1. Joined
    11 Nov '05
    Moves
    43938
    08 Jul '07 07:071 edit
    Originally posted by AThousandYoung
    2*pi*(r+5) - (2*pi*r)
    2*pi*(r+5-r)

    10*pi
    What about if the track resembles a figure eight?
    Does it matter if you have a little loop at one side and a bigger one at the other side? Or does they have to be symmetric?
  2. Standard memberwolfgang59
    Quiz Master
    RHP Arms
    Joined
    09 Jun '07
    Moves
    48793
    08 Jul '07 10:38
    Providing there is no 'figure of 8' or similar the difference in track length will be 2*pi*(track width)

    A figure of eight will exactly even out the inside and outside lanes.
  3. Joined
    28 Jan '06
    Moves
    24604
    08 Jul '07 19:16
    Originally posted by wolfgang59
    Well I guess that spoils it for the non-Mathematicians! I was hoping for some intuitive guesses before the calculators came out!


    What if it were Jupiter and not the Earth though? ;-)
    Oh. Sorry. I was just so excited. It was a surprizing answer to me, so thanks for the question.
  4. Joined
    11 Nov '05
    Moves
    43938
    08 Jul '07 20:12
    Allright, let's complicate a little bit further:

    You have this closed circuit racing track of Scalectric, no eight.

    What about if you have two crosses, where left track become a right track and vice versa? Does this alter the previous answers?
    Where and how should you put them on the track to make it calculable?
  5. Standard memberAThousandYoung
    or different places
    tinyurl.com/2tp8tyx8
    Joined
    23 Aug '04
    Moves
    26660
    10 Jul '07 02:17
    Originally posted by FabianFnas
    Allright, let's complicate a little bit further:

    You have this closed circuit racing track of Scalectric, no eight.

    What about if you have two crosses, where left track become a right track and vice versa? Does this alter the previous answers?
    Where and how should you put them on the track to make it calculable?
    I guess you'd do each loop separately and then add them.
  6. Joined
    11 Nov '05
    Moves
    43938
    10 Jul '07 06:38
    Originally posted by AThousandYoung
    I guess you'd do each loop separately and then add them.
    Actually, if you put them at places with some certain properties you don't have to calculate anything.
    Yes, I played with Scalectric too when I was young.
  7. Joined
    02 Jun '07
    Moves
    2926
    15 Jul '07 21:18
    i've passed math courses through college calculus, so I don't dispute the math. However, it seems extraordinarily counterintuitive to me that 1 meter more ribbon around a golf ball's equator remains "magically suspended" by the same height as 1 meter more ribbon does around Jupeter's equator. This intuitive feeling is intensified while I am moving on a long journy over the surface of the earth. Therefore, I have postulated what physics would be like if pi was a function of diameter (if pi was greater for very small diameters (compared to human scale) and smaller for very large diameters) and I think that very small spheres might seem to behave like standing waves to observers of human scale, and that apparent red shifts and time dilations might manifest themselves when human scale observers scan the overall universe. Anyone care to comment on my absurd thought experiment?
  8. B is for bye bye
    Joined
    09 Apr '06
    Moves
    27526
    15 Jul '07 22:15
    Originally posted by m3mnbgl5n79
    i*** Therefore, I have postulated what physics would be like if pi was a function of diameter (if pi was greater for very small diameters (compared to human scale) and smaller for very large diameters)... Anyone care to comment on my absurd thought experiment?
    I wouldn't want to stand in any buildings that you had a hand in.
  9. Joined
    11 Nov '05
    Moves
    43938
    16 Jul '07 09:471 edit
    Originally posted by FabianFnas
    Actually, if you put them at places with some certain properties you don't have to calculate anything.
    Yes, I played with Scalectric too when I was young.
    Okay, you have non-eight track and you want the left end right track equally long. Then you put two cross-overs somewhere. Where do you put them?

    Answer: You put the two cross-overs anywhere as long they are parallell to eachother but in opposite directions. Then you are sure that the two tracks are equally long.
Back to Top

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.I Agree