| Topic: |
Science > Physics |
| User: |
"exponent137" |
| Date: |
11 Jan 2005 12:14:03 PM |
| Object: |
quantum rocket |
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
aproximately uniform acceleration of rocket.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
The question is: what is real acceleration, how to calculate it, or how
the acceleration is distributed under small time intervals.
Regards exponent137
.
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| User: "Old Man" |
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| Title: Re: quantum rocket |
11 Jan 2005 08:30:21 PM |
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This is a multi-part message in MIME format.
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charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
"exponent137" <exponent137@hotmail.com> wrote in message =
news:1105467243.057247.243720@f14g2000cwb.googlegroups.com...
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
aproximately uniform acceleration of rocket.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
=20
The question is: what is real acceleration, how to calculate it, or =
how
the acceleration is distributed under small time intervals.
Regards exponent137
With this ship, one could go any where in the universe in much less=20
than one lifetime. Presented below, are numerical results for a=20
(relativistic) constant acceleration (1 g) rocket trip wherein fuel mass =
usage is taken into account. In this calculation, the rocket mass=20
decreases exponentially with proper(ship)time. For propulsion, a=20
100% efficient laser is assumed,=20
whereby mass is converted to a non diverging light beam with 100%=20
efficiency. For a constant acceleration of 1 g, with initial ship mass, =
m0, the ship mass, m, declines as
m / m0 =3D exp(-g*T / c)
dt =3D dT*gamma
gamma =3D 1 / [1 - (v / c)^2]^(1 /2)
where T is proper (ship) time, t is earth time, and v is the ship speed. =
Conservation of energy requires, at any final ship speed
E =3D m0*c^2 =3D E(light) + E(ship)
=3D -p(light)*c + (m*c^2)*gamma
But, conservation of momentum requires=20
-p(light)=3D p(ship) =3D m*v*gamma,=20
and therefore,=20
v / c =3D [1 - (m / m0)^2] / [1 + (m / m0)^2]
It is clear that ship velocity and travel distance do not depend on the=20
value of the initial ship mass and that constant ship acceleration can=20
be maintained for unlimited time. However payload mass decreases=20
exponentially with time and is finite for all finite ship times, and =
payload=20
mass approaches, but never, reaches zero. From these equations, for=20
example, at 5 years ship time, the ship has traveled 85.2 light years =
and=20
ship speed is 0.9999*c. 99.44% of the initial ship mass has gone to=20
propellant energy and 0.56% remains as payload. Quantities for other=20
ship times are given in the table below. =20
Ship Time Earth Time Distance
T(years) t(years) V/C X(ly) Mass(m / m0)
0.0000 0.0000 0.0000 0.0000 1.0000
1.0000 1.1892 0.7767 0.5667 0.3545
2.0000 3.7762 0.9689 2.9331 0.1257
3.0000 10.8020 0.9960 9.8807 0.0445
4.0000 30.5254 0.9995 29.5764 0.0158
5.0000 86.1309 0.9999 85.1721 0.0056
6.0000 242.9817 1.0000 242.0194 0.0020
7.0000 685.4528 1.0000 684.4892 0.0007
8.0000 1933.6603 1.0000 1932.6963 0.0002
9.0000 5454.8478 1.0000 5453.8836 0.0001
10.0000 15388.1027 1.0000 15387.1385 0.0000
11.0000 43409.7728 1.0000 43408.8086 0.0000
12.0000 122458.7857 1.0000 122457.8215 0.0000
13.0000 345455.7113 1.0000 345454.7471 0.0000
14.0000 974529.1311 1.0000 974528.1669 0.0000
[Old Man]
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charset="iso-8859-1"
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2900.2523" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY>
<DIV>"exponent137" <<A=20
href=3D"mailto:exponent137@hotmail.com">exponent137@hotmail.com</A>> =
wrote in=20
message <A=20
href=3D"news:1105467243.057247.243720@f14g2000cwb.googlegroups.com">news:=
1105467243.057247.243720@f14g2000cwb.googlegroups.com</A>...</DIV>
<DIV> </DIV>
<DIV>>I ask for the following question:<BR>> We have rocket on =
photonic=20
propulsion. Photons fly out and they gives<BR>> aproximately uniform=20
acceleration of rocket.<BR>> In one case<BR>> 1. these photons are =
from=20
the laser<BR>> and in second case<BR>> 2. these photons are from =
the black=20
body.<BR>> In both cases direction of these photons is only in one=20
angle.<BR>> <BR>> The question is: what is real acceleration, how =
to=20
calculate it, or how<BR>> the acceleration is distributed under small =
time=20
intervals.<BR>> Regards exponent137</DIV>
<DIV><BR> </DIV>
<DIV>With this ship, one could go any where in the universe in much less =
