Potentially painful

[robin.pain@tesco.net]

Biking is much more fun than electronics, especially as summer is
icumen in, but do you youngsters *know* about it?

E.g. you know that its a lot more fun discharging a small 370V
capacitor than a *huge* 5V one because energy is proportional to the
voltage *squared* (because if you double voltage then the current
doubles too and energy is their product)because
Energy = CV^2

I
-------------
| |
| |
V | Energy |
| |
| |
| |
-------------

The same thing applied to speed. If you double the speed you quaduple
the energy because Energy = MV^2

So if you hit a massive artic' while doing 1 mph (the lorry is
stationary) on yer bike then (assuming you weight "one" unit) you will
dissipate one unit of energy.

If you hit the same vehicle at 101 mph when it is doing 100 mph then,
ignoring the lorry's slight speed increase, you will dissipate the
difference between you doing 101 and 100 i.e.

101*101 - 100*100 == ~200

That's nearly 200 hundred times more dangerous.

Wazzat? ya don't believe it? (The local boy racer is reading this over
my shoulder, he is snorting disbelief). Well if you prefer a belief
system to a mathematical system then look at it this way: How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

Cheers

Robin Pain (ouch)

 

[Severi Salminen]

How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

Yeah, and since the Earth is moving at 30km/s around the Sun, we have so
much energy stored in us that on every step we explode like a nuclear
bomb. I can't imagine what should happen if we take the speed of our
solar system into account, not to mention the galaxy...

 

[Bob Masta]

On 8 Mar 2005 01:28:40 -0800, robin.pain@tesco.net
(robin.pain@tesco.net) wrote:

Biking is much more fun than electronics, especially as summer is
icumen in, but do you youngsters *know* about it?

E.g. you know that its a lot more fun discharging a small 370V
capacitor than a *huge* 5V one because energy is proportional to the
voltage *squared* (because if you double voltage then the current
doubles too and energy is their product)because
Energy = CV^2

I
-------------
| |
| |
V | Energy |
| |
| |
| |
-------------

The same thing applied to speed. If you double the speed you quaduple
the energy because Energy = MV^2

So if you hit a massive artic' while doing 1 mph (the lorry is
stationary) on yer bike then (assuming you weight "one" unit) you will
dissipate one unit of energy.

If you hit the same vehicle at 101 mph when it is doing 100 mph then,
ignoring the lorry's slight speed increase, you will dissipate the
difference between you doing 101 and 100 i.e.

101*101 - 100*100 == ~200

That's nearly 200 hundred times more dangerous.

Wazzat? ya don't believe it? (The local boy racer is reading this over
my shoulder, he is snorting disbelief). Well if you prefer a belief
system to a mathematical system then look at it this way: How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

Cheers

Robin Pain (ouch)

The relative speed difference is all that matters.
Consider that both of you are zipping along at
over 1000 mph on the surface of the Earth,
and the Earth is whizzing along at 19 miles
per second, etc, etc, none of which matters here.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

 

[James Beck]

In article <bd24a397.0503080128.77ba467e@posting.google.com>,
robin.pain@tesco.net says...

Biking is much more fun than electronics, especially as summer is
icumen in, but do you youngsters *know* about it?

E.g. you know that its a lot more fun discharging a small 370V
capacitor than a *huge* 5V one because energy is proportional to the
voltage *squared* (because if you double voltage then the current
doubles too and energy is their product)because
Energy = CV^2

I
-------------
| |
| |
V | Energy |
| |
| |
| |
-------------

The same thing applied to speed. If you double the speed you quaduple
the energy because Energy = MV^2

So if you hit a massive artic' while doing 1 mph (the lorry is
stationary) on yer bike then (assuming you weight "one" unit) you will
dissipate one unit of energy.

If you hit the same vehicle at 101 mph when it is doing 100 mph then,
ignoring the lorry's slight speed increase, you will dissipate the
difference between you doing 101 and 100 i.e.

101*101 - 100*100 == ~200

That's nearly 200 hundred times more dangerous.

Wazzat? ya don't believe it? (The local boy racer is reading this over
my shoulder, he is snorting disbelief). Well if you prefer a belief
system to a mathematical system then look at it this way: How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

Cheers

Robin Pain (ouch)

I think the rest of the universe might not go along with your theory.

