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Notices by Don Romano 🍹 (thj@mastodon.cloud), page 13

  1. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 18:19:07 EDT Don Romano 🍹 Don Romano 🍹

    So why did the camera break when setting the same position twice?

    Because the vector calculations that control the camera orbit take two motion endpoints and a target point, calculates the surface normal and corner angle of the resulting triangle, and uses that as an axis of rotation for a vector that pivots around the target in an arc passing between the two other corners, and you can't take a surface normal for a triangle with an angle of zero, because that's not a triangle.

    In conversation Tuesday, 03-Sep-2019 18:19:07 EDT from mastodon.cloud permalink
  2. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 18:11:40 EDT Don Romano 🍹 Don Romano 🍹

    When the image in the 3D engine flickered, he thought it was because I wasn't using double buffering, but that's the default in WebGL, and the actual problem was numerical instability in a floating point equation in a vertex shader. He has no clue how any of this works.

    It's not merely that he doesn't know things. It's that he doesn't know that he doesn't know them. He knows so little that he doesn't see how limited his knowledge is, and how utterly unhelpful his suggestions are.

    In conversation Tuesday, 03-Sep-2019 18:11:40 EDT from mastodon.cloud permalink
  3. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 18:08:04 EDT Don Romano 🍹 Don Romano 🍹

    I don't tell people these things face to face.

    I just want people to respect me a little, and to understand that when I choose to do things a certain way, there's usually a very good reason for it, and this reason will become apparent later.

    My co-developer told me to ignore corner cases. He then complained when setting the camera to the same position twice broke it.

    He has no idea what counts as a corner case in a 3D engine. He thinks it's simple and is impatient because things take time.

    In conversation Tuesday, 03-Sep-2019 18:08:04 EDT from mastodon.cloud permalink
  4. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 17:57:59 EDT Don Romano 🍹 Don Romano 🍹

    Why be so uncompromising on professional standards of software development?

    The reason is simple: Why did I spend close to 30 years developing my skills?

    People take jobs that are unrelated to their degrees all the time. However, it only takes a few years to get a degree.

    I have dedicated myself to this craft for longer than some of my younger colleagues have been alive, and I never stopped learning.

    Tell me again why I should lower my standards...

    In conversation Tuesday, 03-Sep-2019 17:57:59 EDT from mastodon.cloud permalink
  5. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 17:43:43 EDT Don Romano 🍹 Don Romano 🍹

    When someone doesn't understand you, and feels insecure about it, you are put in a paradoxical situation where explaining it again will only serve to make them feel even less secure.

    The only thing that will work on the human level is to tell him that his way is better, even if it won't lead to the best outcome.

    There's another human in the equation, though, who isn't a very good liar, and will compromise on everything *but* his professional standards.

    In conversation Tuesday, 03-Sep-2019 17:43:43 EDT from mastodon.cloud permalink
  6. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 17:36:06 EDT Don Romano 🍹 Don Romano 🍹

    When you can plainly see that your work output maintains a higher standard than someone else's, and you're a terrible liar, and you need to have meetings with that person, that's not a good situation to be in.

    I told him that I think we are having some communication issues, but the problem is more that he doesn't understand me, and feels very insecure about it.

    Maybe he's uncomfortable because he's older than me and has worked there for much longer.

    In conversation Tuesday, 03-Sep-2019 17:36:06 EDT from mastodon.cloud permalink
  7. Don Romano 🍹 (thj@mastodon.cloud)'s status on Tuesday, 03-Sep-2019 17:10:46 EDT Don Romano 🍹 Don Romano 🍹

    But again, the fact that ugly code is all you need doesn't mean that I have to be content with that.

    I'm not sure what to do. The ambition level at work isn't high and the pay isn't great, but I need to work there for a few years to show prospective employers that I can actually hold down a job.

    I wish I wasn't in the situation of only being able to do this now, four years away from age 40. I'm a bit old to be climbing the ladder.