</DIV>
<DIV>than one lifetime. Presented below, are numerical results for =
a=20
</DIV>
<DIV>(relativistic) constant acceleration (1 g) rocket trip wherein fuel =
mass=20
</DIV>
<DIV>usage is taken into account. In this calculation, the rocket mass =
</DIV>
<DIV>decreases exponentially with proper(ship)time. For =
propulsion, a=20
</DIV>
<DIV>100% efficient laser is assumed, <BR>whereby mass is converted to a =
non=20
diverging light beam with 100% <BR>efficiency. For a constant =
acceleration=20
of 1 g, with initial ship mass, </DIV>
<DIV>m0, the ship mass, m, declines as</DIV>
<DIV> </DIV>
<DIV>m / m0 =3D exp(-g*T / c)<BR>dt =3D dT*gamma<BR>gamma =3D 1 / [1 - =
(v / c)^2]^(1=20
/2)</DIV>
<DIV> </DIV>
<DIV>where T is proper (ship) time, t is earth time, and v is the ship =
speed.=20
<BR>Conservation of energy requires, at any final ship speed</DIV>
<DIV> </DIV>
<DIV>E =3D m0*c^2 =3D E(light) +=20
E(ship)<BR> =
=3D=20
-p(light)*c + (m*c^2)*gamma</DIV>
<DIV> </DIV>
<DIV>But, conservation of momentum requires </DIV>
<DIV> </DIV>
<DIV>-p(light)=3D p(ship) =3D m*v*gamma, </DIV>
<DIV> </DIV>
<DIV>and therefore, </DIV>
<DIV> </DIV>
<DIV>v / c =3D [1 - (m / m0)^2] / [1 + (m / m0)^2]</DIV>
<DIV><BR>It is clear that ship velocity and travel distance do not =
depend on the=20
<BR>value of the initial ship mass and that constant ship acceleration =
can=20
</DIV>
<DIV>be maintained for unlimited time. However payload mass =
decreases=20
</DIV>
<DIV>exponentially with time and is finite for all finite ship times, =
and=20
payload </DIV>
<DIV>mass approaches, but never, reaches zero. From these =
equations, for=20
</DIV>
<DIV>example, at 5 years ship time, the ship has traveled 85.2 light =
years and=20
</DIV>
<DIV>ship speed is 0.9999*c. 99.44% of the initial ship mass has =
gone to=20
</DIV>
<DIV>propellant energy and 0.56% remains as payload. Quantities =
for other=20
</DIV>
<DIV>ship times are given in the table below. </DIV>
<DIV> </DIV>
<DIV> Ship Time Earth=20
Time &nb=
sp; =20
Distance<BR> T(years) =20
t(years) =20
V/C X(ly) =
Mass(m /=20
m0)<BR> =
0.0000 =20
0.0000 =
0.0000 =20
0.0000 =
1.0000<BR> =20
1.0000 =
1.1892 =20
0.7767 =
0.5667 =20
0.3545<BR> =
2.0000 =20
3.7762 =
0.9689 =20
2.9331 =
0.1257<BR> =20
3.0000 10.8020 =20
0.9960 =
9.8807 =20
0.0445<BR> 4.0000 =20
30.5254 0.9995 =20
29.5764 =
0.0158<BR> =20
5.0000 86.1309 =20
0.9999 85.1721 =20
0.0056<BR> 6.0000 =20
242.9817 1.0000 =20
242.0194 =
0.0020<BR> =20
7.0000 685.4528 =20
1.0000 684.4892 =20
0.0007<BR> 8.0000 =20
1933.6603 1.0000 =20
1932.6963 =
0.0002<BR> =20
9.0000 5454.8478 =
1.0000 =20
5453.8836 =
0.0001<BR> =20
10.0000 15388.1027 1.0000 =20
15387.1385 =
0.0000<BR> =20
11.0000 43409.7728 1.0000 =20
43408.8086 =
0.0000<BR> =20
12.0000 122458.7857 1.0000=20
122457.8215 =
0.0000<BR> =20
13.0000 345455.7113 1.0000=20
345454.7471 =
0.0000<BR> =20
14.0000 974529.1311 1.0000=20
974528.1669 0.0000</DIV>
<DIV> </DIV>
<DIV>[Old Man]</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV></BODY></HTML>
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| User: "tj Frazir" |
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| Title: Re: quantum rocket |
11 Jan 2005 10:32:46 PM |
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Your trying to push a train with a fly .
good luck.
If you do however ,, we nead anoter rower..can you row a boat ??
xue01.jpg
Address:http://www.gakei.com/xue/xue01.jpg Changed:8:34 AM on Sunday,
October 31, 2004
Hu Longg Winded
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| User: "Uncle Al" |
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| Title: Re: quantum rocket |
11 Jan 2005 03:58:37 PM |
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exponent137 wrote:
I ask for the following question:
We have rocket on photonic propulsion.
Note the perceptible background sucking sound.
Photons fly out and they gives
aproximately uniform acceleration of rocket.
Already in deep mud. Do you have a goal? Are you going to approach
that goal along a geodesic or mimimum action path?
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
Learn about collmation, Bose-Eisntein condensate, coherence in time
and space. What is the temp of a laser population inversion? How hot
a blackbody can you contain? How do you direct gamma photons in the
uninteresting 2(pi) steradians of blackbody emission direction?
You'd better learn how to take a cosine.
The question is: what is real acceleration, how to calculate it, or how
the acceleration is distributed under small time intervals.
Momentum is conserved according to Newton. Nothing else matters until
relativistic speeds.
