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Doubts If ATO will be working

Posted by: Ekkehard Augustin - Sun Sep 04, 2005 11:37 am
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Doubts If ATO will be working 
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Post    Posted on: Wed Nov 16, 2005 9:14 pm
Remember that the maximum L/D value will be preserved regardless at the optimum angle of attack and I can't see good reasons to spend much time at other angles.

The drag coefficient won't change drastically with angle of attack except for slender wings and at most your going to be a small factor out if it does - which in the scheme of things isn't that important.

[EDIT] The Wiki article seems to be correct, I think things may be arranged so that all values are functions of angle of attack.


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Post    Posted on: Wed Nov 16, 2005 10:38 pm
nihiladrem, I think you misunderstood my idea. It's probably my fault, though. What I was getting at was that we need to find out the magnitude of the upward forces of aerodynamic lift, centrifugal force, and bouyancy at different altitudes, given a hypothetical ship w/ certain drag, thrust, and aerodynamic lift charicteristics. That's a reasonable thing to do, and it will tell us whether lift can replace bouyancy up to a great enough altitude to get some real orbital velocity up.

I wasn't thinking about working things out in terms of speed, but in terms of altitude. Sure, you need to get to orbital velocity, but you can do that once you get higher up, IF you can get enough lift between all sources to get to that altitude. As you said, the increments probably don't need to be as fine as a mile, but I certainly think that a tabular computation would be the only way to go. For me, anyway. YOU might be able to get away w/ some advanced math, but I'm just taking calculus one right now!

[Edit: Nilihadrem, one more thing about your post: I assume you get that 60% of the weight figure by subtracting the amount of centrifugal force, which would be 40%. What altitude is that for, and what about bouyancy at that altitude? Those are going to be important; thus my reason for wanting a table.]


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Post    Posted on: Thu Nov 17, 2005 12:21 am
LukeSkyWalker wrote:
I assume you get that 60% of the weight figure by subtracting the amount of centrifugal force, which would be 40%.
Yes, and if you start calculating for a high speed that's the first thing you want to work out.
LukeSkyWalker wrote:
What altitude is that for, and what about bouyancy at that altitude? Those are going to be important; thus my reason for wanting a table.

Altitude really depends on the details, but with slightly more realistic parameters (you can't fly with fixed cylindrical wings) it was somewhere north of 150km. I don't have great atmospheric data at these altitudes, and the formula I used would have underestimated the density.

The bouyancy is zero, if you are trying to run at as low drag as possible the bouyancy is vanishingly low by the time you're above the speed of sound.


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Post    Posted on: Thu Nov 17, 2005 8:16 am
Drag coefficient won't change drastically with angle of attack? That doesn't make sense. If you don't believe me, fold a paper plane and try throwing it with its tip pointing to the ceiling.

Also, why wouldn't cylindrical wings work? I agree that they probably wouldn't be very efficient, but especially at these speeds the whole idea is just to push an object through the air at an angle, so that it deflects the air downward, right?

Last night I figured out that Al is probably the total area of wing that has airflow attached to it. When you get into a stall, the airflow over the wing breaks down and you lose lift. Now I just need to know whether Ad = Al, or Ad = Al sin AoA.

Anyway, I've found some more aerodynamics theory here, so I'll be reading that tonight to try and see whether I can make some more sense of it.


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Post    Posted on: Thu Nov 17, 2005 5:16 pm
Lourens wrote:
Drag coefficient won't change drastically with angle of attack?

If you change the frontal area to compensate (I more or less suggested drag be approximated using true frontal area with a non-optimal nosecone) while the drag is increasing, the drag coefficient will not change drastically. Even if you make a factor of three error in drag, as long as your L/D remains accurate, much of the physics remains unchanged with just a few changes (to altitude at a given speed etc). That's ok for most of the calculations concerning feasibility. If you want to model things away from maximum L/D (which is important if you gain speed rapidly) more complex analysis might be helpful.

I went too far - fixed cylindrical wings can just about work if they are swept back, but without being swept back they don't even have an angle of attack and can't generate lift. Even when swept back at ~60 degrees they make truly horrible wings for supersonic flight.


