> Volkswagen just build the world's most aerodynamic road car ever.
A drag coefficient of 0.158 is not quite setting a record.
Aptera's road-legal car has a coefficient of only 0.13, while Hyundai achieved a drag coefficient of 0.144 with the Aero Challenge concept car [1] (but it's not clear whether that vehicle is road-legal, or even what it looks like).
Aptera is getting pretty close to shipping first customer cars this year, or at least seems to be, while this is a one-off research prototype - I think your comment is not quite fair.
Yes, it's a prototype. But it's from a company that has built countless production cars, and which also produced a road-legal limited production run of a pretty similar car (just with ICE instead of electric): https://en.wikipedia.org/wiki/Volkswagen_1-litre_car
I'm pretty sure they could bring this to market almost immediately, if there was a sizeable market for this kind of vehicle.
Aptera on the other hand has been around for a long time now, changing their product strategy every now and then, making announcements... I wish them all the best, I really do, but until they actually ship something, I will also stay skeptical.
> I still keep in a drawer an old OnePlus 5 I bought in 2017 with a pristine Android system.
How pristine is that when the last supported Android version is 10?
> This phone is sometimes necessary. Like when a known bug in /e/OS prevented my banking app from working for several months.
This surprises me. Until very recently I was using a phone I bought in 2017, which similarly supports no Android beyond version 10.
Since early this year, my bank app started insisting that I upgrade my Android to 11 or higher, and hence stopped working.
More recently, Gmail also started insisting I upgrade Android. As a result I had to get a new phone, which in many ways is worse than my old phone (still 7.2" but worse aspect ratio, no more full fingerprint sensor and much worse battery life).
So the idea of keeping an old phone around in order to bank seems rather odd to me.
> The attack above depends on conditions that ordinary Bitcoin keys do not meet.
> Ordinary Bitcoin keys are not affected by any of this
> A properly generated private key is spread across roughly 2 to the power of 256 possibilities. The square root of that is 2 to the power of 128, which is not attackable by anyone, with any budget, ever.
Note that this is not an extension of the pure λ-calculus.
Abstraction (λx.e) and application (f a) are missing,
although the let construct "let x = e1 in e2" is equivalent to their combination
((λx.e2) e1).
The paper has few details on the higher-level specification language in which users specify desired behaviour:
> Specification Language. Specifications are written as relations between input and output ports, expressed using linear expressions. On their own, specifications are not λ _λ programs. It is the job of the synthesizer to find λ _λ programs that realize a given specification. For example, a simple switching behavior can be specified as output[i] = input[j], while a 2x2 AllReduce operation can be written as output[1] = (input[1] + input[2])/sqrt(2) and output[2]= (input[1] - input[2])/sqrt(2).
> So in 2018, an international team of scientists created a BEC out of strontium atoms, and hit one of those atoms with a carefully tuned laser, exciting its outermost electron and turning it into a Rydberg atom. Several other atoms from the BEC were caught within between that outer electron’s inflated orbital.
This is not so much putting 170 atoms in one, but inflating one beyond the size of its 170-atom neighbourhood.
The hard part being to cool the neighbourhood to such an extremely low temperature that the inflated atom doesn't ionize.
The outer orbital is being treated as the last layer of an onion and the atoms are "inside" that. But orbitals are not spherical so our geometrical intuitions kinda fail.
Edit: Sorry for unclear wording: well, s-orbitals are spherical overlapping probability clouds, but they are not shells like an onion. Nor are they spheres as we commonly think of the geometrical object. To say something is "inside" a probability cloud gets difficult. Disclaimer: my memory/understanding of physics/chemistry/probability is poor. Edit 2: when orbitals of atoms overlap for molecules, don't the orbitals get funkier shaped probability distributions? Surely the same thing happens here and the cloud shapes would change - orbitals are not independent things you can just smoosh together or overlap willy-nilly due to qwuantum blah blah.
A little bit of hand waving, but I don't mind it much. A cloud of steam rising from a boiling pot doesn't have a well defined border or consistent shape, but a human looking at a nearby object can usually make a clear and reasonable decision about whether or not that object is "inside" the steam cloud.
So I thought of a nice question: how close-in does the hydrogen in Rubidium Hydride sit. For gaseous RbH and thinking in solid spheres instead of clouds:
The bond length of RbH is ~4.5 units.
The Hydrogen proton is *inside* the empty 5s radius (~4.7 units).
If we added an electron ("an antibonding electron") it would go into the 5s orbital. But that stretches the bond making the distance between the nuclei ~4.8 units.
In RbH, the Rb atom gives up its single outer 5s electron to the H atom, so there is no populated 5s electron shell in a Rubidium Hydride molecule.
That's the Bohr model. It works surprisingly well for hydrogen. Then there's the probability-cloud model, which works much better. And then there's the “strictly speaking, the orbital isn't actually a probability distribution at all; it's a wavefunction, Ψ, and ∣Ψ∣^2 gives you the probability density” model.
It gets weird. Like, maybe-there's-only-one-electron-in-the-universe weird.
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