"The Starflight Handbook," Mallove and Matloff, ISBN 0-471-61912-4
Chapter 5, Technical Note 5-2, "The Pressure of Laser Photons," p. 75.
I'd give you the stated equation and its derivation for starship
accleration, but then it would have no value to you. Look it up, git,
or derive it yourself. If you care enough to know then you care
enough to find out. Theory of continuous rocket guidance was used by
Fischer Black, Myron Scholes, and Robert Merton ((building on Kiyosi
Ito) in their Nobel Prize-winning option pricing economic theory
(except for the stiff - no postumous Nobels). Look it up. (The
original Black-Scholes paper - the Nobel stuff - was rejected by all
journals to which it was submitted. Their company thereafter,
Long-Term Capital Management, made 24% profit ($312 million) its first
year, 1994.)
A rocket's trajectory is precisely tracked and necessary minute
adjustments made. It is not enough to know where a rocket is every
second or so. One needs to know exactly where a rocket is
continuously. A rocket's trajectory is computed in an uninterrupted
stepless continuum.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "Sam Wormley" |
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| Title: Re: quantum rocket |
11 Jan 2005 12:30:15 PM |
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exponent137 wrote:
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
approximately uniform acceleration of rocket.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
The question is: what is real acceleration, how to calculate it, or how
the acceleration is distributed under small time intervals.
Conservation of Momentum
http://scienceworld.wolfram.com/physics/ConservationofMomentum.html
Photon momentum = p = h/lambda
Can you count photons?
.
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| User: "" |
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| Title: Re: quantum rocket |
11 Jan 2005 05:05:56 PM |
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In article <XWUEd.1429$OF5.629@attbi_s52>, Sam Wormley <swormley1@mchsi.com> writes:
exponent137 wrote:
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
approximately uniform acceleration of rocket.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
The question is: what is real acceleration, how to calculate it, or how
the acceleration is distributed under small time intervals.
Conservation of Momentum
http://scienceworld.wolfram.com/physics/ConservationofMomentum.html
Photon momentum = p = h/lambda
Can you count photons?
There is no need for counting, no more so than one needs to count
molecules in standard rocket exhaust.
The emitted energy and momentum are related by
p = E/c
thus
dp/dt = (1/c)*dE/dt
Since dp/dt is the force (thrust) F, and dE/dt is the rediated power P,
you have
F = P/c
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "John Schoenfeld" |
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| Title: Re: quantum rocket |
11 Jan 2005 08:45:45 PM |
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exponent137 wrote:
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
aproximately uniform acceleration of rocket.
To levitate a 5000kg grounded device you require approximately 50 000 N
of gravitationally opposing force. The reactive force of your rocket is
the speed of light divided by the emitted radiation power reducing the
net electrical power per newton of net force to roughly 300 megawatts.
The power to sustain levitation of this device is not more than 150 000
terrawatts (roughly equivalent to the daily output of 70 million
medium-sized nuclear power plants). Since your rocket would vaporize
all known observers at lift-off your rocket can never fly.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
The question is: what is real acceleration, how to calculate it, or
how
the acceleration is distributed under small time intervals.
Regards exponent137
.
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| User: "Old Man" |
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| Title: Re: quantum rocket |
11 Jan 2005 10:26:28 PM |
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"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
news:1105497945.497595.14800@z14g2000cwz.googlegroups.com...
exponent137 wrote:
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
aproximately uniform acceleration of rocket.
To levitate a 5000kg grounded device you require approximately 50 000 N
of gravitationally opposing force. The reactive force of your rocket is
the speed of light divided by the emitted radiation power reducing the
net electrical power per newton of net force to roughly 300 megawatts.
The power to sustain levitation of this device is not more than 150 000
terrawatts (roughly equivalent to the daily output of 70 million
medium-sized nuclear power plants). Since your rocket would vaporize
all known observers at lift-off your rocket can never fly.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
The question is: what is real acceleration, how to calculate it, or
how
the acceleration is distributed under small time intervals.
Regards exponent137
Put some calculation where your big mouth is.
With this ship, one could go any where in the universe in much less
than one lifetime. Presented below, are numerical results for a
(relativistic) constant acceleration (1 g) rocket trip wherein fuel mass
usage is taken into account. In this calculation, the rocket mass
decreases exponentially with proper(ship)time. For propulsion, a
100% efficient laser is assumed,
whereby mass is converted to a non diverging light beam with 100%
efficiency. For a constant acceleration of 1 g, with initial ship mass,
m0, the ship mass, m, declines as
m / m0 = exp(-g*T / c)
dt = dT*gamma
gamma = 1 / [1 - (v / c)^2]^(1 /2)
where T is proper (ship) time, t is earth time, and v is the ship speed.