    In conversation Tuesday, 03-Sep-2019 17:10:46 EDT from mastodon.cloud permalink
  8. Don Romano 🍹 (thj@mastodon.cloud)'s status on Monday, 02-Sep-2019 02:01:12 EDT Don Romano 🍹 Don Romano 🍹
    • Byllgrim

    @byllgrim The Hard Problem of Consciousness kind of forces you to consider such options. I can't explain why I notice that I'm here. One could imagine a biomachine resembling a human and being equally capable, but incapable of noticing that it exists. A philosophical zombie. And we clearly aren't. So who's "looking"?

    In conversation Monday, 02-Sep-2019 02:01:12 EDT from mastodon.cloud permalink
  9. Don Romano 🍹 (thj@mastodon.cloud)'s status on Monday, 02-Sep-2019 01:58:12 EDT Don Romano 🍹 Don Romano 🍹
    • Byllgrim

    @byllgrim Or, as a physical body with a part of the being attached, and that part is what we call a soul. And much like an ant doesn't realise that it's a part of the anthill megsorganisn, humans don't realise they're a part of the universe megaorganism.

    In conversation Monday, 02-Sep-2019 01:58:12 EDT from mastodon.cloud permalink
  10. Don Romano 🍹 (thj@mastodon.cloud)'s status on Monday, 02-Sep-2019 01:54:56 EDT Don Romano 🍹 Don Romano 🍹
    • Byllgrim

    @byllgrim It's essentially just saying that humans are the universe looking at itself. You can believe in that in a literal sense without getting overly metaphysical, since matter and energy is what the universe is made of, and we are matter and energy looking at matter and energy. In the Buddhist sense, it's more spiritual, with the universe more like a large being and humans more like sentinels sent out by that being to keep an eye on itself.

    In conversation Monday, 02-Sep-2019 01:54:56 EDT from mastodon.cloud permalink
  11. Don Romano 🍹 (thj@mastodon.cloud)'s status on Monday, 02-Sep-2019 01:22:46 EDT Don Romano 🍹 Don Romano 🍹

    Gjertrud vil bli pensjonist i en buss?

    Men Gjertrud er da allerede pensjonist?

    Sist *Æ* sjækka va det bare *EIN* ting ho Gjertrud ønska sæ, og det va *FOOD* prosæssor. Bættre, sjøh!

    In conversation Monday, 02-Sep-2019 01:22:46 EDT from mastodon.cloud permalink
  12. Don Romano 🍹 (thj@mastodon.cloud)'s status on Sunday, 01-Sep-2019 13:58:46 EDT Don Romano 🍹 Don Romano 🍹

    This is by far the best explanation of Buddhism I've ever seen:

    https://www.youtube.com/watch?v=h6fcK_fRYaI

    It's exactly the sort of thing of thing Alan Watts would talk about in great detail.

    In conversation Sunday, 01-Sep-2019 13:58:46 EDT from mastodon.cloud permalink
  13. Don Romano 🍹 (thj@mastodon.cloud)'s status on Sunday, 01-Sep-2019 07:41:24 EDT Don Romano 🍹 Don Romano 🍹

    If you have a parametric function in polar form

    P = ( r, θ )

    where r and θ are functions of t, and you convert it to Cartesian coordinates and call it the derivative

    C' = ( r cos θ, r sin θ )

    of a function C, then it's not exactly trivial to find that function through integration, unless r and θ are extremely simple.

    In conversation Sunday, 01-Sep-2019 07:41:24 EDT from mastodon.cloud permalink
  14. Don Romano 🍹 (thj@mastodon.cloud)'s status on Sunday, 01-Sep-2019 06:43:48 EDT Don Romano 🍹 Don Romano 🍹

    Cubic Hermite splines are a lot easier to control than cubic Bezier splines.

    With the cubic Bezier function, you set 2 endpoints and 2 control points along the x axis, and the control points signify, uh, whatever.

