Another Larry Brasfield Self-Aggrandizing Put-Down Anecdote

[Active8]

On Fri, 25 Mar 2005 08:04:29 GMT, Fred Bloggs wrote:

Active8 wrote:
On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen wrote:

John Larkin wrote:

On Thu, 24 Mar 2005 19:42:44 -0800, Robert Monsen

snip

"A canoeist in a still pond can reach shore by jerking sharply on the
rope attached to the bow of the canoe. How do you explain this? (yes she
can! - it's true)" (question 11, chap 9, "Physics, Part I", Halliday and
Resnick, 3rd edition).

That's the book I had in HS and used at U of PA. I bought it when I
moved to a new school. And that's a damned good question. Thanks,
I'll remember to look at those noodle twisters when I'm bored.

Sadly, I don't know the answer to this, other than some hand wavings
about friction of the water on the boat. The chapter is entitled
"Conservation of Linear Momentum", so it probably has something to do
with conservation of momentum... Maybe somebody with a few more
intact brain cells left over from college can explain it.

I think it's because the energy she uses to jerk her hand is
internal to her [strike lewd remark] body. If she stopped the motion
on her own, an oppositely directed energy from inside her body would
negate that. But the rope stops her motion and that is external to
her body.

It has to do with F=M*dV/dt which for small time intervals (jerk)
becomes F*dt=(Vf-Vi) or in this simple case, Vf(velocity imparted to
canoe)=F*dt, where F is impulse force she applied to rope during the
jerk of duration dt- so many pounds*milliseconds.

Damn Fred. Right after I posted that, I went and had some sardines
and was thinking that if I were to try to quatify that, I'd look at
impulse and momentum. The key is that her jerk/motion energy is
internal to her and the stopping energy is imparted by the canoe.

--
Best Regards,
Mike

 

[Fred Bloggs]

Active8 wrote:

On Fri, 25 Mar 2005 08:04:29 GMT, Fred Bloggs wrote:


Active8 wrote:

On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen wrote:


John Larkin wrote:


On Thu, 24 Mar 2005 19:42:44 -0800, Robert Monsen

snip

"A canoeist in a still pond can reach shore by jerking sharply on the
rope attached to the bow of the canoe. How do you explain this? (yes she
can! - it's true)" (question 11, chap 9, "Physics, Part I", Halliday and
Resnick, 3rd edition).

That's the book I had in HS and used at U of PA. I bought it when I
moved to a new school. And that's a damned good question. Thanks,
I'll remember to look at those noodle twisters when I'm bored.


Sadly, I don't know the answer to this, other than some hand wavings
about friction of the water on the boat. The chapter is entitled
"Conservation of Linear Momentum", so it probably has something to do
with conservation of momentum... Maybe somebody with a few more
intact brain cells left over from college can explain it.

I think it's because the energy she uses to jerk her hand is
internal to her [strike lewd remark] body. If she stopped the motion
on her own, an oppositely directed energy from inside her body would
negate that. But the rope stops her motion and that is external to
her body.

It has to do with F=M*dV/dt which for small time intervals (jerk)
becomes F*dt=(Vf-Vi) or in this simple case, Vf(velocity imparted to
canoe)=F*dt, where F is impulse force she applied to rope during the
jerk of duration dt- so many pounds*milliseconds.


Damn Fred. Right after I posted that, I went and had some sardines
and was thinking that if I were to try to quatify that, I'd look at
impulse and momentum. The key is that her jerk/motion energy is
internal to her and the stopping energy is imparted by the canoe.

I am assuming that 1) the rope is tied off to a pier, and 2) she remains
fixed to the canoe, but in the case of some women you can't be too sure
about that

 

[Rich Grise]

On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen wrote:

It was similar to the 'Questions' section at the end of chapters in my
first college physics book. They were generally qualitative puzzlers
like this, that were intended to make you think about what could be
happening.

Here is an example:

"A canoeist in a still pond can reach shore by jerking sharply on the
rope attached to the bow of the canoe. How do you explain this? (yes she
can! - it's true)" (question 11, chap 9, "Physics, Part I", Halliday and
Resnick, 3rd edition).

