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Simple gravity concept question

  1. Jul 21, 2007 #1
    When an aircraft begins to nose up the effects of gravity begin to slow its velocity.

    By dividing 100% by 90 degrees you get the number 1.11. So for each 1 degree incline the force of gravity increases by 1.11%?

    So at

    1 degree the force of gravity is 1.11%
    45 degrees the force of gravity is 50%
    90 degress the force of gravity is 100%

    -90 degress the force of gravity is -100% meaning it is accelerating the aircraft in the dive.

    So at 1 degree the effect would be .0111*Ma?
    at 45 degrees it would be .5*Ma?
    at 90 degrees it would be 1.0*Ma?

    Diving straight down at -90 degrees would be -1.0*Ma?

    I just want to make sure I am mathematically and theoretically correct. I don't need to use a sine wave or anything like that? Meaning the Sin of the angle determines the percentage effect.
     
  2. jcsd
  3. Jul 21, 2007 #2

    xez

    User Avatar

    Well the sin of the angle of climb does determine the
    force component of gravity that is directed
    'backward' relative to the nose to tail orientation.

    There's no real reason to divide X degrees by X percent
    to get an interpolatory linear estimate sort of thing
    when you can just say that:

    F_rearward = sin(theta)
    F_broadside_bottomward=cos(theta)

    e.g. in level flight gravity pulls downward SORT-OF
    at 90 degrees to the flight direction, and there is no
    "tail-ward" gravity component if the head-to-tail line is
    taken to be horizontal.

    Anyway it's all a little confused and somewhat meaningless
    since to achieve level flight you STILL have to have an
    force upward counteracting the force of gravity, so you
    have an angle of attack and lift effect of the wings to the
    airstream that is giving you 9.81*Mass_kg newtons of lift
    in level flight to balance the 9.81*Mass_kg newtons of
    gravity force.

    You have your Lift/Drag ratio that is then the
    transformation of the Lift value to a Drag value which
    *is* of course a rearward force (Drag) so even in level
    flight at constant speed there is so much force
    in the FORWARD direction that is needed to overcome
    the Drag which, of course is converted to a vertical force
    by the wings' lifts/angles of attack to equal the downward
    gravity force.

    So it seems unnatural and unnecessary to distinguish
    the drag, vs. downward vs. forward vs. upward etc.
    forces in the way you've mentioned since you can't really
    discount the gravitational and drag effects even in level
    flight or under any other circumstance...
     
  4. Jul 21, 2007 #3

    xez

    User Avatar

    i.e. the aircraft is *always* 'nose up' to 'fight' gravity
    due to the angle of attack of the wings, even if the 'nose'
    may have horizontal orientation while the wings have
    a positive angle of attack.
     
  5. Jul 21, 2007 #4
    xez, thanks a ton!

    So a pure percentage i.e. 45 degress = 50% gravity effect is wrong. Wow, that seems so counterintuitive to me. The Sine of the angle determines the effect of gravity... wow. So according to my calc at 45 degrees the effect of gravity is about 71%. I know you are right but doesn't that sound crazy? I'm glad I asked because I would have never understood that on my own. The only reason I asked about sine was because I saw it in a different equation with gravity. I thought it must have been an error or weird exception.


    As for the other forces I know you're right. I'm just talking about a simple, theoretical world with as few variables as possible. Any more difficult and my mind will explode.
     
  6. Jul 21, 2007 #5

    xez

    User Avatar

    Well it's non-intuitive to think of sin/cos as relating to
    forces if you're thinking in terms of sine waves versus
    linear slopes.

    It will all be much clearer if you look back on the
    geometric definition of sin, cos, tan:

    Draw a circle of radius 1 circling the origin [x=0,y=0].
    X is your horizontal axis.
    Y is your vertical axis.
    Theta is the the angle of any given point on the circle
    starting measurement from the positive X axis, i.e.
    the circle contains point [x=+1,y=0] and that's the
    zero-degree point.

    X = COS theta
    Y = SIN theta

    If you recall that the circle has a RADIUS of 1, that means
    there's a length one vector from any point on the
    circle to the origin [0,0].

    If you recall the pythagorean theorem for the length
    of the sides of a right triangle you get:
    a^2 + b^2 = c^2 where c is the hypotenuse, and a
    and b are the lengths of the other two sides.

    So at 45 degrees on the unit-radius circle you have a
    line of length 1 to the origin at 45 degrees.
    The value of X is the COS, and the value of Y is the SIN.
    X = SIN 45 deg = sqrt(2)/2,
    Y = COS 45 deg = sqrt(2)/2,
    so the length of the triangles sides formed between
    the origin, the circle point at 45 degrees, and the
    X or Y axis is:
    hypotenuse^2 = x^2 + y^2
    1^2 = (sqrt(2)/2)^2 + (sqrt(2)/2)^2
    1 = 2/4 + 2/4
    1 = 1.

    So the reason it's 0.707.... is just because that's the
    length of the side of the triangle that's needed to reach
    a circle of radius 1 at 45 degrees.

    If you look at it graphically it'll make sense how to use
    SIN and COS in the context of an angle as part of a
    circle.

    TAN = Sin(theta) / Cos(theta) = Y / X.

    So at 45 degrees the X component of the force and the
    Y component of the force (SIN and COS) are equal,
    but their values aren't 0.5 just because of the radius of
    circle used to define SIN, COS, etc.
     
  7. Jul 22, 2007 #6
    Thanks xez

    I read your post several times and decided to do a test. I used a graphics program and built a 45 degree angle and measured the distances. Holy smokes you are right. I uploaded it here

    http://imageigloo.com/images/72145degree.gif

    It sounds counterintuitive and yet it is absolutely true.

    Thanks a TON!!!! I would have never understood that concept without your explanation. :biggrin:
     
  8. Jul 23, 2007 #7

    xez

    User Avatar

    You're quite welcome; I'm glad that helped you
    visualize the geometric aspects of the angles and vector
    force components!

    I'm sorry that I haven't been able to give a really simple
    answer to your overall desire for a simplified model of
    thrust vs. drag vs. lift vs. final speed vs. time etc.
    I kind of see what you're wanting, and it's not that complex
    but then again planes aren't rockets. It's easier to look
    at rockets since they have no 'lift' and no 'stall' or
    variable lift vs. angle vs. speed issues etc. etc.

    Take a look at this site; I'm sure they'll go into more
    detail than you're interested in in terms of the
    physics / mechanics of the flight dynamics, but perhaps
    some of the ways of looking at things will resonate with you
    and you can find new ways to simplify or think about the
    sorts of questions you have vs. the information you have
    available about the variables.

    http://www.av8n.com/how/
     
  9. Jul 24, 2007 #8
    Thanks xez,

    That site is gold! I'll put it in my favorites.

    As for my equation I'm looking for something just that simple, i.e. the rocket equation. I'm trying to create something theoretical with just a few variables. Don't worry though, I'm hot on the trail of that equation. Sadly, I've been sworn off working on it until next week.

    Thanks again!!!
     
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