Conservation of energy requires, at any final ship speed
E = m0*c^2 = E(light) + E(ship)
= -p(light)*c + (m*c^2)*gamma
But, conservation of momentum requires
-p(light)= p(ship) = m*v*gamma,
and therefore,
v / c = [1 - (m / m0)^2] / [1 + (m / m0)^2]
It is clear that ship velocity and travel distance do not depend on the
value of the initial ship mass and that constant ship acceleration can
be maintained for unlimited time. However payload mass decreases
exponentially with time and is finite for all finite ship times, and payload
mass approaches, but never, reaches zero. From these equations, for
example, at 5 years ship time, the ship has traveled 85.2 light years and
ship speed is 0.9999*c. 99.44% of the initial ship mass has gone to
propellant energy and 0.56% remains as payload. Quantities for other
ship times are given in the table below.
Ship Time Earth Time Distance
T(years) t(years) V/C X(ly) Mass(m / m0)
0.0000 0.0000 0.0000 0.0000 1.0000
1.0000 1.1892 0.7767 0.5667 0.3545
2.0000 3.7762 0.9689 2.9331 0.1257
3.0000 10.8020 0.9960 9.8807 0.0445
4.0000 30.5254 0.9995 29.5764 0.0158
5.0000 86.1309 0.9999 85.1721 0.0056
6.0000 242.9817 1.0000 242.0194 0.0020
7.0000 685.4528 1.0000 684.4892 0.0007
8.0000 1933.6603 1.0000 1932.6963 0.0002
9.0000 5454.8478 1.0000 5453.8836 0.0001
10.0000 15388.1027 1.0000 15387.1385 0.0000
11.0000 43409.7728 1.0000 43408.8086 0.0000
12.0000 122458.7857 1.0000 122457.8215 0.0000
13.0000 345455.7113 1.0000 345454.7471 0.0000
14.0000 974529.1311 1.0000 974528.1669 0.0000
[Old Man]
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| User: "John Schoenfeld" |
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| Title: Re: quantum rocket |
12 Jan 2005 12:59:43 PM |
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Old Man wrote:
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
To levitate a 5000kg grounded device you require approximately 50
000 N
of gravitationally opposing force. The reactive force of your
rocket is
the speed of light divided by the emitted radiation power reducing
the
net electrical power per newton of net force to roughly 300
megawatts.
The power to sustain levitation of this device is not more than 150
000
terrawatts (roughly equivalent to the daily output of 70 million
medium-sized nuclear power plants). Since your rocket would
vaporize
all known observers at lift-off your rocket can never fly.
Put some calculation where your big mouth is.
With this ship, one could go any where in the universe in much less
than one lifetime.
If technology able to produce a photon drive was known, chances are
that a physics allowing much more efficient modes of propulsion would
also be known.
Presented below, are numerical results for a
(relativistic) constant acceleration (1 g) rocket trip wherein fuel
mass
usage is taken into account. In this calculation, the rocket mass
decreases exponentially with proper(ship)time. For propulsion, a
100% efficient laser is assumed,
whereby mass is converted to a non diverging light beam with 100%
efficiency. For a constant acceleration of 1 g, with initial ship
mass,
m0, the ship mass, m, declines as
m / m0 = exp(-g*T / c)
dt = dT*gamma
gamma = 1 / [1 - (v / c)^2]^(1 /2)
where T is proper (ship) time, t is earth time, and v is the ship
speed.
Conservation of energy requires, at any final ship speed
E = m0*c^2 = E(light) + E(ship)
= -p(light)*c + (m*c^2)*gamma
But, conservation of momentum requires
-p(light)= p(ship) = m*v*gamma,
and therefore,
v / c = [1 - (m / m0)^2] / [1 + (m / m0)^2]
It is clear that ship velocity and travel distance do not depend on
the
value of the initial ship mass and that constant ship acceleration
can
be maintained for unlimited time. However payload mass decreases
exponentially with time and is finite for all finite ship times, and
payload
mass approaches, but never, reaches zero. From these equations, for
example, at 5 years ship time, the ship has traveled 85.2 light years
and
ship speed is 0.9999*c. 99.44% of the initial ship mass has gone to
propellant energy and 0.56% remains as payload. Quantities for other
ship times are given in the table below.
Ship Time Earth Time Distance
T(years) t(years) V/C X(ly) Mass(m / m0)
0.0000 0.0000 0.0000 0.0000 1.0000
1.0000 1.1892 0.7767 0.5667 0.3545
2.0000 3.7762 0.9689 2.9331 0.1257
3.0000 10.8020 0.9960 9.8807 0.0445
4.0000 30.5254 0.9995 29.5764 0.0158
5.0000 86.1309 0.9999 85.1721 0.0056
6.0000 242.9817 1.0000 242.0194 0.0020
7.0000 685.4528 1.0000 684.4892 0.0007
8.0000 1933.6603 1.0000 1932.6963 0.0002
9.0000 5454.8478 1.0000 5453.8836 0.0001
10.0000 15388.1027 1.0000 15387.1385 0.0000
11.0000 43409.7728 1.0000 43408.8086 0.0000
12.0000 122458.7857 1.0000 122457.8215 0.0000
13.0000 345455.7113 1.0000 345454.7471 0.0000
14.0000 974529.1311 1.0000 974528.1669 0.0000
[Old Man]
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| User: "Old Man" |
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| Title: Re: quantum rocket |
12 Jan 2005 02:09:46 PM |
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"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
news:1105556383.362685.240050@f14g2000cwb.googlegroups.com...
