Arc of impossibility?

Discussion in 'Math' started by AlbertHall, Sep 17, 2018.

  1. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,181
    1,119
    Hi,

    Thanks for the reply.

    Oh in your previous post you were talking about some vector operations so i thought you could show that work.
    I would really like to see this other approach as it differs from mine.
    I understand your most recent approach and the explanation is better than before i think.

    I'll try to post mine either tonight or tomorrow sometime. I've been busy watching the judge confirmation hearings today so didnt do much else.

    For a quick explanation though, the logic is that if we have two points and some radius, no matter what that radius is we have two possible centers which could in the degenerate case be the same but in the more general case are not the same, and because the radius is a maximum constraint that means the the two centers can not be farther apart then 2*R. The result of this approach has been very interesting and with a surprise twist which i will show. The main point i was trying to make though is that this constitutes a verbal description while the mathematical proof would be more towards a set of equations that automatically spit out the result. I'll be sure to post this very soon.
     
  2. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,181
    1,119
    Hi,

    Thanks for your reply.

    [joking ON] It's my secret forever :) [joking OFF]

    I'll try to post tonight or tomorrow as soon as possible.
     
  3. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,181
    1,119
    Hello again,

    Here is an alternate proof in the form of a derivation.

    Distance between the two given points:
    d=sqrt((x1-x2)^2+(y1-y2)^2)
    or alternately:
    d=sqrt(dx^2+dy^2)

    Centers (ha,ka) and (hb,kb) of the possible two circles:
    ha=(x2+x1)/2+(sqrt(r^2-d^2/4)*(y1-y2))/d
    ka=(y2+y1)/2+(sqrt(r^2-d^2/4)*(x2-x1))/d
    hb=(x2+x1)/2-(sqrt(r^2-d^2/4)*(y1-y2))/d
    kb=(y2+y1)/2-(sqrt(r^2-d^2/4)*(x2-x1))/d

    Squared distance between those centers:
    (ha-hb)^2+(ka-kb)^2

    which comes out to:
    (4*(r^2-d^2/4)*(y1-y2)^2)/d^2+(4*(r^2-d^2/4)*(x2-x1)^2)/d^2

    and noting the two definitions of 'd' far above and seeing that we have d^2 in the numerator and denominator when factored:
    ((dy^2+dx^2)*(2*r-d)*(2*r+d))/d^2

    we get a simplification:
    (2*r-d)*(2*r+d)

    and since this is the squared distance between centers, taking the square root:
    sqrt(4*r^2-d^2)

    This last expression is real when 4*r^2>=d^2 but imaginary otherwise.
    When the expression is imaginary the circle centers are farther apart then 2*r. Thus we can say that 4*r^2 must be greater than or equal to d^2. This also allows us to calculate a minimum radius for two given points (which of course not that hard to do anyway).

    Lastly, if we calculate the imaginary part of the above we get as one of the *factors*:
    sin(atan2(0,4*r^2-d^2)/2)

    and this factor is either 1 or 0 or undefined, but the undefined result is only when 4*r^2-d^2 is zero. Thus any other case (other than when the two centers are exactly 2*r apart) this function spits out a binary result :)
    Of course as usual the numerical stability must be examined before use too.

    So the results match, and the last thing is just a little bit of a trick.
     
    Last edited: Sep 28, 2018
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