Sadly, I don't know the answer to this, other than some hand wavings
about friction of the water on the boat. The chapter is entitled
"Conservation of Linear Momentum", so it probably has something to do
with conservation of momentum... Maybe somebody with a few more
intact brain cells left over from college can explain it.

There's nothing to it. They even did it with the Enterprise once when
they were stuck in some kind of negative energy vortex thing. They hit
the thrusters with one pulse, and coasted out. It's the same thing with
the canoe. The energy of the jerk is imparted to the canoeist/canoe
system, and the canoe simply coasts to shore. Imagine you're standing on
the dock, holding the rope, which is attached to the canoe. Could you
jerk that rope sharply and have the canoe come to you? Of course. Same
thing. I was stuck on a raft in the middle of a pond without a paddle
once, and didn't want to get wet because Mom would kill me for playing
around in the pond (actually a backed-up drainage ditch), so I kind of
rocked it asymmetrically, turning the whole raft into sort of a paddle.
I made it back to shore.

What I wonder about is the assertion that a rocket's speed, relative
to some imaginary fixed frame of reference, can't ever be faster
than the speed of the exhaust relative to the rocket. I wonder if
that's really true?

Thanks,
Rich

 

[Active8]

On Tue, 22 Mar 2005 18:06:05 -0500, Active8 wrote:

On Tue, 22 Mar 2005 11:45:04 GMT, Fred Bloggs wrote:

Look at the total schizoid fake make more stuff up- a real zero- this is
getting so old:

From: "Larry Brasfield" <donotspam_larry_brasfi...@hotmail.com
snip

"Robert Monsen" <rcsurn...@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...
No, as I recall, kinetic energy is 1/2 * m * v^2.

I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

The instantaneous kinetic E obeys .5*m*v^2, but the total solution
would require a partial diff eq.

Better yet, if you know the mass *and* speed of the ejected fuel,
conservation of momentum is the easiest solution.
--
Best Regards,
Mike

 

[Active8]

On Tue, 22 Mar 2005 18:50:15 -0500, Active8 wrote:

On Tue, 22 Mar 2005 18:06:05 -0500, Active8 wrote:

On Tue, 22 Mar 2005 11:45:04 GMT, Fred Bloggs wrote:

Look at the total schizoid fake make more stuff up- a real zero- this is
getting so old:

From: "Larry Brasfield" <donotspam_larry_brasfi...@hotmail.com
snip

"Robert Monsen" <rcsurn...@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...
No, as I recall, kinetic energy is 1/2 * m * v^2.

I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

And WTF? The equation for E is not a claim, it's a fact. The last 3
sentences above are so indicative of flawed logic ( or trolling )
that they defy comment.

If I shit a pound per day and grow shit^2 pounds of tomatoes a year
for shit^3 kcal of energy... I'm shitting linearly, growing
quadratically, and consuming cubically. I shit you not, larry.
You're my favorite turd this week.

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

The instantaneous kinetic E obeys .5*m*v^2, but the total solution
would require a partial diff eq.

Better yet, if you know the mass *and* speed of the ejected fuel,
conservation of momentum is the easiest solution.

M_r = mass of rocket and fuel, pre-burn
V_r = velocity of rocket and fuel, pre-burn
m_r = mass of rocket, post burn.
v_r = velocity of rocket, post burn
v_f = velocity of ejected fuel, post burn
m_f = mass of ejected fuel, post burn

M_r * V_r = m_r * v_r + m_f * v_f

A stumped physics teacher does not a smart larry make.
--
Best Regards,
Mike

 

[John Larkin]

On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen
<rcsurname@comcast.net> wrote:

My issue (and Fred's, I'm gussing) wasn't really about the puzzler
itself, it was with the "Let's see if we can show this guy up" attitude
it was originally posted with.

He posted another, challenging us to explain why a transistor's Ic
goes down when you zener the b-e junction. This one also has a trick
answer: it doesn't.

