity (of anything) is
Constant,unless a Force acts on the body.
Newton's Second Law tells us how quickly the velocity of the body
differs from constant, when a force does act on the body. We can see
from Newton's Second Law that the rate of change of velocity of
"Acceleration" is inversely proportional to the mass of the body,
which is to say that big things change slow, while little things
can change their velocity very quickly indeed. So unlike the first
law, this second law accords with common experience.
Any questions on Newton's Second Law of Motion ?
I'm sure all the girls here would love to hear anything further you can
tell us about Newton's Second Law of Motion, Mr Bodkin. Pay attention
girls. Mr.Bodkin has deep knowledge of physics (he has read a lot of
books on the subject), and his skill in the subject is about as good
as a would be that of a cabinet maker who had read a lot of books
on "How to Make Cabinets" eh! :)
This is not new stuff, Mr Stein. Though he wrote Principia in Latin, the
English translation “motion” that Newton used corresponds to what is now
called momentum.
The net force is the time derivative of momentum.
As you might well understand (or maybe not), the actual statement is
broader than the one you stated, and moreover it is STILL correct
relativistically.
How about YOU cast it the way Newton actually described it, Mr Bodkin.
Odd Bodkin
2021-02-24 18:56:28 UTC
Reply
Permalink
Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
Newton's 2nd Law of Motion
Force = Mass * Acceleration.
That is the very definition of "Force", and it is measured in "Newtons".
GRAVITY is a FORCE DOWN due to THE ACCELERATION OF GRAVITY of 9.8 m/s/s
In general ACCELERATION is RATE OF CHANGE OF VELOCITY.
So by Newton's First Law of Motion Velocity (of anything) is
Constant,unless a Force acts on the body.
Newton's Second Law tells us how quickly the velocity of the body
differs from constant, when a force does act on the body. We can see
from Newton's Second Law that the rate of change of velocity of
"Acceleration" is inversely proportional to the mass of the body,
which is to say that big things change slow, while little things
can change their velocity very quickly indeed. So unlike the first
law, this second law accords with common experience.
Any questions on Newton's Second Law of Motion ?
I'm sure all the girls here would love to hear anything further you can
tell us about Newton's Second Law of Motion, Mr Bodkin. Pay attention
girls. Mr.Bodkin has deep knowledge of physics (he has read a lot of
books on the subject), and his skill in the subject is about as good
as a would be that of a cabinet maker who had read a lot of books
on "How to Make Cabinets" eh! :)
This is not new stuff, Mr Stein. Though he wrote Principia in Latin, the
English translation “motion” that Newton used corresponds to what is now
called momentum.
The net force is the time derivative of momentum.
As you might well understand (or maybe not), the actual statement is
broader than the one you stated, and moreover it is STILL correct
relativistically.
How about YOU cast it the way Newton actually described it, Mr Bodkin.
I just did.

In English translation: The alteration of motion is ever proportional to
the motive force impressed; and is made in the direction of the right line
in which that force is impressed.”

The “alteration of motion”, as elaborated elsewhere in the Principia, is
the time derivative of momentum.

It is unfortunate that you were so badly educated in the basics that you
cannot state the laws accurately, attribute the ideas to the right man, nor
apply the ideas to answer simple conceptual questions.
--
Odd Bodkin -- maker of fine toys, tools, tables
Keith Stein
2021-02-23 21:56:44 UTC
Reply
Permalink
Post by Keith Stein
3. To every action there is an equal and opposite reaction.
Any questions on Newton's 3rd Law of Motion ?

Self explatory this one i would have thought,
so perhaps we just have time for a bit more
pointeless hechling from the Python or the Bodkin eh! :)

keith stein
.
Wilburn Neubecker
2021-02-23 22:00:39 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
3. To every action there is an equal and opposite reaction.
Any questions on Newton's 3rd Law of Motion ?
Self explatory this one i would have thought,
so perhaps we just have time for a bit more pointeless hechling from the
Python or the Bodkin eh!
You too, can you do zero-math science?
Keith Stein
2021-02-23 22:21:55 UTC
Reply
Permalink
Post by Wilburn Neubecker
Post by Keith Stein
Post by Keith Stein
3. To every action there is an equal and opposite reaction.
Any questions on Newton's 3rd Law of Motion ?
Self explatory this one i would have thought,
so perhaps we just have time for a bit more pointeless hechling from the
Python or the Bodkin eh!
You too, can you do zero-math science?
If one is serious about physics, Mr Neubecker, i believe it is essential
to learn how to program a computer. The computer does the maths eh! :)

keith stein
Keith Stein
2021-02-25 20:26:43 UTC
Reply
Permalink
Post by Keith Stein
...... can you do zero-math science?
If one is serious about physics, Mr Neubecker, i believe it is essential
to learn how to program a computer. The computer does the maths eh!  :)
Seriously Wilburn, LEARN TO PROGRAM,
Not necessarily any particular language, but you must understand in
principle how the computer operates..It's just numbers in boxes which
we can label P,T,x,y,z,whatever, and manipulate according to any
equation we chose. If we can write it down, the computer can work it
out. No problem.

There is more to physics than putting numbers in equations of course, Mr
Neubecker, but not a lot eh! :)

keith stein
Jim Bunce
2021-02-25 21:12:42 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
...... can you do zero-math science?
If one is serious about physics, Mr Neubecker, i believe it is
essential to learn how to program a computer. The computer does the
maths eh!
Seriously Wilburn, LEARN TO PROGRAM,
Not necessarily any particular language, but you must understand in
principle how the computer operates..It's just numbers in boxes which we
can label P,T,x,y,z,whatever, and manipulate according to any equation
we chose. If we can write it down, the computer can work it out. No
problem.
There is more to physics than putting numbers in equations of course, Mr
Neubecker, but not a lot eh!
That's not how computers works. Are not doing mathematics. They are just
a tool in doing modeling in physics. Which supposes you know their
limitation, otherwise you can't trust the garbage they spit out. Stop
being incompetent.
Keith Stein
2021-02-25 22:29:30 UTC
Reply
Permalink
Post by Jim Bunce
There is more to *physics* than putting numbers in equations of course, Mr
Neubecker, but not a lot eh!
That's not how computers works.
No, that IS how *physicists* work Mr Bunce.
Post by Jim Bunce
Are not doing mathematics. They are just
a tool in doing modeling in physics. Which supposes you know their
limitation, otherwise you can't trust the garbage they spit out.
But can we trust the garbage you spit out Mr Bunce ?
Post by Jim Bunce
Stop being incompetent.
Has that command ever been obeyed, anywhere, ever ?
Stop with the incompetent orders Mr Bunce. :)

keith stein
Odd Bodkin
2021-02-23 22:37:52 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
3. To every action there is an equal and opposite reaction.
Any questions on Newton's 3rd Law of Motion ?
Self explatory this one i would have thought,
so perhaps we just have time for a bit more
pointeless hechling from the Python or the Bodkin eh! :)
keith stein
.
Here’s a question for you, Mr Stein? If there is a reaction equal and
opposite to every action, how does there end up being a net force on any
body? Don’t action and reaction cancel out?

Here’s another question, Mr Stein. A horse starts walking on the packed
snow. What propels the horse forward, Mr Stein?
--
Odd Bodkin -- maker of fine toys, tools, tables
Keith Stein
2021-02-23 23:49:16 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
3. To every action there is an equal and opposite reaction.
Any questions on Newton's 3rd Law of Motion ?
Self explatory this one i would have thought,
so perhaps we just have time for a bit more
pointeless hechling from the Python or the Bodkin eh! :)
keith stein
.
Here’s a question for you, Mr Stein? If there is a reaction equal and
opposite to every action, how does there end up being a net force on any
body? Don’t action and reaction cancel out?
Good question Mr.Bodkin. No they don't cancel out, because the action
acts on one of the bodies, and the reaction acts on the other eh!
Neither body experiences both the "action" and the "reaction" but if
they did then, as you say Mr.Bodkin, they would indeed "cancel out".
Post by Odd Bodkin
Here’s another question, Mr Stein. A horse starts walking on the packed
snow. What propels the horse forward, Mr Stein?
Fear of freezing to death in the snow i should think Mr. Bodkin.

keith stein
Odd Bodkin
2021-02-24 15:23:55 UTC
Reply
Permalink
Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
3. To every action there is an equal and opposite reaction.
Any questions on Newton's 3rd Law of Motion ?
Self explatory this one i would have thought,
so perhaps we just have time for a bit more
pointeless hechling from the Python or the Bodkin eh! :)
keith stein
.
Here’s a question for you, Mr Stein? If there is a reaction equal and
opposite to every action, how does there end up being a net force on any
body? Don’t action and reaction cancel out?
Good question Mr.Bodkin. No they don't cancel out, because the action
acts on one of the bodies, and the reaction acts on the other eh!
Neither body experiences both the "action" and the "reaction" but if
they did then, as you say Mr.Bodkin, they would indeed "cancel out".
Very good Mr Stein.
Post by Keith Stein
Post by Odd Bodkin
Here’s another question, Mr Stein. A horse starts walking on the packed
snow. What propels the horse forward, Mr Stein?
Fear of freezing to death in the snow i should think Mr. Bodkin.
So now it is clear that you do not understand even high school physics,
enough to be able to answer a series of simple conceptual problems. So what
point might you have for trotting out high school physics that you don’t
well understand?
Post by Keith Stein
keith stein
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-24 16:52:13 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Post by Odd Bodkin
Here’s a question for you, Mr Stein? If there is a reaction equal and
opposite to every action, how does there end up being a net force on any
body? Don’t action and reaction cancel out?
Good question Mr.Bodkin. No they don't cancel out, because the action
acts on one of the bodies, and the reaction acts on the other eh!
Neither body experiences both the "action" and the "reaction" but if
they did then, as you say Mr.Bodkin, they would indeed "cancel out".
Very good Mr Stein.
Thank you Mr.Bodkin, that is HIGH PRAISE INDEED! coming from you,
my previous best score from you was either "very bad" or "very poor",i
don't remember exactly but my previous best from you was no where near
"good" and i never even dreamed i'd ever see a "very good". :)
Post by Odd Bodkin
Post by Keith Stein
Post by Odd Bodkin
Here’s another question, Mr Stein. A horse starts walking on the packed
snow. What propels the horse forward, Mr Stein?
Fear of freezing to death in the snow i should think Mr. Bodkin.
So now it is clear that you do not understand even high school physics,
This is more like the Odd Bodkin of Old,
But it is not me who does not understand "physics", Mr.Bodkin,
but rather it is you who does not understand "horses" eh!

keith stein
Odd Bodkin
2021-02-24 18:50:37 UTC
Reply
Permalink
Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Post by Odd Bodkin
Here’s a question for you, Mr Stein? If there is a reaction equal and
opposite to every action, how does there end up being a net force on any
body? Don’t action and reaction cancel out?
Good question Mr.Bodkin. No they don't cancel out, because the action
acts on one of the bodies, and the reaction acts on the other eh!
Neither body experiences both the "action" and the "reaction" but if
they did then, as you say Mr.Bodkin, they would indeed "cancel out".
Very good Mr Stein.
Thank you Mr.Bodkin, that is HIGH PRAISE INDEED! coming from you,
my previous best score from you was either "very bad" or "very poor",i
don't remember exactly but my previous best from you was no where near
"good" and i never even dreamed i'd ever see a "very good". :)
Post by Odd Bodkin
Post by Keith Stein
Post by Odd Bodkin
Here’s another question, Mr Stein. A horse starts walking on the packed
snow. What propels the horse forward, Mr Stein?
Fear of freezing to death in the snow i should think Mr. Bodkin.
So now it is clear that you do not understand even high school physics,
This is more like the Odd Bodkin of Old,
But it is not me who does not understand "physics", Mr.Bodkin,
but rather it is you who does not understand "horses" eh!
keith stein
I notice you still can’t answer the basic physics question. And it’s high
school physics.
--
Odd Bodkin -- maker of fine toys, tools, tables
Keith Stein
2021-02-25 03:59:55 UTC
Reply
Permalink
Bodkin's problem is that he reads a lot, and understands nothing,
not horses, not girls, and not physics .... Best ignore him.

