# The First Weightless Wedding

Back in the early nineties, when I unsuccessfully tried to get a weightless experience business started, this was one of the markets for it. It took a lot longer than I hoped or expected back then, and I didn’t make it happen, but I’m glad that someone did. And I see that she used one of the designs that Misuzu had the contest for as a wedding dress.

Of course, as usual, almost everyone gets this wrong:

The idea of these flights seems to be that the plane makes 16 huge dips from 36,000 feet to 24,000 feet to simulate zero gravity.

The weightless effect doesn’t just occur on the descent — it occurs through the entire parabola, up and down. If it only happened on the way down, the weightless period would only be half the time that it actually is.

## 8 thoughts on “The First Weightless Wedding”

1. Bill White says:

Hmmm . . .

From the Zero G Corporation website:

Aboard our specially modified Boeing 727, G-FORCE ONE, weightlessness is achieved by doing aerobatic maneuvers known as parabolas. Specially trained pilots perform these aerobatic maneuvers which are not simulated in any way. ZERO-G’s passengers experience true weightlessness. Before starting a parabola, G-FORCE ONE flies level to the horizon at an altitude of 24,000 feet. The pilots then begins to pull up, gradually increasing the angle of the aircraft to about 45° to the horizon reaching an altitude of 34,000 feet. During this pull-up, passengers will feel the pull of 1.8 Gs. Next the plane is “pushed over” to create the zero gravity segment of the parabola. For the next 20-30 seconds everything in the plane is weightless. Next a gentle pull-out is started which allows the flyers to stabilize on the aircraft floor. This maneuver is repeated 12-15 times, each taking about ten miles of airspace to perform.

During the ascent, the passengers experience 1.8 gees and during the descent they experience weightlessness for between 10 and 20 seconds.

Rand is this quote a test, or something?

The weightless effect doesn’t just occur on the descent — it occurs through the entire parabola, up and down. If it only happened on the way down, the weightless period would only be half the time that it actually is.

2. It’s poorly worded and misleading (one of the reasons that so many journalists get it wrong). “Push over” in this context doesn’t mean go to a negative flight path angle. It means it’s reduced from forty-five degrees and has started into the parabola. The first ten to fifteen seconds are ascending, the second are descending, until it reaches forty-five degrees negative, at which point it pulls out (another high-gee maneuver). It implies that 34,000 feet is the peak altitude, but if so, it’s simply wrong, because at peak altitude the aircraft is flying level.

If I were in charge of Zero G, I’d rewrite this to make it more clear what’s going on, because it doesn’t make sense as written. You don’t have to take my word for it — this is just basic kinematics and dynamics. Suborbital flights are parabolic (at least approximately, for short ones –longer ones are more accurately modelled as a section of an ellipse) so they are symmetrical. And a parabola in an aircraft is in essence a suborbital flight within the atmosphere.

3. The best way to think of it is as firing a cannonball, with the “explosion” being the end of the pullup under thrust, after which the engines are idled back to just the level needed to balance the drag, resulting in a ballistic trajectory. The forty-five degree angle (as Galileo determined) maximizes range (and presumably, though I’d have to pull a derivative, time).

4. mpthompson says:

Rand is correct. Zero Gs are indeed experienced during a portion of the ascent.

While the aircraft transitions from level flight to the ascent the passengers will experience 1.8 Gs, but that will quickly return to 1 G as the plane assumes a steady climb. Zero Gs occur when the pilot then pushes the nose forward to change the vertical acceleration of the aircraft to -9.8 m/s^2 — matching the acceleration of gravity in the vertical direction. The aircraft is still ascending during this phase while the passengers experience zero Gs, but its vertical velocity is falling rapidly. Eventually the aircraft reaches a vertical velocity of 0 m/s (the peak of the parabolic trajectory) and continues to accelerate towards the ground at -9.8 m/s^2. Ultimately the pilot must pull out of the downward acceleration and begin the next parabolic cycle.

BTW, the length of time in zero G flight is entirely dependent upon the vertical velocity the aircraft can achieve. The faster the vertical velocity the longer the weightless ride will be. Outfitting the Concorde for zero G flight would have been fun :-).

5. …the length of time in zero G flight is entirely dependent upon the vertical velocity the aircraft can achieve.

Right. It’s not obvious to me why they would do exactly a forty-five degree angle in and out — I’d think that’s a function of the Mach limit of the airframe and how many gees the structure/passengers can take.

6. Leland says:

I dunno, but that looked fairly umcomfortable. Brides hitting her head on the ceiling or falling on the ground. Didn’t hear audio, so maybe she was ok with it.

It will be better when they can do such a thing in a sub-orbital flight rather than parabolic flight.

7. Didn’t hear audio, so maybe she was ok with it.

She didn’t seem to complain, at least in that story.

It will be better when they can do such a thing in a sub-orbital flight rather than parabolic flight.

As I said, there really is no distinction between the two, except that one doesn’t leave the atmosphere (which isn’t why the weightless period is longer — that’s a function of the vertical speed component, as Mike said).

8. notanexpert says:

The best way to think of it is as firing a cannonball, …

I wonder if a better analogy might be to think about what a baby feels if you were to toss it up in the air. It feels weightless from the time it leaves your hands (moving upward) to the time it (hopefully) returns to your loving arms.