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#1 2006-03-05 04:55:14
- MarsDog
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- From: vancouver canada
- Registered: 2004-03-24
- Posts: 852
Re: 30 days to Mars = Earth Geo to Phobos
30 days = 2,592,000 seconds
56,000,000 million kilometers
Earth Geo to Mars Phobos transfer
21.6 km/second + 2.7 = 24.3 Delta-V
http://www.pma.caltech.edu/~chirata/deltav.html
Using beamed microwave for Exhaust of 8 km/sec:
http://monolith.caltech.edu/Papers/ParkinThruster.pdf
http://monolith.caltech.edu/Papers/ParkinLauncher.pdf
Rocket equation:
http://en.wikipedia.org/wiki/Tsiolkovsk … t_equation
initial mass / final mass = e ^ (Delta-V/Vexhaust) = e ^(24.3 / 8) = e^3.04 = 20.9
initial takeoff mass = 83,600 kg == 4,000 kg to Phobos
Second stage needed to slow it down == leaving only 191 kg
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#2 2006-03-05 08:45:09
- GCNRevenger
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- From: Earth
- Registered: 2003-10-14
- Posts: 6,056
Re: 30 days to Mars = Earth Geo to Phobos
"initial takeoff mass = 83,600 kg == 4,000 kg to Phobos
Second stage needed to slow it down == leaving only 191 kg"
Oh yeah, that'll work reaaal good. So you only need a 90 metric tonne vehicle to deliver 200kg of payload, not counting tank mass? To store some 83,000kg of liquid hydrogen fuel, you need between 4,000 and 8,000kg of fuel tank, which gives you negative payload when you subtract that out. And thats without the mass of the engine itself.
Your math also doesn't take into account the time required to accelerate or decelerate, which can be considerable. One third of a VASIMR ship's mission is spent spiraling out and back into orbit despite its superhigh efficiency.
[i]"The power of accurate observation is often called cynicism by those that do not have it." - George Bernard Shaw[/i]
[i]The glass is at 50% of capacity[/i]
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#3 2006-03-08 14:40:54
- publiusr
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- Registered: 2005-02-24
- Posts: 682
Re: 30 days to Mars = Earth Geo to Phobos
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#4 2006-03-08 16:13:54
- MarsDog
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- From: vancouver canada
- Registered: 2004-03-24
- Posts: 852
Re: 30 days to Mars = Earth Geo to Phobos
Your math also doesn't take into account the time required to accelerate or decelerate, which can be considerable.
from equation (3):
http://hypertextbook.com/physics/mechan … equations/
v^2 = v0^2 + 2a(x − x0) (3)
V^2 = 2 * accelearation * distance
Distance=V^2/(2*acceleration)
If acceleration=20 m/sec^2 = 0.02 km/sec^2 or around twice Earth surface gravity
and needed speed of 25 km per second
distance spaceship is accelerating = 25^2/(0.04)= 15,625 km
v=1/2 a*t or t=2v/a = 2*25/0.02 = 2500 seconds = 42 minutes
============================
There are mistakes, but the 42 minutes at each end is just an inaccuracy.
Even if ignoring forces from planets, need to consider deceleration from the Sun.
Starting at 25km/sec from Earth and arriving Mars how fast ? etc
Idea is to accelerate near Earth, to be in range of beamed propulsion.
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#5 2006-03-08 19:53:24
- GCNRevenger
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- Registered: 2003-10-14
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Re: 30 days to Mars = Earth Geo to Phobos
So you want your microwave-beam rocket to accelerate at double normal gravity?? (2Gs!) You would need a very large engine and a truely mind-boggling beam to generate that kind of thrust, maybe up into the terrawatt region. That kind of thrust is clearly impractical, and even dangerous if it were to suddenly be interrupted.
What you need is less acceleration, lots less, as little as you can get away with because the less acceleration you need the less power and the smaller the engine you need.
[i]"The power of accurate observation is often called cynicism by those that do not have it." - George Bernard Shaw[/i]
[i]The glass is at 50% of capacity[/i]
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#6 2006-03-08 22:44:18
- MarsDog
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- From: vancouver canada
- Registered: 2004-03-24
- Posts: 852
Re: 30 days to Mars = Earth Geo to Phobos
You would need a very large engine and a truely mind-boggling beam to generate that kind of thrust, maybe up into the terrawatt region. That kind of thrust is clearly impractical, and even dangerous if it were to suddenly be interrupted.
http://www.pha.jhu.edu/~jlotz/aoptics/node2.html
telescope resolution equation : target/distance = 1.22 λ/Diameter
140 Gig: λ = 0.0021 meters
distance= 15,625,000 meters
Assume target on rocket =10 meters
Diameter needed = 1.22 X 0.0021 X 15,625,000 / 100 = 4,000 meters
============================
In first second. Power into exhaust
momentum mv exhaust = mv rocket
Mexhaust X 8000 = 100,000 X 20m/sec gives 250 kg propellant
0.5mv^2 = 0.5 X 250 X 8000^2 = 8,000,000,000 watts or 8 Gigawatts
into rocket 0.5 X 100,000 X 20^2 = 20,000,000 watt or 20 MW
http://monolith.caltech.edu/Papers/ParkinLauncher.pdf
100 Megawatt is given for 1 ton reuseable craft
Scaled for 100 ton gives 10 Gigawatts
============================
Can get velocity as function of distance from Sun: Vf^2=Vi^2 - 2GM/r
But then how to get time to travel ?
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#7 2006-03-11 10:38:29
- GCNRevenger
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- Registered: 2003-10-14
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Re: 30 days to Mars = Earth Geo to Phobos
15,000km range may still not be enough, and you are talking about building a microwave generator four kilometers in diameter?? And, for this kind of small beam size, you are going to need high frequency microwaves, which in turn requires you put the thing on top of a mountain where the air is thin and dry. Do you know of any mountains with a flat summit two or three miles in diameter?
Also, because fuel is used up and the vehicle gets lighter, the vehicle will be exposed to nineteen G of acceleration, enough to liquify humans and obliterate anything sensitive.
10 Gigawatts, with a 20% penalty for transmission and atmospheric losses, amounts to a dozen full size nuclear power plants!
[i]"The power of accurate observation is often called cynicism by those that do not have it." - George Bernard Shaw[/i]
[i]The glass is at 50% of capacity[/i]
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#8 2006-03-14 07:31:47
- cIclops
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- Registered: 2005-06-16
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Re: 30 days to Mars = Earth Geo to Phobos
10 Gigawatts, with a 20% penalty for transmission and atmospheric losses, amounts to a dozen full size nuclear power plants!
Or only 5,000 2MW wind turbines; make sure you check the weather forecast before launch :>
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