Newton's Three Laws of Motion.

1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.

2. Force = Mass * Acceleration.

3. To every action there is an equal and opposite reaction.

and NEWTON'S LAW OF GRAVITATION

F = G*m1*m2/r^2

where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters

Next week we work out how much weight one can lose by relocating to the
moon. You should already know that you can't lose any mass that way eh!
Odd Bodkin
2021-02-25 15:34:28 UTC
Reply
Permalink
Post by Keith Stein
Bodkin's problem is that he reads a lot, and understands nothing,
not horses, not girls, and not physics .... Best ignore him.
Well, I gather you are not happy with being informed that your grip of
first year physics is thin and inaccurate, though you would like to take
pride in it.

So, you’d like to continue with your crappy Cliff Notes version of first
year physics?
Post by Keith Stein
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
and NEWTON'S LAW OF GRAVITATION
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters
Next week we work out how much weight one can lose by relocating to the
moon. You should already know that you can't lose any mass that way eh!
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-26 16:58:35 UTC
Reply
Permalink
Post by Keith Stein
Next week we work out how much weight one can lose by relocating to the
moon.
We are going to need to know exactly how heavy you are to answer that of
course, but i think we can get the percentage reduction without the need
for you to disclose any personal information, so let's go for that.

I don't think we even need to know Newton's Gravitational Constant, and
we could do this immediately if we knew the gravitational acceleration
on the moon, which is probably available on the net, but lets do this
the hard way.

F = G*m1*m2/r^2

where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters

So Weight on moon = G*mGirl*mMoon/rMoon^2
And Weight on Earth = G*mGirl*mEarth/rEarth^2

Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2

Google kindly provided this:

Moon Ratio (Moon/Earth)
Mass (10^24 kg) 0.07346 0.0123
Equatorial radius (km) 1738.1 0.2725

Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
= 0.0123*(1/0.2725)^2
~= 0.16

Witch means, if i didn't make no mistakes, you'd lose about 84% of your
weight when you relocate to the moon, but not even a milligram of your
mass eh!
Odd Bodkin
2021-02-26 17:09:43 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
Next week we work out how much weight one can lose by relocating to the
moon.
We are going to need to know exactly how heavy you are to answer that of
course, but i think we can get the percentage reduction without the need
for you to disclose any personal information, so let's go for that.
I don't think we even need to know Newton's Gravitational Constant, and
we could do this immediately if we knew the gravitational acceleration
on the moon, which is probably available on the net, but lets do this
the hard way.
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters
So Weight on moon = G*mGirl*mMoon/rMoon^2
And Weight on Earth = G*mGirl*mEarth/rEarth^2
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
Moon Ratio (Moon/Earth)
Mass (10^24 kg) 0.07346 0.0123
Equatorial radius (km) 1738.1 0.2725
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
= 0.0123*(1/0.2725)^2
~= 0.16
Witch means, if i didn't make no mistakes, you'd lose about 84% of your
weight when you relocate to the moon, but not even a milligram of your
mass eh!
Might I suggest you just post pictures of a nearby children’s science
museum’s exhibits?
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-26 17:44:44 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
Next week we work out how much weight one can lose by relocating to the
moon.
We are going to need to know exactly how heavy you are to answer that of
course, but i think we can get the percentage reduction without the need
for you to disclose any personal information, so let's go for that.
I don't think we even need to know Newton's Gravitational Constant, and
we could do this immediately if we knew the gravitational acceleration
on the moon, which is probably available on the net, but lets do this
the hard way.
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters
So Weight on moon = G*mGirl*mMoon/rMoon^2
And Weight on Earth = G*mGirl*mEarth/rEarth^2
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
Moon Ratio (Moon/Earth)
Mass (10^24 kg) 0.07346 0.0123
Equatorial radius (km) 1738.1 0.2725
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
= 0.0123*(1/0.2725)^2
~= 0.16
Witch means, if i didn't make no mistakes, you'd lose about 84% of your
weight when you relocate to the moon, but not even a milligram of your
mass eh!
Might I suggest you just post pictures of a nearby children’s science
museum’s exhibits?
Thank you Mr.Bodkin, and
the next time i want advice
,(on how to make a table),
i shall come to you eh!
Odd Bodkin
2021-02-26 18:36:02 UTC
Reply
Permalink
Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Post by Keith Stein
Next week we work out how much weight one can lose by relocating to the
moon.
We are going to need to know exactly how heavy you are to answer that of
course, but i think we can get the percentage reduction without the need
for you to disclose any personal information, so let's go for that.
I don't think we even need to know Newton's Gravitational Constant, and
we could do this immediately if we knew the gravitational acceleration
on the moon, which is probably available on the net, but lets do this
the hard way.
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters
So Weight on moon = G*mGirl*mMoon/rMoon^2
And Weight on Earth = G*mGirl*mEarth/rEarth^2
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
Moon Ratio (Moon/Earth)
Mass (10^24 kg) 0.07346 0.0123
Equatorial radius (km) 1738.1 0.2725
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
= 0.0123*(1/0.2725)^2
~= 0.16
Witch means, if i didn't make no mistakes, you'd lose about 84% of your
weight when you relocate to the moon, but not even a milligram of your
mass eh!
Might I suggest you just post pictures of a nearby children’s science
museum’s exhibits?
Thank you Mr.Bodkin, and
the next time i want advice
,(on how to make a table),
i shall come to you eh!
What is the point of posting high school physics tidbits on a relativity
newsgroup? Why not post about levers and pulleys?
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-26 19:18:02 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Thank you Mr.Bodkin, and
the next time i want advice
,(on how to make a table),
i shall come to you eh!
What is the point of posting high school physics tidbits on a relativity
newsgroup? Why not post about levers and pulleys?
Might i suggest that you return to your important debates with Ken Seto,
or Mitch, or Mr.Wozniak, or something. Surely you have more important
things to do than wasting your time on this "high school" trivia,
Mr.Bodkin. Maybe you could read a book on "How to Ride a Bike", and just
keep reading it until you sre as good at bike riding as you are at
physics. Shouldn't take long eh! :)

keith stein
Odd Bodkin
2021-02-26 22:15:36 UTC
Reply
Permalink
Post by Keith Stein
Post by Odd Bodkin
Post by Keith Stein
Thank you Mr.Bodkin, and
the next time i want advice
,(on how to make a table),
i shall come to you eh!
What is the point of posting high school physics tidbits on a relativity
newsgroup? Why not post about levers and pulleys?
Might i suggest that you return to your important debates with Ken Seto,
or Mitch, or Mr.Wozniak, or something. Surely you have more important
things to do than wasting your time on this "high school" trivia,
Mr.Bodkin. Maybe you could read a book on "How to Ride a Bike", and just
keep reading it until you sre as good at bike riding as you are at
physics. Shouldn't take long eh! :)
keith stein
So you wish to be left alone posting off-topic material in a relativity
group as you see fit?
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-27 11:39:29 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
Next week we work out how much weight one can lose by relocating to
the moon.
We are going to need to know exactly how heavy you are to answer that of
course, but i think we can get the percentage reduction without the need
for you to disclose any personal information, so let's go for that.
I don't think we even need to know Newton's Gravitational Constant, and
we could do this immediately if we knew the gravitational acceleration
on the moon, which is probably available on the net, but lets do this
the hard way.
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
r is the distance between the masses in meters
So  Weight on moon = G*mGirl*mMoon/rMoon^2
And Weight on Earth = G*mGirl*mEarth/rEarth^2
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth/rMoon)^2
Moon    Ratio (Moon/Earth)
Mass (10^24 kg)            0.07346            0.0123
Equatorial radius (km)    1738.1            0.2725
Therefore (Wt on Moon / Wt on Earth) = (mMoon/mEarth)*(rEarth)^2
=  0.0123*(1/0.2725)^2
~=  0.16
Witch means, if i didn't make no mistakes, you'd lose about 84% of your
weight when you relocate to the moon, but not even a milligram of your
mass eh!
So MASS is the quantity of matter in a body, and is a property which
travels everywhere with the body. Your MASS does not change no matter
where you go, so it is the same on Mars, or on the Earth, or on the moon.

Your WEIGHT on the other hand is the FORCE acting on your body due to
its attraction to the planet, or moon, on which you happen to be.

We know from Newton's Second Law that:
FORCE = MASS * ACCELERATION
Now i know, because i just Googled it, that gravity on the moon is 1.6
m/s/s, and i also know that gravity on this Earth is 9.8 m/s/s, because
physicists just know that.

Witch means that the strength of gravity on the moon is about 16% of
it's value here on Earth, and that means i didn't make no mistake when
we worked this out the hard way. So you can indeed lose about 84% of
your WEIGHT by relocating to the moon. That's physics, and it offers a
lot more than can be achieved with any diet eh! :)
Keith Stein
2021-02-24 00:15:58 UTC
Reply
Permalink
Who knew these men on s.p.r. were such "bitches" ?
Keith Stein
2021-02-26 11:18:58 UTC
Reply
Permalink
" EQUATIONS IS EASY "

Don't be SCARED of EQUATIONS girls. Nothing in any equation that you
can't solve, with a little help from me and your computer....
but first YOU DO NEED TO understand that
if x=1 and y=2 then x+y=3
And really there is no equation any more scary than that,
not when you use a computer because the computer also knows;
if x=1 and y=2
then
x*y = 2, and x-y =-1, and x^2 + y^2 = 5,and ,,,,,,,,,,

but you don't need the computer to tell you that, i hope, although
you would probably need at least a calculator if say x=6.7346213E-07,
although to your computer that is no more difficult than when x=1, of
course.

Anyway i repeat, don't be scared of equations. EQUATIONS IS EASY eh!
Keith Stein
2021-02-24 12:12:02 UTC
Reply
Permalink
Post by Keith Stein
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
No one completed the set Lesson 1 Homework, but that could be my fault
since i haven't mentioned NEWTON'S LAW OF GRAVITATION
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
and r is the distance between the masses in meters
and you need to learn all that too, but i can promise you
that in the end it will all be worth the effort. You will
with the aid of Newton's Laws learn how easy it is for anyone
to loose weight, by moving to the moon eh! :)
Keith Stein
2021-02-24 13:19:51 UTC
Reply
Permalink
Post by Keith Stein
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
No one completed the set Lesson 1 Homework, but that could  be my fault
since i haven't mentioned NEWTON'S LAW OF GRAVITATION
F = G*m1*m2/r^2
where F is the FORCE of attraction in Newtons
G is Newton's Gravitation Constant in m.k.s units
m1 and m2 the MASSES attracting each other in kilograms
and r is the distance between the masses in meters
and you need to learn all that too, but i can promise you
that in the end it will all be worth the effort. You will
with the aid of Newton's Laws learn how easy it is for anyone
to loose weight, by moving to the moon eh!   :)
Witch brings me to my point, the difference between MASS and WEIGHT.
It can be confusing because whereas MASS is measured in kilograms,
WEIGHT is strictly a force measured in Newtons, but is in practice
more usually and usefully measured in kilogram-weight.
You won't lose so much as a milligram from your MASS by going to
the moon, or even in space, where it is easy to lose 100% of WEIGHT,
so what you need to decide is, "Is it MASS or is it WEIGHT that you want
to lose, and then you'll be a physicist who DOES KNOW THE DIFFERENCE eh!
Paul B. Andersen
2021-02-24 14:14:06 UTC
Reply
Permalink
Post by Keith Stein
Witch brings me to my point, the difference between MASS and WEIGHT.
It can be confusing because whereas MASS is measured in kilograms,
WEIGHT is strictly a force measured in Newtons, but is in practice
more usually and usefully measured in kilogram-weight.
You won't lose so much as a milligram from your MASS by going to
the moon, or even in space, where it is easy to lose 100% of WEIGHT,
so what you need to decide is, "Is it MASS or is it WEIGHT that you want
to lose, and then you'll be a physicist who DOES KNOW THE DIFFERENCE eh!
Good grief!

