Results 1 to 4 of 4

Math Help - trigonometrical heighting??

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    5

    Exclamation trigonometrical heighting??

    as part of a measurement scheme to monitor subsidence a target at P2 was mounted on the face of an embankment and observed from a theodolite at P1.
    Height of Collimation of theodolite at P1
    zenith angle P1-P2 95d30m00s
    Coefficient of refraction (K) -0.28
    R=6375km

    From the above compute the elevation of the target at P2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by jane?? View Post
    as part of a measurement scheme to monitor subsidence a target at P2 was mounted on the face of an embankment and observed from a theodolite at P1.
    Height of Collimation of theodolite at P1
    zenith angle P1-P2 95d30m00s
    Coefficient of refraction (K) -0.28
    R=6375km

    From the above compute the elevation of the target at P2
    More information is required:

    [1] The slope distance from P1 to P2.

    [2] The elevation of P1.

    You are probably aware that the coefficent of curvature and the coefficient of refraction have opposite signs.

    Curvature & Refraction become important when the distance from P1 to P2 is larger than 300 metres.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    5
    this is all that was givin i also left out the value for the height of colimation of the theodolite at P1 is 100m
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by jane?? View Post
    this is all that was givin i also left out the value for the height of colimation of the theodolite at P1 is 100m
    The slope (or line of sight) distance from P1 to P2 is required to
    compute the Curvature & Refraction correction for elevation determination.

    The
    coefficient of refraction = -0.28
    doesn't fit the normally acceptable range using miles or kilometers.


    The polar radius of the earth is approx: 6356.9 kilometers = 3950.0 miles
    The equatorial radis of earth is approx: 6378.4 kilometers = 3963.4 miles

    The R given as 6375km is within that range.
    The (C)orrection (vertical adjustment) required to compute the difference
    from a true horizontal line to a level line, at a (D)istance on the earth
    with a (R)adius is:
    C = R - sqrt( R^2 - D^2)
    C = correction in units given
    R = Radius of the earth
    D = Horizontal distance from observer

    If using kilometers:
    C = 6375 - sqrt( 6375^2 - 1^2 ) = 0.00007843
    and multiply by 1000 to get the correction in meters
    The curvature correction from a true horizontal line to a level line
    at 1 kilometer is 0.07843 meters


    If using miles:
    C = 3961.241 - sqrt( 3961.241 - 1^2 ) = 0.00012622
    and multiply by 5280 to get the correction in feet
    The curvature correction from a true horizontal line to a level line
    at 1 mile is 0.66644 feet

    The refraction correction (in earth's atmosphere) is approximately 1/7 the curvature correction.
    If using kilometers the coeffiecient for refraction would be close to
    0.07843/7 = 0.0112 per kilometer squared
    If using miles it would be 0.0952 per mile squared.
    The given coefficient does not fit either.

    If instead it had stated:
    "THE REFRACTION CORRECTION IS -0.28"
    The the horizontal (or slope) distance could be computed.

    Pedantic Note:
    The Earth curvature formula works uniformly well for
    a horizontal distance out to 32 kilometers (approx 20 miles). Distances greater than
    that require additional correcting tweaks.
    HOWEVER:
    The correction for refraction will vary (very noticeably) with air density. As the
    air temperature and barometric pressure change, so will the coefficient of refraction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometrical equation
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: September 23rd 2011, 03:13 AM
  2. trigonometrical
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 1st 2011, 01:12 PM
  3. Where did the trigonometrical functions come from???
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: June 29th 2011, 10:53 AM
  4. Prove the following trigonometrical identity-
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: February 3rd 2010, 01:38 AM
  5. trigonometrical qn
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 4th 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum