Wednesday, December 26, 2012


Action Causes Reaction.

Let a small mass of high velocity exhaust particles 

exit a rocket in one direction;

subsequent momentum increases speed

of the rocket's large mass in opposite direction.

Momentum is mass times velocity, m × v.

Momentum is a conserved quantity.
Within a closed system of interacting objects,
total momentum of entire system remains constant.

When object A collides with object B in an isolated system,
total momentum of the two objects before the collision
is the same as the total momentum after the collision.
Any momentum lost by object A goes to object B.
Recall momentum equilibrium:
P = M × V = m × v = p
Product of large mass × slow velocity will equal product of small mass × fast velocity.

This momentum relationship between fast moving fuel particles and a slow moving spaceship can be expressed:
Mship × Vship = mfuel × vfuel
Thought Experiment (TE) assumes following items:
1) For every second of powered flight, space vessel's speed increases by 10 meters per second (m/s) . Thus, Vship is a constant 10 m/s. This value coincides with g, acceleration due to gravity near Earth’s surface. Thus, TE assumes our notional spaceship can expel enough high speed particles to provide gravity equivalence throughout the flight. This simulates Earth gravity for ship and contents.
2) Let the fuel consumption be one gram of particles for every second of powered space flight. I.e., set mfuel to a constant 1.0 gram.
3) Assume spaceship's propulsion uses an onboard particle accelerator to give exhaust fuel particles extremely high speeds. Since current devices easily accelerate particles to 99% of light speed, our thought experiment will eventually sustain fuel exhaust velocity, vfuel, to near light speed, c.
4) Rename the vfuel term as "vExh" to more clearly indicate exhaust velocity of particles exiting vessel. We choose to independently vary vExh by incrementally increasing from row to row in the following tables. Thus, vExh is our Independent Variable (IV).
MShip × VShip = mfuel × vExh
Given following terms:
  • Mship, ship mass, in grams
  • Vship,  ship velocity, in meters/sec
  • mfuel, exhaust particles, in grams.
  • vExh,  exhaust particle speed in m/s
For consistency, TABLE-1 uses grams and meters per second (m/s) on both sides of momentum exchange equation.
1,000,000 gm10 m/s1.0 gm10,000,000 m/s
2,000,000 gm
10 m/s
1.0 gm
20,000,000 m/s
3,000,000 gm
10 m/s
1.0 gm
30,000,000 m/s

Solve for Mship  to determine how much ship's mass can be propelled 10 m/s by a given mass of high speed fuel particles in opposite direction. Since Mship  depends on the value of vExh, Mship is our Dependent Variable (DV).
Mship = mfuel × vExh


Mship = 1.0 gm × 30,000,000 m/s

10 m/sec
Mship =
3,000,000 gm
Momentum Exchange.
Thus, total momentum of a system of objects is "conserved"; total amount of mass times velocity is static. To further understand momentum conservation, consider Newton's Law of Momentum Conservation. When two objects collide, the forces acting between the two objects are equal in magnitude and opposite in direction. In equation form, m*v = -M * V.
The forces between the two objects are equal in magnitude and opposite in direction, which leads to Newton's Law of Momentum Conservation. In a collision, the momentum change of object A is equal and opposite to the momentum change of object B. That is, the momentum lost by object A is equal to the momentum gained by object B. In a collision between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object A loses 75 units of momentum, then object B gains 75 units of momentum. Yet, the total momentum of the two objects (object A plus object B) is the same before the collision as it is after the collision; the total momentum of the system (the collection of two objects) is conserved.
For any collision occurring in an isolated system, momentum is conserved - the total amount of momentum of the collection of objects in the system is the same before the collision as after the collision.
This exchange can be readily observed by watching billiard balls collide. 
Table-1 calculations are straight forward;
however, two terms, vExh and MShip, have a lot of inconvenient digits. simplify this process by mixing some units.

Express vExh as dc × c
"decimal c" times light speed
For convenience, mix different velocity units. In particular, we'd like to continue expressing ship's velocity increase, VShip, as 10 meters per second (m/s); but we'd like to change vExh units from meters per second to decimal light speed, dc c, where c remains the light speed constant, c = 299,792,458 m/sec; and dc is the variable decimal component.