Why do you state these obvious trivialities?
--
Paul

https://paulba.no/
Keith Stein
2021-02-24 16:11:07 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Keith Stein
Witch brings me to my point, the difference between MASS and WEIGHT.
It can be confusing because whereas MASS is measured in kilograms,
WEIGHT is strictly a force measured in Newtons, but is in practice
more usually and usefully measured in kilogram-weight.
You won't lose so much as a milligram from your MASS by going to
the moon, or even in space, where it is easy to lose 100% of WEIGHT,
so what you need to decide is, "Is it MASS or is it WEIGHT that you
want to lose, and then you'll be a physicist who DOES KNOW THE
DIFFERENCE eh!
Good grief!
Why do you state these obvious trivialities?
I do it in the hope that the girls will appreciate that relocating to
the moon, while it will invariable cure any girls weight problem, may
not be really what she is trying to achieve, and there are more
important things than losing weight, like learning physics.

Perhaps you would care to explain to the girls how they can use Newtons
Law of Gravitation to work out exactly how much lighter they would be
on the moon than they are at home on their bathroom scales, which i am
recommending they should take with them to the moon.

Do hope you can spare the time to give the girls a lesson Paul...

Sincerely,

keith eh!
Paul B. Andersen
2021-02-24 18:35:59 UTC
Reply
Permalink
Post by Keith Stein
Do hope you can spare the time to give the girls a lesson Paul...
I am not a sexist, so I leave that to you.

Women have taught me a good deal physics and astronomy.
--
Paul

https://paulba.no/
Keith Stein
2021-02-25 11:01:39 UTC
Reply
Permalink
Post by Paul B. Andersen
Women have taught me a good deal physics and astronomy.
Then you have been more fortunate than me Mr.Andersen, for I have learnt
very little "physics" from women myself, and virtually no "astronomy"
Post by Paul B. Andersen
In the star catalogues you will find thousands of stars with
detailed information about their position, their velocity,
their spectral class, their distance.....
I was hoping i might persuade you to explain how one uses the numbers in
the star catalogues to determine where to point a telescope, Mr.Andersen
bearing in mind that this must vary throughout the night, and throughout
the year eh!

keith stein
P.S. It was a woman,Margaret, who taught me how to program a computer,
Paul, and really one knows "NOTHING!", 'till one knows how to program.
Paul B. Andersen
2021-02-25 14:22:56 UTC
Reply
Permalink
Post by Keith Stein
Post by Paul B. Andersen
Women have taught me a good deal physics and astronomy.
Then you have been more fortunate than me Mr.Andersen, for I have learnt
very little "physics" from women myself, and virtually no "astronomy"
from anyone,
You could learn a lot astronomy from Henrietta Swan Leavitt
even if she is dead.
I have.
Post by Keith Stein
Post by Paul B. Andersen
In the star catalogues you will find thousands of stars with
detailed information about their position, their velocity,
their spectral class, their distance.....
I was hoping i might persuade you to explain how one uses the numbers in
the star catalogues to determine where to point a telescope, Mr.Andersen
bearing in mind that this must vary throughout the night, and throughout
the year eh!
Don't worry, Keith, your ignorance is of no consequence.
If you wish to remedy it, read a book or two.
I am not going to teach you.

All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted.

An equatorial mount makes it a lot simpler to make the telescope
track the star, because then the constant rotation 360 degrees
per sidereal day is around one axis only (parallel to the Earth axis).

Not used on big telescopes, though. They have to be rotating
around three axes.

But all they have to do nowadays is to enter the stars RA, DEC
and proper motion into the computer that controls the telescope.
Post by Keith Stein
keith stein
P.S. It was a woman,Margaret, who taught me how to program a computer,
Paul, and really one knows "NOTHING!", 'till one knows how to program.
I have programmed computers since 1968, and know that it is worth
"NOTHING" if you don't know the physics of the systems the computer
should control.

You would know "NOTHING" about how to make the computer control
the telescope. Knowing a programming language isn't knowing
how to program the computer to do a specific job.
--
Paul

https://paulba.no/
Keith Stein
2021-02-25 15:56:25 UTC
Reply
Permalink
On 25/02/2021 14:22, Paul B. Andersen wrote:
......... can't point the telescope to were
Post by Paul B. Andersen
the star was when the light was emitted.
That is in fact exactly where they DO point the telescope, Mr.Andersen,
and the reason that i know this and you don't is because i know that
light travels relative to the medium, not the observer eh! And i also
know how to program a computer to correctly plot the straight line
joining the telescope to WHERE THE STAR WAS WHEN THE LIGHT LEFT THE
STAR, witch you apparently do not, in spite of all my tuition eh!
Post by Paul B. Andersen
But all they have to do nowadays is to enter the stars RA, DEC
and proper motion into the computer that controls the telescope.
Well i had been hoping that you'd show us how to write the program
which steers the telescope, Mr. Andersen, but sure any mug can type
coordinates into the computer.
Post by Paul B. Andersen
Post by Keith Stein
P.S. It was a woman,Margaret, who taught me how to program a computer,
Paul, and really one knows "NOTHING!", 'till one knows how to program.
I have programmed computers since 1968, and know that it is worth
"NOTHING" if you don't know the physics of the systems the computer
should control.
That was about the same time as my introduction to programming, but i
wasn't using it to control anything myself Paul. I was using it to solve
differential equations. And yes i did know what the solutions should
look like, and yes it is very important to stay grounded in the reality
of experiment.
Post by Paul B. Andersen
You would know "NOTHING" about how to make the computer control
the telescope.
Which is why i was asking you to explain to how to do that Mr.Andersen,
and sadly i still don't know even after all your help eh! :)
Post by Paul B. Andersen
Knowing a programming language isn't knowing
how to program the computer to do a specific job.
Exactly Mr.Andersen, and that's why we needed YOU
to show us how to write the program witch controls the telescope,
Mr.Andersen, these girls expect to do more than just type in the
coordinates :(

Never mind for the present anyway Paul. What you did give was actually
quite helpful,thanks, and the computing skills of most of our audience
is not yet sufficient to enable them to appreciate it even if you had
typed out the entire control program (in every last ugly bracket of C++,
or whatever). As you say, if i wanna know something about "Astronomy",
better buy a book on it eh! Never thought of that! Thanks Paul.

keith
Paul B. Andersen
2021-02-25 20:19:30 UTC
Reply
Permalink
Post by Keith Stein
......... can't point the telescope to were
Post by Paul B. Andersen
the star was when the light was emitted.
That is in fact exactly where they DO point the telescope, Mr.Andersen,
and the reason that i know this and you don't is because i know that
light travels relative to the medium, not the observer eh! And i also
know how to program a computer to correctly plot the straight line
joining the telescope to WHERE THE STAR WAS WHEN THE LIGHT LEFT THE
STAR, witch you apparently do not, in spite of all my tuition eh!
We have been through this before, Keith, I wrote it to provoke you.
I succeeded!

As you know, during one year a star will appear to move along
an ellipse with semi-major axis 40.98". The real position
of the star when the light was emitted is at the centre of
that ellipse.

If the distance to the star is 100 light years, then you
believe that 100 years ago, the star was moving around
an ellipse with semi-major axis 0.02 light years.

I am not going to quarrel about this again.

Cranks never learn, even when the obvious is shoved in their face.
--
Paul

https://paulba.no/
Keith Stein
2021-02-25 21:47:53 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Keith Stein
......... can't point the telescope to were
Post by Paul B. Andersen
the star was when the light was emitted.
That is in fact exactly where they DO point the telescope, Mr.Andersen,
and the reason that i know this and you don't is because i know that
light travels relative to the medium, not the observer eh! And i also
know how to program a computer to correctly plot the straight line
joining the telescope to WHERE THE STAR WAS WHEN THE LIGHT LEFT THE
STAR, witch you apparently do not, in spite of all my tuition eh!
We have been through this before, Keith, I wrote it to provoke you.
I succeeded!
As you know, during one year a star will appear to move along
an ellipse with semi-major axis 40.98".
And as you know Paul, the position of the star is not changing during
the year. What is actually changing is the frame of reference of the
observer. And in every inertial frame of reference the path of the ray
of light IS a straight line from where the star was when the light left
the star, to where it is received in your observatory on Earth (i am
neglecting any refraction effects due to changes in refractive index, of
course)

The real position
Post by Paul B. Andersen
of the star when the light was emitted is at the centre of
that ellipse.
Where the star WAS or IS depends on the FRAME OF REFERENCE OF THE
OBSERVER, and in the frame of reference of ANY EARTHBOUND OBSERVER
the star was never at the center of the ellipse, in EVERY frame
of reference the telescope points at where the star was when the
light left the star, but i have told you this before Mr.Andersen.
Post by Paul B. Andersen
If the distance to the star is 100 light years, then you
believe that 100 years ago, the star was moving around
an ellipse with semi-major axis 0.02 light years.
I have told you and told you again and again, but STILL YOU DON'T GET
IT! IT IS NOT THE STAR WHICH MOVES, rather it is the frame of reference
of YOU which is changing during the year. The star DOES NOT MOVE Paul.
Post by Paul B. Andersen
I am not going to quarrel about this again.
Why should we quarrel Paul, we both know i'm right eh! :)

keith stein
Paul B. Andersen
2021-02-26 09:19:35 UTC
Reply
Permalink
Post by Keith Stein
Post by Paul B. Andersen
Post by Keith Stein
......... can't point the telescope to were
Post by Paul B. Andersen
the star was when the light was emitted.
That is in fact exactly where they DO point the telescope, Mr.Andersen,
and the reason that i know this and you don't is because i know that
light travels relative to the medium, not the observer eh! And i also
know how to program a computer to correctly plot the straight line
joining the telescope to WHERE THE STAR WAS WHEN THE LIGHT LEFT THE
STAR, witch you apparently do not, in spite of all my tuition eh!
We have been through this before, Keith, I wrote it to provoke you.
I succeeded!
As you know, during one year a star will appear to move along
an ellipse with semi-major axis 40.98".
And as you know Paul, the position of the star is not changing during
the year. What is actually changing is the frame of reference of the
observer. And in every inertial frame of reference the path of the ray
of light IS a straight line from where the star was when the light left
the star, to where it is received in your observatory on Earth (i am
neglecting any refraction effects due to changes in refractive index, of
course)
The real position
Post by Paul B. Andersen
of the star when the light was emitted is at the centre of
that ellipse.
Where the star WAS or IS depends on the FRAME OF REFERENCE OF THE
OBSERVER, and in the frame of reference of ANY EARTHBOUND OBSERVER
the star was never at the center of the ellipse, in EVERY frame
of reference the telescope points at where the star was when the
light left the star, but i have told you this before Mr.Andersen.
Post by Paul B. Andersen
If the distance to the star is 100 light years, then you
believe that 100 years ago, the star was moving around
an ellipse with semi-major axis 0.02 light years.
I have told you and told you again and again, but STILL YOU DON'T GET
IT!  IT IS NOT THE STAR WHICH MOVES, rather it is the frame of reference
of YOU which is changing during the year. The star DOES NOT MOVE Paul.
Post by Paul B. Andersen
I am not going to quarrel about this again.
Why should we quarrel Paul, we both know i'm right eh!       :) >
Post by Paul B. Andersen
Cranks never learn, even when the obvious is shoved in their face.
Thanks for the confirmation of my words.
--
Paul

https://paulba.no/
Thomas 'PointedEars' Lahn
2021-02-25 22:46:13 UTC
Reply
Permalink
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted.
Parallax has *nothing* to do with the speed of light.