John

 

[Active8]

On Fri, 25 Mar 2005 00:46:13 -0500, Active8 wrote:

Better clean that up. My fumbling fingers typed "=" instead of "-"
in two places.

let D be the round d

k = fuel mass expenditure wrt time
M = initial mass of rocket

m = M - kt instantaneous mass of rocket

1
KE = --- (M - kt)a^2 t^2
2

expand that

DE 1 2
-- = - v
Dm 2

DE 2
-- = mat - kat
Dv

1 2 2
DE = - v dm + mat dv - kat dv
2


dE 1 2 dm dv 2 dv
-- = - v -- + mv -- - kat --
dt 2 dt dt dt

dE 1 2 2
-- = - kv + mav - kv
dt 2

1 2 2
dE = - kv dt + Fv dt - kv dt
2

integrate
1 2 2
E = - kv t + Fvt - kv t
2

1 2
E = - mv + Fx - kxv
2

.
.. . Total energy = KE + thrust energy - exhaust energy

QED
--
Best Regards,
Mike

 

[Larry Brasfield]

"Active8" <reply2group@ndbbm.net> wrote in message
newsxi1f69au1tb$.dlg@ID-222894.news.individual.net...

On Tue, 22 Mar 2005 16:53:22 -0800, Larry Brasfield wrote:
"Active8" <reply2group@ndbbm.net> wrote in message
news:1undbh2xrxhm0$.dlg@ID-222894.news.individual.net...
Brasfield once wrote:
I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

And WTF? The equation for E is not a claim, it's a fact. The last 3
sentences above are so indicative of flawed logic ( or trolling )
that they defy comment.

Well, your "fact" is not accepted physics today. It is a very
good approximation for velocities well below the speed of
light, but it cannot be defended as gold plated Truth.
If you doubt this, look here:
http://www.kineticbooks.com/physics/17467/17516/sp.html
or search on words: "kinetic energy" formula relativistic .

Rockets travel well below the speed of light, last I checked. So the
above is irrelevant to the discussion.

It was offered only as evidence to refute the assertion:
"The equation for E is not a claim, it's a fact"
As I recall, that recently became part of the discussion.

That said, I will admit to phrasing my doubt of the matter
in a provocative way. If that makes me a troll, so be it.
But I posted it because I thought it was an interesting
puzzler.

It is, but it looks more like you were trying to make your physics
teacher look dumb and you smart by posing a question you couldn't
answer.

You assume facts not in evidence. What leads you to
believe this happened in front of a classroom of kids?
As a matter of fact, it occured after school during one
of many half-hour sessions he and I spent. How could
a teenager expect that his teacher would not be able to
answer such a question? What evidence supports the
allegation that it was anything but a curious kid asking
an expert a question that puzzled him?

....

I posted to the "Twist You Noodle" forum and
someone pulled my head out of my ass for me.

Looking for more of your favorites, were you?

The physics are fairly simple, (as you have
suggested), but only if you consider the right set of
objects. The puzzle is one that took me in as a high
school student, along with a smart friend and my high
school physics teacher who was dedicated and good
at what he did. So, I thought it had a good chance of
puzzling folks who were following the thread up to then.
All in good fun, and not a typical troll.

I'm not a bit puzzled. The flawed logic is the assumption that a
linear decrease in fuel mass is contrary to a quadratic increase in
kinetic energy.

Yes, that is the key. There is an implicit (but incorrect)
reliance on the notion of energy conservation, while
paying attention to only 2 of the 3 players. The burnt
fuel and the rocket are visible, the exhaust is invisible.
(If you ever do this on a whiteboard for someone else's
puzzlement, be sure to not draw an exhaust plume!)

And that *instantaneous E" or better yet

|
E = .5 * m * V^2 |
|m=k

variable mass is another story.

And way beyond anybody's attention if actually dealt with.

And to hint that it breaks E = .5*m*v^2 is incorrect.

It could hardly be a puzzle without that dilemma.

