varyingExpertise

joined 2 months ago
[–] varyingExpertise 5 points 23 hours ago

Hier SpongeBobs Chef mit Dollarzeichen in den Augen einfügen. Dummfug bringt empörte Klicks.

[–] varyingExpertise 7 points 2 days ago

It really uses a lot of energy and it doesn't have V2L or thermal preconditioning for the battery or super fast CCS charging or any of the other niceties that more modern platforms have.

That said, all of that is just fine for me, I just pull my trailer when working in the forest and very occasionally go on longer trips which are fine as well, to be honest. Chilling a bit at the charger every 2 hours is okay, usually my dog tags along so we're taking a break anyway.

At home it gets charged from my solar panels for about half the year, so that's cheap driving at least in summer.

Nah, you're right, it's a nice but not very modern thing that has been overtaken by time. I like it though, but I also use an IBM Model M, so take that as you will.

[–] varyingExpertise 4 points 2 days ago (1 children)

This is an ex leasing car. It was driven by the owner of a agriculture machine repair shop who apparently went everywhere his customers went as well, judging by the buckets of fine, sandy, northern german soil I have dug out from being the fender liners.

I think, Audi offered them for very low leasing rates for a while and therefore the market is now flooded with lots of these. They aren't very efficient and they are quite bad on the software side, so dealers are glad when they get them off the lot.

I just like having an electric car with all wheel drive that can be raised enough to travel the forest roads around here without worrying and it fills that specific need quite well.

[–] varyingExpertise 1 points 2 days ago

Right? This is a thing for me now as well.

[–] varyingExpertise 4 points 2 days ago

No, not at all. This is my first nice car and I made use of the general uncertainty around used electric cars and the eTron being a relatively old concept in this fast moving market. Cost around 30k€ for a bottom right configuration.

[–] varyingExpertise 9 points 2 days ago (2 children)

Yeah, bought it used. They are pretty cheap because they are really bad electric cars but nice, you know, cars. So I treated myself to a bad electric car which is a nice car for me anyway.

 

I don't know how I never got that idea. I've always complained that a panorama roof is mainly a feature for the rear passengers - well, this is actually rather excellent.

[–] varyingExpertise 2 points 5 days ago* (last edited 5 days ago)

Oh, wir (also ich und meine Nachbarn) hören heute also die alten Thunderdome Compilations. So ein Morgen wird das also.

"Amsterdam sucks, New York is a waste of time, to hell with London and certainly fuck Frankfurt.... THIS IS THE THUNDERDOME."

[–] varyingExpertise 2 points 5 days ago (1 children)

Ja, klar. Am Ende war die Frage eigentlich nur "wie wirkt sich das Seitenverhältnis auf den Vorfaktor aus". Und die Antwort war "eigentlich gar nicht, im Sinne einer Abschätzung", da hat @captain_unicode den deutlich kürzeren Weg zu einem sinnvollen Ergebnis gewählt.

[–] varyingExpertise 2 points 5 days ago (3 children)

Yes, aber du berücksichtigst dabei nicht, dass ein Bildschirm nicht quadratisch ist. In erster Näherung ist das so natürlich trotzdem plausibel und wäre dafür auch gut.

[–] varyingExpertise 4 points 5 days ago* (last edited 5 days ago) (5 children)

Naja. Dinge die mehr Fläche mit der gleichen Leuchtdichte hell machen sollen brauchen mehr Leistung. Die Fläche hängt quadratisch von der Diagonale ab. Mehr Diagonale ist quadratisch mehr Leistung. Sind die 30% für 10" plausibel? Schaun mer mal:

Für ein 16:9-Display gilt:

- Bezeichnen wir die Diagonale mit x, die Breite mit w und die Höhe mit h
- Wir wissen w/h = 16/9
- Die Diagonale x folgt dem Satz des Pythagoras: x² = w² + h²

1. Aus dem Seitenverhältnis: w = (16/9) × h
2. In Pythagoras einsetzen: x² = ((16/9) × h)² + h²
3. Vereinfachen: x² = (256/81) × h² + h² = (337/81) × h²
4. Nach h auflösen: h = x × (81/337)
5. Dann w: w = (16/9) × x × sqrt(81/337)

Die Fläche a ist Breite × Höhe, also:

a = w × h = (16/9) × x × sqrt(81/337) × x × sqrt (81/337)

Vereinfacht:

a = (16/9) × x² × (81/337) ≈ 0.4273 × x²

70 Zoll: 2093 square inch
80 Zoll: 2734 square inch

Also, etwa 30% mehr Fläche. Passt.

Eigentlich irgendwie so neunte Klasse Geometrie, wenn ich mich recht erinnere, ich würds schöner finden, wenn man annehmen könnte, dass sowas jeder selbst irgendwie überschlagen kann. Aber dann hätten wir eine Menge Probleme nicht.

[–] varyingExpertise 2 points 5 days ago

Hab die Tage noch mit einem Bauleiter von nem Windpark hier geredet. Anscheinend ist die Straßeninfrastruktur für Anlieferungen auch ne Katastrophe.

[–] varyingExpertise 1 points 6 days ago

Alright, not very original but I'll take it. Now do a)!

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