Re: starship-design: Re: FTL travel

[STAR1SHIP@aol.com]

In a message dated 4/1/00 7:10:37 AM Pacific Daylight Time, 
Shealiak@XS4ALL.nl writes:

> Hi Tom,

Hello Timothy,

>  Some remarks on different letters.
>  
>  >Does the group have any thought, ideas, methods or machines to
>  >solve the problem or know of others attempt or solution to answer
>  >the question "How far is that star?"
>  
>  You mentioned the parallax method. You didn't like the idea to use the
>  starship to deviate from the straight line towards the destination star. I
>  suggest to sent out probes at right angles to the ship's direction of
>  travel. These probes should contain similar equipment as the Hipparcos
>  satellite that Steve mentioned.
>  The probes may be expendible or not. They may even have their own smaller
>  engine system, so that they can accelerate with the main starship.

Good thinking. Probes perpendicular to the line of travel is not a bad idea 
as long as the base of the triangle formed is long enough to permit accurate 
trigonometric calculations from the measured angle difference. With angles to 
close to 90 degrees and the distance calculation from trig tables of say 4 
light years will actually measure 4 light years plus or minus 3 light years. 

Suggested probes sent to the star beforehand face the same problem of needing 
the distance beforehand before the energy requirements can used to launch the 
probe. I am not willing to wait for this type probe to send me back the data 
in 4 years before I can launch.  Both type probes are not satisfactory to my 
way of thinking for when I reach the first star and find no suitable place to 
land I want to continue on to another star. Selection of which is limited by 
the number of probes launched from the limited mass of my ship.

As I recall 50 years + or -  20 years ago the distance to the stars published 
in star tables was found to be off by a factor of two from a bad formula. I 
do not remember how the current star tables were derived by calculation or 
measured directly. I was looking for the information or a mechanical range 
finder I could depend on. Any more suggestions.

>  
>  - -  -   -   -     -      -      -
>  
>  >Taking for example the general equation MVof exhaust=MVof payload.
>  >to accelerate a given payload mass of 5 tons twice light speed requires I 
>  >accelerate 100 tons of exhaust to 1/10 light speed (at 1 g acceleration).
>  >Using the equation E(kinetic)=1/2 MV we get the energy required.
>  
>  It actually is E(kinetic) = 1/2 m v^2  (Thus v squared.)

You are correct. It has been years since I made the calculation of 1/2 ton 
needing to be converted to energy to propel 100 tons of propellant  to 
one-tenth light speed. I probably used the correct formula then and did not 
redo my calculation while currently discussing it. I then used the energy 
value calculated and plugged the value into E=Mc^2 to get the mass converted. 
I recommend any using my data to check my arithmetic and my formulas as I do 
on construction or test time.

>  
>  (The momentum in classical physics indeed is defined as p = m v)
>  
>  - -  -   -   -     -      -      -

That is why the gamma calculation of .89 C times 5 tons of payload is not 
equal to 1/10 C times 100 tons of exhaust and therefore violates conservation 
of momentum as the momentum of the payload does not equal to the momentum of 
the exhaust even when gamma is used on the exhaust velocity as the gamma 
effect is a curve and not linear on a graph. At such times I throw out the 
formula out rather than the universal law of conservation of momentum as 
directed by Einstein. That formula is one of the gamma Lorenz transforms that 
Einstein made later corrections for with his derivations that I use in my 
calculations as Lorenz knew nothing of time dilation discovered by Einstein.

>  
>  >Cancellation of the gamma factor is common with the general rocket
>  >equation as (gamma times mass times velocity)of payload=(gamma
>  >times mass times velocity) of propellant. Although the respective
>  >gamma variable are different letters the values calculated are the
>  >same so as gamma subscript payload (gp)=gamma subscript(ge) exhaust
>  >then  ge/gp = 1 and makes the cancellation valid.
>  
>  >Gamma is a complex variable with mass, velocity, time and spatial
>  >dimensions.
>  
>  The gamma used most often only depends on the variable called velocity.
>  Furthermore it so simple that even I can remember it.
>  
>  gamma = 1/Sqrt[1-v^2/c^2]

I can probably show you six or seven gamma formulas from the equation fields 
(field defined as a plane surface like paper where all applicable sets of 
equations are listed available for problem solving). The correct ones 
uncommonly used are only the partial derivatives of the completed gamma I 
stated which when graphed on E Vs V axis and is an exponential curve with 
exponent of two and asymptote at the imaginary vertical line c.


>  
>  The velocity of the exhaust(propellant) is not at all equal to the velocity
>  of the payload. Hence the two gammas ge and gp should be calculated with
>  different v. You'll see that the ge and gp will not be the same, nor cancel
>  each other out.
>  
>  - -  -   -   -     -      -      -

I repeat I can multiply both sides of an equation by the same variable or 
constant and the equation remains equal to solve the equation. As the 
complete gamma is a complex variable all variables must be filled not just 
velocity to conserve momentum.
immediately after multiplying both sides by gamma I can cancel by dividing 
both sides by gamma canceling them out completely. Cancellation is understood 
to be 1(unity or singularity) and not zero and meaningless as thought by 
Steve. The equation returns to Me*Ve=Mp*Vp with 1 not shown but understood as 
Gp/Ge =1

>  
>  >I have no desire to teach the mistaught and misbehaving
>  
>  Somehow I get the idea that this is told *to you* at several times in your
>  life. Otherwise you might have realized that misunderstanding does not have
>  to be a deliberate act (of misbehavior).

Well said. No offense was meant. True any time in past I made the claim mass 
can exceed light speed the majority assumed it was because I did not know 
something they were taught and then try and proceed to teach me as they were 
taught without even listening to what I was taught by Einstein.

The phrase mistaught I use quickly circumvents the common train of thought 
therefore changes the response and then both sides of the question can be 
examined reasonably.

The phrase misbehaving I use as a last resort, when the debater failing to 
convince or provided credible source or proof resorts to name calling, 
character assassination and making unsubstantiated claims as I present only 
credible source and proofs.

Misunderstanding I have no problem with so use and answer questions as best I 
can to clarify that misunderstood by the examiner or examinee even if it be 
me or they not understanding.

Best regards,
Tom


>  
>  
>  Timothy
>