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Re: Physic help
- To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)
- Subject: Re: Physic help
- From: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)
- Date: Wed, 8 May 1996 07:59:50 -0500
- Cc: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com, neill@foda.math.usu.edu, 101765.2200@compuserve.com, MLEN3097@Mercury.GC.PeachNet.EDU
At 12:20 AM 5/8/96, Timothy van der Linden wrote:
>>>Yes, but that still means that the momentum and energy are not necessary
>>>equally distributed. Rex has given a 2 particle calculation but I wonder if
>>>that makes much sense, since all energy/momentum is finally directed in one
>>>direction namely backwards.
>>
>>I guess it would even out in the plasma. Collisions and such.
>
>Yes, something like this I had in mind.
>
>>>What isn't mentioned in your fusion reactions is that within the engine
>>>probably many photons are generated too. You should see these photons just
>>>as a form of energy and not as an ingredient of the reaction.
>>
>>Supposedly not. Thats why the reactions have virtually no radiation, and
>>virtually all the energy can be converted to electricity.
>
>I'm a bit amazed by that, maybe not initially but as soon as the particles
>start colliding with each other I think photons may be created. I don't know
>when or why photons are created, but I assume that something like blackbody
>radiation will certainly create a bunch of them. Rex, do you know about that?
The info was taken from a paper by Bussard on reactors using these fuels.
The reactors converted virtuall all (99+%) of the fueles energy directly to
electricity. No shielding or cooling system included.
I was impressed.
Probably the random collisions will cause heat and light. But not until
its clear of the system (or the energy is converted to electricity).
>>>Didn't I show this a while ago? It shows the fuel to ship ratios.
>>>
>>> End velocity -->
>>> +------+-------+-------+------+-------+--------+
>>> | 0.20 | 0.30 | 0.40 | 0.50 | 0.60 | 0.70 |
>>> +-----+------+-------+-------+------+-------+--------+
>>> f | 200 | 7.6 | 22.2 | 69.5 | 244 | 1032 | 5906 |
>>> | 250 | 9.7 | 31.9 | 114.6 | 467 | 2338 | 16422 |
>>> || | 300 | 12.0 | 44.4 | 180.0 | 839 | 4896 | 41401 |
>>> || | 350 | 14.6 | 60.2 | 272.7 | 1439 | 9662 | 96907 |
>>> || | 400 | 17.6 | 79.8 | 401.4 | 2376 | 18191 | 213876 |
>>> \/ | 450 | 21.0 | 104.1 | 577.2 | 3805 | 32958 | 449882 |
>>> | 500 | 24.7 | 133.8 | 813.9 | 5941 | 57820 | 908988 |
>>> +-----+------+-------+-------+------+-------+--------+
>>>
>>>Where f is c^2 / 2.058E14 = 439 (c=3E8 m/s)
>>
>>Having no idea how to tie F to acctual fuels. This didn't help much.
>
>I'm sorry, I hope you have an idea by now: f=c^2/(J/kg)
>
>>>Our best reaction is 2H + 3He --> 4He +1H + 18.3 MeV
>>>with 3.51E14 J/kg so that f=257
>>
>>??? That would give you a fuel mass ration of under 40 acording to your
>> table.
>
>Yes that's right, but keep in mind that these ratios are just for
>acceleration or deceleration, if you want to accelerate and decelerate
>without refilling in between then square the numbers (40^2=200).
>
>Tim
I was only calculating the ratio in the explorer craft for deceleration
(acceleration fuel isn't carried by the ship). With the
velocities/specific impulse that was giving the fuel ration was far higher.
???
Kelly
----------------------------------------------------------------------
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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