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Post    Posted on: Thu Nov 17, 2005 6:13 pm
I am no aerodynamics expert, but I do know that there are two kinds of drag. There is parasitic drag and induced drag. Parasitic drag is the regular kind everybody thinks of and induced drag is a byproduct of lift generation and is related to angle of attack.
Here is a web reference I googled up.
http://142.26.194.131/aerodynamics1/Drag/Page6.html

By the way, I don't think you should assume the wings will be cylindrical. We don't know what shape JP will really use and I see no reason to limit our simulations to a cylinder.


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Post    Posted on: Thu Nov 17, 2005 9:13 pm
Thanks to all who are writing posts in this thread...fasinating stuff. While I don't have much technical expertise in these areas or higher math skills I'm still getting a pretty fair idea of where you are going.

While I really like the ATO concept I've always had doubts about how they are actually going to do it.

One other thing. I'm assuming that provided the craft can get to orbit that the flight characteristics you are trying to model would be used in reverse to re-enter? Would a large amount of whatever fuel would be used be needed for a slow de-orbit?


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Post    Posted on: Thu Nov 17, 2005 9:51 pm
I don't think so. Drag would take the place of thrust, but buoyancy and aerodynamic lift should be the same. You would need a small amount of power to initiate reentry though, just like the shuttle.


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Post    Posted on: Thu Nov 17, 2005 10:15 pm
Pivotal here is what you mean by a a slow de-orbit. If you mean much slower than a shuttle reentry then you would most definitely need lots of fuel/power, doing reentry very slowly is *almost* as hard as getting into orbit very slowly. That said there's no benefit from slowing down reentry using fuel, if you can survive reaching orbit slowly, you should be able to survive reentry without using thrust.


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Post    Posted on: Thu Nov 17, 2005 10:22 pm
What JP has said they mean is 5 days to orbit and 5 days back down.


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Post    Posted on: Thu Nov 17, 2005 10:27 pm
This was one of the things that suggested assumptions of extreme L/D...

nihiladrem wrote:
LukeSkyWalker wrote:
What altitude is that for, and what about bouyancy at that altitude? Those are going to be important; thus my reason for wanting a table.
Altitude really depends on the details, but with slightly more realistic parameters (you can't fly with fixed cylindrical wings) it was somewhere north of 150km. I don't have great atmospheric data at these altitudes, and the formula I used would have underestimated the density.

Managed to drop a factor of ten in that calculation. Altitude now seems to be nearer 130km.


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Post    Posted on: Fri Nov 18, 2005 11:28 pm
I don’t recommend the L/D ratio as an accurate predictor of ATO performance. Its use involves assuming constraints on the ATO that do not apply.

Let me clarify that.

To derive this particular relationship between lift, drag, and thrust, start with the only universal condition for sustained flight, which is that the sum of vertical forces (Fy) be equal to or greater than zero:

Fy > 0

This can be expanded to a list of all vertical forces on the vehicle, but only aerodynamic lift (L), buoyancy (B), centrifugal force (Iy), and vertical drag (Dy) are large enough to be of interest to us here:

Fy = L + B + Iy + Dy – W

In order to make the math easier, an effective weight or catchall weight (Wc) can be introduced to stand in for every vertical force that isn’t aerodynamic lift:

Let Wc = W – B – Iy – Dy

Fy = L – Wc

Which implies that:

L – Wc > 0

L > Wc

Aerodynamic lift and horizontal drag (Dx) can be expressed in terms of horizontal relative velocity (vrel) by assigning descriptive lift and drag coefficients (cl and cd):

L = 0.5 * Rho * A * cl * vrel^2

Dx = 0.5 * Rho * A * cd * vrel^2

L / Dx = cl / cd

This allows a statement of aerodynamic lift in terms of horizontal drag:

L = Dx * cl / cd

Which means:

Dx > Wc * cd / cl

All this so far is completely applicable to the ATO system.

At this point, the usual next steps in the derivation are to assume that the vehicle is traveling at approximately terminal velocity (vrel = vt) and that the corresponding horizontal drag is equal to the horizontal thrust (Dx = Fx). However, neither or these two assumptions is strictly true in the case of the ATO system. vrel = vt can be justified if the vehicle fails to ascend, and proving vrel = vt remains a simple way to exclude parameter sets as impractical. However, JP Aerospace swears up and down that the ATO will accelerate all the way to orbital speed, so vrel = vt isn’t a valid assumption if the system actually works.

Also, more fundamentally, there is potentially a horizontal inertial force component (Ix) contributing to Dx, so that:

Dx = Fx + Ix

Ix is orthogonal to Fy, and thus not subject to operations on Wc. Also, Ix is determined by the vehicle’s absolute velocity, vabs, and not by vrel. So, it can vary independently of L.