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)
Paul Alsing
2021-02-26 03:03:13 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted.
Parallax has *nothing* to do with the speed of light.
True enough, but for a close star, if you don't take parallax into account, you might not be unable to accurately point your telescope!
Paul B. Andersen
2021-02-26 09:34:43 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted.
Parallax has *nothing* to do with the speed of light.
PointedEars
Who said it had?

https://paulba.no/pdf/Stellar_aberration.pdf
--
Paul

https://paulba.no/
Thomas 'PointedEars' Lahn
2021-02-26 10:50:37 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
^^^^^^^^^^^^ ^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
the star was when the light was emitted.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Parallax has *nothing* to do with the speed of light.
Who said it had?
You did.
Post by Paul B. Andersen
https://paulba.no/pdf/Stellar_aberration.pdf
Yeah, thanks, I am a third-year Astronomy student :-D

PointedEars
--
Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)
Jim Bunce
2021-02-27 02:17:02 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
the star was when the light was emitted.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
That's not speed, idiot.
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Parallax has *nothing* to do with the speed of light.
Who said it had?
You did.
Post by Paul B. Andersen
https://paulba.no/pdf/Stellar_aberration.pdf
Yeah, thanks, I am a third-year Astronomy student :-D
homepage designer.
Paul B. Andersen
2021-02-27 10:29:59 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
^^^^^^^^^^^^ ^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
the star was when the light was emitted.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Parallax has *nothing* to do with the speed of light.
Who said it had?
You did.
You can't be serious! :-D

You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angels RA and DEC give
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.

To get the star position at a later time, you must
correct for the proper motion of the star.

So "where the star was when the light was emitted"
simply means the position of the star charted in the star
catalogues.

(I had a reason for expressing it that way, it was
a response to Keith Stein who claims that you always
have to point the telescope to where the star was
when the light was emitted.)

So my statement is equivalent to:
"they can't point the telescope in the direction given
by the RA and DEC in the position of the star, they have
to correct for stellar aberration and parallax"

(The parallax will almost always be so small compared
to the field of the telescope that it can be ignored.
But if the position of the Earth isn't on the line
Sun - star, it is a difference due to parallax, however small.)

How you can make my statement to mean that parallax
depend on the speed of light beats me.

Was it the statement "where the star was when the light
was emitted" that trigged you to make the blunder?
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
https://paulba.no/pdf/Stellar_aberration.pdf
Yeah, thanks, I am a third-year Astronomy student :-D
So you should have known better!
Post by Thomas 'PointedEars' Lahn
PointedEars
--
Paul

https://paulba.no/
Keith Stein
2021-02-27 18:52:45 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax.....
You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angels RA and DEC give
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.
To get the star position at a later time, you must
correct for the proper motion of the star.
So "where the star was when the light was emitted"
simply means the position of the star charted in the star
catalogues.
Not quite Mr.Andersen. From what you have just told us, the star
catalogues give the position at which the star would be observed on
January 1 2000 in the inertial frame of reference of the Sun. But the
position at which a star is seen depends on the frame of reference of
the observer, which is why on Earth we must correct for the time of day,
(rotation of Earth),and the time of the year (aberration and parallax).
Post by Paul B. Andersen
(I had a reason for expressing it that way, it was
a response to Keith Stein who claims that you always
have to point the telescope to where the star was
when the light was emitted.)
Which is indeed true Mr.Andersen, and if you point the telescope
anywhere else you ain't going to see the star. But of course you
must point at where the star was IN YOUR FRAME OF REFERENCE, not
in the frame of reference of the Sun, as given in the star catalogues.
Post by Paul B. Andersen
"they can't point the telescope in the direction given
by the RA and DEC in the position of the star, they have
to correct for stellar aberration and parallax"
(The parallax will almost always be so small compared
to the field of the telescope that it can be ignored.
But if the position of the Earth isn't on the line
Sun - star, it is a difference due to parallax, however small.)
Thank you Paul, but tell us what lies in the direction RA = 0, DEC = 0.
Anything of interest ?

keith stein
Paul B. Andersen
2021-02-27 20:21:27 UTC
Reply
Permalink
Post by Keith Stein
Thank you Paul, but tell us what lies in the direction RA = 0, DEC = 0.
Anything of interest ?
keith stein
The vernal point.
--
Paul

https://paulba.no/
Paul Alsing
2021-02-28 02:49:41 UTC
Reply
Permalink
Post by Keith Stein
Post by Paul B. Andersen
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax.....
You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angels RA and DEC give
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.
To get the star position at a later time, you must
correct for the proper motion of the star.
So "where the star was when the light was emitted"
simply means the position of the star charted in the star
catalogues.
Not quite Mr.Andersen. From what you have just told us, the star
catalogues give the position at which the star would be observed on
January 1 2000 in the inertial frame of reference of the Sun. But the
position at which a star is seen depends on the frame of reference of
the observer, which is why on Earth we must correct for the time of day,
(rotation of Earth),and the time of the year (aberration and parallax).
Post by Paul B. Andersen
(I had a reason for expressing it that way, it was
a response to Keith Stein who claims that you always
have to point the telescope to where the star was
when the light was emitted.)
Which is indeed true Mr.Andersen, and if you point the telescope
anywhere else you ain't going to see the star. But of course you
must point at where the star was IN YOUR FRAME OF REFERENCE, not
in the frame of reference of the Sun, as given in the star catalogues.
Post by Paul B. Andersen
"they can't point the telescope in the direction given
by the RA and DEC in the position of the star, they have
to correct for stellar aberration and parallax"
(The parallax will almost always be so small compared
to the field of the telescope that it can be ignored.
But if the position of the Earth isn't on the line
Sun - star, it is a difference due to parallax, however small.)
Thank you Paul, but tell us what lies in the direction RA = 0, DEC = 0.
Anything of interest ?
keith stein
Mr. Stein, I'm sorry, but you don't have the brains given to a mud fence! You REALLY need to find a new hobby!
Maciej Wozniak
2021-02-28 07:49:08 UTC
Reply
Permalink
Post by Paul Alsing
Post by Keith Stein
Post by Paul B. Andersen
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax.....
You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angels RA and DEC give
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.
To get the star position at a later time, you must
correct for the proper motion of the star.
So "where the star was when the light was emitted"
simply means the position of the star charted in the star
catalogues.
Not quite Mr.Andersen. From what you have just told us, the star
catalogues give the position at which the star would be observed on
January 1 2000 in the inertial frame of reference of the Sun. But the
position at which a star is seen depends on the frame of reference of
the observer, which is why on Earth we must correct for the time of day,
(rotation of Earth),and the time of the year (aberration and parallax).
Post by Paul B. Andersen
(I had a reason for expressing it that way, it was
a response to Keith Stein who claims that you always
have to point the telescope to where the star was
when the light was emitted.)
Which is indeed true Mr.Andersen, and if you point the telescope
anywhere else you ain't going to see the star. But of course you
must point at where the star was IN YOUR FRAME OF REFERENCE, not
in the frame of reference of the Sun, as given in the star catalogues.
Post by Paul B. Andersen
"they can't point the telescope in the direction given
by the RA and DEC in the position of the star, they have
to correct for stellar aberration and parallax"
(The parallax will almost always be so small compared
to the field of the telescope that it can be ignored.
But if the position of the Earth isn't on the line
Sun - star, it is a difference due to parallax, however small.)
Thank you Paul, but tell us what lies in the direction RA = 0, DEC = 0.
Anything of interest ?
keith stein
Mr. Stein, I'm sorry, but you don't have the brains given to a mud fence! You REALLY need to find a new hobby!
Mr. Alsing, I'm sorry, but you don't have the brains given to a mud fence!
You have also no dignity, honesty and responsibility. Surely teaching
dogs how to bark is a right hobby for you.
Thomas 'PointedEars' Lahn
2021-03-05 02:57:00 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
^^^^^^^^^^^^ ^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
the star was when the light was emitted.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Parallax has *nothing* to do with the speed of light.
Who said it had?
You did.
You can't be serious! :-D
See the markings above. There is no point denying what you did, though it
*might* have been just unfortunate wording. Your reaction below belies that
assumption, though: You appear to actually believe that parallax would have
to do with the speed of light.
Post by Paul B. Andersen
You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angels RA and DEC give
Angels have nothing to do with it, trust me :)
Post by Paul B. Andersen
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.
My research indicates that this NOT in general so. Some catalogs include
some types of stellar aberration and even parallax, others do not:

<http://star-www.rl.ac.uk/docs/sun67.htx/sun67se4.html>

And none of this has inherently to do with parallax, the point that I am
making which you are still missing.

Parallax is the *apparent* change of position of a (celestial) object (in
the sky) against the background (e.g. fixed stars sky) because it is viewed
from different positions. It does NOT depend on the speed of the observer
relative to the observed object (and therefore NOT on the speed of light),
only on the position of the observer. The assumption there is that the
peculiar motion of stars relative to the observer can be neglected as within
a year the difference is too small), and that stellar aberration is already
considered.

You can observe parallax easily with our own eyes if you focus on an object
with one eye and then with the other eye. If the speed of light would be
infinitely fast¹, parallax would be *exactly the same*:

* *.* * * * *. * *
\ .---. /
'.___.'
\p: /
\:/
*
/:\
/ :p\
/ : \
/ : \ d
/ : \
/ : \
/ __:___ \
.-''' : '''-.
T S-------T
`-..______...-'

[Annual parallax: The position of a star (*) in front of the star
background (* * *) appears to shift as Terra (T) moves in its orbit and
the star is observed from different positions in the orbit. The farther
away the star, the smaller this shift. For stars that are close enough,
then, trigonometry can be used to estimate their distance

d ≈ (1 AU)/(sinp) ≈ (1 AU)/p.

Surveying missions like Gaia use the same principle on a larger scale due
to much better image resolution.]
Post by Paul B. Andersen
[tl;dr]
PointedEars
___________
¹ surprisingly, there is no way to refute *experimentally* that the speed of
light would not be half-as-fast as expected in one direction and
infinitely fast in the opposite direction
--
Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)
Tom Roberts
2021-03-05 05:11:14 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
¹ surprisingly, there is no way to refute *experimentally* that the speed of
light would not be half-as-fast as expected in one direction and
infinitely fast in the opposite direction
This is a footnote, so I'm not sure whether you intend this specifically
to apply to the earlier test, or in general. As the earlier text is
"infinitely fast", I don't see how it could relate to that at all.

Note that the observed accuracy of the GPS shows that the one-way speed
of light from satellite to receiver cannot vary with direction by as
much as a few parts per million -- such variation would require the
satellites to follow paths inconsistent with orbital mechanics.

Tom Roberts
Maciej Wozniak
2021-03-05 05:47:25 UTC
Reply
Permalink
Post by Tom Roberts
Post by Thomas 'PointedEars' Lahn
¹ surprisingly, there is no way to refute *experimentally* that the speed of
light would not be half-as-fast as expected in one direction and
infinitely fast in the opposite direction
This is a footnote, so I'm not sure whether you intend this specifically
to apply to the earlier test, or in general. As the earlier text is
"infinitely fast", I don't see how it could relate to that at all.
Note that the observed accuracy of the GPS shows that the one-way speed
of light from satellite to receiver cannot vary with direction
Lyin shit Tom, of course, doesn't mention that GPS is applying
only one frame. Even if he gladly emhasize it on other occasions.
Thomas 'PointedEars' Lahn
2021-03-06 05:08:15 UTC
Reply
Permalink
Post by Tom Roberts
Post by Thomas 'PointedEars' Lahn
¹ surprisingly, there is no way to refute *experimentally* that the speed
of light would not be half-as-fast as expected in one direction and
infinitely fast in the opposite direction
This is a footnote, so I'm not sure whether you intend this specifically
to apply to the earlier test, or in general. As the earlier text is
"infinitely fast", I don't see how it could relate to that at all.
My argument was that the speed of light does not matter for parallax, and
this footnote is pointing out that even if this hypothetical scenario where
the one-way speed of light is infinite would be true, it would not matter
for parallax.
Post by Tom Roberts
Note that the observed accuracy of the GPS shows that the one-way speed
of light from satellite to receiver cannot vary with direction by as
much as a few parts per million -- such variation would require the
satellites to follow paths inconsistent with orbital mechanics.
Thank you. I agree that there is no indication of a preferred direction
globally; however, that says nothing about what happens in single case, like
a reflection.

I wonder what you think about the video where I found this:

PointedEars
--
Q: Why is electricity so dangerous?
A: It doesn't conduct itself.