Old Man wrote:
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
To levitate a 5000kg grounded device you require approximately 50
000 N
of gravitationally opposing force. The reactive force of your
rocket is
the speed of light divided by the emitted radiation power reducing
the
net electrical power per newton of net force to roughly 300
megawatts.
The power to sustain levitation of this device is not more than 150
000
terrawatts (roughly equivalent to the daily output of 70 million
medium-sized nuclear power plants). Since your rocket would
vaporize
all known observers at lift-off your rocket can never fly.
Put some calculation where your big mouth is.
With this ship, one could go any where in the universe in much less
than one lifetime.
If technology able to produce a photon drive was known, chances are
that a physics allowing much more efficient modes of propulsion would
also be known.
We already have the technology. A laser isn't required.
All one needs is a rather large parabolic reflector with
a heat source, such as a nuclear reactor, at it's focus:
F = P / c, regardless of wavelength.
Except for the small loss of neutrino energy and of beam
divergence (the dimensions of the reactor must be much
smaller than those of the reflector), that's near 100 %
conversion efficiency of mass (nuclear binding energy)
to directed electromagnetic radiation.
[Old Man]
(relativistic) constant acceleration (1 g) rocket trip wherein fuel
mass
usage is taken into account. In this calculation, the rocket mass
decreases exponentially with proper(ship)time. For propulsion, a
100% efficient laser is assumed,
whereby mass is converted to a non diverging light beam with 100%
efficiency. For a constant acceleration of 1 g, with initial ship
mass,
m0, the ship mass, m, declines as
m / m0 = exp(-g*T / c)
dt = dT*gamma
gamma = 1 / [1 - (v / c)^2]^(1 /2)
where T is proper (ship) time, t is earth time, and v is the ship
speed.
Conservation of energy requires, at any final ship speed
E = m0*c^2 = E(light) + E(ship)
= -p(light)*c + (m*c^2)*gamma
But, conservation of momentum requires
-p(light)= p(ship) = m*v*gamma,
and therefore,
v / c = [1 - (m / m0)^2] / [1 + (m / m0)^2]
It is clear that ship velocity and travel distance do not depend on
the
value of the initial ship mass and that constant ship acceleration
can
be maintained for unlimited time. However payload mass decreases
exponentially with time and is finite for all finite ship times, and
payload
mass approaches, but never, reaches zero. From these equations, for
example, at 5 years ship time, the ship has traveled 85.2 light years
and
ship speed is 0.9999*c. 99.44% of the initial ship mass has gone to
propellant energy and 0.56% remains as payload. Quantities for other
ship times are given in the table below.
Ship Time Earth Time Distance
T(years) t(years) V/C X(ly) Mass(m / m0)
0.0000 0.0000 0.0000 0.0000 1.0000
1.0000 1.1892 0.7767 0.5667 0.3545
2.0000 3.7762 0.9689 2.9331 0.1257
3.0000 10.8020 0.9960 9.8807 0.0445
4.0000 30.5254 0.9995 29.5764 0.0158
5.0000 86.1309 0.9999 85.1721 0.0056
6.0000 242.9817 1.0000 242.0194 0.0020
7.0000 685.4528 1.0000 684.4892 0.0007
8.0000 1933.6603 1.0000 1932.6963 0.0002
9.0000 5454.8478 1.0000 5453.8836 0.0001
10.0000 15388.1027 1.0000 15387.1385 0.0000
11.0000 43409.7728 1.0000 43408.8086 0.0000
12.0000 122458.7857 1.0000 122457.8215 0.0000
13.0000 345455.7113 1.0000 345454.7471 0.0000
14.0000 974529.1311 1.0000 974528.1669 0.0000
[Old Man]
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| User: "John Schoenfeld" |
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| Title: Re: quantum rocket |
12 Jan 2005 03:43:13 PM |
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Old Man wrote:
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
news:1105556383.362685.240050@f14g2000cwb.googlegroups.com...
Old Man wrote:
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
To levitate a 5000kg grounded device you require approximately
50
000 N
of gravitationally opposing force. The reactive force of your
rocket is
the speed of light divided by the emitted radiation power
reducing
the
net electrical power per newton of net force to roughly 300
megawatts.
The power to sustain levitation of this device is not more than
150
000
terrawatts (roughly equivalent to the daily output of 70 million
medium-sized nuclear power plants). Since your rocket would
vaporize
all known observers at lift-off your rocket can never fly.
Put some calculation where your big mouth is.
With this ship, one could go any where in the universe in much
less
than one lifetime.
If technology able to produce a photon drive was known, chances are
that a physics allowing much more efficient modes of propulsion
would
also be known.
We already have the technology. A laser isn't required.
All one needs is a rather large parabolic reflector with
a heat source, such as a nuclear reactor, at it's focus:
F = P / c, regardless of wavelength.
In that case, generating propellant photons is trivially achieved with
a yagi antenna. However, the difference between that and a photon drive
is the difference between an Aristotlean bicycle and a scramjet. As
stated previously, radiative levitation of a 5 tonne vehicle requires
the power of 70 000 000 nuclear power plants.
