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Boeing-Backed, Hybrid-Electric Commuter Plane To Hit Market In 2022 (reuters.com)
An anonymous reader quotes a report from Reuters: A Seattle-area startup, backed by the venture capital arms of Boeing and JetBlue announced plans on Thursday to bring a small hybrid-electric commuter aircraft to market by 2022. The small airliner is the first of several planes planned by Zunum Aero, which said it would seat up to 12 passengers and be powered by two electric motors, dramatically reducing the travel time and cost of trips under 1,000 miles (1,600 km). Zunum's plans and timetable underscore a rush to develop small electric aircraft based on rapidly evolving battery technology and artificial intelligence systems that avoid obstacles on a road or in the sky. In a separate but related development, Boeing said on Thursday it plans to acquire a company that specializes in electric and autonomous flight to help its own efforts to develop such aircraft. Zunum's planes would fly from thousands of small airports around big cities to cut regional travel times and costs.
I've read about how planes are required to have certain amounts of fuel at certain points relative to their trip.
You need to have 100 or 200 nmi of fuel left once you reach for destination to divert if necessary. Depends on the Part and certificate under which you operate.
More to my point I have heard that this is why planes often have to dump unused jet fuel (usually conveniently done over less-desirable neighborhoods near the airport) before landing.
Planes have to have a provision to reduce weight if the plane must land before the destination. Generally, if the maximum takeoff weight (MTOW) is greater than 105% of the maximum landing weight (MLW), there will be a fuel dump option. It is not used often, only during an abnormality or declared emergency, if the type is so equipped. Otherwise, the plane will circle, if practical, to burn fuel to reduce weight for landing. If that isn't possible, an overweight landing will happen with the air frame taken out of service for a D-level inspection.
If your fuel instead is primarily batteries, how will that change these regulations?
The air frame will be designed such that MTOW = MLW. This isn't an unusual design criterion at all.
Quadcopter don't scale. I assume that's what you meant - virtually all of the toy and hobby "drones" are quadcopters.
The power produced by a propeller is proportional to it's length.
The weight of a craft, however, is proportional to it's length X width X height.
Suppose we have a toy that's 1 foot X 1 X 1. It's one cubic foot. Perhaps it weighs one pound. The 1 foot prop needs to make 1 pound of thrust.
Now we scale that "ten times bigger". Now the dimensions are 10x10x10. That's 1,000 cubic feet! "Ten times the size" is about a THOUSAND times the weight. But our prop is only ten times as long, so it makes ten times the thrust, enough to lift TEN pounds, not a thousand pounds.
In other words, as the size of craft increases, weight increases with roughly the size (length) CUBED. Prop thrust only increases directly proportional to size (length).
It's therefore therefore relatively easy to lift a small craft with props, but the power requirements go up real fast as the size increases, until you basically hit a wall of impossible physics. The largest helicopters that can be physically built carry about 40 people, whereas an A380 plane seats 853 people.
For an actuator disk of area A, with induced velocity v at the rotor disk, and with p as the density of air, the mass flow rate m through the disk area is:
m =pAv
By conservation of mass, the mass flow rate is constant across the slipstream both upstream and downstream of the disk (regardless of velocity). Since the flow far upstream of a helicopter in a level hover is at rest, the starting velocity, momentum, and energy are zero. If the homogeneous slipstream far downstream of the disk has velocity w, by conservation of momentum the total thrust T developed over the disk is equal to the rate of change of momentum, which given zero starting velocity is:
T=mw
Because tip velocity can't exceed c, w is limited to far below the transonic regime. Therefore w can't be increased beyond an easily achievable value. Meaning thrust T is limited to a (roughly) constant factor times m, mass air flow. Recall mass air flow is pAv. p, air density, we can't change. v is limited to far subsonic, so we can increase thrust T only proportionally to A, the area of the rotor disk. By middle school geometry the area of the disc is pi 2 r. Pi and 2 being constants, the area, and therefore the thrust are directly proportional to r, the radius (the length of the rotor blade),
Yes, of course you're right. It's the AREA of the disk, multiplied by the velocity, which determines how much air is moved at what rate, which equals how much thrust is created (Newton's third law). I typed it as circumference rather than area.
What I didn't go into was a consideration with real props which brings it closer to directly proportional, so the actual real-life performance is in between the area and the circumference. The inner part of the blade is of course moving slower than the tips, in terms of linear speed. In other words, if the rotor tips are moving through the air at 400MPH, halfway toward the hub it's only slicing through the air at 200MPH. So the portion of the rotor which is delivering maximum power is at the ends, measured by the circumference.
Does that make sense? The total area is proportional to r^2, but the area of blade working at full speed is proportional to just r. Meaning the total thrust is between X * r and X * r^2.
Your correction points out why large rotorcraft normally use one lifting rotor rather than multiple as hobby craft often do. A single 40 foot rotor has twice the swept area of the two 20-foot rotors which would fit on the same airframe. Of course, when you need rotor too big to build and affix to the airframe, such as a heavy-lift copter, you may end up with tandem rotors simply because a 120' rotor is Impractical.