Community > Forum > Technology & Science > Maglev launch

Maglev launch

Posted by: Garnetstar - Thu Sep 11, 2003 1:24 am
Post new topic Reply to topic
 [ 98 posts ] 
Go to page Previous  1, 2, 3, 4, 5, 6, 7  Next
Maglev launch 
Author Message
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Thu Jan 11, 2007 2:20 pm
Enthusiast wrote:
Launch a 1 meter sphere...
A sphere would have higher drag than the proposed shape, which is this:
Image
Which is from this web site: http://www.launchpnt.com/Space_Launch.32.0.html

Still, you calculations agree with my intuition, the deceleration would make it unworkable. On the other hand, LaunchPoint is a company that knows what they are doing and I must assume we are not smarter than they are. They must have thought about this and believe that they can solve it.


Back to top
Profile WWW
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Thu Nov 10, 2005 7:15 pm
Posts: 40
Location: Houston
Post    Posted on: Thu Jan 11, 2007 2:30 pm
Ekkehard,

Well, although I calculated the weight and cross-sectional area based on a sphere, I actually used the coefficient of drag for a blunt cone. (Cd =0.6 vs. 0.9 for a sphere). So imagine the projectile as being a blunt cone with the same cross-sectional area as a sphere of the same weight. The blunt cone is a favored shape for reentry, at least, because it reduces the heat transfer. Apparently, at hypersonic speeds, the main source of heating is not air friction, believe it or not, but compression heating of the air column by the shock wave in front of the projectile. At very high Mach numbers the air is heated to thousands of degrees this way. Heat is conducted and/or radiated back to the projectile from this superheated shockwave. The heat flow depends on the temperature of the shock wave and its distance from the projectile. Blunt shapes cause the shock wave to form further away (in front of) the object, which reduces the heat transferred. I found several articles online that discuss this qualitatively, but I couldn't find one that had the equations.

I'm not sure how relevant your comments are about radiating heat at higher altitudes. For this type of launch, the heat transfer problem you need to worry about is at low altitude, not high altitude. At low altitude the air is the thickest and you are going the fastest, both of which tend to increase the heat generated.

_________________
Enthusiast


Back to top
Profile
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Thu Nov 10, 2005 7:15 pm
Posts: 40
Location: Houston
Post    Posted on: Thu Jan 11, 2007 2:36 pm
Campbelp,

Thanks for pointing that diagram out from their website. I think a needle shaped projectile would reduce drag further, but at the expense of greater heating, as I mentioned to Ekkehard above. If anybody can come up with some equations, let's try to calculate the heating.

BTW, did you look at the g forces on that thing? We're talking artillery shells, not spaceships! And don't forget the centripetal force while it's being accelerated in a ring. If you assume an exit velocity of 5000 m/s and a 1 km diameter for the ring, you'd be pulling 5000 g's just going around in circles.

_________________
Enthusiast


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Thu Jan 11, 2007 4:57 pm
Yeah, heating would be a problem; but I am not even considering that yet. First I have to get past the obvious obstacle of a thousand G deceleration right out of the launch ramp. If it slows down so quickly at launch, how would it ever make it to orbit? As to heating and G forces, I have to assume they are planning some exotic material and sacrificing the tip (expecting it to vaporize) and launching non-delicate payloads only, like fuel or water.


Back to top
Profile WWW
Moderator
Moderator
avatar
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745
Location: Hamburg, Germany
Post    Posted on: Fri Jan 12, 2007 8:49 am
Hello, Enthusiast,

in one of the threads about SSO in this section I was told that SSO handles the heating by the large surface over which the heat can be distribute when I asked for something like that.

So waht about involving the surface into your calculations?



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


Back to top
Profile
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Thu Nov 10, 2005 7:15 pm
Posts: 40
Location: Houston
Post    Posted on: Fri Jan 12, 2007 2:34 pm
I haven't tried calculating the heat loads yet. I'm still looking for the appropriate set of equations. Surface area is already figured into the drag calculation, though. Cross-sectional area, anyway. You can see the equations on this Wikipedia entry:

http://en.wikipedia.org/wiki/Drag_%28physics%29

I'm using the formula for "drag at high velocity".