(from: WolframAlpha)
Paul B. Andersen
2021-03-05 14:18:45 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
^^^^^^^^^^^^ ^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
the star was when the light was emitted.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Parallax has *nothing* to do with the speed of light.
Who said it had?
You did.
You can't be serious! :-D
See the markings above. There is no point denying what you did, though it
*might* have been just unfortunate wording. Your reaction below belies that
assumption, though: You appear to actually believe that parallax would have
to do with the speed of light.
So you ARE serious. But I still can't understand how it is
possible to interpret what I said that way.

Why not simply admit that you made a minor blunder?
Why is it so important to you to insist that I
said that parallax depend on the speed of light,
which I never said?
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angles RA and DEC give
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.
You do now this, don't you?

It is quite simple.
When you observe a star, it will, due to stellar aberration,
during one year appear to move along an ellipse. The major axis
of this ellipse will be parallel to the ecliptic plane.
The major axis will be:
arcsin(2v/c) radians ~= 40.98 arcseconds
The minor axis will be:
40.98*cos(ecliptic latitude) arcsecs

(It will be a very small distortion of the ellipse due
to the eccentricity of Earth orbit.)

The parallax will distort this ellipse slightly. The distortion
will always be towards the point in the centre of the ellipse.
If the maximum parallax is p, then the major axis will be
shortened by 2p, while the minor axis will be shortened
by 2p*cos(ecliptic latitude)

But the parallax will always be small relative to 41 arcsecs.
The parallax of the closest star, Proxima Centauri, is p = 0.77".
And for a star 100ly away p = 0.033". It's a reason why
the parallax is given in mas (milli-arcsecs) in the star catalogues.

So the parallax will for most stars be very small relative to
the stellar aberration, and because of its symmetry, it will
not affect the position of the centre of the ellipse.

And the important point is:
The position of the star charted in the star catalogues
is the RA and DEC of the centre of that ellipse.
And this point is where we would see the star from the Sun.

And remember the obvious:
The position charted in the star catalogues is
where the star was relative to the Sun when the light
was emitted.
Post by Thomas 'PointedEars' Lahn
My research indicates that this NOT in general so. Some catalogs include
The position of the star, given as RA and DEC doesn't
include stellar aberration and parallax, but there may
be differences between the catalogues due to different
reference systems. Finding the correct position of the vernal
point and the correct angle of the equatorial plane isn't
easy, since both are changing with time due to the precession
and nutation of Earth's axis, and converting from one EPOCH to
another may not be trivial.

But this is nitpicking, and irrelevant to the issue:

Did I claim that stellar parallax depend on the speed of light?

Generally what I said above is correct, and I think you know it.
Post by Thomas 'PointedEars' Lahn
<http://star-www.rl.ac.uk/docs/sun67.htx/sun67se4.html>
And none of this has inherently to do with parallax, the point that I am
making which you are still missing.
Parallax is the *apparent* change of position of a (celestial) object (in
the sky) against the background (e.g. fixed stars sky) because it is viewed
from different positions. It does NOT depend on the speed of the observer
relative to the observed object (and therefore NOT on the speed of light),
only on the position of the observer. The assumption there is that the
peculiar motion of stars relative to the observer can be neglected as within
a year the difference is too small), and that stellar aberration is already
considered.
You can observe parallax easily with our own eyes if you focus on an object
with one eye and then with the other eye. If the speed of light would be
* *.* * * * *. * *
\ .---. /
'.___.'
\p: /
\:/
*
/:\
/ :p\
/ : \
/ : \ d
/ : \
/ : \
/ __:___ \
.-''' : '''-.
T S-------T
`-..______...-'
[Annual parallax: The position of a star (*) in front of the star
background (* * *) appears to shift as Terra (T) moves in its orbit and
the star is observed from different positions in the orbit. The farther
away the star, the smaller this shift. For stars that are close enough,
then, trigonometry can be used to estimate their distance
d ≈ (1 AU)/(sinp) ≈ (1 AU)/p.
Surveying missions like Gaia use the same principle on a larger scale due
to much better image resolution.]
I know what parallax is.

This is irrelevant to the issue:
Did I claim that stellar parallax depend on the speed of light?

Since the position charted in the stellar catalogues is where
the star was relative to the Sun when the light from star
was emitted, my statement:

To point the star in the right direction to see a star
"they have to correct for stellar aberration and parallax
because they can't point the telescope to were the star
was when the light was emitted."

Or with slightly different words:
"they have to correct for stellar aberration and parallax
because they can't point the telescope to the RA and DEC
given in the star catalogues."

There is no way this statement can be interpreted to
mean that aberration depend on the speed of light.

And I still can't understand how you managed to interpret
it in such a way.

So I can only repeat my question:
Was it the statement you marked: "where the star was when
the light was emitted" that trigged you to make the strange
blunder that this means that parallax depend on the speed of light?
--
Paul

https://paulba.no/
Keith Stein
2021-03-05 15:34:30 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Paul B. Andersen
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
All astronomers, amateurs as well as professionals, know how
to make the telescope point in the right position when
the right ascension, declination and time are known.
They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
^^^^^^^^^^^^                                        ^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
the star was when the light was emitted.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Post by Paul B. Andersen
Post by Thomas 'PointedEars' Lahn
Parallax has *nothing* to do with the speed of light.
Who said it had?
You did.
You can't be serious! :-D
See the markings above.  There is no point denying what you did,
though it
*might* have been just unfortunate wording.  Your reaction below
belies that
assumption, though: You appear to actually believe that parallax would have
to do with the speed of light.
So you ARE serious. But I still can't understand how it is
possible to interpret what I said that way.
Why not simply admit that you made a minor blunder?
Why is it so important to you to insist that I
said that parallax depend on the speed of light,
which I never said?
Post by Paul B. Andersen
You know of course that when the position of
a star at EPOCH J2000 is right ascension RA and
declination DEC, then the angles RA and DEC give
the direction from the Sun to where the star was when
the light that reached the Sun at January 1, 2000,
was emitted.
You do now this, don't you?
It is quite simple.
When you observe a star, it will, due to stellar aberration,
during one year appear to move along an ellipse. The major axis
of this ellipse will be parallel to the ecliptic plane.
arcsin(2v/c) radians ~= 40.98 arcseconds
40.98*cos(ecliptic latitude) arcsecs
(It will be a very small distortion of the ellipse due
to the eccentricity of Earth orbit.)
The parallax will distort this ellipse slightly. The distortion
will always be towards the point in the centre of the ellipse.
If the maximum parallax is p, then the major axis will be
shortened by 2p, while the minor axis will be shortened
by 2p*cos(ecliptic latitude)
But the parallax will always be small relative to 41 arcsecs.
The parallax of the closest star, Proxima Centauri, is p = 0.77".
And for a star 100ly away p = 0.033". It's a reason why
the parallax is given in mas (milli-arcsecs) in the star catalogues.
So the parallax will for most stars be very small relative to
the stellar aberration, and because of its symmetry, it will
not affect the position of the centre of the ellipse.
The position of the star charted in the star catalogues
is the RA and DEC of the centre of that ellipse.
And this point is where we would see the star from the Sun.
The position charted in the star catalogues is
where the star was relative to the Sun when the light
was emitted.
My research indicates that this NOT in general so.  Some catalogs include
The position of the star, given as RA and DEC doesn't
include stellar aberration and parallax, but there may
be differences between the catalogues due to different
reference systems. Finding the correct position of the vernal
point and the correct angle of the equatorial plane isn't
easy, since both are changing with time due to the precession
and  nutation of Earth's axis, and converting from one EPOCH to
another may not be trivial.
Did I claim that stellar parallax depend on the speed of light?
Generally what I said above is correct, and I think you know it.
<http://star-www.rl.ac.uk/docs/sun67.htx/sun67se4.html>
And none of this has inherently to do with parallax, the point that I am
making which you are still missing.
Parallax is the *apparent* change of position of a (celestial) object (in
the sky) against the background (e.g. fixed stars sky) because it is viewed
from different positions.  It does NOT depend on the speed of the
observer
relative to the observed object (and therefore NOT on the speed of light),
only on the position of the observer.  The assumption there is that the
peculiar motion of stars relative to the observer can be neglected as within
a year the difference is too small), and that stellar aberration is already
considered.
You can observe parallax easily with our own eyes if you focus on an object
with one eye and then with the other eye.  If the speed of light would be
* *.* * * * *. * *
\ .---. /
'.___.'
\p: /
\:/
*
/:\
/ :p\
/  :  \
/   :   \ d
/    :    \
/     :     \
/    __:___   \
.-'''  :   '''-.
T       S-------T
`-..______...-'
[Annual parallax: The position of a star (*) in front of the star
background (* * *) appears to shift as Terra (T) moves in its
orbit and
the star is observed from different positions in the orbit.  The
farther
away the star, the smaller this shift.  For stars that are close
enough,
then, trigonometry can be used to estimate their distance
d ≈ (1 AU)/(sinp) ≈ (1 AU)/p.
Surveying missions like Gaia use the same principle on a larger
scale due
to much better image resolution.]
I know what parallax is.
Did I claim that stellar parallax depend on the speed of light?
Since the position charted in the stellar catalogues is where
the star was relative to the Sun when the light from star
To point the star in the right direction to see a star
"they have to correct for stellar aberration and parallax
because they can't point the telescope to were the star
was when the light was emitted."
"they have to correct for stellar aberration and parallax
because they can't point the telescope to the RA and DEC
given in the star catalogues."
There is no way this statement can be interpreted to
mean that aberration depend on the speed of light.
BUT "aberration" DOES depend on the speed of light,Mr.Andersen,
as I know YOU know Mr.Andersen :) You meant, of course, that you never
suggested "that *parallax* depend on the speed of light", and
no you never suggested it did, Paul. Perhaps our Thomas Lahn was
just creating an opportunity to give us his lesson on "PARALLAX".
(Maybe he just learnt about it in his Astronomy course eh! :) )

keith stein
Post by Paul B. Andersen
And I still can't understand how you managed to interpret
it in such a way.
Was it the statement you marked: "where the star was when
the light was emitted" that trigged you to make the strange
blunder that this means that parallax depend on the speed of light?
Paul B. Andersen
2021-03-06 19:39:56 UTC
Reply
Permalink
Post by Keith Stein
Post by Paul B. Andersen
To point the star in the right direction to see a star
"they have to correct for stellar aberration and parallax
because they can't point the telescope to were the star
was when the light was emitted."
"they have to correct for stellar aberration and parallax
because they can't point the telescope to the RA and DEC
given in the star catalogues."
There is no way this statement can be interpreted to
mean that aberration depend on the speed of light.
BUT "aberration" DOES depend on the speed of light,Mr.Andersen,
as I know YOU know Mr.Andersen :) You meant, of course, that you never
suggested "that *parallax* depend on the speed of light", and
no you never suggested it did, Paul.
Right. Stupid blunder!
Good catch.
--
Paul

https://paulba.no/
Thomas 'PointedEars' Lahn
2021-03-05 18:35:17 UTC
Reply
Permalink
Post by Paul B. Andersen
Why not simply admit that you made a minor blunder?
Pot calling the kettle black.

You wrote

“They know that they have to correct for […] parallax because they can't
point the telescope to were the star was when the light was emitted.“

This is WRONG.

PointedEars
--
Q: How many theoretical physicists specializing in general relativity
does it take to change a light bulb?
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)
Keith Stein
2021-03-05 19:11:00 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Why not simply admit that you made a minor blunder?
Pot calling the kettle black.
“They know that they have to correct for […] parallax because they can't
point the telescope to were the star was when the light was emitted.“
This is the unedited version of that:

"They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted."

