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Re: New idea Laser launcher/scoop systems

At 3:11 PM 3/12/96, Timothy van der Linden wrote:
>To Kelly,
>>>OK, unfortunately (as you stated below) the paper didn't show all there was,
>>>your numbers only showed a decrease of mass for lower exhaust-velocities.
>>>I found out that for high exhaust speed the mass goes up because of the
>>>extra kinetic energy (could be overcome by beaming) and for lower speeds the
>>>mass goes up too, because of the waste of energy (ie. using fusion fuel as
>>>reaction mass).
>>I'm not sure I follow this.  By waste energy are you implying the same
>>specific impulse ratio could be developed with lower fuel to reaction mass
>I assume that with "impulse ratio" you mean "impulse velocity". Than the
>answer is yes.
>So after "extracting" the energy of the fusion fuel, you could better dump
>the excess mass without accelerating it! (Because if you accelerate it the
>other mass could be accelerated less)

Actually no.  Specific impulse is a term with rockets that tells how much
thrust you can get out of a mass of reaction mass or fuel.  (in a normal
rocket they are the same.)  For example a pound of hydrogen and oxegen
pumped into a shuttle motor generates about 455 pound secounds of thrust.
My numbers showed that unless the specific impulse were geting out of the
Explorer classes engines is over 2,000,000 we don't get to go.  So I was
woundering is using the fusion fuel directly would make more or less
performance mass sence then mixing it in with extra reaction mass, or
converting the power to electricity that accelerates a smaller reaction
mass in a lineac?  Given that the energy in the fusion reactors was all
turned into kinetic energy in the exaust plasma, I had assumed using it
directly would be most efficent.

>>>So one needs to find the valley between the (ever rising) hills.
>>>The formula to do this:
>>>  BestG:=1+1/(f-1);                     (the relativistic gamma)
>>>  BestV:=c*Sqrt(Sqr(BestG)-1)/BestG     (the accompanying velcity)
>>>The optimum exhaust-velocity is only dependent on the mass:energy ratio of
>>>the fuel (thus not on the final velocity).
>>> f   exhaust velocity (in c)
>>>200        0.09987
>>>250        0.08935
>>>300        0.08158
>>>350        0.07554
>>>400        0.07067
>>>450        0.06663
>>>500        0.06321
>>>(f is the mass:energy ratio (about 270 is the best for fusion untill now))
>>>The next table shows the ship:fuel ratios needed to accelerate upto the
>>>final velocities that vary horizontally. Vertically the energy:mass ratios
>>>of the fuel are varied.
>>>Every time the optimal exhaust speed is used (this can be looked up in the
>>>previous table).
>>>                  End velocity -->
>>>      0.10      0.20      0.30      0.40    0.50    0.60    0.70
>>>200     2.7       7.6      22.2      69.5     244    1032    5906
>>>250     3.1       9.7      31.9     114.6     467    2338   16422
>>>300     3.4      12.0      44.4     180.0     839    4896   41401
>>>350     3.8      14.6      60.2     272.7    1439    9662   96907
>>>400     4.1      17.6      79.8     401.4    2376   18191  213876
>>>450     4.5      21.0     104.1     577.2    3805   32958  449882
>>>500     4.9      24.7     133.8     813.9    5941   57820  908988
>>> ^
>>> +--- Energy:mass ratio of the (fusion) fuel
>>Mass energy ration of what?  Kinetic energy to mass?  Thermal energy to
>>mass? etc...
>* Mass:energy ratio shows what part of the mass of a fuel can be turned into
>energy. For a matter & anti-matter mix all mass can be turned into energy
>and thus gives a ratio 1:1. For fusion fuel only about 1/300 part of the
>total mass can be converted to energy so that means a ratio of 1:300.
>Note. I've defined f as total mass divided by energy mass (f=mass/energy),
>this way you always get ratios bigger than or equal to 1.
>* Fuel:Ship ratio is clear I think, it shows what part of the total mass of
>the starship is used for fuel and what is used for habitation etc.
>>Your table apears to show that I could get lower ship to fuel mass ratios
>>than I expected.
>Yes, that what I'm trying to tell everybody for the last week. :)
>If we content ourselves with a final velocity of 0.2c, we can accelerate and
>decelerate having a total fuel:ship ratio of 12*12=144. (f=300) (Ofcourse we
>still need to mine fuel for the homeward trip)
>If we assume we are beamed away (and have to decelerate by ourselves) while
>using the same fuel:ship ratio of 144 then we can decelerate from a velocity
>of 0.385c (not in the tables)

Great, 144 is probably impractical.  But it sounds like stoping from
..33c-.36c is possible.  Assuming you can find the fuel!



Kelly Starks                       Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)