Maybe wings in orbit aren't such a stupid idea after all.

The reason for a two stage to orbit design are that you want to end up
in orbit with the minimum amount of excess stuff. Since you can't
throw away much else, that primarily means you want minimum engine and
minimum tankage.

Tankage is fairly straightforward -- just throw a big tank away, like
the shuttle does. Engine is harder. As I've posted earlier, the burn
to orbit generally takes around 430 seconds. With ballistic rockets,
it's hard to get the engine to work longer, so it's hard to end up
with less engine weight in orbit, unless you dump this most expensive
part of the vehicle on the way up.

To get into orbit at all with a propane-lox level Ve (3300 m/s), you
really need to get the ballistic upper stage started at about 2000 -
2500 m/s, depending on your altitude. If you want the engine that
takes you all the way to orbit to deliver that velocity, you have to
(a) have a big tank of gas, but you needed that anyway
(b) negate gravity losses
(c) keep it from hitting the ground for around 1000 seconds
(d) not have too much drag

(c) might not require as large wings as I had previously thought. The
last 430 seconds is completely ballistic, and during the first 570 the
vehicle weight is dropping and the velocity is rising. At some point
the air density is dropping fast too, of course, but at that point the
vehicle mass has decreased enough that we can ignore

So the launcher idea I'm thinking of is an air-dropped thing with
wings that uses a single rocket motor all the way to orbit, and dumps
it's gas tank on the way up. The wings are not large enough to
support the craft at launch. Basically, once dropped the thing dives,
ignites the motor, and accelerates until it goes supersonic, at which
point it has enough lift to pull out and start climbing.

A ballistic second-stage rocket taking it's main engine all the way to
orbit takes about 430 seconds to burn and starts at 2000-2500 m/s.
When it starts burning, the thrust is less than 1G and so for a short
while the thing is actually slowing down. The Space Shuttle does this
once the SRBs seperate, IIRC. Fire that rocket for another 570
seconds, and the starting acceleration (not counting gravity) will be
something like 0.25 - 0.3 G. If the supersonic L/D of the wing can be
made better than 4, and the wing doesn't mass too much, it's a win.

The Concorde wing had a L/D of 8 when supersonic. Sounds good.

So the launcher idea I'm thinking of is an air-dropped thing with
wings that uses a single rocket motor all the way to orbit, and dumps
it's gas tank on the way up.
Like an SS1 with a drop tank? Just brainstorming but if SS1 had a twin
strapped under it, that was just a fuel tank, that might hold 30+ times the
fuel of a single SS1. The tank could be dropped when empty and then flown back
by computer and remote control, keeping the cost down.
^
//^\\
~~~ near space elevator ~~~~
~~~members.aol.com/beanstalkr/~~~

Allen Meece wrote:
So the launcher idea I'm thinking of is an air-dropped thing with
wings that uses a single rocket motor all the way to orbit, and dumps
it's gas tank on the way up.
Like an SS1 with a drop tank? Just brainstorming but if SS1 had a twin
strapped under it, that was just a fuel tank, that might hold 30+ times the
fuel of a single SS1. The tank could be dropped when empty and then flown back
by computer and remote control, keeping the cost down.

I'm assuming you'r getting "30+" times from the fact that it it needs some
30* the speed.
However, that doesn't work, as you've also got to accellerate the fuel,
so it becomes 900*, but then the engines, and ... can't cope with that,
so it gets orribly complex.

"Ian Stirling" wrote in message
...
Allen Meece wrote:
So the launcher idea I'm thinking of is an air-dropped thing with
wings that uses a single rocket motor all the way to orbit, and dumps
it's gas tank on the way up.
Like an SS1 with a drop tank? Just brainstorming but if SS1 had a twin
strapped under it, that was just a fuel tank, that might hold 30+ times
the
fuel of a single SS1. The tank could be dropped when empty and then
flown back
by computer and remote control, keeping the cost down.

I'm assuming you'r getting "30+" times from the fact that it it needs some
30* the speed.
However, that doesn't work, as you've also got to accellerate the fuel,
so it becomes 900*, but then the engines, and ... can't cope with that,
so it gets orribly complex.

..and even if you could get all the way to orbit, you had better improve the
TPS a fair bit or it will get mighty warm on the way back down. ;-)

It will be most interesting to see what Mr Rutan comes up with next.