Jim

 

[John Larkin]

On 8 Mar 2005 01:28:40 -0800, robin.pain@tesco.net
(robin.pain@tesco.net) wrote:

Biking is much more fun than electronics, especially as summer is
icumen in, but do you youngsters *know* about it?

E.g. you know that its a lot more fun discharging a small 370V
capacitor than a *huge* 5V one because energy is proportional to the
voltage *squared* (because if you double voltage then the current
doubles too and energy is their product)because
Energy = CV^2

I
-------------
| |
| |
V | Energy |
| |
| |
| |
-------------

The same thing applied to speed. If you double the speed you quaduple
the energy because Energy = MV^2

So if you hit a massive artic' while doing 1 mph (the lorry is
stationary) on yer bike then (assuming you weight "one" unit) you will
dissipate one unit of energy.

If you hit the same vehicle at 101 mph when it is doing 100 mph then,
ignoring the lorry's slight speed increase, you will dissipate the
difference between you doing 101 and 100 i.e.

101*101 - 100*100 == ~200

That's nearly 200 hundred times more dangerous.

Wazzat? ya don't believe it? (The local boy racer is reading this over
my shoulder, he is snorting disbelief). Well if you prefer a belief
system to a mathematical system then look at it this way: How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

Cheers

Robin Pain (ouch)

You have just disproven the theory of relativity and thus destroyed
the entire foundation of modern physics. Bad boy!

John

 

[Robert Monsen]

robin.pain@tesco.net wrote:

Biking is much more fun than electronics, especially as summer is
icumen in, but do you youngsters *know* about it?

E.g. you know that its a lot more fun discharging a small 370V
capacitor than a *huge* 5V one because energy is proportional to the
voltage *squared* (because if you double voltage then the current
doubles too and energy is their product)because
Energy = CV^2

I
-------------
| |
| |
V | Energy |
| |
| |
| |
-------------

The same thing applied to speed. If you double the speed you quaduple
the energy because Energy = MV^2

No, as I recall, kinetic energy is 1/2 * m * v^2.

So if you hit a massive artic' while doing 1 mph (the lorry is
stationary) on yer bike then (assuming you weight "one" unit) you will
dissipate one unit of energy.

If you hit the same vehicle at 101 mph when it is doing 100 mph then,
ignoring the lorry's slight speed increase, you will dissipate the
difference between you doing 101 and 100 i.e.

101*101 - 100*100 == ~200

That's nearly 200 hundred times more dangerous.

No, it's 200 times less dangerous if you hit something stationary.
However, you haven't yet hit anything stationary.

The dissipation can also be measured from the lorry's reference frame.
From that frame, the system has only your kinetic energy measured in
relation to it's own velocity (1/2 m * 1^2). When you hit the lorry, the
system has a tiny bit of kinetic energy (it's now moving slightly in the
pre-collision reference frame), and a tiny bit of energy from the
collision (probably in heat, or in radiation). It isn't possible to
determine the relative percentages in general.

From the road's reference frame, the difference in energy you perceived
is translated into that small amount of heat, plus a change in the
kinetic energy of the lorry + you system. You only dissipated a tiny bit
of the energy into heat (which is what hurts); you transferred the rest
of it to the kinetic energy of the lorry + you system.

Wazzat? ya don't believe it? (The local boy racer is reading this over
my shoulder, he is snorting disbelief). Well if you prefer a belief
system to a mathematical system then look at it this way: How much
energy do you have to put it to get ya bike up to 1 mph? Not much, you
just start pushing it and after a couple of strides both of you are up
to 1 mph but what about 100 to 101? First of all you have to *be*
running at 100 *before* you start pushing and *then* you must push and
get it to 101. The ground you have to cover *while* pushing is
evidently a hellofalot more. Does that not seem like a *lot* more
stored energy to you?

No. It's the same amount of energy. Assume you are on a train travelling
at 100 mph. Now, accelerate 1 mph in the direction the train is
travelling. How much energy do you need to expend?

Cheers

Robin Pain (ouch)

--
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.

 

[Larry Brasfield]

"Robert Monsen" <rcsurname@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).

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

 

[Don Klipstein]

In article <d9oXd.34$UW6.2012@news.uswest.net>, Larry Brasfield wrote:

"Robert Monsen" <rcsurname@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 have resolved it. Consider the kinetic energy of the rocket exhaust
along with that of the car. Assuming none of the exhaust loses kinetic
energy due to turbulent mixing with air that converts kinetic energy to
heat, the combined sum of kinetic energies of the car and exhaust will
increase linearly with respect to time during the time period that rocket
exhaust is being generated.