Decimal Component (dc) is the ratio of vExh to c:
dc = vExh / c
EXAMPLE: Let vExh = 60 million meter/second;
then, dc = 6×107 m/s  ÷ 30×107 m/s = .2

Express Mship as metric Tonnes (mT)
For added convenience, mix different mass units. We'll continue using "grams" for fuel mass burned every second of powered flight; however, we'll now express ship's mass in metric Tonnes (mTs).  By definition: 1 mT = 1,000,000 gms; thus, TABLE-1's example value of MShip=3,000,000 gms changes to 3 mT for TABLE-2. (For reference, a large pickup truck is about 3 metric tons.)

1 mT = 1,000 kg = 1,000,000 gm
For convenience, change from consistent units (grams and m/s) to mixed terms.  Thus, express Mship in mT and vExh in decimal c.
Mship = mfuel × dcc


 To compute MShip value in mTs, use following terms:
  • Mass Fuel (mFuel) remains the given constant, 1 gm.
  • Fuel Velocity (vfuel) becomes dc c with c as light speed constant, and decimal component (dc) becomes Independent Variable (IV).
  • Mass Ship (Mship) is Dependent Variable (DV) measured in metric Tonnes (mT)
  • Ship Velocity (Vship) remains constant 10 m/s for every second of powered flight.
  • 300 is resulting conversion constant which replaces "c", the light speed constant.
Mship = 300 × mfuel × dc

3 mT
1.0 gm
10 m/sec
.1 c
6 mT
1.0 gm
10 m/sec
.2 c
9 mT
1.0 gm
10 m/sec
.3 c

Mship = 300 × 1 gm × .3

10 m/s
9 mT
Convert Vship and mfuel to Useful Rates
Recall original ship-fuel momentum exhaust equation:
Mship × Vship= mfuel × vExh
Divide both sides by one second:
Mship × Vship

1 second
= mfuel × vExh

1 second
Rearrange as follows:
Mship × Vship

1 sec
= mfuel

1 sec
× vExh
Since definition of acceleration is velocity increase per time, Vship per second is an acceleration. We've chosen Vship to be 10 m/s for every second of powered flight; this is same value as g, acceleration due to Earth's surface gravity.
Thus, we can substitute term "g" for expression, Vship/sec:
Mship × g=mfuel

1 sec
× vExh
Further, we can introduce new term, fuel flow per sec (ffsec), and substitute for the expression: mfuel/sec . Thus far, we've arbitrarily assigned a constant value of 1 gram per second to m fuel /sec (now "ffsec"); however, this term can be any value.
Thus, we can substitute term "ffsec" for expression, mfuel/sec, and rearrange: 

× vExh
From TABLE-2: "Mixed Units" 
Express exhaust particles velocity, vExh,  as "decimal light speed", dc c.
Substitute above terms as shown:

× dc × c
Rearrange for following:

× dc × ffsec
Express Term, c/g, as a constant.

=299,792,458 m/sec

9.780327 m/sec2
= 30.57×106  sec
NOTE: Precise values.
Mship=30.57×106 sec× dc ×

"Sec"s cancel out; however, use "fuel flow" term (ffsec) to reinforce concept of fuel quantity used in a second; but expressed in grams. (not gm/sec).
Mship=30.57×106 × dc ×ffsec
To state Mship as metric Tonnes (mTs), use following conversion constant:
k=1 mT


= 30.57×106 

Following table shows that both fuel flow quantity
and fuel particle exhaust velocity are independent variables.
Mship= dc × 30.57×106 × ffsec
Mship= dc × 30.57 mega-ffsec
dc × c
3 mT
1.0 gm/sec
.1 c
60 mT
10.0 gm/sec
.2 c
900 mT
100.0 gm/sec
.3 c
With Newtonian mechanics, we propose that expelling one gram of high speed particles (ffsec) at a velocity of .1 c (vExh) is sufficient to propel a 3 metric Tonne (mT) ship (Mship) at 1-g-force for one second.

We now propose that at near light speeds; relativistic effects further increase the propulsion. To quantity this mass increase due to near light speeds, we use the Lorentz Transform.