Can you SEE THE DIFFERENCE, Mr. Lahn ?
Post by Thomas 'PointedEars' Lahn
This is WRONG.
PointedEars
The "difference", is that what Mr.Andersen wrote was NOT wrong, Mr.Lahn
or rather it was WRONG, but NOT for the reason that had been implied by
your selective editing. The actual reason it WAS WRONg, PointedEars,
was because in actual fact that *IS* WHERE THEY *DO* POINT THE TELESCOPE

keith stein

keith stein
Paul B. Andersen
2021-03-06 20:22:33 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul B. Andersen
Why not simply admit that you made a minor blunder?
Pot calling the kettle black.
“They know that they have to correct for […] parallax because they can't
point the telescope to were the star was when the light was emitted.“
Paul B. Andersen wrote:
"They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted."
Post by Thomas 'PointedEars' Lahn
This is WRONG.
PointedEars
Do you really not understand that if you don't correct
for parallax, then you will see the star offset from
the centre of the telescope field by the parallax angle?

A third year astronomy student must know
that my statement above is correct, even if
the parallax might be negligible.

But the issue was:
Did I claim that stellar parallax depend on the speed of light?

I think you better explain how you can interpret the statement:
"They know that they have to correct for stellar aberration
and parallax because they can't point the telescope to were
the star was when the light was emitted."

To mean:
"stellar parallax depend on the speed of light"

Why not simply admit that you made a minor blunder?
--
Paul

https://paulba.no/
Chris M. Thomasson
2021-02-27 19:46:38 UTC
Reply
Permalink
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Is physics for girls different than physics for men? For some reason,
this reminds me of a part in the following video that shows the
Encyclopedia of Numbers:

Here is the relevant part, numbers just for men!

http://youtu.be/rVtHrgdcvZA

;^)
Keith Stein
2021-02-28 08:36:33 UTC
Reply
Permalink
Post by Chris M. Thomasson
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Is physics for girls different than physics for men?
Yes, there's no bullshit in "Physics for Girls" Chris.
Post by Chris M. Thomasson
For some reason,
this reminds me of a part in the following video that shows the
http://youtu.be/rVtHrgdcvZA
That's a rubbish Encyclopedia Mr.Thomasson. There are
billions of numbers not listed, and "Judy Numbers" are
not even mentioned eh!
Post by Chris M. Thomasson
Here is the relevant part, numbers just for men!
http://youtu.be/rVtHrgdcvZA
;^)
Chris M. Thomasson
2021-02-28 20:42:53 UTC
Reply
Permalink
Post by Keith Stein
Post by Chris M. Thomasson
Post by Keith Stein
What i am about to teach you is.....
" Classical Physics ",
and, unlike much of "Modern Physics",
this IS well worth learning. I promise, but you
will have to trust me on this, because you
won't appreciate it until you master it. Whoops !
Sorry i should of said miss it perhaps,
Newton's Three Laws of Motion.
1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.
2. Force = Mass * Acceleration.
3. To every action there is an equal and opposite reaction.
Worry not if you don't understand 2, or 1 either, or 3 even,
just learn NEWTON'S THREE LAWS, like they were a sci.fi script for now,
you can't be a physicist if you don't know them.
And even Anarchists obey Newton's Laws eh!
Homework CLASSICAL PHYSICS Lesson 1 Learn Newton's Laws
Is physics for girls different than physics for men?
Yes, there's no bullshit in "Physics for Girls" Chris.
Well, just make sure girls and boys are being taught the same physics. ;^)
Post by Keith Stein
Post by Chris M. Thomasson
For some reason, this reminds me of a part in the following video that
http://youtu.be/rVtHrgdcvZA
That's a rubbish Encyclopedia Mr.Thomasson. There are
billions of numbers not listed, and "Judy Numbers" are
not even mentioned eh!
Funny! LOL!
Post by Keith Stein
Post by Chris M. Thomasson
Here is the relevant part, numbers just for men!
http://youtu.be/rVtHrgdcvZA
;^)
Keith Stein
2021-02-28 16:21:51 UTC
Reply
Permalink
Physicists associate numbers with just about everything, like position,
time, velocity, mass, force, pressure, temperature, and so on, and if
physicists can't associate a number with it, then physicists are not,
at least professionally, interested in it.

Suppose one of these numbers so beloved of physicists has the value 'P'.
So 'P' is just some number, 13.73 maybe, or whatever, it really does not
matter what it is. Indeed a competent physicist can arrange for the
value of P to have any value you like, simply by changing the "units" in
which P is measured. But let us suppose that our units, and procedure
for for measuring P, have been defined, and it is found that the
parameter P has the value P = 13.73 = P0 = Value of P at time t=0

What physicists often want to know is how this value P changes with
time. An experimental physicist would determine this simply and directly
by measuring the value of P at various times, thus obtaining the values:
at t=0 P=P0, at t=1 P=P1, at t=2 P=P2,at t=3 P=P(3), at t=4...,,, etc.
But this is not the way of the theoretical physicist. A theoretical
physicists needs to obtain this string of values without even leaving
her desk.

She acheives this by "inventing" even more numbers. She defines a new
number, which she calls "the rate of change of P", giving it a new
symbol "rP". Current Physics Text Books would probably use the notation
"dP/dt" to denote the rate of change of P, but it's still just some
number, and is more conveniently stored in the computer using the
notation i am recommending to you here, so dP/dt =rp, and, although we
haven't come to this yet, the notation is easily extended so that the
rate of change of the rate change of P = r2P = d2P/dt^2.

The important point is that both dP/dt (rP) and d2P/dt^2 (r2P) are just
numbers, and although, in principle, your computer couldn't care less
what you call them, in practice most computers will object if you try to
stick with the Newton's original notation. So here's my tip for you
girls that you are not going to find in any physics text book:

dP/dt = rP
and d2P/dt^2 = r2P

The important point to grasp here is that rP and r2P are JUST NUMBERS.

The simplest case to consider is the case where r2P = 0, and rP = 0.
In that case the value P will just remain constant at P0, and the
theoretical physicist can determine, even without the aid of her
computer, that:
at t=0 P=P0, and at t=1 P=P0, and at t=2 P=P0,,, at t = whatever,, P=P0
How easy was that eh!

However most of the problems tackled by theoretical physicists are a
little more tricky than that, so taking the next step in complexity let
us consider the case where r2P remains as before equal to zero, but this
time the rate of change of P (i.e. 'rP') now has some non-zero value,
which we shall denote by 'rP', but in more conventional physics texts is
more likely to appear as 'dP/dt'. First thing you should note is that
since "the rate of change of 'the rate of change of P'" (i.e. r2P) is
still fixed at "zero",the value of rP will remain constant, but not so
the value of P, since "the rate of change of P" (i.e. rP) is NOT zero.

If the rate of change of P is constant at rP, then the value of P at any
future time is obtained very simply from:

P(t) = P0 + rP*t

Homework PHYSICS for Girls.. Week2 Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
Odd Bodkin
2021-02-28 16:29:34 UTC
Reply
Permalink
Post by Keith Stein
Physicists associate numbers with just about everything, like position,
time, velocity, mass, force, pressure, temperature, and so on, and if
physicists can't associate a number with it, then physicists are not,
at least professionally, interested in it.
Suppose one of these numbers so beloved of physicists has the value 'P'.
So 'P' is just some number, 13.73 maybe, or whatever, it really does not
matter what it is. Indeed a competent physicist can arrange for the
value of P to have any value you like, simply by changing the "units" in
which P is measured. But let us suppose that our units, and procedure
for for measuring P, have been defined, and it is found that the
parameter P has the value P = 13.73 = P0 = Value of P at time t=0
What physicists often want to know is how this value P changes with
time. An experimental physicist would determine this simply and directly
at t=0 P=P0, at t=1 P=P1, at t=2 P=P2,at t=3 P=P(3), at t=4...,,, etc.
But this is not the way of the theoretical physicist. A theoretical
physicists needs to obtain this string of values without even leaving
her desk.
She acheives this by "inventing" even more numbers. She defines a new
number, which she calls "the rate of change of P", giving it a new
symbol "rP". Current Physics Text Books would probably use the notation
"dP/dt" to denote the rate of change of P, but it's still just some
number, and is more conveniently stored in the computer using the
notation i am recommending to you here, so dP/dt =rp, and, although we
haven't come to this yet, the notation is easily extended so that the
rate of change of the rate change of P = r2P = d2P/dt^2.
The important point is that both dP/dt (rP) and d2P/dt^2 (r2P) are just
numbers, and although, in principle, your computer couldn't care less
what you call them, in practice most computers will object if you try to
stick with the Newton's original notation. So here's my tip for you
dP/dt = rP
and d2P/dt^2 = r2P
The important point to grasp here is that rP and r2P are JUST NUMBERS.
The simplest case to consider is the case where r2P = 0, and rP = 0.
In that case the value P will just remain constant at P0, and the
theoretical physicist can determine, even without the aid of her
at t=0 P=P0, and at t=1 P=P0, and at t=2 P=P0,,, at t = whatever,, P=P0
How easy was that eh!
However most of the problems tackled by theoretical physicists are a
little more tricky than that, so taking the next step in complexity let
us consider the case where r2P remains as before equal to zero, but this
time the rate of change of P (i.e. 'rP') now has some non-zero value,
which we shall denote by 'rP', but in more conventional physics texts is
more likely to appear as 'dP/dt'. First thing you should note is that
since "the rate of change of 'the rate of change of P'" (i.e. r2P) is
still fixed at "zero",the value of rP will remain constant, but not so
the value of P, since "the rate of change of P" (i.e. rP) is NOT zero.
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2 Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
Now you want to discuss basic calculus on a newsgroup about relativity.
Why?
--
Odd Bodkin — Maker of fine toys, tools, tables
Keith Stein
2021-02-28 16:42:36 UTC
Reply
Permalink
PHYSICS for Girls.. Week2 Just Numbers

Just one tiny typo correction on previous post,
'rp' should have been 'rP'. Sorry 'bout that eh!
Post by Keith Stein
Physicists associate numbers with just about everything, like position,
time, velocity, mass, force, pressure, temperature, and so on, and if
physicists can't associate a number with it, then physicists are not,
at least professionally, interested in it.
Suppose one of these numbers so beloved of physicists has the value 'P'.
So 'P' is just some number, 13.73 maybe, or whatever, it really does not
matter what it is. Indeed a competent physicist can arrange for the
value of P to have any value you like, simply by changing the "units" in
which P is measured. But let us suppose that our units, and procedure
for for measuring P, have been defined, and it is found that the
parameter P has the value P = 13.73 = P0 = Value of P at time t=0
What physicists often want to know is how this value P changes with
time. An experimental physicist would determine this simply and directly
at t=0 P=P0, at t=1 P=P1, at t=2 P=P2,at t=3 P=P(3), at t=4...,,, etc.
But this is not the way of the theoretical physicist. A theoretical
physicists needs to obtain this string of values without even leaving
her desk.
She acheives this by "inventing" even more numbers. She defines a new
number, which she calls "the rate of change of P", giving it a new
symbol "rP". Current Physics Text Books would probably use the notation
"dP/dt" to denote the rate of change of P, but it's still just some
number, and is more conveniently stored in the computer using the
notation i am recommending to you here, so  dP/dt = rP, and, although we
haven't come to this yet, the notation is easily extended so that the
rate of change of the rate change of P = r2P = d2P/dt^2.
The important point is that both dP/dt (rP) and d2P/dt^2 (r2P) are just
numbers, and although, in principle, your computer couldn't care less
what you call them, in practice most computers will object if you try to
stick with the Newton's original notation. So here's my tip for you
dP/dt = rP
and  d2P/dt^2 = r2P
The important point to grasp here is that rP and r2P are JUST NUMBERS.
The simplest case to consider is the case where r2P = 0, and rP = 0.
In that case the value P will just remain constant at P0, and the
theoretical physicist can determine, even without the aid of her
at t=0 P=P0, and at t=1 P=P0, and at t=2 P=P0,,, at t = whatever,, P=P0
How easy was that eh!
However most of the problems tackled by theoretical physicists are a
little more tricky than that, so taking the next step in complexity let
us consider the case where r2P remains as before equal to zero, but this
time the rate of change of P (i.e. 'rP') now has some non-zero value,
which we shall denote by 'rP', but in more conventional physics texts is
more likely to appear as 'dP/dt'. First thing you should note is that
since "the rate of change of 'the rate of change of P'" (i.e. r2P) is
still fixed at "zero",the value of rP will remain constant, but not so
the value of P, since "the rate of change of P" (i.e. rP) is NOT zero.
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2  Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
Paul B. Andersen
2021-02-28 19:10:32 UTC
Reply
Permalink
< snip a lot of words with little content, and which ends with this:>
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2  Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
Very few children are reading this NG, so what's
the point with posting these trivialities here?
--
Paul

https://paulba.no/
Keith Stein
2021-03-03 09:13:42 UTC
Reply
Permalink
Post by Paul B. Andersen
< snip
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2  Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
What do i mean by P(1) ?

