[bathat]

>LEO requires about 7.8 km/s, Skylon's jet engines can go around 1.7 km/s. You are reaching 20% of your orbital velocity without reaction mass.

Unfortunately, kinetic energy scales as speed squared, so 20% of your orbital speed represents less than 5% of your orbital kinetic energy. To put this in perspective, the difference in gravitational potential energy between LEO and the earth's surface represents about 15% of your total on-orbit energy.

Now, it's true that an air-breathing engine doesn't need to carry reaction mass (and, sometimes, oxidizer), but the air engines add considerable complexity (which is always a bad thing) as well as weight (because you still need to carry a conventional rocket to finish orbit insertion). So what you need to do is ask how the weight penalty of the air engine compares to the weight penalty of carrying extra fuel in a conventional rocket (bearing in mind, of course, that there a pernicious positive feedback loops when scaling a booster).

[lmm]

>Unfortunately, kinetic energy scales as speed squared, so 20% of your orbital speed represents less than 5% of your orbital kinetic energy.

Fortunately this is counterbalanced by the Oberth effect. Getting to 20% of your orbital velocity requires expending 20% of your rocket's delta-V. And since delta-V is logarithmic in your propellant mass (rocket equation) that could easily translate to needing half as much fuel.

[bathat]

>Fortunately this is counterbalanced by the Oberth effect. Getting to 20% of your orbital velocity requires expending 20% of your rocket's delta-V. And since delta-V is logarithmic in your propellant mass (rocket equation) that could easily translate to needing half as much fuel.

Well, that doesn't really address my point, which was that you need to compare the weight of the hybrid engine to the weight of the extra fuel. The first problem is that an air-breathing engine is going to be something like 3 to 5% of the initial mass, and you have to carry it with you to orbit[1]. The second problem is how the fuel scales:

v_hybrid = 0.8 * v_conventional

Assume both have similar engines:

ln(m_hybrid-initial/ m_hybrid-final) = ln((m_conventonal-initial / m_conventional-final)^0.8)

m_hybrid-intial = (m_conventional-intial / m_conventional-final)^0.8 * m_hybrid-final

let delta equal the expression in parenthesis

m_h-i = delta^0.8 * (m_payload + m_engine) = delta^0.8 * (m_payload + m_h-i * 0.05)

m_h-i * (1 - 0.05 * delta^0.8) = delta^0.8 * m_payload

So the fuel load in a hybrid is going to be:

m_final = delta^(1-0.2) / (1 - 0.05delta^(1-0.2)) m_payload,

The factor in the denominator is what really kills you, and the hybrid is only going to give you a net benefit for deltas less than about 15. So, not only is there not a factor of 2 fuel savings, there isn't any fuel savings at all! Even if you assume an engine weight of only 3% of initial mass, the benefit is only for delta < 35, which is better than just about all actual (as opposed to paper) launchers. By the way, the Shuttle had a delta of about 85-90 for LEO precisely because its designers made the decision to bring wings (which we neglected above) along for the ride to orbit. That also contributed to the 1 in 50 accident rate of that launch system.

And none of this addresses the fact that you are optimizing the f*ck out of one of the least expensive components of launch cost by introducing all sorts of unnecessary complexity.

[1] Ok, I suppose you don't, but then you have to have some way of recovering it, and that adds an enormous amount of complexity to the system.

[EthanHeilman]

>And none of this addresses the fact that you are optimizing the fck out of one of the least expensive components of launch cost by introducing all sorts of unnecessary complexity.

As I understand it, construction of the rocket is the most expensive part of a launch system. The point of skylon is to create a reusable single stage to orbit space plane. Shouldn't skylon's reusability make it "optimize the fck" out of one of the most expensive components?