# Re: New idea Laser launcher/scoop systems

```At 5:40 PM 3/11/96, Timothy van der Linden wrote:
>>>Kelly, how did you calculate these numbers?
>>>
>>>Tim
>>
>>found the equation.
>>
>>A.1 The rocket equation
>>
>>The "rocket equation" tells you that
>>the ratio of rocket mass when full of fuel to rocket mass after
>>burning the fuel is:
>>
>>          m(i) / m(f) = exp ( dV / v(e) )
>
>Yups, I already found it when you pointed out where to look for it.

Pity I didn't read that until after I sent the equation.  Oh, well at least
we are on the same equation.

>===============================================================================
>
>>>I already see what makes the difference (besides the fact that formula
>>>doesn't take into account what the energy:mass ratio of the fuel is). For
>>>slightly different exhaust speeds (around 0.08c) the ratios are optimal. I
>>>must have overlooked that in Kellies table, where at the top this speed of
>>>25M is used. I don't understand however why all the other numbers are shown
>>>in the table.
>>
>>I wanted to show all mass ratio's with various specific impulse numbers.
>
>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
ratios?  Given that the fusion reactions are listed as releasing all the
fusion energy in the kinetic energy of there particals, I assumed that was
about the best we could do.

>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
>
>
>If anything is unclear, don't hesitate to ask.
>
>Timothy

Mass energy ration of what?  Kinetic energy to mass?  Thermal energy to
mass? etc...

Your table apears to show that I could get lower ship to fuel mass ratios
than I expected.

Kelly

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

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