Results 1 to 6 of 6

Math Help - How do I 'describe' this simply trigonometric relationship?

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    6

    How do I 'describe' this simply trigonometric relationship?

    I'm interested in the decreasing angles between objects at increasing distance from the viewer, as attached.

    I know that the the angle is the arctan of the opposite / adjacent. This is basic high school SOH CAH TOA.

    But how do I describe this relationship in text form? I'm interested in the way that distant things in the horizon appear more and more clumped together, because the more distant things are, the less 'vertical angle' from the observer they get to share.

    So I might say something like the "vertical angle of features decreases with the arctangent of distance over height" but that sounds wrong. How would I express this simply trigonometric concept elegantly?

    The specific example I'm looking at is mountains. When you stand on a high mountain the foreground appears large, but as you look higher and higher towards the horizon things get more and more squished together. Let's ignore earth curvature here for the moment! If anyone can help with the wording I'd be grateful. Thanks



    How do I 'describe' this simply trigonometric relationship?-stickman.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2010
    Posts
    6
    Is this any good?

    "Angular declination below the horizon is greater for closer objects, with the trigonometrical relationship Tan θ= height / distance"
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by jt7747 View Post
    Is this any good?

    "Angular declination below the horizon is greater for closer objects, with the trigonometrical relationship Tan θ= height / distance"
    You see those 5 small isosceles triangles standing on the ground?
    Let's say they are equally spaced..

    What's happening is that the angle at the observer's eye is decreasing as he focuses on more distant pairs of ground objects
    equally spaced.
    You can see it in your sketch.

    If you draw that same angle for closer objects,
    it would cause them to be much closer,
    if you know what I mean.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2010
    Posts
    6
    That's right, they are evenly spaced. I designed the diagram that way.

    Yes - it is helpful to think of the angle subtended between pairs of markers - large for close pairs, increasingly small for distant.

    It's the nature of that reduction with distance. I guess it's exponential, hec anything with a Tan function is exponential. I'm just trying to word it as a mathematician would, ie "varies in proportion to the tan of the bla bla bla please insert correct phrase here!"
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by jt7747 View Post
    That's right, they are evenly spaced. I designed the diagram that way.

    Yes - it is helpful to think of the angle subtended between pairs of markers - large for close pairs, increasingly small for distant.

    It's the nature of that reduction with distance. I guess it's exponential, hec anything with a Tan function is exponential. I'm just trying to word it as a mathematician would, ie "varies in proportion to the tan of the bla bla bla please insert correct phrase here!"
    If someone doesn't do more work on this, I will pick up the thread in the morning.
    Unfortunately, it's naptime here!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    If \theta is the angle the object makes to the eye of the observer
    and the point on the ground that the observer is directly over, then...

    tan\theta=\frac{height}{distance}

    and height may be fixed, while distance allowed to vary.

    (d)tan\theta=h=constant

    tan(0)=0

    tan\left(\frac{{\pi}}{2}\right)\rightarrow\infty

    Therefore there is a non-linear inverse relationship between the angle and the distance.
    Or, the product of "distance" and the "tangent of the angle" is constant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Can I simply e^3x/e^2x ?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 30th 2011, 06:29 AM
  2. [SOLVED] Simply ordered set
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 24th 2011, 11:09 AM
  3. Simply help.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 31st 2009, 12:20 AM
  4. How would you describe this relationship
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: June 15th 2009, 12:09 AM
  5. Simply complex
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: March 30th 2008, 04:25 PM

Search Tags


/mathhelpforum @mathhelpforum