And now I see the mandacity that Jim alluded to. E does not vary
quadratically with time, it varies quadratically with velocity

Under the stated assumptions of the problem, "accellerating at a
constant rate", the velocity varies linearly with time. Therefore
the kinetic energy varies quadratically with time and velocity.
I fail to see how mendacity enters into this.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[John - KD5YI]

Fred Bloggs wrote:

Look at the total schizoid fake make more stuff up- a real zero- this is
getting so old:

From: "Larry Brasfield" <donotspam_larry_brasfi...@hotmail.com
Newsgroups: sci.electronics.basics
References: <bd24a397.0503080128.77ba467e@posting.google.com
uMGdnQ0E54dKlbPfRVn-tw@comcast.com
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"Robert Monsen" <rcsurn...@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...
No, as I recall, kinetic energy is 1/2 * m * v^2.


I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

I do not understand how acceleration can be constant due to constant thrust
when the mass is changing (fuel depletion). This seems to me to be a
contradiction.

John

 

[John - KD5YI]

Larry Brasfield wrote:

"John - KD5YI" <kd5yikes@mindspring.com> wrote in message
news:T5g0e.1972$gI5.1121@newsread1.news.pas.earthlink.net...

Fred Bloggs wrote:

[Irrelevant FB spew cut. Irrelevant and scurrilous FB subject revised.]

"Robert Monsen" <rcsurn...@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...
No, as I recall, kinetic energy is 1/2 * m * v^2.

I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).

I do not understand how acceleration can be constant due to constant thrust when the mass is changing (fuel depletion). This seems
to me to be a contradiction.


Think of it as an engineering approximation.
For an atomic powered, ion rocket engine,
it can be a very close approximation.

The reason to make the approximation is
to keep the math simple. The error is not
relevant to the puzzle's apparent dilemma,
so the extra math would be a distraction.

Oh. They had ion engines way back then?

 

[Active8]

On Fri, 25 Mar 2005 12:54:27 -0500, Active8 wrote:

Ok. I suck at ascii math and my work on paper was sloppy. Why don't
we just build this rocket car thing already?

let D be the round d

k = fuel mass expenditure wrt time dm/dt or m/t
M = initial mass of rocket

m = M - kt instantaneous mass of rocket

1 2
KE = --- (M - kt)v
2

expand that and take the partial derivatives wrt m and v

DE 1 2
-- = - v
DM 2

DE 2
-- = Mat - kat
Dv

1 2 2
dE = - v dM + Mat dv - kat dv
2


dE 1 2 dM dv 2 dv
-- = - v -- + Mv -- - kat --
dt 2 dt dt dt

dE 1 2 2
-- = - kv + Mav - kv
dt 2

1 2 2
dE = - kv dt + Fv dt - kv dt
2

integrate

1 2 2
E = - kv t + Fvt - kv t
2

1 2
E = - mv + Fx - kxv
2

.
.. . Total energy = KE + thrust energy - exhaust energy

QED
--
Best Regards,
Mike

 

[Larry Brasfield]

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message news:22j841db7unbaq3es8p6mniees46cic3ui@4ax.com...

On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen
rcsurname@comcast.net> wrote:

My issue (and Fred's, I'm gussing) wasn't really about the puzzler
itself, it was with the "Let's see if we can show this guy up" attitude
it was originally posted with.

That whole line of thinking is insulting and founded
only on misreading between the lines. How anybody
can take it as more than a mystified kid asking the
question because he wants an answer says more
about the conjecturers as it does about my motive.

He posted another, challenging us to explain why a transistor's Ic
goes down when you zener the b-e junction. This one also has a trick
answer: it doesn't.

You have misconstrued what I said.

I wrote "One 2nd order effect is that the extra
majority carriers reduce the equilibrium density of
minority carriers in the base region, reducing what is
usually thought of as the C-B leakage."

I never claimed it was the only 2nd order effect,
and explicitly said it was not: "I'm only stating the
effect of the 2nd order effect I mentioned. There
are a few others. You have to add them all up if
you're nuts enough to want to predict how they
will show up in real devices."