Except for the small loss of neutrino energy and of beam
divergence (the dimensions of the reactor must be much
smaller than those of the reflector), that's near 100 %
conversion efficiency of mass (nuclear binding energy)
to directed electromagnetic radiation.
It is likely that superior propulsion technology would be available
before such power could be generated by a mobile device.
[Old Man]
(relativistic) constant acceleration (1 g) rocket trip wherein
fuel
mass
usage is taken into account. In this calculation, the rocket mass
decreases exponentially with proper(ship)time. For propulsion, a
100% efficient laser is assumed,
whereby mass is converted to a non diverging light beam with 100%
efficiency. For a constant acceleration of 1 g, with initial ship
mass,
m0, the ship mass, m, declines as
m / m0 = exp(-g*T / c)
dt = dT*gamma
gamma = 1 / [1 - (v / c)^2]^(1 /2)
where T is proper (ship) time, t is earth time, and v is the ship
speed.
Conservation of energy requires, at any final ship speed
E = m0*c^2 = E(light) + E(ship)
= -p(light)*c + (m*c^2)*gamma
But, conservation of momentum requires
-p(light)= p(ship) = m*v*gamma,
and therefore,
v / c = [1 - (m / m0)^2] / [1 + (m / m0)^2]
It is clear that ship velocity and travel distance do not depend
on
the
value of the initial ship mass and that constant ship acceleration
can
be maintained for unlimited time. However payload mass decreases
exponentially with time and is finite for all finite ship times,
and
payload
mass approaches, but never, reaches zero. From these equations,
for
example, at 5 years ship time, the ship has traveled 85.2 light
years
and
ship speed is 0.9999*c. 99.44% of the initial ship mass has gone
to
propellant energy and 0.56% remains as payload. Quantities for
other
ship times are given in the table below.
Ship Time Earth Time Distance
T(years) t(years) V/C X(ly) Mass(m / m0)
0.0000 0.0000 0.0000 0.0000 1.0000
1.0000 1.1892 0.7767 0.5667 0.3545
2.0000 3.7762 0.9689 2.9331 0.1257
3.0000 10.8020 0.9960 9.8807 0.0445
4.0000 30.5254 0.9995 29.5764 0.0158
5.0000 86.1309 0.9999 85.1721 0.0056
6.0000 242.9817 1.0000 242.0194 0.0020
7.0000 685.4528 1.0000 684.4892 0.0007
8.0000 1933.6603 1.0000 1932.6963 0.0002
9.0000 5454.8478 1.0000 5453.8836 0.0001
10.0000 15388.1027 1.0000 15387.1385 0.0000
11.0000 43409.7728 1.0000 43408.8086 0.0000
12.0000 122458.7857 1.0000 122457.8215 0.0000
13.0000 345455.7113 1.0000 345454.7471 0.0000
14.0000 974529.1311 1.0000 974528.1669 0.0000
[Old Man]
.
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| User: "Old Man" |
|
| Title: Re: quantum rocket |
12 Jan 2005 06:33:31 PM |
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|
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
news:1105566193.518684.287530@z14g2000cwz.googlegroups.com...
Old Man wrote:
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
news:1105556383.362685.240050@f14g2000cwb.googlegroups.com...
Old Man wrote:
"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
To levitate a 5000kg grounded device you require approximately
50
000 N
of gravitationally opposing force. The reactive force of your
rocket is
the speed of light divided by the emitted radiation power
reducing
the
net electrical power per newton of net force to roughly 300
megawatts.
The power to sustain levitation of this device is not more than
150
000
terrawatts (roughly equivalent to the daily output of 70 million
medium-sized nuclear power plants). Since your rocket would
vaporize
all known observers at lift-off your rocket can never fly.
Put some calculation where your big mouth is.
With this ship, one could go any where in the universe in much
less
than one lifetime.
If technology able to produce a photon drive was known, chances are
that a physics allowing much more efficient modes of propulsion
would
also be known.
We already have the technology. A laser isn't required.
All one needs is a rather large parabolic reflector with
a heat source, such as a nuclear reactor, at it's focus:
F = P / c, regardless of wavelength.
In that case, generating propellant photons is trivially achieved with
a yagi antenna. However, the difference between that and a photon drive
is the difference between an Aristotlean bicycle and a scramjet. As
stated previously, radiative levitation of a 5 tonne vehicle requires
the power of 70 000 000 nuclear power plants.
In terms of mass-energy, a "photon drive" is much more
efficient than a "mass drive" (thermal ejection of atoms) by
many orders of magnitude. However, If one accelerated
and ejected 50 MeV electrons, the mass-energy efficiency
would be 99 %, but, in order to maintain charge neutrality
and efficiency, 50 MeV positrons would have to be ejected
in the same direction as that of the electrons. The same
results can be derived via 100 GeV hydrogen atoms or
alternatively, 100 GeV / nucleon for heavier atoms.
Except for the small loss of neutrino energy and of beam
divergence (the dimensions of the reactor must be much
smaller than those of the reflector), that's near 100 %
conversion efficiency of mass (nuclear binding energy)
to directed electromagnetic radiation.
It is likely that superior propulsion technology would be available
before such power could be generated by a mobile device.
