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2 yr. ago

  • I suspected, but I'm not about to let facts get in the way of comedy.

    (No really, I feel bad)

  • Maybe, maybe not. Time will tell.

    See you around, hey?

  • I've also been watching this account, and I think it's an agitator organization. The posts are too frequent across several focused topics, and engagement is contemporaneously too cogent across those topics for it to be a single person.

    I call shenanigans.

  • Alternatively: I find I rarely need or want to visit a user's profile, and would be fine with accessing it through the 3 dots menu.

  • Not only does it need blocking on the wheels, but that thing's also apt to be stolen with the hitch just sitting there like that. They might consider parking a car in front of it too.

  • Oh, goddammit.

  • Black eyed peas. I'm not gonna tellya twice.

  • Nuclear energy has insane energy density in terms of MJ/kg (something like 3.9 x 106 ) versus chemical fuels (4.5 x 101), but it's grossly inefficient because most of the output is waste heat and "hot" isotopes-- the last things we need. I don't have hard numbers on hand but I wouldn't say nuclear is more than a few tens of percent efficiency. Then there's the capital costs to build, maintain and operate plants PLUS costs to source, refine, transport, and store the fuel, and then transport and discard (contain) waste product. Not worth it at scale.

    Versus Solar, Wind and Tidal which are far less energy dense per unit mass of working fluid¹, but enjoy up to 80% efficiency, and are relatively easy to scale.

    Nuclear still makes sense, I think, in interior areas like the American Midwest where wind and solar are fickle, and transportation (transmission) costs for tidal would be unsustainable.


    ¹ Not a fair comparison because solar efficiency is quantized on intensity x area / time, while wind and tidal would quantized on flux density, or (mass / area) x velocity (over time?).

  • Sure that's easy. All you need is an HX711 sensor load cell amplifier taped to the floor of the chamber where it sits under an edge of your bottle.

    Full, 5 gal (US) jug or water weights 40lbs nominal. If you have the same bottom feed dispenser I have, it starts drawing air with about 16-20oz of water left in the bottom.

    You'd want to calibrate first for the sensor without weight (tare), and then for a full bottle's weight (40lb) accounting for cosine error of the bottle being slightly tilted by the thickness of the sensor. Or, tape a couple of shims at 120° offset from the sensor to level out the bottle.

    Finally, note the weight of an "empty" bottle with that 18ish oz of water in it, which will probably be close to the zero value.

    Now you can calculate % full by looking at the ratio of the current weight between your full/empty value as:

    %Water = [ (W_full - (W-18oz -W_tare) ) - W_now ] / (W_full - (W_18oz - W_tare) ) x 100

  • Did we get to the disappointment and broken promises yet? The new class induction into the Graveyard of Good Ideas is always my favorite talk.

  • I guess technically that'd be the email account at my employer where I started some decades ago. Before Google and their Gmail was a thing, anyway. Or, livejournal, I guess. ...MySpace... Oof, AOL... Usenet, maybe...

    Gah. I'm old.