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SpaceShipOne Flight Test

Soft writes "Scaled Composites' entry for the X-Prize, the SpaceShipOne, has had a successful first (unpowered) flight test. The spacecraft was dropped from the White Knight carrier aircraft at 47,000 ft (14 km) and 105 kt (194 km/h, 120 mph) and touched down after a 1.1-hour glide at Mojave airport. Photos are available."

5 of 175 comments (clear)

  1. Flight Time? by FreeLinux · · Score: 5, Informative

    Flight Time: 1.1 hours / 19 minutes

    The post refers to a 1.1 hour flight, which shocked me as a rather long glide from 47,000 feet, but after reading the article it seems that total flight duration was 1.1 hours and actual glide time was a more understandable 19 minutes. 19 minutes is still great from that altitude as Nasa's shuttle has a much higher sink rate, despite its greater weight.

  2. Re:Can it handle re-entry? by blufive · · Score: 5, Informative
    I have to wonder how it could possibly handle the heat and stresses of atmospheric re-entry.

    Re-entry from orbit involves hitting the atmosphere at almost-orbital speeds - about 17,000+ mph.

    SSO is designed to fly SUB-orbital. Its re-entry will be MUCH slower. Scaled Composites' website quotes a maximum speed of about 2,500 mph. Kinetic heating shouldn't be a major problem at that sort of speed.

  3. SpaceDev's engine is ready by Anonymous Coward · · Score: 5, Informative

    Spacedev completed the last and full scale test of it's rocket motor for SpaceShip One last week. So, it has a way go up now. Here's the link...
    SpaceDev Performs Successful Rocket Motor Test

  4. The cool thing about all of this by chadamir · · Score: 5, Interesting

    is that it's sort of like living in the 1950s and experiencing all of this new space stuff for the first time. We are lucky to be living in interesting times.

  5. Re:Can it handle re-entry? by WegianWarrior · · Score: 5, Informative

    First, the X-price is about creating a sub-orbital craft, not an orbital one. Still, it's a valid question you ask.

    Rockets to slow the capsule / spaceship down for rentry purposes has been used on every single manned spaceship I know of. They are called retro-rockets and are employed to initiate re-entry at the proper time and place to put the capsule / spaceship down where it's supposed to come down. The alternative is to stay in orbit until it dacays naturaly, and then who knows how long you will stay up there or where you will come down.

    That said, I assume you knew that already and are wondering about rocketengines / other engines that can be used continualy for a logner period of time to brake the craft faster than purely aerodynamic braking can achive?

    In theory it is nothing stopping you from trying that - apart from the weight of both engines and fuel. Not only does the rule of thumb tells us that for every kilogram you want to take into orbit, you'll burn ten kilograms of fuel to get it there, but as the engiens and fuel will have to be protected against the heat of re-entry, you nead a larger (thus heavier) heatshield as well as a larger (heavier) craft overall. And that in turns means - you guessed it - that you'll have to burn even more fuel getting it up there.

    On the other hand, if you're simply suggesting dropping the relative groundvelocity of the craft to zero before it re-enetered the atmosphere, so it would drop straight down, I see two problems. Firstly, you would have to do it fast (since loss in speed means loss in altitude - thus meeting the atmosphere), which means an allmighty kick in the pants for the poor astronauts (very hight G). Secondly, the heatpulse would be about the same anyway - the craft will have a whooping huge potential energy from simply beeing that high, and that will be converted to kinetic energy (read; speed) on the way down. Remember Epot = mgh while Ekin = 1/2mv, and if we assume that all the potential energy is transformed into kinetic energy (which it ain't, a whole lot will turn into heat), we find that Epot = Ekin, thus mgh=1/2mv. Simplify, and you see that the speed (v) equals the square root of two times the height multiplied with the gravitatinal pull (v=sq(2gh) ). Thus, if we set the height to 100 km (100000 meter) and we assume that g is constant at 9.82ms, we find that the speed of the craft as it reaches the surface is no less than 1401.42 meter a second, equal to 5045kmh, equal to 3136mph, or about 4.25 Mach. So to summarise, you won't save anything by 'stopping' in your orbital tracks.

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