--------------------------------------------------

Back to the original: If a car moving 1 MPH hits a stationary one, or a
car moving 101 MPH rear-ends one doing 100, the amount of kinetic energy
loss from causing a tiny dent or a "bump" noise from the bumper or slight
heating of any cushioning in the bumper is the same. But if the cars
doing 101 and 100 go out of control, their remaining kinetic energy is a
big problem if they hit enything else. If their wheels lock, the
remaining kinetic energy can abrade a lot of rubber from the tires.
Although I doubt a car doing 101 tapping the rear bumper of one doing 100
will result in any catostrophe unless at least one driver is incompetent
in ways besides one tapping another at high speeds.

- Don Klipstein (don@misty.com)

 

[Rich Grise]

On Tue, 08 Mar 2005 23:10:12 +0000, Don Klipstein wrote:

Back to the original: If a car moving 1 MPH hits a stationary one, or
a
car moving 101 MPH rear-ends one doing 100, the amount of kinetic energy
loss from causing a tiny dent or a "bump" noise from the bumper or
slight heating of any cushioning in the bumper is the same. But if the
cars doing 101 and 100 go out of control, their remaining kinetic energy
is a big problem if they hit enything else. If their wheels lock, the
remaining kinetic energy can abrade a lot of rubber from the tires.
Although I doubt a car doing 101 tapping the rear bumper of one doing
100 will result in any catostrophe unless at least one driver is
incompetent in ways besides one tapping another at high speeds.

I like the one where he asks, "if you're going 100 MPH, and you meet

another car head-on that's coming the other way at 100 MPH, how much
energy is expended?" the point being that hitting a 100 MPH car isn't
like a 200 MPH crash at all - it's exactly the same as hitting a brick
wall, assuming all the standard assumptions.

Cheers!
Rich

 

[Jon Hoyle]

Hello!

Just wanna add a couple of my thoughts after reading this one. Please feel
free to correct things, lol.

'Accelerating at a constant rate because its thrust is constant' True or
false?
I think false. Does drag increase with speed up until terminal velocity
is reached?
If thrust is constant, will drag eventually equal the thrust once a
certain speed is reached?
Therefore does the thrust need to increase to match the increase in
drag?

'The kinetic energy of the rocket car is allegedly M * V^2 / 2, so it is
increasing quadratically versus time' True or false?
False?
Does 'M' get smaller as the rocket car uses fuel?
What shape would the graph be if it does?

'But the fuel consumed increases only linearly with time' True or false?
False?
As pressure and temperature builds inside a combustion chamber, does the
fuel burn rate (thrust) also increase?

Well that was fun. Please tell me my understanding of physics is phlawed

Jon


"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message
news:d9oXd.34$UW6.2012@news.uswest.net...

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).

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

 

[Robert Monsen]

Larry Brasfield wrote:

"Robert Monsen" <rcsurname@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.

You've put that behind you then.

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?

Beats me. May have something to do with the exhaust, as Don says. Might
also have to do with the heat generated in the rocket. Probably
something to do with reference frames, and the velocity of the fuel
exhaust changing with regards to a stationary observer. That is all it
really can be, since the closed system is rocket, fuel, exhaust. Inside
the system, the energy is constant. However, working it out is another
matter.

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

Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'

--
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 Tue, 8 Mar 2005 12:44:56 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"Robert Monsen" <rcsurname@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).

At low velocities, a rocket is a very inefficient source of
propulsion; at near-zero velocity, it's using its usual amount of fuel
but hardly delivering any kinetic energy to the vehicle. As velocity
increases, efficiency improves (or rather becomes less terrible) and
vehicle energy accumulates faster. That trend continues until you run
out of fuel.

In a system that goes from extremely inefficient to only rather
inefficient, it's not hard to shape the efficiency curve into a
quadratic. A cog railway can be nearly 100% efficient, so it will need
increasing amounts of fuel if it accelerates at constant gees, but
anywhere on the path it will be a lot more efficient than a rocket.

That make sense?

John

 

[Robert Monsen]

John Larkin wrote:

On Tue, 8 Mar 2005 12:44:56 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:


"Robert Monsen" <rcsurname@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).



