Re: more physics (short)

```At 6:37 PM 6/2/96, Timothy van der Linden wrote:
>>I went over the Bussard paper and found 14E6 as the upper exaust vel for p
>>+ 11B.  Which obviously isn't what I'm getting.  Eiather I miscalculated
>>the watts per Kg (I sent my equations for that in another letter) or I
>>misread Bussards paper.  (I suppose I could write him.)  I'ld hate to write
>>off about 15% of spec impulse if I didn't need to.  (See table below.)
>
>>If you know how to get the exaust speed from Mev numbers let me know.  I'm
>>sure those numbers from the papers are correct.
>
>Watch carefully:
>
>The reaction as found in my tables book:
>
>H + 11B --> 3 4He + 8.7 MeV
>
>Let's assume energy conversion is 100% and ALL reaction products will be
>accelerated in the 100% effective lineac.
>
>First turn MeV in to Joules:
>
>8.7 MeV = 8.7E6 * 1.602E-19 = 1.394E-12 Joule
>
>Then determine the weight of the particles in kg:
>
>H + 11B is approx 12 u (u=atomic mass unit)
>
>12 u = 12 * 1.661E-27 = 1.993E-26 Kg
>
>Thus the energy per kg is:
>
>1.394E-12 / 1.993E-26 = 6.995E13 J/Kg

So we used the same numbers (thou in a different order) and got about the
same number.

One suprize, I had that the mass of Protons and neutrons as about 1.673
E-27 kg, where you have 1.661E-27.  Is that a typo?  We came up with the
same velocity, which seems odd if we were using a different constant.

Eiather way (unless Rex has an alternate opinion) it looks like the lower
numbers I just computed are the ones to use.

<Sigh>

>
>(Don't use much more significant numbers, if you do use them you need to
>define some things more accurate and cannot simply think that 11 H weighs as
>much as 11B.)
>
>Using E=0.5 m v^2 I still get v=11.8E6 m/s, so I really wonder how Bussard
>can get a higher velocity, unless he dumps some of the reaction mass instead
>of accelerating it.

No the numbers were not based on an engineering output.  That was supposed
to be the direct output from the reactions.  Could be I misread something
and the 14E6 number was for something else (though I can't see what).

>>I was working up the following table of the fuel mass ratios needed to get
>>to or from certain speeds given the exaust velocities.
>>
>>NEW numbers
>>
>>Fuel --> Exhaust
>>Vexh                75E6m/s     100 E6m/s  125 E6m/s 150 E6m/s
>>
>>p + 11B --> 3 4He
>>  11,800,000 m/s       576.0     4,790.0    39,900    332,000
>>
>>6Li + 6Li --> 3 4He
>> (Combined)
>>  17,800,000 m/s        67.6       275.0     1,120      4,570
>>
>>3He + 3He --> 4He + 2 p
>> 20,300,000 M/s         42.5       138.0       472      1,620
>>
>>De + 3He --> 4He + p
>>  26,500,000 m/s        16.9        43.5       112        287
>
>Of course ;) I get completly different numbers (at least for higher Vend)
>
>  674  2,680  82,000  1,200,000
>   73    181   1,728     10,000
>   44     97     708      3,377
>   18     34     154        509
>
>I assume this time relativistics is the origin (for 0.5c gamma=1.15 which
>means its not almost equal to 1 anymore) for our differences

??
Why are you geting larger (often much larger) numbers for 75E6m/s, 125
E6m/s, and 150 E6m/s, but smaller for 100 E6m/s?  Also the deltas don't
seem even.  Hum.  I suppose I should rerun my numbers with relativistic
equations.  With our luck I'll come up with a third set of numbers.

thou.  I normally wouldn't have pushed them that far.  But De + 3He was
looking suprizingly good.

>>Amazing how touchy the fuel ratios are to changes in exaust vel/specific
>>impulse.  Look at the difference between 6Li + 6Li and 3He + 3He!  ;)
>
>Well yes, that is because of the energy per kilogram of fuel. Which is 3
>times higher for 3He + 3He. (This is what my "magic" number f says)
>
>
>Timothy

Pity you gave no way to use you magic number.  (Like the value for given
fuels?)  Also the F number table didn't give the numbers for the fuels,
which naturally  didn't have even F numbers.

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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