The "mobile device" Old Man had in mind would have
the mass of a small moon, or possibly, the Moon itself.
Look at the table that Old Man provided:
With a final ship mass (payload) at 0.01 % of initial
mass, 9 years ship time gets 5454 light years, but
you still have to slow down. So, after 18 years ship
time, and 10,908 light years, the payload is 1 x 10^(-8)
that of the initial ship mass.
That sounds terrible, but a thermal mass drive wouldn't
get 1 light year in a 100 years ship time.
[Old Man]
[Old Man]
(relativistic) constant acceleration (1 g) rocket trip wherein
fuel
mass
usage is taken into account. In this calculation, the rocket mass
decreases exponentially with proper(ship)time. For propulsion, a
100% efficient laser is assumed,
whereby mass is converted to a non diverging light beam with 100%
efficiency. For a constant acceleration of 1 g, with initial ship
mass,
m0, the ship mass, m, declines as
m / m0 = exp(-g*T / c)
dt = dT*gamma
gamma = 1 / [1 - (v / c)^2]^(1 /2)
where T is proper (ship) time, t is earth time, and v is the ship
speed.
Conservation of energy requires, at any final ship speed
E = m0*c^2 = E(light) + E(ship)
= -p(light)*c + (m*c^2)*gamma
But, conservation of momentum requires
-p(light)= p(ship) = m*v*gamma,
and therefore,
v / c = [1 - (m / m0)^2] / [1 + (m / m0)^2]
It is clear that ship velocity and travel distance do not depend
on
the
value of the initial ship mass and that constant ship acceleration
can
be maintained for unlimited time. However payload mass decreases
exponentially with time and is finite for all finite ship times,
and
payload
mass approaches, but never, reaches zero. From these equations,
for
example, at 5 years ship time, the ship has traveled 85.2 light
years
and
ship speed is 0.9999*c. 99.44% of the initial ship mass has gone
to
propellant energy and 0.56% remains as payload. Quantities for
other
ship times are given in the table below.
Ship Time Earth Time Distance
T(years) t(years) V/C X(ly) Mass(m / m0)
0.0000 0.0000 0.0000 0.0000 1.0000
1.0000 1.1892 0.7767 0.5667 0.3545
2.0000 3.7762 0.9689 2.9331 0.1257
3.0000 10.8020 0.9960 9.8807 0.0445
4.0000 30.5254 0.9995 29.5764 0.0158
5.0000 86.1309 0.9999 85.1721 0.0056
6.0000 242.9817 1.0000 242.0194 0.0020
7.0000 685.4528 1.0000 684.4892 0.0007
8.0000 1933.6603 1.0000 1932.6963 0.0002
9.0000 5454.8478 1.0000 5453.8836 0.0001
10.0000 15388.1027 1.0000 15387.1385 0.0000
11.0000 43409.7728 1.0000 43408.8086 0.0000
12.0000 122458.7857 1.0000 122457.8215 0.0000
13.0000 345455.7113 1.0000 345454.7471 0.0000
14.0000 974529.1311 1.0000 974528.1669 0.0000
[Old Man]
.
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| User: "tj Frazir" |
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| Title: Re: quantum rocket |
12 Jan 2005 01:13:28 AM |
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What do you call the white and back painted squares that spin in a bulb
via sunlight ?
A laser shinning on the white side would allso spin it.
Spin a 50 foot frozzen disk of compressed steel up to 1/2 over a few
years because motor controles cant spin a disk as fast as a laser can .
Suspend the disk as a solid supperconductor over liquid n . and
spin it with 2 laser beams to hold its position.. like the sound
berior as the
disk spins its taking up more space making the low lower at the same
time centrifical force is trying to pull it apart..to point ,,at 1/2 C
the gravity is so intence the centrifica force CANT pull it apart and
it has no moe friction.
Once its past that point you can hold it in a vice and squeeze and
controle it.
Evrything will fall twards the low.
lower than the earth ,,dirt falls up to hit it.
place is a the center of mass on a ship and it wount move ,,but place
it near the top and more of the mass of the ship is falling up than is
falling down ,,evrything falls twards the low.
Its a fact of quantom physics ,,a disk at atoms speeds.
Dont call ME WRONG unless your ready to back it up.
I will use quantom ,,spin an atom faster and gain mass is mass taking
up more space in motion so is a larger low.
Its the ame damb thing in prorptions tat would hurt a PC let alone yer
head.
goo forgot p0000246.jpg
Address:http://dell1500sc.nmefc.gov.cn/product/jidi/tupian/images/s-xuelong/p0000246.jpg
Changed:2:23 PM on Tuesday, September 24, 2002
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| User: "tj Frazir" |
|
| Title: Re: quantum rocket |
12 Jan 2005 01:12:51 AM |
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|
What do you call the white and back painted squares that spin in a bulb
via sunlight ?
A laser shinning on the white side would allso spin it.
Spin a 50 foot frozzen disk of compressed steel up to 1/2 over a few
years because motor controles cant spin a disk as fast as a laser can .
