I am in Florida for a meeting this week, having flown from Albany to Ft. Lauderdale. Due to the vagaries of the air travel system, though, this required a change of planes in Orlando. The Orlando-Ft. Lauderdale flight is sufficiently short that I like to think of it as a ballistic route– you’re not cruising at altitude for any significant time, but start going down practically as soon as you’re finished going up.
Let’s imagine that we have a commercial enterprise– let’s call it Angry Birds Airlines– that wants to fling objects from Orlando to Ft. Lauderdale. What would that require? That is, could you build some sort of giant catapult to fling people from Orlando to Ft. Lauderdale? And could you estimate the reasonableness of this using physics from first-year classical mechanics?
For the sake of argument, let’s imagine that the giant slingshot launcher and the glass-and-steel arrival terminal the passengers will be crashing into are at the same elevation. We can further simplify the problem by ignoring the effect of air resistance. That’s not remotely a reasonable assumption, of course, but it will do for these purposes– if the idea isn’t feasible without air resistance, there’s no way it will work with air resistance.
Given those assumptions, it’s easy to find the range for a projectile launched at speed v and at an angle θ with the horizontal. The result, cribbing the equation graphic from that Wikipedia page, is:
Here, d is the distance traveled and g is the acceleration due to gravity, 9.8m/s2. The maximum range for a given launch velocity occurs at an angle of 45 degrees, so let’s just assume that our angry but entrepreneurial avians have set their passenger launch system at that angle.
Now, we know the start and end points for our journey, namely Orlando and Ft. Lauderdale. Google Maps says these are separated by about 200 miles (give or take), so what we want to know is what speed you would need to launch a projectile at in Orlando to have it land in Ft. Lauderdale. We can re-arrange our equation to get:
v = (gd)1/2
that is, the speed we want is the square root of the distance between the cities multiplied by the acceleration of gravity.
A distance of 200 miles is 300 km at the precision we’re using, and g is close to 10, so we have:
v = (3,000,000 m2/s2)1/2 = 1,700 m/s
That’s 5 and a bit times the speed of sound. So, yeah, that’s not going to be a real successful money maker for our furious feathered friends. On the bright side, though, the trip would only take about four minutes…
So, if you’ve ever wondered what physicists do to pass the time during long airport layovers, now you know.
Not to mention the sudden deceleration on arrival in FLL…
I dunno; air resistance might help make the landing survivable. It was certainly helpful in the suborbital Mercury-Redstone flights, because it meant the astronauts were not pounded to pemmican on arrival. On the other hand, the Redstone is a ballistic *rocket*, and what you’re describing is a launcher that stays on the ground — a super-trebuchet, a railgun, a cannon, whatever — and that means the acceleration to 1,700 m/s has to be pretty darn near instantaneous. If you’re still bored, you could calculate the G load and try to estimate the condition (and perhaps recognizability) of the passengers after launch. 😀
Reminds me a little of HARP, the Canadian project to ultimately launch pico satellites using gun-launched rockets. The idea was that the gun would serve as the first stage, allowing quite small sounding rockets to achieve orbit. They got to a nice proof-of-concept but lost funding before they could scale up to an orbital launch. One of the limiting factors was that the satellites would have needed to be quite robust in order to survive the launch.
But you are saying that this would work perfectly for luggage right?
The device for this has been built before, the Paris gun
http://en.wikipedia.org/wiki/Paris_Gun
so with air resistance you only get about 80 miles range, with a top of the arc being at 25 miles. While not practical for a trip Orlando-Ft. Lauderdale, it would be great for Tucson-Phoenix (talk about a BAD flight during monsoon season, it never makes it past 6000 ft) or to connect the NY airports.
There was once a proposed design for air travel that had the airliners sweeping in above airports and picking up and delivering passenger compartments by way of a cable while remaining in flight. This was planned to eliminate the need for the airplanes to land to take on or put out passengers or cargo. On the field the passenger boxes would be brought up to matching speed on a truck before the connection was made and the compartment hoisted away.
Sounded like quite a ride.
This was simply an enlargement of the very real technique developed in the early 60s of retrieving people/loads from the ground by having a cargo aircraft pick up a cable held aloft by a balloon.
The key factor in both cases is that a load suspended by a cable tends to balance the forces of gravity and air resistance so that the initial relative movement is up. A great benefit if you intend to avoid obstacles on the ground. (Sounds to me like a dynamic worth mathematical exploration.)
Don’t you also need to take into consideration the curvature of the Earth? Or, for really precise calculations, the non-spherical nature of the Earth.
Also, do Coriolis Forces come into play?
Factoring those effects into the equation should keep you busy on a Florida to California flight. 😉
Dave
That is, could you build some sort of giant catapult to fling people from Orlando to Ft. Lauderdale?
Depending what part of Florida you’re in, there might be a market for a device that can fling people back to Orlando. People who live down there, as I once did, often develop a love-hate relationship with tourists, some of whom the locals would quite frankly prefer not to wander beyond Disney World. (Those would be the ones who allow their brains to go on vacation along with their bodies.)
There is a huge flaw in your analysis: you have assumed an instantaneous launch (and landing) with infinite G forces. How did I figure this out? I tried calculating how long the “throw” would have to be if you wanted to launch with a constant a = 3g. It takes over 48 km to get up to 1700 m/s!
Just like your aircraft flight, much of the trip will be spent during “takeoff” and “landing”. Unlike your plane trip, tourists would probably pay extra just to take that E-ticket Ride to the beach and back.
I’ve flown on a route like that a lot in the past. “We’ve reached our cruising altitude and begun our descent to ….” Back in the ancient past, it was quite a trick to hand out drinks and snacks when the plane never really leveled off for more than a few minutes.
Just like your aircraft flight, much of the trip will be spent during “takeoff” and “landing”.
However, it would avoid an annoying feature of some flights I have been on: where pushback to wheels up takes more time than wheels up to wheels down. That’s even normal for some flights: not just puddlejumper hops over mountain ranges like Denver-Aspen (not a good idea if you are susceptible to motion sickness, especially for late afternoon/early evening departures–on that DEN-ASE flight the flight attendant passed out the water bottles *before* we pushed back since she knew they wouldn’t be able to do in-flight service), but mainline routes like Newark-Boston. And while I haven’t flown that particular route, it may be true of Orlando-FLL as well.
Recalculate for operations on the moon. Reaction mass could be very expensive there. You would still need a rocket engine for landing and such. But a runway and catapult system could probably save a lot of reaction mass.
“So, if you’ve ever wondered what physicists do to pass the time during long airport layovers, now you know.”
How do you like your new smartphone?
More importantly why didn’t you take the Southwest direct flight from Albany to Ft. Lauderdale?
More importantly why didn’t you take the Southwest direct flight from Albany to Ft. Lauderdale?
Because there’s only one direct flight a day, and it’s at a time I wouldn’t be able to make after my morning class. And, this week, the direct flight was cancelled due to snow, while the indirect flight made it out.
The comment about taxi and ground hold time reminded me of something you young folks may never experience: taking off on a puddle-jumper flight and immediately going into a holding pattern for your destination. Today you can’t take off until you have a clearance for a landing slot. Back then (before the traffic control strike in the Reagan era) you didn’t get a landing clearance until you were in the air.