I mean the value of P when t = 1, which you can obtain by substituting
t=1 in the above equation, thus obtaining:

P(1) = P0 + rP*1

But you can do better than that, because if you read the question you
will see that i also gave you the values of P0 and rP, and to acheive a
better grade you must also substitute the values P0=13.73 and rP=0, thus
obtaining:

P(1) = 13.73 + 0.1*1

Which is better, but still not good enough to get you full marks, for
you should also evaluate the RHS (Right Hand Side) of the above
equation, thus obtaining the required answer of:

P(1) = 13.83

And by using the values t = 2,3,4 and 100 you should now see that:
P(2) = 13.93
P(3) = 14.03
P(4) = 14.13
P(100) = 23.73

If you got all those right then "well done", and if you didn't get them
all right then go back and do them all again, and again, and again 'till
you do get them all right, and make sure you can work out P(5), P(6),
P(7), and P(22) too eh!
Post by Paul B. Andersen
Very few children are reading this NG,
And that Mr.Andersen is your fault, for if you write something
witch IS of interest to children, then you can be sure that
(eventually) a child will read it, and find it interesting eh!
Post by Paul B. Andersen
so what's the point with posting these trivialities here?
Instead being such a grumpy old nay sayer Mr.Andersen, you should
write a post which will encourage children to read this newsgroup.
You could give a good lesson if you put your mind to it Paul, but
do please remember that this thread is strictly on:
"Classical Physics"
i don't want you corrupting my girls with any of that relativity
bullshit eh!

keith stein
Paul B. Andersen
2021-03-03 12:34:30 UTC
Reply
Permalink
Post by Keith Stein
Post by Paul B. Andersen
Very few children are reading this NG,
so what's the point with posting these trivialities here?
< snip
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2  Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
What do i mean by P(1) ?
I mean the value of P when t = 1, which you can obtain by substituting
P(1) = P0 + rP*1
But you can do better than that, because if you read the question you
will see that i also gave you the values of P0 and rP, and to acheive a
better grade you must also substitute the values P0=13.73 and rP=0, thus
P(1) = 13.73 + 0.1*1
Which is better, but still not good enough to get you full marks, for
you should also evaluate the RHS (Right Hand Side) of the above
P(1) = 13.83
P(2) = 13.93
P(3) = 14.03
P(4) = 14.13
P(100) = 23.73
If you got all those right then "well done", and if you didn't get them
all right then go back and do them all again, and again, and again 'till
you do get them all right, and make sure you can work out P(5), P(6),
P(7), and P(22) too eh!
And that Mr.Andersen is your fault, for if you write something
witch IS of interest to children, then you can be sure that
(eventually) a child will read it, and find it interesting eh!
Instead being such a grumpy old nay sayer Mr.Andersen, you should
write a post which will encourage children to read this newsgroup.
You could give a good lesson if you put your mind to it Paul, but
"Classical Physics"
i don't want you corrupting my girls with any of that relativity
bullshit eh!
keith stein
Very few children are reading this NG,
so what's the point with posting these trivialities here?
--
Paul

https://paulba.no/
Paparios
2021-03-03 12:45:33 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Keith Stein
i don't want you corrupting my girls with any of that relativity
bullshit eh!
keith stein
Very few children are reading this NG,
so what's the point with posting these trivialities here?
--
Old timers turn to behave like children, using diapers and all that stuff!!!!
Chris M. Thomasson
2021-03-03 13:18:22 UTC
Reply
Permalink
Post by Paparios
Post by Paul B. Andersen
Post by Keith Stein
i don't want you corrupting my girls with any of that relativity
bullshit eh!
keith stein
Very few children are reading this NG,
so what's the point with posting these trivialities here?
--
Old timers turn to behave like children, using diapers and all that stuff!!!!
Reminds me of the old riddle:

What starts out walking on all fours, then two, then three?
Keith Stein
2021-03-03 19:07:27 UTC
Reply
Permalink
Post by Chris M. Thomasson
What starts out walking on all fours, then two, then three?
What starts out walking on all fours, then two, then three,
then six,then wheels, but could still manage with six for
short distances ?

Reminds me of My Old Mother eh!
Chris M. Thomasson
2021-03-05 07:02:26 UTC
Reply
Permalink
Post by Keith Stein
Post by Chris M. Thomasson
What starts out walking on all fours, then two, then three?
What starts out walking on all fours, then two, then three,
then six,then wheels, but could still manage with six for
short distances ?
Reminds me of My Old Mother eh!
No shit! Bless her soul :^)
Odd Bodkin
2021-03-03 14:02:22 UTC
Reply
Permalink
Post by Keith Stein
Post by Paul B. Andersen
< snip
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2  Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
What do i mean by P(1) ?
I mean the value of P when t = 1, which you can obtain by substituting
P(1) = P0 + rP*1
But you can do better than that, because if you read the question you
will see that i also gave you the values of P0 and rP, and to acheive a
better grade you must also substitute the values P0=13.73 and rP=0, thus
P(1) = 13.73 + 0.1*1
Which is better, but still not good enough to get you full marks, for
you should also evaluate the RHS (Right Hand Side) of the above
P(1) = 13.83
P(2) = 13.93
P(3) = 14.03
P(4) = 14.13
P(100) = 23.73
If you got all those right then "well done", and if you didn't get them
all right then go back and do them all again, and again, and again 'till
you do get them all right, and make sure you can work out P(5), P(6),
P(7), and P(22) too eh!
Post by Paul B. Andersen
Very few children are reading this NG,
And that Mr.Andersen is your fault, for if you write something
witch IS of interest to children, then you can be sure that
(eventually) a child will read it, and find it interesting eh!
Post by Paul B. Andersen
so what's the point with posting these trivialities here?
Instead being such a grumpy old nay sayer Mr.Andersen, you should
write a post which will encourage children to read this newsgroup.
Why, in your learned opinion, should children be encouraged to visit a
newsgroup about relativity, to find a lesson about high school physics and
mathematics?

Do you also think it’s worthwhile to encourage children to visit JPL to
learn about simple machines like levers and screws?
Post by Keith Stein
You could give a good lesson if you put your mind to it Paul, but
"Classical Physics"
i don't want you corrupting my girls with any of that relativity
bullshit eh!
keith stein
--
Odd Bodkin -- maker of fine toys, tools, tables
Keith Stein
2021-03-03 20:55:16 UTC
Reply
Permalink
Post by Keith Stein
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t.............(1)
Surprisingly that equation will hold even if rP is not constant, at
least it holds to a very good approximation, which becomes perfect
when t = 0

P(0) = P0

This is my error. For consistency i should of called 'P0' 'P(0), but no
matter, and saved us from the tautology of P(0) = P(0), but whatever
what you should note is that equation 1 is exactly true when t=0,
and exactly true when rp is constant, and a good approximation if
t is a very short amount of time.

Last week i introduced the rather strange concept of "the rate of change
of the rate change", and you've probably forgotten but we even went so
far as to give it its own special symbol.

Can anyone remember what we decided to call the rate of change of the
rate of change of P ?

We decided to call it "r2P", but more conventionally it would appear
as "d2P/dt^2", but you must not be scared by that, IT'S JUST A NUMBER.

There are problems where r2P is not constant, but there are plenty of
physics probems where r2P is constant, so let's not get ahead of
ourselves. For now we will assume that r2P is constant.
P is just any parameter, but it might be easier if we consider a
particular example of a parameter, let's say DISTANCE, witch would
normally be given the symbol x rather than P, so lets do that. So:
DISTANCE = X
RATE OF CHANGE OF DISTANCE = dX/dt = rX = velocity
RATE OF CHANGE OF VELOCITY = d2X/dt^2 = r2X = acceleration

Don't worry if that all seems very confusing right now. Really
ALL JUST NUMBERS IN BOXES. We just have to tell our computer
which number to put in witch box. It's easy when you know how.
Odd Bodkin
2021-03-03 21:14:11 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t.............(1)
Surprisingly that equation will hold even if rP is not constant, at
least it holds to a very good approximation, which becomes perfect
when t = 0
P(0) = P0
This is my error. For consistency i should of called 'P0' 'P(0), but no
matter, and saved us from the tautology of P(0) = P(0), but whatever
what you should note is that equation 1 is exactly true when t=0,
and exactly true when rp is constant, and a good approximation if
t is a very short amount of time.
Last week i introduced the rather strange concept of "the rate of change
of the rate change", and you've probably forgotten but we even went so
far as to give it its own special symbol.
Can anyone remember what we decided to call the rate of change of the
rate of change of P ?
We decided to call it "r2P", but more conventionally it would appear
as "d2P/dt^2", but you must not be scared by that, IT'S JUST A NUMBER.
There are problems where r2P is not constant, but there are plenty of
physics probems where r2P is constant, so let's not get ahead of
ourselves. For now we will assume that r2P is constant.
P is just any parameter, but it might be easier if we consider a
particular example of a parameter, let's say DISTANCE, witch would
DISTANCE = X
RATE OF CHANGE OF DISTANCE = dX/dt = rX = velocity
RATE OF CHANGE OF VELOCITY = d2X/dt^2 = r2X = acceleration
Don't worry if that all seems very confusing right now. Really
ALL JUST NUMBERS IN BOXES. We just have to tell our computer
which number to put in witch box. It's easy when you know how.
Ken’s grasp of relativity seems to be confined to first year physics, a
little calculus, and how to program a computer to iteratively solve a
second order differential equation. Woo. Hoo.
--
Odd Bodkin -- maker of fine toys, tools, tables
Prokaryotic Capase Homolog
2021-03-04 02:19:45 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Keith Stein
Don't worry if that all seems very confusing right now. Really
ALL JUST NUMBERS IN BOXES. We just have to tell our computer
which number to put in witch box. It's easy when you know how.
Ken’s grasp of relativity seems to be confined to first year physics, a
little calculus, and how to program a computer to iteratively solve a
second order differential equation. Woo. Hoo.
"Ken"???

:-)
Odd Bodkin
2021-03-04 03:05:53 UTC
Reply
Permalink
Post by Prokaryotic Capase Homolog
Post by Odd Bodkin
Post by Keith Stein
Don't worry if that all seems very confusing right now. Really
ALL JUST NUMBERS IN BOXES. We just have to tell our computer
which number to put in witch box. It's easy when you know how.
Ken’s grasp of relativity seems to be confined to first year physics, a
little calculus, and how to program a computer to iteratively solve a
second order differential equation. Woo. Hoo.
"Ken"???
:-)
Keith. Over the two mixed up.
--
Odd Bodkin -- maker of fine toys, tools, tables
Keith Stein
2021-03-04 16:36:34 UTC
Reply
Permalink
Post by Keith Stein
Post by Keith Stein
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t.............(1)
Surprisingly that equation will hold even if rP is not constant, at
least it holds to a very good approximation, which becomes perfect
when t = 0
P(0)  = P0
What this equation says is very profound and very simple.
Nothing changes in zero time, SO when t=0 all variables in
the computer are at their "INITIAL VALUES". i.e " known "

What are the values of these " variables "at the time t ?

In the simple case where r2P = 0 we've already answer that, it's
Equation (1), witch was......

