# more physics (short)

```Sorry I was doing a bit more physics calculations for the Web pages and ran
into a problem.  So if you can verify my new numbers vs the old ones I'ld
appreciate it.

Kelly

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Using the old equation p=1/2 M V^2  we get the
power (in watts) = 1/2 Mass (in Kg) * (exaust velocity in meters/sec)^2

Given the fuel numbers show watts per Kg the equation shuffels around to:

Exaust velocity (in Meters per secound) =sqrt (2 * Power)

>> Did I mess this up?  Yes I'm ignoring relativistic <<

Fuel --> Exhaust        Watts /
kg

p + 11B --> 3 4He       6.926 E13      11,800,000 m/s
De + 3He --> 4He + p    3.505 E14      26,500,000 m/s
*
6Li + 6Li --> 3 4He
(Combined)             1.596 E14      17,800,000 m/s
**
3He + 3He --> 4He + 2 p
2.059 E14       20,300,000 M/s
***

This all seems pretty straight forward, but then I noticed we had
previously come up with exaust velopcities of:

================================================================
Fuel type                   Exhaust speed            Specific Impulse
(Isp)
p + 11B --> 3 4He          14,000,000 M/sec     1,400,000.  seconds

De + 3He --> 4He + p

6Li + 6Li --> 3 4He       21,000,000 M/sec     2,100,000.  seconds

3He + 3He --> 4He + 2 p   24,000,000 M/sec     2,400,000.  seconds

So which numbers are corect, the first group or secound?

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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