At low velocities, a rocket is a very inefficient source of
propulsion; at near-zero velocity, it's using its usual amount of fuel
but hardly delivering any kinetic energy to the vehicle. As velocity
increases, efficiency improves (or rather becomes less terrible) and
vehicle energy accumulates faster. That trend continues until you run
out of fuel.

In a system that goes from extremely inefficient to only rather
inefficient, it's not hard to shape the efficiency curve into a
quadratic. A cog railway can be nearly 100% efficient, so it will need
increasing amounts of fuel if it accelerates at constant gees, but
anywhere on the path it will be a lot more efficient than a rocket.

That make sense?

John

Here is a web page written by somebody who has been driven insane by
this problem:

http://nov55.com/enrgy.html

His solution is that kinetic energy is really mv, not 1/2 mv^2. He has
equations and everything. Pretty wierd stuff. Who knows?

--
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.

 

[sergiometra]

Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'

if there is not resistance in circuit and cap, circuit will oscillate
forever;
otherwise energy dissipate in heat and oscillation eventually end
regards
sergio

 

[sergiometra]

Here is a web page written by somebody who has been driven insane by
this problem:

http://nov55.com/enrgy.html

His solution is that kinetic energy is really mv, not 1/2 mv^2. He has
equations and everything. Pretty wierd stuff. Who knows?

I think he demonstrate the exact opposite;
if you jump from 2nd floor, speed at arrival is not double that a jumping
from 1st floor as it take less time from floor 1 to 0 than from 2 to 1 , but
energy is double as you need double energy to go up 2 floor than 1 , so
energy increase more rapidly than speed i/e Energy>mv;
regards
sergio

 

[Larry Brasfield]

"John Larkin" <jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote
in message news:sa3t2113mifbj3e38eovdn74qc546dni4i@4ax.com...

On Tue, 8 Mar 2005 12:44:56 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"Robert Monsen" <rcsurname@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).


At low velocities, a rocket is a very inefficient source of
propulsion; at near-zero velocity, it's using its usual amount of fuel
but hardly delivering any kinetic energy to the vehicle. As velocity
increases, efficiency improves (or rather becomes less terrible) and
vehicle energy accumulates faster. That trend continues until you run
out of fuel.

In a system that goes from extremely inefficient to only rather
inefficient, it's not hard to shape the efficiency curve into a
quadratic. A cog railway can be nearly 100% efficient, so it will need
increasing amounts of fuel if it accelerates at constant gees, but
anywhere on the path it will be a lot more efficient than a rocket.

That make sense?

Yes. I should have known this little puzzle would
not take you folks long to sort out. The key is to
consider that starting condition. All the energy ends
up in the exhaust at the limit of zero rocket speed.
The problem gets a little easier to see if it involves
chucking cannonballs from a railway car.

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

 

[Larry Brasfield]

"Jon Hoyle" <NOSPAMbrumsoftNOSPAM@blueSPAMyonder.co.uk> wrote
in message news:vhvXd.27926$y25.8386@fe3.news.blueyonder.co.uk...

Hello!
Hi.
Just wanna add a couple of my thoughts after reading this one. Please feel free to correct things, lol.

'Accelerating at a constant rate because its thrust is constant' True or false?
I think false. Does drag increase with speed up until terminal velocity is reached?
If thrust is constant, will drag eventually equal the thrust once a certain speed is reached?
Therefore does the thrust need to increase to match the increase in drag?

You have assumed something into the problem
that complicates it without affecting its nature.
The "terminal velocity" concept requires some
kind of liquid or gaseous medium. No such
medium was postulated in the original problem.

'The kinetic energy of the rocket car is allegedly M * V^2 / 2, so it is increasing quadratically versus time' True or false?
False?
Does 'M' get smaller as the rocket car uses fuel?
What shape would the graph be if it does?

This is a reasonable point, but again a complication
that does not change the nature of the problem.
One can imagine an ion engine whose mass usage
is so low that it can be neglected for a limited time
that still fully exposes the quadratic/linear "dilemma".

'But the fuel consumed increases only linearly with time' True or false?
False?
As pressure and temperature builds inside a combustion chamber, does the fuel burn rate (thrust) also increase?

Another complication. The original problem is not
about startup characteristics of rocket motors.

Well that was fun. Please tell me my understanding of physics is phlawed

I'm not ready to say "flawed", but I wonder if you
plan to go into engineering. If so, you will have to
learn how to neglect certain phenomena and how
to decide that such neglect is appropriate.

Jon


"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message news:d9oXd.34$UW6.2012@news.uswest.net...
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?

