"Half of all flights in the world are under 2 hours." Then they give examples of Fjord-town hopping in Norway, or Island hopping in Hawaii, which seem very niche.
Will they be able to compete in any of the busiest routes? Could it take me Melbourne-Sydney in 1 - 2 hours for < $100? I'd sacrifice some time and $ for an eco-friendly option, but obviously there's a limit.
https://en.wikipedia.org/wiki/List_of_busiest_passenger_flig...
Boston, Massachusetts (1h) Washington, D.C. (1h) Philadelphia, Pennsylvania (45m) Baltimore, Maryland (1h) Toronto, Ontario (1h 45m) Montreal, Quebec (1h 30m) Cleveland, Ohio (1h 45m) Pittsburgh, Pennsylvania (1h 15m) Buffalo, New York (1h 10m) Providence, Rhode Island (45m) Portland, Maine (1h 10m) Bermuda (1h 50m) Nantucket, Massachusetts (1h 15m) Martha’s Vineyard, Massachusetts (1h 25m) Ottawa, Ontario (1h 30m) Quebec City, Quebec (1h 45m) Rochester, New York (1h 5m) Syracuse, New York (1h) Manchester, New Hampshire (1h) Bangor, Maine (1h 25m) Burlington, Vermont (1h 10m) Richmond, Virginia (1h 10m) Norfolk, Virginia (1h 15m) Raleigh/Durham, North Carolina (1h 25m) Charlotte, North Carolina (1h 45m) Greensboro, North Carolina (1h 30m) Charleston, South Carolina (2h) Myrtle Beach, South Carolina (1h 45m) Savannah, Georgia (2h)
If it gets a little better than current 2 hour flight time then Detroit Chicago and Atlanta are in range - very significant.
The range is apparently 125 miles electric, and 500 miles as a hybrid. That's not enough for Melbourne Sydney. Give it a couple doubling-times though (5 years total?) and it will be.
Meanwhile aircraft take forever to develop and there should be enough of a market in shorter hall flights to occupy a scaling company in the meantime anyways.
The only way to reduce current is to put batteries in parallel, which for a given voltage doubles the weight of the battery pack.
You could go with lower voltage, but that means for a given, power, you need more current, which drops your efficiency.
This is supposed to be able to handle "up to" 2-hour flights, and (for jets at least) takeoff power is about 3x cruise power, so about 1.5 watts per watt-hour of battery, or a 1.5C peak discharge rate. What chemistries are you thinking of that cannot handle this?
The full power system for a prop looks like this: you have a battery of a specific voltage, which then runs a motor, which then runs a gearbox, which then turns a prop.
The motor and the gearbox can be considered as one unit - an electric motor has two factors, KV(RPM/volt) and KT(torque/amp). The higher the KV, the lower the KT. A high KV motor spins fast, but draws a lot of current for the same torque - putting it through a reduction gearbox turns it into a low KV, high KT motor. Naturally, low KV motors or (low KV setups) are more efficient because they draw less current for a given torque, and the heating power loss varies with current^2.
The prop needs to spin at certain RPM for max aerodynamic efficiency. Given the slider for motor/gearbox selection between high KV/low KT and the opposites, you generally want to have as high voltage as possible, so that you can run a low KV/high KT setup, which means that for the given torque, the current is minimal. I.e you have a motor spinning really fast, through a large reduction gear, driving a prop at the necessary speed and torque without much load on the motor.
So lets say you determine that you want a certain voltage, which requires a stack of cells in series. The only way to get more capacity is to duplicate that stack and put them in parallel. So your weight becomes quantized by the number of stacks you have in parallel. And the more stacks you have in series, the higher the weight jumps between parallel stacks counts.
Subtracting cells from stacks doesn't work well. Lets say you have a single stack of 10 cells 10s1p. If you do something like 8s1p, you lower the output voltage, which means you need to have slightly higher gear ratio to spin the prop at the same efficient rpm, which means you draw more current, which means the extra capacity in the cells doesn't really matter if you are drawing more current.
If your motors draw W Watts at peak, and you have N cells at V volts then the peak per-cell current will be (approximately -- there are internal losses) W/(NV) regardless of the geometry of your stacks.
If N is small, your designs may be restricted, but I'll hazard a guess that N will be large for a 30 passenger plane capable of 2 hour flights (e.g. a Tesla model 3 is almost 3000 cells; 31s96p).
The overall point that Im trying to make is that slight battery specific energy density improvements don't matter when compared against the power losses which are proportional to square root of the current.
An airplane under typical operation has a much narrower ranges of power than a Tesla in high power mode. Takeoff power is a low single-digit multiple of cruise power. For my math earlier I used a 3:1 ratio. A Model S Plaid is more like 50:1, or 30:1 for the "regular" Model S.
> The overall point that Im trying to make is that slight battery specific energy density improvements don't matter when compared against the power losses which are proportional to square root of the current.
You've completely missed making that point, since I still haven't seen an argument so far that the current must be very high.
There's already an Australian electric trucking company trailing plug n'replace heavy battery packs (with a forklift).
That allows for fast turnover times while used batteries recharge at terminals.
High altitude where the air is less dense is better for speed due to less drag. This is also power source agnostic.
When you size a gas engine for a plane, you design around the power needed to takeoff on a given runway, which is fine because in cruise, you will throttle down and use less fuel.
When you design an electric vehicle, if you design it the same way, you end up carrying effectively dead battery weight, which also affects power consumption during cruise (you either go faster, or use use more wing, which creates more induced drag).
IF you design it for cruise, you end up being horribly inefficient at takeoff.
The only 2 solutions are
1. 10 mile long runway where you can take your time accelerating, to reduce the torque requirement on the motor. You would basically limit the current draw and the plane would accelerate slowly, and eventually you get to cruise speed
2. Changing your flight profile, where you get to high altitude ASAP where you can be more efficient,