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Cake day: June 29th, 2023

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  • I went up to the Lake Champlain area where there was some high altitude cloud cover. Fortunately, it didn’t affect the viewing basically at all. A cool side effect of the clouds/related atmospheric conditions though was that the sun had a 22° halo. I wish that 1) I’d had a camera that could capture it and that 2) I’d had the presence of mind to pay attention to what happened to it in the moments before and after totality.




  • ornery_chemist@mander.xyztoxkcd@lemmy.worldxkcd #2907: Schwa
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    6 months ago

    The point about stress is interesting. I’ve been playing with pronouncing the phrase, and almost everything tends toward [ɐ] when I speak the syllables one at a time, even the ones I marked with and pronounce as a schwa in normal speech. The notable exceptions are the final schwas in “obstruction” and “onions”, which tend toward [ɪ], and the -nel of “tunnel”, which is something like [nɫ] (vocalic ɫ) ~ [nəɫ].


  • ornery_chemist@mander.xyztoxkcd@lemmy.worldxkcd #2907: Schwa
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    6 months ago

    It helps when most of the vowels are the same and most other letters match their English counterparts lol.

    In case you get the urge to learn sooner:

    Here are some quick refs for consonants and vowels in English (RP = received pronunciation (a standardized form of English from the UK), GA = General American). Wikipedia pages for specific English dialects (e.g., Australian English) also contain a bunch of word/IPA pairs. Here are audio charts for vowels and consonants.



  • ornery_chemist@mander.xyztoxkcd@lemmy.worldxkcd #2907: Schwa
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    6 months ago

    Thank you for reminding me of this channel, I’d forgotten about it.

    Interesting about the merging. Schwa has always been weird for me because in my dialect it can be many sounds. I grew up saying “obstruction” as [ʌbstɹʌkʃɪn] like those around me. Then I hit grade school and was told by a straight-faced teacher that both the first and last syllables in this and similar words were schwas while pronouncing them differently :)



  • So many solver solutions that day, either Z3 or Gauss-Jordan lol. I got a little obsessed about doing it without solvers or (god forbid) manually solving the system and eventually found a relatively simple way to find the intersection with just lines and planes:

    1. Translate all hailstones and their velocities to a reference frame in which one stone is stationary at 0,0,0 (origin).
    2. Take another arbitrary hailstone (A) and cross its (rereferenced) velocity and position vectors. This gives the normal vector of a plane containing the origin and the trajectory of A, both of which the thrown stone must intersect. So, the trajectory of the thrown stone lies in that plane somewhere.
    3. Take two more arbitrary hailstones B and C and find the points and times that they intersect the plane. The thrown stone must strike B and C at those points, so those points are coordinates on the line representing the thrown stone. The velocity of the thrown stone is calculated by dividing the displacement between the two points by the difference of the time points of the intersections.
    4. Use the velocity of the thrown stone and the time and position info the intersection of B or C to determine the position of the thrown stone at t = 0
    5. Translate that position and velocity back to the original reference frame.

    It’s a suboptimal solution in that it uses 4 hailstones instead of the theoretical minimum of 3, but was a lot easier to wrap my head around. Incidentally, it is not too hard to adapt the above algorithm to not need C (i.e., to use only 3 hailstones) by using line intersections. Such a solution is not much more complicated than what I gave and still has a simple geometric interpretation, but I’ll leave that as an exercise for the reader :)