_________________
Enthusiast


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Fri Jan 12, 2007 4:51 pm
I don't think heat load is the place to spend all your efforts. From a conservation of energy viewpoint, all the kinetic energy lost due to deceleration is converted to heat. If the deceleration is less, the heat will be less. If the heating is so high that it can't be handled with existing technology, like shuttle tiles or high temperature alloys or whatever, then the deceleration is great enough to slow the projectile far below orbital velocity before it ever escapes the atmosphere.

So if the dart shape has aerodynamic characteristics that allow it to escape the atmosphere before slowing down below orbital velocity, then it automatically does not heat up too much. Conversely, if it does heat up too much, then it must decelerate too quickly to ever make it to orbit.

Now I am talking overall heating, not heating at a certain location. The dart shape may very well have low enough drag to make it out of the atmosphere without decelerating too much, at the expense of concentrating all the heat load on the very tip of the projectile, but computing that would require expensive CFD software I think. If that is their plan, they could just make the tip out of tungsten or other high melting point material and allow it to vaporize. The projectile would make it to orbit with much less overall heating and much less velocity lost to atmospheric drag, but with the tip burnt off.


Back to top
Profile WWW
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Thu Nov 10, 2005 7:15 pm
Posts: 40
Location: Houston
Post    Posted on: Fri Jan 12, 2007 7:18 pm
Aerodynamic heating at hypersonic speeds is mainly discussed in the context of reentry. So the problem to be solved is a bit different, and the atmospheric density is much, much less, but I think the physics is similar. One article I read on reentry indicates that only a tiny fraction of the heat dissipated is absorbed by the spacecraft. Presumably you leave behind a trail of hot air. (Well not you personally...at least, I don't know you well enough to say so... :lol: )

The other thing they say about reentry heating is that a blunt shape is crucial for limiting the heat flux to the vehicle. In fact, it is stated that "the heat load experienced by an entry vehicle was inversely proportional to the drag coefficient". This has to do with the shape of the shock wave and its distance form the vehicle. The vehicle is primarily heated by being in close proximity to the hot plasma of the shock wave. At the highest speeds, radiative heating dominates. At lower (but still hypersonic) speeds, conduction is important.

Unlike reentry, our goal is not to slow down. So to get into space this way in the first place, you are going to want to reduce the drag coefficient. But that makes the heating problem worse. Which is the bigger problem for a maglev launcher? Without some actual numbers in front of us, I think it's hard to say.

_________________
Enthusiast


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Fri Jan 12, 2007 9:08 pm
I don’t think you are understanding what I am getting at.

As you said, the goal here is to get up to speed and stay there, not slow down. That means low drag is a must, and all the low drag shapes are sharp. So step 1 is to see if that sharp shape has low enough drag, totally ignoring heating. After you confirm low drag, step 2 is to see if heating is a show stopper. You don’t do step 2 if step 1 says it won’t work.


Back to top
Profile WWW
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Wed Jan 10, 2007 6:36 pm
Posts: 47
Location: Texas
Post    Posted on: Fri Jan 12, 2007 9:37 pm
I have a fair knowledge of physics, but will defer to the actual rocket scientists for this:

How much velocity can a track that is five kilometers long impart on an object at 5g's worth of acceleration?

now assume that track starts at about 1km above sea level at the equator, and rises 1 km during the run.

How much acceration would then be needed to get 1 kilogram into low earth orbit after being launched from such a track?

Now given a fairly inefficient rocket (pick a cheap fuel) how big would a rocket have to be to launch about 100 tons into orbit at that height (2km) and moving at that velocity (whatever the calculation from above is)?

_________________
I hope to leave the world better off for my having been here.


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Sat Jan 13, 2007 2:13 am
For constant acceleration starting from zero speed, use this formula:
V^2=2*A*X
5 Gees is about 50 meters per second squared and 5 km is 5000 meters, so
V^2=2*50*5,000
V=SQRT(500,000)
V=707 meters per second or 0.7 km/s.
Orbital speed is a little over 7 km/s, so you only got 10% of that speed from your track.
Orbital altitude is at least 200 km, so you got only 1% of that from your track.
The shuttle reaches that speed and altitude within the first minute of flight.
So I don’t think the track would be any better than standard air launch ideas, which start at about a half or a third the speed but 5 or 10 times the altitude.