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

 

[Robert Monsen]

sergiometra wrote:

Energy is a funny thing. You have to be careful about how you count it up.

The classic riddle here on s.e.b is 'given a capacitor charged up to V
volts, the energy is 1/2 * C * V^2. If you connect an equal uncharged
cap in parallel, the charge will equalize such that the voltage is 1/2
what it was. Thus, the energy is now 2 * (1/2 * C * (V/2)^2) = 1/2 what
it was before. Where did the energy go?'



if there is not resistance in circuit and cap, circuit will oscillate
forever;
otherwise energy dissipate in heat and oscillation eventually end
regards
sergio

Where is the oscillation coming from again? I know that an LC circuit
near absolute zero will oscillate, but I didn't say anything about an
inductor.

But you are right, the transfer of energy requires some dissipation.
That is the answer, the extra energy is dissipated. It's easier to see
if you put in resistors instead of wires between the caps (but wires are
just resistors with lower values, right?). Then, the integral of power
through the resistor from 0 to infinity gives the solution, which is 1/2
the original energy.

Start with cap A charged to V, connected to an uncharged cap B through
one terminal. The other terminal has a switch and a resistor. B is
uncharged. To the circuit, the two resistors are simply series caps, so
the total is 1/2 C. Thus, the voltage across the resistor at t is

V(t) = V * exp (-t*2/RC)

Thus, instantaneous power through the resistor is

P(t) = V(t)^2/R

The integral of this from 0 to infinity is

C*V^2/4 = 1/2 * (1/2 C * V^2)

--
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.

 

[Robert Monsen]

sergiometra wrote:

Here is a web page written by somebody who has been driven insane by
this problem:

http://nov55.com/enrgy.html

His solution is that kinetic energy is really mv, not 1/2 mv^2. He has
equations and everything. Pretty wierd stuff. Who knows?



I think he demonstrate the exact opposite;
if you jump from 2nd floor, speed at arrival is not double that a jumping
from 1st floor as it take less time from floor 1 to 0 than from 2 to 1 , but
energy is double as you need double energy to go up 2 floor than 1 , so
energy increase more rapidly than speed i/e Energy>mv;
regards
sergio

You didn't read the page. His thesis is that energy should be measured
as mv, not 1/2 mv^2. He claims that mv, and not 1/2mv^2 is conserved. He
uses rockets in a stationary reference frame to prove this (but doesn't
seem to take the kinetic energy of the fuel into account).

--
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.

 

[Robert Monsen]

Larry Brasfield wrote:

"John Larkin" <jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote
in message news:sa3t2113mifbj3e38eovdn74qc546dni4i@4ax.com...

On Tue, 8 Mar 2005 12:44:56 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:


"Robert Monsen" <rcsurname@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).


At low velocities, a rocket is a very inefficient source of
propulsion; at near-zero velocity, it's using its usual amount of fuel
but hardly delivering any kinetic energy to the vehicle. As velocity
increases, efficiency improves (or rather becomes less terrible) and
vehicle energy accumulates faster. That trend continues until you run
out of fuel.

In a system that goes from extremely inefficient to only rather
inefficient, it's not hard to shape the efficiency curve into a
quadratic. A cog railway can be nearly 100% efficient, so it will need
increasing amounts of fuel if it accelerates at constant gees, but
anywhere on the path it will be a lot more efficient than a rocket.

That make sense?



Yes. I should have known this little puzzle would
not take you folks long to sort out. The key is to
consider that starting condition. All the energy ends
up in the exhaust at the limit of zero rocket speed.
The problem gets a little easier to see if it involves
chucking cannonballs from a railway car.

Ok. I guess I'm feeling thick today, but it sounds like your solution is
to attach a reference frame to the accelerating rocket. I don't know
enough physics to determine if that's reasonable.

The rate of change of kinetic energy of the rocket when the rocket isn't
moving is still the time derivative of 1/2mv^2, which is

F^2*t/M

where M is the (assumed unchanging) mass of the rocket, and F is thrust.
This is a function of t, meaning that the rate of energy at t=0 is 0, so
all the energy goes into the fuel. That is nice, but what happens at
t=1? We are still increasing our 'rate of change' of energy...

If you were to say that, for example, the rate of change of energy of
the thrust (from the point of view of a stationary observer) was
decreasing with t, and the sum of these was constant, I'd buy that.

--
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.