Subject:
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Re: Transit Time to Mars
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Thu, 16 Dec 1999 00:54:12 GMT
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Reply-To:
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lpieniazek@novera.com(nomorespam)
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Viewed:
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175 times
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Jasper Janssen wrote:
> The thing to remember is that maintaining an acceleration of 1g takes
> a tremendous amount of fuel.
No it doesn't. It takes a tremendous amount of CHEMICAL fuel, but you
need to use something with a much higher specific impulse. The problem
is that your chemical exhaust is going WAY too slow, hence you're not
transferring much momentum.
> The Shuttle does several g's for a few
> seconds with its _entire_ solid-fuel boosters, and then several g's
> for another few dozen minutes burning up its entire external fuel
> droptank.
>
> That is a _bloody_ huge amount of material gone in a very short while.
> You'd need to maintain that not for half an hour, but for 30 hours.
> And remember that this grows exponentially - every time you double the
> acceleration time or the weight, you near-double the fuel weight,
> which means you have to near-double the amount of fuel again, etc.
>
> A doubling in payload or burn time can easily mean one or two extra
> stages to your missile, and 10 times the amount of fuel.
Again, you need different technology. Ion drive engines, while
admittedly not capable of 1G acceleration *today*, are MUCH more
efficient in terms of acceleration imparted per g of reaction mass.
Orders of magnitude more efficient. Your most efficient reaction drive
engines will accelerate their reaction mass to a very significant
fraction of c. Chemical gets what, 20,000 meters per second? Bah.
positively puny specific impulse.
Get Matthew Verdier involved in this conversation. I'm playing a rocket
scientist here(1), but he IS a rocket scientist. Lives at the Cape and
works for NASA.
1 - I actually filled out and submitted an application to NASA for the
astronaut corps at one point. Just dreaming but I do know a bit about
propulsion systems. Chemical is so... so... last millenium. And the
shuttle is late 60's era technology, after all.
--
Larry Pieniazek larryp@novera.com http://my.voyager.net/lar
- - - Web Application Integration! http://www.novera.com
fund Lugnet(tm): http://www.ebates.com/ ref: lar, 1/2 $$ to lugnet.
NOTE: Soon to be lpieniazek@tsisoft.com :-)
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Message is in Reply To:
 | | Re: Transit Time to Mars
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| (...) Assume that a mile is 1.6 km, because dammit, you don't calculate these things in imperial (you'd need g in miles/sec. Ugh.) (...) All right. x == v0 * t + 0.5 * a * t x == 57.6e9 m v0 == 0 (this means I'm calculating from reaching orbit, (...) (25 years ago, 16-Dec-99, to lugnet.off-topic.geek)
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