And the posed question said nothing about Ic going
down. On that, it said only "What happens to the
current as indicated by the meter? Now, why?

You have already seen that the effects are small,
being 4+ orders of magnitude down. Why do you
insist that what you see is contrary to what I've
stated? I only claimed that there would be small
effects. The fact that I named one whose sign
differs from what you saw does not contradict
the thrust of my prediction, which is that there
is no 1st order (ie "large") effect.

And to deal with another matter, there is no
"trick" answer. There are answers founded
on facts and reasoning, and all the rest.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Active8]

On Wed, 23 Mar 2005 15:40:35 GMT, John - KD5YI wrote:

Fred Bloggs wrote:
Look at the total schizoid fake make more stuff up- a real zero- this is
getting so old:

From: "Larry Brasfield" <donotspam_larry_brasfi...@hotmail.com
Newsgroups: sci.electronics.basics
References: <bd24a397.0503080128.77ba467e@posting.google.com
uMGdnQ0E54dKlbPfRVn-tw@comcast.com
Subject: Re: Potentially painful
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"Robert Monsen" <rcsurn...@comcast.net> wrote in message
news:uMGdnQ0E54dKlbPfRVn-tw@comcast.com...
No, as I recall, kinetic energy is 1/2 * m * v^2.

I recall seeing that claim from a high school physics teacher when
I was a smart-ass twerp. I posed the following puzzle to him:
A rocket car starts at rest, accellerating at a constant rate
because its thrust is constant. It is burning fuel at a constant
rate to produce that constant thrust. The kinetic energy of
the rocket car is allegedly M * V^2 / 2, so it is increasing
quadratically versus time. But the fuel consumed increases
only linearly with time. How can this be?

I would be interested in your take on this. My physics teacher
could not resolve it, (but, to his credit, that bothered him).


I do not understand how acceleration can be constant due to constant thrust
when the mass is changing (fuel depletion). This seems to me to be a
contradiction.

That's a good candidate for the "mendacity hidden in his problem
description." that JT warned me about.
--
Best Regards,
Mike

 

[Robert Monsen]

Active8 wrote:

On Fri, 25 Mar 2005 06:00:46 GMT, Fred Bloggs wrote:


Active8 wrote:

On Thu, 24 Mar 2005 13:48:06 -0800, Larry Brasfield wrote:


"Active8" <reply2group@ndbbm.net> wrote in message news:3ohe0rdjzmmf.dlg@ID-222894.news.individual.net...


On Thu, 24 Mar 2005 11:31:41 -0800, Larry Brasfield wrote:

What did Socrates do? I don't recall that.

That p.o.s. doesn't know. He only brought that up because I first
referred to it in a response JF regarding Brasfield's pedophile-like
technique of capturing the attention of juvenile punk OPs.


This must be a clue:

http://www.mindspring.com/~mfpatton/sclinic.htm

I got the gist after a few paras. I don't know in what context you
used the phrase, but I'll venture that Larry meant that his teacher
just kept on until he gave in.

Socratic method is a teaching scheme which uses questions to prompt
students toward a particular answer, rather than just telling them what
to think. The basic idea it was based on was that you already know
everything (in the 'ideal plane'), but have forgotten some of it due to
your existence here in the real world. The questions simply prompt you
to remember what you've forgotten. It's annoying, and doesn't really
work all that well, but educators seem to like it because it gives them
a sense of superiority. They get to watch the student squirm while
trying to come up with an answer the teacher wants.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[John Fields]

On Fri, 25 Mar 2005 10:36:41 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message news:22j841db7unbaq3es8p6mniees46cic3ui@4ax.com...
On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen
rcsurname@comcast.net> wrote:

My issue (and Fred's, I'm gussing) wasn't really about the puzzler
itself, it was with the "Let's see if we can show this guy up" attitude
it was originally posted with.

That whole line of thinking is insulting and founded
only on misreading between the lines. How anybody
can take it as more than a mystified kid asking the
question because he wants an answer says more
about the conjecturers as it does about my motive.