The actual relationship between L, Dx, F and W then becomes:

Fx > Wc * cd / cl – Ix

This equation is just as applicable to the ATO as the equation Fx > W * L/D would be to a conventional airplane, and can be used to exclude parameters. However, cases with:

Wc < 0

are trivial solutions – they convey nothing useful. Also, cases are physically possible where Fx < W * L/D but the craft still ascends because Fx > Wc * cd / cl – Ix remains true. The relationship between Fx and Ix over time is more complex than a simple ratio. The value of Ix depends on which way the ship is moving at the time. Thus, navigation becomes as important a factor as raw engine power.

Requiring only Fy > 0 (the only universal condition for sustained flight) will give you useful data all the way up, even under conditions where L/D fails to do so. Trying to use L/D to simulate the ATO’s flight path at all points is simply too much sugar for a dime.

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Post    Posted on: Sat Nov 19, 2005 5:25 am
WhoaH! That's a lot of good stuff. I'll have to study it. In the mean time, I'd like to say one more thing to nilihadrem.

Let me see if I follow your thinking correctly: In order for the ATO to gain altitude, it needs to go forward and develop lift (as well as drag). In order to develop lift at higher altitude, where there will be less bouyancy and ALSO less air mass flowrate over the ATO 'wings,' you will need to fly the thing faster, but the decrease in drag due to the thinner atmosphere won't make up for having to support more wieght by aerodynamic lift AND the increase in drag (relative to lift) caused by flying at super or hypersonic velocities.

But I detect one assumption you make, which I am not sure is valid. You keep returning to the idea that the balloon-ship will need to go super or hypersonic, against non-negligible densities of air. My thinking is that you might not have to. It may be that you can get all the lift you need to get up to 80 km, from which the ship can begin to accelerate and displace lift with centrifugal force as it rises, and not get to super and hyper sonic speeds until it is very near orbital altitude and the atmosphere becomes negligible. So your assumption of going high speed at lower altitudes and incurring huge drag losses may not be correct. I think in order to know for sure, though, we'd need to do some real, solid aerodynamic analysis rather than conjectures.

As for JP possibly withholding some crucial information, that seems unlikely.

One note about thrust and ISP: ISP is the opposite of energy efficiency for thrust. ISP is reaction mass efficiency. To get more thrust from a limited solar input, you want to decrease the exhaust velocity (ISP) and hence energy per thrust, while increasing the amount of propellant used. Up to a certain point, that is, where you'd have problems w/ using too much fuel. But suppose JP were to use on-board compressors to provide nitrogen for low-ISP electric thrusters of some sort? Or what if he used some sort of atmospheric electric thruster, w/ the ion engines saved for the orbital insertion burn, if you will? And who says props are out of the question?

Hmm. Atmospheric ion propulsion. Now, there's a smart way and a stupid way to do that. Stupid way: try to get an exhaust stream, like any other engine. Unless you can influence a huge area of air around the ship, which seems unlikely do to the nature of electric feild lines. Smart way: move only the air directly around the ship, so that you effectively use your energy to overcome friction, and your ship will accelerate to whatever speed you can make the flowin sheath sustain. YOu can't get alot of thrust, but you can cut friction.


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Post    Posted on: Sat Nov 19, 2005 4:02 pm
JP has got more competition, the airforce have managed to fly an airship up to 74,000 feet witha payload of 60lb. Still a long way from orbital altitude but getting closer.

http://www.spaceref.com/news/viewpr.html?pid=18343

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Post    Posted on: Sat Nov 19, 2005 4:19 pm
LukeSkywalker wrote:
As for JP possibly withholding some crucial information, that seems unlikely.


:) Good one!

Regarding engine power, if you know how much electrical power the ATO has to work with, you can estimate its available thrust for a given exhaust velocity (or vice versa).

The kinetic power released by the exhaust (Pe) is related to the reaction mass (me) and the exhaust velocity (ve) by some overall efficiency (e):

Pe = 0.5 * e * me * ve^2

The engine thrust (F) is:

F = me * ve

So, you can just substitute that back into the power equation:

Pe / e = 0.5 * Fe * ve

where Pe / e is the necessary power consumption of the engine. This allows a better estimate of the necessary engine performance than just picking your favorite class of electric engine out of a hat.

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