Surely you can't of all forgotten:

P(t) = P0 + rP*t.............(1)

witch you mastered/missed in week2 homework..you remember how
Post by Keith Stein
If the rate of change of P is constant at rP, then the value of P at any
P(t) = P0 + rP*t
Homework PHYSICS for Girls.. Week2 Just Numbers
Given that r2P = 0, and rP= 0.1, and P0 = 13.73 then
determine the values of P(1), P(2), P(3), P(4) and P(100)
But what about those cases were rP (the rate of change of P) is NOT
constant ? In that case still no problem for us, we must make it into
another variable, for which we shall need the initial value [ rP(0) ]
,and the rate of change of the rate of change,r2P, which we assume
constant for now.
rP(t) = rP(0) + r2P*t

r2P*t is the change in the value of rP in time,
and what the equation above says is:
"NEW VALUE OF rP = OLD VALUE OF rP + CHANGE IN VALUE of rP"
witch is pretty obvious, if you think about it.

Don't worry, you'll getta da hanga of ita
i promiss ya. It's very easy whena youa
getta the hanga of it,. witch YOU certainly WILL,
if you don't give up first eh! so STICK WITH IT,
This is as bad as it gets. Here's something to cheer you all up:

NO HOMEWORK THIS WEEK.

HorrraaaaaH eh!
Keith Stein
2021-03-05 09:01:12 UTC
Reply
Permalink
PHYSICS for Girls.." NEW VALUE = OLD VALUE + CHANGE IN VALUE "

P(t) = P(0) + rP * t .......1

rP(t) = rP(0) + r2P * t ..... 2

Equation 1 is valid when rP is constant
Equation 2 is valid when r2P is constant

Both equations 1 and 2 say:

NEW VALUE = OLD VALUE + CHANGE IN VALUE

Witch is obvious, if you think about it.
Keith Stein
2021-03-05 12:16:10 UTC
Reply
Permalink
"Newton's First and Second Laws".

1. Every body continues in a state of rest, or uniform motion in a
straight line, unless it is acted upon by an external force.

[ You should note that Law 1 is valid only when Force = 0 ]

2. Force = Mass * Acceleration.

[You should see note that when Force = 0, Acceleration = 0]
Post by Keith Stein
P(t) = P(0) + rP * t .......1
Equation 1 is valid when rP is constant [i.e.Newton's First Law!]
Equation 2 is valid when r2P is constant[ g = -9.8 m/s/s ]
rP(t) = rP(0) + r2P * t ..... 2
NEW VALUE = OLD VALUE + CHANGE IN VALUE
Witch is obvious, if you think about it.
Keith Stein
2021-03-06 12:34:50 UTC
Reply
Permalink
On 05/03/2021 12:16, Keith Stein wrote:

NEW VALUE = OLD VALUE + CHANGE IN VALUE
If the rate of change of P is constant then:

P(t) = P(0) + rP * t .......1

where P(t) is the new value of P at time t = t
P(0) is the old value of P at time t = 0
rP is the rate of change of P

P is just the numeric value of any parameter whatsoever, and so
equation 1 is a very general truth which enables us to determine the
new value of the parameter P(t) when we know the old value P(0) and the
rate of change of the parameter rP. This equation will be exactly true
when the rate of change of the parameter (rP) is constant, and will
also be a good approximation whenever we are considering very short
intervals of time, because nothing changes (much) in a very short period
of time, and this later fact often enables equation 1 to be employed in
computer simulations, even when rP is not constant!, but we are getting
ahead of ourselves here, so let us for now just consider cases where rP
is constant.

Example:

If Temperature is 7C at 7am when i put the central heating on,
and the temperature then rises at a steady 3 degrees per hour,
what is the temperature at 9am ?
Well if the temperature rises by 3 degrees per hour (rP=3), then
after 2 hours (t=2) it will have risen 6 degrees, and if we add
that to our starting temperature of 7C, we will see the answer is
that when the staff arrive at 9am the temperature will be a barmy 13C,
and of course you didn't need any computer, or equations, to work that
one out, but nevertheless you should see how the calculation we have
just carried out is a very simple application of Equation 1.

P(t) = P(0) + rP * t .......1
= 7 + 3 * 2
= 7 + 6
= 13

Yes that probably looks like a laborious way to do a simple sum witch
you can do immediately in your head, but by learning how this simple
sum is performed by computer you'll soon be able to tackle problems
which would have been way beyond the capabilities of Newton, or even
the mighty Maxwell!

For cases where rp is not constant we must make it into a variable rP(t)
,and by applying our equation 1 to this new variable we obtain our
Equation which is valid where r2P is constant, and also a good
approximation whenever t is very short:

rP(t) = rP(0) + r2P * t ........ (2)

Can you see that there is nothing new in Equation 2 really ?
Equation 2 is just Equation 1 applied to the new variable 'rP'.
Don't worry if you don't see that, I mention that more for the
benefit of any kibitzers who may be watching, rather than anything
you girls need to worry about.
Keith Stein
2021-03-06 20:09:30 UTC
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Post by Keith Stein
NEW VALUE = OLD VALUE + CHANGE IN VALUE
P(t)   =    P(0)   +    rP * t   .......1
the computer does not need to be told what the varios symbols stand for,
it really doesn't care, it simply takes the number in the box with that
lable, does with it whatever asked for in the equation, and sticks the
answer whereever we tell it eh!

Equation 1 applies where the rate of change is constant.
Equation 2 applies where the rate of change is changing!
Post by Keith Stein
rP(t) =  rP(0)   +   r2P * t   ........ (2)
Note that equations 1 and 2 will apply to every variable, and will, i
think, ALWAYS be correct, whatever the variable represents, BECAUSE
Equations 1 and 2 are expressing a truth about NUMBERS which can be
applied to ANY variables in the computer with equal justification,
irrespective of what the variables stand for!

So P is just any parameter, but it might be easier if we consider a
particular example of a parameter, let's say DISTANCE, witch would
normally be given the symbol x rather than P, so lets do that:
DISTANCE = X
RATE OF CHANGE OF DISTANCE = dX/dt = rX = velocity = v

So Equation 1 becomes:

X(t) = X(0) + v * t ............ 1'

RATE OF CHANGE OF VELOCITY = d2X/dt^2 = r2X = acceleration = a
and Equation 2 becomes:

v(t) = v(0) + a * t ............ 2'

Those may look a little more familiar to you, but if not worry not,
there is nothing difficult here.Don't try to memorise them. Try to
UNDERSTAND THEM eh!
Keith Stein
2021-03-07 19:31:50 UTC
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Permalink
y
DISTANCE = X
RATE OF CHANGE OF DISTANCE = dX/dt = rX = velocity = v
RATE OF CHANGE OF VELOCITY = d2X/dt^2 = r2X = acceleration = a
X(t) = X(0)  +  v * t ............ 1
v(t) = v(0)  +   a * t ............ 2
Car Problem:

How far will a car travel in 3 hours at a steady speed of 60 mph ?
Of course you don't need any equation or computer to answer this,
but if you learn how very simple problems like this can be solved
by computer then you will soon find that you can solve problems that
would have been way beyond the cababilities of even an Einstein eh!

Velocity is constant so we can use Equation 1:

X(t) = X(0) + v * t ............ 1

t = time = 3 hours
v = velocity = 60 mph
X(0) = Distance traveled at time 0 = 0
X(t) = X(3) = Distance traveled after 3 hours = What we want to know.

Substituting these values in Equation 1 gives:

X(3) = 0 + 60 * 3
= 180
= Distance traveled after 3 hours

and yes it would have been much easier to do it in your head,
but you won't be able to do "Rocket Science" that way eh!

So moving on to Equation 2 which we must use when the velocity
is not constant. This equation is valid if the acceleration "a"
is constant. The acceleration due to gravity of Earth is constant
at 9.8 m/s/s, so falling under gravity is a good example of constant
acceleration.

Falling Stone Problem:
Suppose a stone is dropped down a hole and is heard to hit the bottom
after 2 seconds. How fast is the stone traveling when it hits the
bottom? Here we have a constant acceleration so we use Equation 2.

v(t) = v(0) + a * t ............ 2

Giving: v(2) = 0 + 9.8 * 2
= 19.6 m/s

So the stone is traveling at 19.6 m/s when it hits the bottom, and
again you can probably do that with any equations. But suppose we
want to know how deep this hole is. Now we have a problem because
it's Equation 1 that gives us distance (X), but Equation 1 applies
where the velocity is constant, which is not the case here. However
Equation 1 will be valid if we use the correct AVERAGE VELOCITY for
the journey being considered, and it is easy to obtain the average
velocity for this stone falling down the whole. We know that it started
from 0 m/s and accelerated smoothly to 19.6 m/s when it his the bottom.

Therefore: AVERAGE VELOCITY = (0 + 19.6)/2 m/s
= 9.8 m/s

Using this average value for v in Equation 1:
X(t) = X(0) + v * t ............ 1
Distance traveled after 2 seconds = X(2) = 0 + 9.8 * 2
= 19.6 m

So the hole is 19.6 meters deep, which in this particular
case happens to be numerically equal to the speed of the
stone when it hits the bottom, but that's just unfortunate.
It wouldn't usually be so, which you should check by reworking
this falling stone example for the case where the stone takes
3 seconds to fall from the top to the bottom.

Homework:
If a stone takes 3 seconds to fall down a deep hole, how fast is
it traveling when it hits the bottom, and HOW DEEP IS THE HOLE ?
Odd Bodkin
2021-03-07 20:28:05 UTC
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Permalink
Post by Keith Stein
y
DISTANCE = X
RATE OF CHANGE OF DISTANCE = dX/dt = rX = velocity = v
RATE OF CHANGE OF VELOCITY = d2X/dt^2 = r2X = acceleration = a
X(t) = X(0)  +  v * t ............ 1
v(t) = v(0)  +   a * t ............ 2
How far will a car travel in 3 hours at a steady speed of 60 mph ?
Of course you don't need any equation or computer to answer this,
but if you learn how very simple problems like this can be solved
by computer then you will soon find that you can solve problems that
would have been way beyond the cababilities of even an Einstein eh!
X(t) = X(0) + v * t ............ 1
t = time = 3 hours
v = velocity = 60 mph
X(0) = Distance traveled at time 0 = 0
X(t) = X(3) = Distance traveled after 3 hours = What we want to know.
X(3) = 0 + 60 * 3
= 180
= Distance traveled after 3 hours
and yes it would have been much easier to do it in your head,
but you won't be able to do "Rocket Science" that way eh!
So moving on to Equation 2 which we must use when the velocity
is not constant. This equation is valid if the acceleration "a"
is constant. The acceleration due to gravity of Earth is constant
at 9.8 m/s/s, so falling under gravity is a good example of constant
acceleration.
Suppose a stone is dropped down a hole and is heard to hit the bottom
after 2 seconds. How fast is the stone traveling when it hits the
bottom? Here we have a constant acceleration so we use Equation 2.
v(t) = v(0) + a * t ............ 2
Giving: v(2) = 0 + 9.8 * 2
= 19.6 m/s
So the stone is traveling at 19.6 m/s when it hits the bottom, and
again you can probably do that with any equations. But suppose we
want to know how deep this hole is. Now we have a problem because
it's Equation 1 that gives us distance (X), but Equation 1 applies
where the velocity is constant, which is not the case here. However
Equation 1 will be valid if we use the correct AVERAGE VELOCITY for
the journey being considered, and it is easy to obtain the average
velocity for this stone falling down the whole. We know that it started
from 0 m/s and accelerated smoothly to 19.6 m/s when it his the bottom.
Therefore: AVERAGE VELOCITY = (0 + 19.6)/2 m/s
= 9.8 m/s
X(t) = X(0) + v * t ............ 1
Distance traveled after 2 seconds = X(2) = 0 + 9.8 * 2
= 19.6 m
So the hole is 19.6 meters deep, which in this particular
case happens to be numerically equal to the speed of the
stone when it hits the bottom, but that's just unfortunate.
It wouldn't usually be so, which you should check by reworking
this falling stone example for the case where the stone takes
3 seconds to fall from the top to the bottom.
If a stone takes 3 seconds to fall down a deep hole, how fast is
it traveling when it hits the bottom, and HOW DEEP IS THE HOLE ?
You must have a LOT of idle time, Keith, to be posting high school physics
to no one in particular in a relativity newsgroup. Is there really nothing
more worthwhile you could be doing?
--
Odd Bodkin — Maker of fine toys, tools, tables
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