Lorentz was awarded the Nobel Prize in 1902, but his subsequent "Lorentz Transformation" (LT, published 1904) was based on electromagnetic forces between charges slightly changing due to their motion, causing moving bodies to slightly contract. Among other things, LT paved the way for Einstein's special theory of relativity, which was published 1905, the year after Lorentz published his famous transform. (Einstein’s Nobel Prize awarded 1921). Of course, relativistic effects are due to nature of matter discovered by Einstein’s famous e=mc2 equation. These relativistic effects can be quantified and measured by using the equation associated with the Lorentz Transform (LT).
For example, if a small original mass (mo) could accelerate in a particle accelerator to a significant portion of c, speed of light; its size would grow to relativistic mass (mr) as shown by equation at left, Lorentz Transform.
For very small values of v (say speed of bullets, 1 km/s, or even slower speed of automobile, 100 km/hr = .028 km/s), v is so much less then c, 300,000 kilometers per sec, that LT's radical expression is extremely close to one (1.0) with virtually no mass increase. Thus, for everyday speeds, moving mass equals original mass.

Use LT relation for following:
Express particle exhaust velocity as decimal light speed: 
vExh = dcc
Square both sides:
(vExh)2 = (dc × c)2 = dc2c2
Let mo = 1.
mr = 1/(1- vExh2

Substitute as shown:

mr = 1/(1- dc2c2

Simplify by reducing the ratio dc2c2/c2 to dc2.
Thus, we obtain mr identity:
mr =1

(1 - dc2)
mr2 =1

1 - dc2
dc2 =-1

dc2 =mr2 - 1

Above steps transform mr identity to restate dc identity:
dc =(mr2 - 1)


Thought Experiment (TE) assumes 100% fuel flow efficiency
For simplicity, TE artificially assumes every particle introduced into particle accelerator as ffsec will successfully travel through many kilometers of pathways and exit the exhaust to propel the spacecraft. Thus, relativistic effects will increase original fuel flow per second ffsec to exhaust fuel flow, ffExh:


Consequently, TE assumes there is a relativistic increase in exhaust particle mass; furthermore, TE assumes this mas increase further increases propulsion effect.
To quickly realize significance of Lorentz Transform (LT), arbitrarily pick a speed much higher then what we observe everyday; for example, one half the speed of light or .5 c for following equations.
After particle accelerates to .5c, the mass increases an observable 15%. TABLE-5 shows even greater increases for even higher speeds.
Alter momentum equilibrium equation
to reflect fuel flow mass of exhaust particles:
Reflect "rates" from T-3, Momentum Exhaust:
Mship × g= ffExh × dcc
Solve for Mship:
MShip =
ffExh × dcc

Make relevant fuel flow substitutions:
MShip =
mrffsec × dcc

MShip = mr × dc × c

× ffsec
Rearrange mr identity, to express dc:
dc =

Thus, alternate equation: 
MShip = (mr2- 1) ×c

× ffsec
Replace term, c/g, with conversion constant 30.57×106.
MShip = mrdc×30.57 mega-ffsec
 Fuel flow (ffSec) is consumed at non-relativistic rate; 
TABLE-4 maintains ffSec at constant 1 gm/sec.

 Fuel particle exhaust (ffExh) is expelled at relativistic rate
to increase both 
momentum and propulsion capability.
ffExh =
m× ffsec

TABLE-5: Greater Exhaust Speeds

dc × mr


(1 - dc2)
(mr2-1)30.57 dc mr
In brief, accelerating particles to near light speed increases their mass:
  • 86.6% of light speed, c, doubles their mass.
  • 94.3% of c increases their mass three fold.
  • 96.8% increases it 4 times!!!
Thus far, we've taken the Momentum Conservation equation and applied it for a duration of one second.
For space travel to become routine, technology will have to accomplish two things:
  1. Micro-scale. Spread the amount of fuel ejected in a smooth continuous manner throughout the entire second.
  2. Macro-scale. Reliably repeat above item for each of the 86,400 seconds during each day of space flight

Both above items are easily done now in various modes of transportation, (for example, the typical plane, train, and automobile have done this for many years). However, can micro/macro scale energy conversion be done for exhaust particles accelerated to large portions of c, speed of light???

Note: for current state of nuclear propulsion, see excellent book,
Nuclear Space Power and Propulsion Systems (Progress in Astronautics and Aeronautics) by Claudio Bruno(Editor).


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