    With the cubic Hermite function, you specify the first derivatives (tangents) at the endpoints.

    "What are the slopes at the endpoints?" makes a hell of a lot more sense than "Which two waypoints should we halfheartedly approach as we travel between the endpoints?"

    In conversation Sunday, 01-Sep-2019 06:43:48 EDT from mastodon.cloud permalink
  15. Don Romano 🍹 (thj@mastodon.cloud)'s status on Sunday, 01-Sep-2019 06:10:23 EDT Don Romano 🍹 Don Romano 🍹
    • Mike, First of His Name

    @mike For my part, it's Sunday noon and I'm messing with vectors and differential equations.

    In conversation Sunday, 01-Sep-2019 06:10:23 EDT from mastodon.cloud permalink
  16. Don Romano 🍹 (thj@mastodon.cloud)'s status on Sunday, 01-Sep-2019 04:47:43 EDT Don Romano 🍹 Don Romano 🍹

    Wow... The Aztecs had a fucked up origin myth... A city offers them hospitality, and they say thanks by sacrificing a woman, flaying her and making a boy dance in her skin?

    In conversation Sunday, 01-Sep-2019 04:47:43 EDT from mastodon.cloud permalink
  17. Don Romano 🍹 (thj@mastodon.cloud)'s status on Saturday, 31-Aug-2019 16:37:27 EDT Don Romano 🍹 Don Romano 🍹

    The reason that this problem interests me is because I have a vertex shader and I want to bend a plane while preserving its area.

    Because shaders execute in parallel, and loops are tricky, it would be better to integrate.

    It would be enough if I could simply integrate x'(t), y'(t) over an interval t = [a, b] given φ'(t) = c, where c is a constant. This boils down to rotation around an axis and isn't too hard to hack together, but is far less general.

    In conversation Saturday, 31-Aug-2019 16:37:27 EDT from mastodon.cloud permalink
  18. Don Romano 🍹 (thj@mastodon.cloud)'s status on Saturday, 31-Aug-2019 16:16:23 EDT Don Romano 🍹 Don Romano 🍹

    Math thing that currently interests me...

    Suppose you had 2D parametric function

    x(t), y(t)

    with the property that differential vector is always a unit vector

    | x'(t), y'(t) | = 1

    You then convert that to polar coordinates

    r'(t), φ'(t)

    Since r'(t) is always 1, it can be dismissed, and φ'(t) alone describes the traced shape.

    Problem: Given φ'(t), how to find the original 2D shape?

    Analogy: Find the position of a car moving at constant speed, given the steering wheel angle over time.

    In conversation Saturday, 31-Aug-2019 16:16:23 EDT from mastodon.cloud permalink
  19. Don Romano 🍹 (thj@mastodon.cloud)'s status on Saturday, 31-Aug-2019 14:46:08 EDT Don Romano 🍹 Don Romano 🍹
    • regines

    @regines Yes, and I guess the point I was trying to make is this: Many third-world countries have this problem too. On paper, there is enough food and money to give the citizens a decent life, but the government is corrupt or inept, so you get charities helping people on the ground instead.

    In conversation Saturday, 31-Aug-2019 14:46:08 EDT from mastodon.cloud permalink
  20. Don Romano 🍹 (thj@mastodon.cloud)'s status on Saturday, 31-Aug-2019 12:55:32 EDT Don Romano 🍹 Don Romano 🍹

    When I take Vyvanse, I tend to get in this mood where I want everything I do to be absolutely perfect.

    Granted, I'm already prone to that, but the Vyvanse seems to accentuate it.

    Ever tried to formulate something so it doesn't sound rude, and it's really important that you say it, but you can't find a way of saying it nicely, but you don't want to be blunt, so you just run around in circles and drive yourself mad?

    It's a bit like that, except with source code.

    In conversation Saturday, 31-Aug-2019 12:55:32 EDT from mastodon.cloud permalink
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