Welcome to the NavList Message Boards.

NavList:

A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

Compose Your Message

Message:αβγ
Message:abc
Add Images & Files
    or...
       
    Reply
    Re: HR-1 working re-creation and new questions
    From: Brad Morris
    Date: 2013 Feb 16, 02:11 -0500


    Confirming the result another way, we have:

    Let R = radius of earth
    Let angle = 0.103957108 degrees

    Then height above surface is = R/cos(angle) - R
    = 251103108.1953 / cos(0.103957108) - 251103108.1953
    = 413.3194096555 inches

    Circumference of earth is 2*Pi*R
    = 1577727359.99983 inches

    Arc Length = (360 - 2*angle)/360 * circumference of earth
    = 1576816160.14673 inches

    Length of string is then = 2*sin(angle)*R/cos(angle) + arc length.
    = 2*tan(angle)*R + arc length
    = 1577727360.99976

    Subtract circumference of earth from length of string to find delta.
    Delta is ~1" to an error of less than 1/10,000".  It appears that my angle is off by a tiny bit. 

    Brad











    On Feb 15, 2013 8:52 PM, "David Fleming" <d.l.fleming.1@gmail.com> wrote:

    Brad,

    Let me begin by saying I get numbers that agree with your result.

    Let me define:

    s = half arc length of earth where string is above ground
    a= half string lenth not touching earth
    R= radius of ball
    h= heigth of string at max above earth

    hopefully you can draw appropriate picture for these variables.

    1) delta = 2(a-s) the amount string exceeds sphere circumference is delta

    2) theta = s/R 1/2 angle at center of sphere

    3) tan(theta) = a/R

    4) a^2 + R^2 = (R+h)^2 Pythagorean Theorem


    Given above:

    expand tan(theta)= theta + (1/3)*theta^3 +... = a/r

    ignore other than first two terms on LHS and transpose first term to RHS and substitue 2) for theta on RHS

    (1/3)*theta^3 = (1/R)*(a-s) = delta/2R

    thus given delta and R we compute theta.

    then a = R tan(theta) we compute a

    from 4) we have h = sqroot(a^2 + R^2) - R

    or h = (1/2)*a^2/R ( factor R from sqroot and expand sq root in power series)

    using your value for radius of earth you'll get h = 4.13.36 inches for delta=1".


    It is interesting that in first problem which shoud read for string having circular symmetry height does not depend on size of the ball. A string one inch longer than a cue ball or one inch larger than the earth is 1/2Pi above the surface. While for second case which has mirror symmetry about the point you pull on the string the answer depends on the size of the sphere. And the answer is surprisingly large.

    Dave F
    ----------------------------------------------------------------
    NavList message boards and member settings: www.fer3.com/NavList
    Members may optionally receive posts by email.
    To cancel email delivery, send a message to NoMail[at]fer3.com
    ----------------------------------------------------------------

    View and reply to this message: http://fer3.com/arc/m2.aspx?i=122373

       
    Reply
    Browse Files

    Drop Files

    NavList

    What is NavList?

    Join NavList

    Name:
    (please, no nicknames or handles)
    Email:
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
    Email:

    Email Settings

    Posting Code:

    Custom Index

    Subject:
    Author:
    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site