---
Actually, it says a a lot about _you_. You're the one who's telling
the story in the present, and you relate it in a vein which is
designed to show the teacher up as incompetent and you as, well, once
again, Larry must stay on top of it all... That's what's behind most
everything you've posted, and seems to be the driving force in your
life.

Why? Was your childhood so painful that you've had to spend the
greater part of your life proving to yourself that you're _not_
worthless by trying to prove that everyone else is?
---

He posted another, challenging us to explain why a transistor's Ic
goes down when you zener the b-e junction. This one also has a trick
answer: it doesn't.

You have misconstrued what I said.

I wrote "One 2nd order effect is that the extra
majority carriers reduce the equilibrium density of
minority carriers in the base region, reducing what is
usually thought of as the C-B leakage."

I never claimed it was the only 2nd order effect,
and explicitly said it was not: "I'm only stating the
effect of the 2nd order effect I mentioned. There
are a few others. You have to add them all up if
you're nuts enough to want to predict how they
will show up in real devices."

And the posed question said nothing about Ic going
down. On that, it said only "What happens to the
current as indicated by the meter? Now, why?

You have already seen that the effects are small,
being 4+ orders of magnitude down. Why do you
insist that what you see is contrary to what I've
stated? I only claimed that there would be small
effects. The fact that I named one whose sign
differs from what you saw does not contradict
the thrust of my prediction, which is that there
is no 1st order (ie "large") effect.

And to deal with another matter, there is no
"trick" answer. There are answers founded
on facts and reasoning, and all the rest.

---
As usual, someone (of course) misconstrued what you said and you have
to jump in there and spend an inordinate amount of time defending your
former "position", but not without introducing an out or two for next
time.

"What is _usually_ thought of as the C-B leakage"? (emphasis mine)

To contradict that point would take an inordinate amount of study just
for the purpose of blowing your statement off and would also provide
you with a convenient jumping-off place to start yet another thread
for the purpose of taking the heat off of you by diverting the focus
from the point at hand.

I believe JL asked you in another post whether you were going to do
the testing you claimed you were going to do or whether you were going
to talk the thing to death. Which is it going to be?

--
John Fields

 

[John Woodgate]

I read in sci.electronics.design that Rich Grise <richgrise@example.net>
wrote (in <pan.2005.03.25.17.07.27.620478@example.net&gt about 'Another
Larry Brasfield Self-Aggrandizing Put-Down Anecdote', on Fri, 25 Mar
2005:

What I wonder about is the assertion that a rocket's speed, relative to
some imaginary fixed frame of reference, can't ever be faster than the
speed of the exhaust relative to the rocket. I wonder if that's really
true?

Not true, if you really mean 'arbitrary'. The rocket can't go thataway
faster than the exhaust is going thisaway, though, if you think about
it. But relative to a distant quasar, as the fixed frame of reference,
both are going at a reasonable fraction of the speed of light.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Robert Monsen]

Active8 wrote:

Actually, the puzzle is that the power is constant (a constant rate of
fuel burn) but the rocket's kinetic energy is increasing quadratically


How many times... how many different ways... can I say it. There *is
no puzzle*. If a linear fuel mass expenditure really gave a constant
acceleration or constant power, that's fine. Check your old text.
They start by showing that v = at and then a = F/m. Then they show
that KE varies quadratically with v. If you can accept that, there's
no puzzle.

Pulease...

When I say *puzzle*, perhaps I should have said "apparent discrepancy",
or "poser", or something like that. It's not puzzling if you know the
answer, of course.

When one assumes that a linear change of one dimension should
produce a linear change in a derived unit it's called flawed logic.
At least that's what I called it.

Right, that's the trick with puzzlers, you lead people down the path,
and they arrive someplace unexpected. Then, their job is to figure out
why, or to find the flawed logic.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Jim Thompson]

On Fri, 25 Mar 2005 15:55:38 -0800, Robert Monsen
<rcsurname@comcast.net> wrote:

Active8 wrote:

Actually, the puzzle is that the power is constant (a constant rate of
fuel burn) but the rocket's kinetic energy is increasing quadratically


How many times... how many different ways... can I say it. There *is
no puzzle*. If a linear fuel mass expenditure really gave a constant
acceleration or constant power, that's fine. Check your old text.
They start by showing that v = at and then a = F/m. Then they show
that KE varies quadratically with v. If you can accept that, there's
no puzzle.


