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New SpaceShip One Photos Online

Posted by timothy on from the with-a-fisheye-lens-I'm-in-space-now dept.

Alex Edwards writes "Scaled now have the latest photos from their last 200,000-ft. trip online. The earth curvature on this one shows just how close they got to space."

3 of 49 comments (clear)

  1. 200000 feet = 38 miles up by Picass0 · · Score: 5, Informative

    Mt. Everest = 6 miles
    Edge of Space = 75 miles.
    Space Shuttle orbit ~200 miles (typical)
    International Space Station = 228 miles

    1. Re:200000 feet = 38 miles up by PurpleFloyd · · Score: 5, Informative
      Another few figures of interest:
      • MiG-25 (high speed, high altitude interceptor): ~90,000 feet, and the highest most civilians can go. A few entrepreneurs bought these interceptors from ex-Soviet republics hard up for money. If you've got the money, they'll take you up to about 17 miles and Mach 3.
      • U-2 (spy plane): The US military won't say anything but 'greater than 90,000 feet.' I've personally heard from fairly reliable sources that the real figure is between 90,000 and 95,000 feet.
      • North American X-15 (research/rocket plane): 354,200 ft. Same mission as the X-2. Still holds the altitude record for a piloted aircraft (although I have a feeling that will be broken soon). This plane actually broke the US Air Force's 50-mile definition of space 12 times and the international FAI (Federation Aeronautique Internationale) definition of 100 km twice. It is the closest thing the Space Shuttle has to a direct ancestor.
      I don't know where you got your definition of 75 miles for the edge of space. The USAF awards astronaut wings to any pilot who goes past 50 miles, and the FAA, the FAI, and most importantly the X-Prize backers consider space to start at 100 km, or roughly 62 miles. While Spaceship One does have a good long ways to go, it's not quite as long as you describe it.
      --

      That's it. I'm no longer part of Team Sanity.
  2. Re:Earth curvature my ass by NearlyHeadless · · Score: 5, Informative
    I think the earth curvature is still there, albeit exaggerated by the fisheye lens...
    The distance to the horizon is 549 miles =sqrt(2*r*h+h*h). Or, approximately 546 miles along the surface of the Eath =r*acos(r/(r+h)).

    If I do the math correctly, the distance of the airplane from the (geometric) plane containing the cirle of the horizon is 75.4 miles, and the radius of the circle in the plane is 544 miles. So you would get a roughly equivalent view of circle on the ground that has a radius that is 7.2 times the distance of your eyes above the ground (7.2 = 544/75.4). This works out to about 39 feet for me (12 meters). So, yes, I think you could see the curvature.