Suspend the disk as a solid supperconductor over liquid n . and
spin it with 2 laser beams to hold its position.. like the sound
berior as the
disk spins its taking up more space making the low lower at the same
time centrifical force is trying to pull it apart..to point ,,at 1/2 C
the gravity is so intence the centrifica force CANT pull it apart and
it has no moe friction.
Once its past that point you can hold it in a vice and squeeze and
controle it.
Evrything will fall twards the low.
lower than the earth ,,dirt falls up to hit it.
place is a the center of mass on a ship and it wount move ,,but place
it near the top and more of the mass of the ship is falling up than is
falling down ,,evrything falls twards the low.
Its a fact of quantom physics ,,a disk at atoms speeds.
Dont call ME WRONG unless your ready to back it up.
I will use quantom ,,spin an atom faster and gain mass is mass taking
up more space in motion so is a larger low.
Its the ame damb thing in prorptions tat would hurt a PC let alone yer
head.
.
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| User: "Mark Martin" |
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| Title: Re: quantum rocket |
11 Jan 2005 12:30:46 PM |
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exponent137 wrote:
I ask for the following question:
We have rocket on photonic propulsion. Photons fly out and they gives
aproximately uniform acceleration of rocket.
In one case
1. these photons are from the laser
and in second case
2. these photons are from the black body.
In both cases direction of these photons is only in one angle.
The question is: what is real acceleration, how to calculate it, or
how
the acceleration is distributed under small time intervals.
Momentum is conserved, so whatever momentum is carried by the exhaust
photons, is balanced by that of the spacecraft. In the case of the
laser, the momenta of all the photons will be, in principle, equal. For
the blackbody there will be an average momentum per photon. These
momenta can be calculated from easily Googled formulae.
-Mark Martin
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| User: "exponent137" |
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| Title: Re: quantum rocket |
11 Jan 2005 12:53:07 PM |
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In classical physics the acceleration is uniform. (Mass of the photons
is very small to mass of the rocket). But photons fly out
statistically. It can fly in one moment, there is some moment nothing
and then it fly again.
So they can give uniform acceleration, if we look at large time
intervals. But what happens in small time intervals? Let us imagine
that photons fly out very rarely.
In clasicall physics there is linear dependence of velocity to time
(for rocket).
Is maybe in quantum physis there stairs shape of curve (velocity to
time), where every stair is differently long.
As first, it is not important for me, if there is laser of black body.
What is shape of velocity to time curve?
.
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| User: "Mark Martin" |
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| Title: Re: quantum rocket |
11 Jan 2005 01:08:38 PM |
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exponent137 wrote:
In classical physics the acceleration is uniform. (Mass of the
photons
is very small to mass of the rocket). But photons fly out
statistically. It can fly in one moment, there is some moment nothing
and then it fly again.
So they can give uniform acceleration, if we look at large time
intervals. But what happens in small time intervals? Let us imagine
that photons fly out very rarely.
In clasicall physics there is linear dependence of velocity to time
(for rocket).
Is maybe in quantum physis there stairs shape of curve (velocity to
time), where every stair is differently long.
As first, it is not important for me, if there is laser of black
body.
What is shape of velocity to time curve?
This is true also for classically based rockets. They burn chemicals,
and what shoots out as exhaust is a large number of discrete masses,
the molecules yielded in the reaction between fuel & oxidizer. This
isn't really all that hard a thing to figure out. Just say to yourself,
"Each photon carries a unit of momentum. Each photon to eject out the
rear imparts that much momentum to the spacecraft, an impulse. How
often the spacecraft receives a unit impulse is how often a photon is
ejected." Take it from there.
-Mark Martin
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| User: "exponent137" |
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| Title: Re: quantum rocket |
11 Jan 2005 01:26:07 PM |
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So the next question is:
Does this photon gives larger momentum to rocket in time zero or in
time, which is function of wave lenght. If it gives momentum in time
zero then we obtain (velocity to time curve) in a shape of stairs.
So the next simplified form of question is: We have rocket, from which
fly one photon. This photon give momentum to the rocket. What is curve
velocity to time (or momentum to time)
Next simplification of question is: we have standstill free electron,
which absorb part of energy of one photon. (This is compton effect.)
What is time dependence of velocity to time for this electron in the
moments of absorbtion?
Regards Janko
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| User: "LingChow" |
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| Title: Re: quantum rocket |
11 Jan 2005 03:35:08 PM |
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"exponent137" <exponent137@hotmail.com> wrote in message
news:1105471567.002547.302030@z14g2000cwz.googlegroups.com...
So the next question is:
Does this photon gives larger momentum to rocket in time zero or in
time, which is function of wave lenght. If it gives momentum in time
zero then we obtain (velocity to time curve) in a shape of stairs.
So the next simplified form of question is: We have rocket, from which
fly one photon. This photon give momentum to the rocket. What is curve
velocity to time (or momentum to time)
you have before, and then after. (the photon fly out) you can calculate
both of these.
the time inbetween, or as the photon leaves the Black Body, is very small
about 0.001 pico seconds
Next simplification of question is: we have standstill free electron,
which absorb part of energy of one photon. (This is compton effect.)
What is time dependence of velocity to time for this electron in the
moments of absorbtion?
how did this come from Rockets?
Regards Janko
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