Pulease...

When I say *puzzle*, perhaps I should have said "apparent discrepancy",
or "poser", or something like that. It's not puzzling if you know the
answer, of course.

When one assumes that a linear change of one dimension should
produce a linear change in a derived unit it's called flawed logic.
At least that's what I called it.


Right, that's the trick with puzzlers, you lead people down the path,
and they arrive someplace unexpected. Then, their job is to figure out
why, or to find the flawed logic.

Just return to fundamentals when confronted by posers/shamans...

F =/= ma, except for CONSTANT mass

F = d/dt(mv) http://www.grc.nasa.gov/WWW/K-12/airplane/newton2.html

Sir Isaac Newton first presented his three laws of motion in the
"Principia Mathematica Philosophiae Naturalis" in 1686. His second law
defines a force to be equal to the differential change in MOMENTUM per
unit time as described by the calculus of mathematics, which Newton
also developed.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Robert Monsen]

Rich Grise wrote:

What I wonder about is the assertion that a rocket's speed, relative
to some imaginary fixed frame of reference, can't ever be faster
than the speed of the exhaust relative to the rocket. I wonder if
that's really true?

No, that is a consequence of doing the computation with a fixed mass
rocket. If you let the mass of the rocket vary, then the rocket can
achieve any velocity. My old physics book claims the final velocity is
related by

Mf/Mo = exp(-vf/Vt)

where vf is the final velocity, Mf is the final mass, Mo is the initial
mass, and Vt is the velocity of the thrust with respect to the rocket. Thus,

vf = Vt * ln(Mo/Mf)

Since Mf can get as small as you want, the final velocity can get as
large as you want. Clearly, the more Vt you can get, the better off the
situation is.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[John Larkin]

On Fri, 25 Mar 2005 10:36:41 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in
message news:22j841db7unbaq3es8p6mniees46cic3ui@4ax.com...
On Thu, 24 Mar 2005 23:12:10 -0800, Robert Monsen
rcsurname@comcast.net> wrote:

My issue (and Fred's, I'm gussing) wasn't really about the puzzler
itself, it was with the "Let's see if we can show this guy up" attitude
it was originally posted with.

That whole line of thinking is insulting and founded
only on misreading between the lines. How anybody
can take it as more than a mystified kid asking the
question because he wants an answer says more
about the conjecturers as it does about my motive.

He posted another, challenging us to explain why a transistor's Ic
goes down when you zener the b-e junction. This one also has a trick
answer: it doesn't.

You have misconstrued what I said.

I wrote "One 2nd order effect is that the extra
majority carriers reduce the equilibrium density of
minority carriers in the base region, reducing what is
usually thought of as the C-B leakage."

I never claimed it was the only 2nd order effect,
and explicitly said it was not: "I'm only stating the
effect of the 2nd order effect I mentioned. There
are a few others. You have to add them all up if
you're nuts enough to want to predict how they
will show up in real devices."

Well, I measured the zero-order effect, the initial leakage. And I
measured the first-order effect, the linear increase in Ic with Ie. If
there are higher-order effects, I didn't have the resolution or
enthusiasm to resolve them. The linear effect was clearly the biggie.

And the posed question said nothing about Ic going
down. On that, it said only "What happens to the
current as indicated by the meter? Now, why?

Well, the only answer you provided was that it would go down.

You have already seen that the effects are small,
being 4+ orders of magnitude down. Why do you
insist that what you see is contrary to what I've
stated? I only claimed that there would be small
effects. The fact that I named one whose sign
differs from what you saw does not contradict
the thrust of my prediction, which is that there
is no 1st order (ie "large") effect.

"First order" doesn't mean "large"; it means linear on input. Or maybe
you're using a different kind of polynomial than the ones I'm used to.

John