Back to top
Profile WWW
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Wed Jan 10, 2007 6:36 pm
Posts: 47
Location: Texas
Post    Posted on: Thu Jan 18, 2007 9:46 pm
campbelp2002 wrote:
For constant acceleration starting from zero speed, use this formula:
V^2=2*A*X
5 Gees is about 50 meters per second squared and 5 km is 5000 meters, so
V^2=2*50*5,000
V=SQRT(500,000)
V=707 meters per second or 0.7 km/s.
Orbital speed is a little over 7 km/s, so you only got 10% of that speed from your track.
Orbital altitude is at least 200 km, so you got only 1% of that from your track.
The shuttle reaches that speed and altitude within the first minute of flight.
So I don’t think the track would be any better than standard air launch ideas, which start at about a half or a third the speed but 5 or 10 times the altitude.


This leads to the next question: How much smaller must your rocket then be if you have gotten the first 10% of your energy outside the capsule?

i.e. how much fuel weight did you save?

_________________
I hope to leave the world better off for my having been here.


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Fri Jan 19, 2007 2:54 pm
Well that is a much more complicated problem, because there are many factors that have to be considered, such as launch angle and air resistance. I will ignore such complications and just consider this.
An object that falls from 2km in a vacuum at 1G will hit the ground at 200 m/s, so I will assume you avoid 0.2 km/s of gravity drag by launching at that altitude. That may be an incorrect way to calculate gravity drag but it was all I could think of. If there is a better way, will someone point it out? Orbital speed is about 7 km/s, but the usual deltaV needed to launch into that orbit from the ground is 9 or 10 km/s to take into account gravity and air drag. I’ll be nice and use 9. We save 0.7 from the velocity of the magnetic launch and another 0.2 from the 2 km altitude, so it is a savings of 0.9 km/s, or 10%, which would require the remaining 8.1 km/s from the rocket. Using the rocket equation we can compute the launch mass, and so the propellant used, to reach orbit.
The rocket equation says m0=m1*EXP(Dv/Ev), where m0 is the launch mass, m1 is the mass to orbit, which includes the empty rocket and the payload, Dv is the deltaV required and Ev is the exhaust velocity of the engine. A “cheapâ€


Back to top
Profile WWW
Space Station Commander
Space Station Commander
avatar
Joined: Wed Mar 09, 2005 1:25 am
Posts: 887
Post    Posted on: Fri Jan 19, 2007 7:40 pm
I think a lot about Glushkos statement--that went something to the effect of a payload being a square as a brick is fine--if you have enough rocket.

I think big and simple beats all the talk of mass-efficency gurus and aerodynamic purists---and their over-thinking the plumbing, as it were.


Back to top
Profile
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Wed Jan 10, 2007 6:36 pm
Posts: 47
Location: Texas
Post    Posted on: Mon Jan 29, 2007 6:37 pm
[quote="campbelp2002"]Well that is a much more complicated problem, because there are many factors that have to be considered, such as launch angle and air resistance. I will ignore such complications and just consider this.
An object that falls from 2km in a vacuum at 1G will hit the ground at 200 m/s, so I will assume you avoid 0.2 km/s of gravity drag by launching at that altitude. That may be an incorrect way to calculate gravity drag but it was all I could think of. If there is a better way, will someone point it out? Orbital speed is about 7 km/s, but the usual deltaV needed to launch into that orbit from the ground is 9 or 10 km/s to take into account gravity and air drag. I’ll be nice and use 9. We save 0.7 from the velocity of the magnetic launch and another 0.2 from the 2 km altitude, so it is a savings of 0.9 km/s, or 10%, which would require the remaining 8.1 km/s from the rocket. Using the rocket equation we can compute the launch mass, and so the propellant used, to reach orbit.
The rocket equation says m0=m1*EXP(Dv/Ev), where m0 is the launch mass, m1 is the mass to orbit, which includes the empty rocket and the payload, Dv is the deltaV required and Ev is the exhaust velocity of the engine. A “cheapâ€

_________________
I hope to leave the world better off for my having been here.


Back to top
Profile
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 98 posts ] 
Go to page Previous  1, 2, 3, 4, 5, 6, 7  Next
 

Who is online 

Users browsing this forum: No registered users and 22 guests


© 2014 The International Space Fellowship, developed by Gabitasoft Interactive. All Rights Reserved.  Privacy Policy | Terms of Use