Cameron:-)

I'm assuming you'r getting "30+" times from the fact that it it needs some
30* the speed.
Someone said that orbit would take 30 times the *energy* of a vertical
flight to 100 km. Since then I've also heard *20* times the energy.
SS1 went Mach 3 on it's record flight and orbital speed is Mach 24. That
means SS1 needs to go 8 times faster, to make orbit. The Mach 3 was achieved
in vertical flight and the orbital speed will be done in mostly horizontal
flight which takes less fuel.
I doubt it will take 8 times the fuel to go 8 times the speed. At 2 gee it
should take 4 times the fuel to go 8 times the speed plus a little to boost the
extra fuel weight.
So that might require about 5 times at much fuel to obtain orbit?
^
//^\\
~~~ near space elevator ~~~~
~~~members.aol.com/beanstalkr/~~~

Allen Meece wrote:
I'm assuming you'r getting "30+" times from the fact that it it needs some
30* the speed.
Someone said that orbit would take 30 times the *energy* of a vertical
flight to 100 km. Since then I've also heard *20* times the energy.
SS1 went Mach 3 on it's record flight and orbital speed is Mach 24. That
means SS1 needs to go 8 times faster, to make orbit. The Mach 3 was achieved
in vertical flight and the orbital speed will be done in mostly horizontal
flight which takes less fuel.
I doubt it will take 8 times the fuel to go 8 times the speed. At 2 gee it
should take 4 times the fuel to go 8 times the speed plus a little to boost the
extra fuel weight.
So that might require about 5 times at much fuel to obtain orbit?

Nope.
Taking this in order.
At 100Km, the potential energy of an object at rest is mass*gravity*height,
or about 1MJ/Kg.
To get to 8000m/s (orbital speed, neglecting drag) you need
1/2*mass*velocity^2 = 32MJ/Kg.
24^2/3^2 = 64 times the energy at mach 24, compared to mach 3.
Gravity is irrelevant, other than to make the trajectory less efficiant,
as you have to spend some fuel on avoiding hitting the atmosphere
at high speeds.
It takes the same amount of fuel to accellerate by a given amount
however fast you do it.
SS1 is (about) 40% fuel by mass.
So, increasing the fuel you need by 8 times will increase the mass
significantly.
Now, you have to run the fuel consumption numbers on the whole mass, so
it no longer looks quite so good.

Not to mention that you need a new carrier.

Ian Stirling wrote in message news:40f1c112$0$967
Nope.
Taking this in order.
At 100Km, the potential energy of an object at rest is mass*gravity*height,
or about 1MJ/Kg.
To get to 8000m/s (orbital speed, neglecting drag) you need
1/2*mass*velocity^2 = 32MJ/Kg.
24^2/3^2 = 64 times the energy at mach 24, compared to mach 3.
Gravity is irrelevant, other than to make the trajectory less efficiant,
as you have to spend some fuel on avoiding hitting the atmosphere
at high speeds.
It takes the same amount of fuel to accellerate by a given amount
however fast you do it.
SS1 is (about) 40% fuel by mass.
So, increasing the fuel you need by 8 times will increase the mass
significantly.
Now, you have to run the fuel consumption numbers on the whole mass, so
it no longer looks quite so good.

Not to mention that you need a new carrier.

Just using the basic energy equation does not account for the change
in mass over time as the fuel is used. The actual rocket equation is
logarithmic. It can be found at:
http://web.media.mit.edu/~sibyl/proj...ce/rocket.html

The actual equation is:
V(t) = Vx*ln(dM/dt) -gt

Where V is the velocity, Vx the exhaust velocity, dM/dt the change in
mass of the rocket over time (ie: the fuel being consumed), and g*t
the effects of gravity over time. This basic formula does not include
the effect of drag, but if your starting already in space, as the
original poster described, then drag will basically be zero anyways.

To determine how much energy would be need to reach altitude in the
first place (but with zero lateral velocity), I found this equation on
the 'net:

Max altitude: [-M /(2*k)]*ln([T -M*g -k*v^2] / [T - M*g])
Velocity at burnout: q*[1-exp(-x*t)] / [1+exp(-x*t)]

q = sqrt([T-M*g]/k) and
x = 2*k*q /M
k is the air resistance.

I won't even begin to try to explain *these* equations
Best I just direct you to the site I obtained them from:
http://my.execpc.com/~culp/rockets/rckt_eqn.html#Method