1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Go home airplane, you are drunk! (Another Airplane Problem Thread)

  1. Sep 17, 2014 #1
    An introductory picture.

    1. The problem statement, all variables and given/known data
    Its number in my physics exercise book is 1.147. (just for reference, though completely unimportant). The actual problem description is written below.
    An airplane is flying from point A to the point B which is 400 km west from the point A. Determine how long was it flying if:
    • there was no wind;
    • a wind was blowing from the east;
    • a wind was blowing from the south?
    Wind's speed is 20 m/s, and airplane's speed with regard to air (wind) is 800 km/h.

    2. Relevant equations
    Just as in The Classic Airplane Problem, laws of sines, cosines, tangents and cotangents, and the Pythagorean theorem.

    3. The attempt at a solution
    Distance of 400 km between points A and B is assigned to variable dist, wind's speed has a variable name of v_wg (meaning velocity, wind-ground), airplane's speed with regard to air (wind) goes by the name of v_aw (airplane-wind velocity), and the resultant velocity is named v_ag (velocity airplane-ground).
    • No wind, hence time t is equal to the dist / v_aw ratio, where v_aw when converted to metres per second equals 2000/9 m/s. The final result for this part is t = 1800 s.
    • There is a wind, and it's contributing to the airplane's speed, thus:
      v_ag = v_aw + v_wg.
      v_wg is 20 m/s, so the sum is 2180/9 m/s ... and t = dist / v_ag ≈ 1651,3761 s.
    • There is a wind, and it's blowing from the south. I used two triangles to visualise and try to solve this problem. The right angle is placed in the upper right corner. On the top is the upper cathetus whose value is dist (or s_aw), on the hypotenuse is s_ag, and on the right leg (cathetus) is s_wg. Same naming convention and layout applies to the second, "velocity" triangle.

      1. First (still the c part of this exercise), we don't have much informations about the distance part, so let's go on the velocity triangle. Calculating the hypotenuse by the Pythagorean theorem, v_ag is approximately 223,1204 m/s.
      2. Calculating the angle between v_aw and v_ag by the cosine relation, got approximately 5,1428 degrees.
      3. Going back to the distance triangle. Now that we've got the angle between s_aw and s_ag, we can calculate s_ag. Using the cosine relation, the result is about 401616,7551 m.
      4. t = s_ag / v_ag ≈ 1800 s.
    The book tells me that the a and b parts are solved correctly, while the c part is messed up horribly.
    The problem lies in the fact that I've set v_ag as the upper cathetus, but according to the book's logic, or at least as I see it, it should be the hypotenuse. Swapping v_ag and v_aw on the velocity triangle and using the dist distance gives the correct value book author specified (30,12 minutes), or by dividing s_ag calculated in step 3 of the c part with the v_aw value. I can simply swap the values and do so from now on, but I'd like to understand the reasoning behind it. Is it because it's really going along the upper cathetus (the dist path), but at a speed of the hypotenuse or ... I don't know. Help!
     
  2. jcsd
  3. Sep 17, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    For (c) think of it as the plane having to deploy a portion of its velocity to counteract the effects of the wind. That is, while the plane can move at some speed Vp with respect to the air, the air itself is moving in bulk at some velocity Vw at right angles to the desired direction. If you rotate Vp (keeping its magnitude the same!) so that its vertical component exactly cancels Vw, what's the horizontal component (aimed directly at destination B)?
     
  4. Sep 17, 2014 #3

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I suggest you draw a diagram of the velocities. What you know is the following:

    - Aircraft speed relative to ground.
    - Velocity of wind.
    - Direction of total velocity relative to ground (due west in order to come to B).
    - Total velocity is the vector sum of wind velocity and velocity of aircraft relative to wind.
     
  5. Sep 17, 2014 #4
    Thank you for replying so fast!

    gneill, I understood the fact (well, somewhat I did) that the airplane has to deploy a part of its velocity to compensate for the wind's force. However, what bothered me was its calculation, what goes where? So ... the horizontal component should be, if the vertical is equal to v_wg and the Vp is under a (for now) unknown angle, approximately 221.3204 m/s?

    Orodruin, as I've mentioned above, what should go where, and why so?
     
  6. Sep 17, 2014 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Start by drawing the wind velocity as this is something you know completely. Then draw a horizontal line from the base of the arrow representing wind velocity to represent the total velocity since you know that it has to be due west. Now you have to draw the aircraft velocity relative the wind. It has to be an arrow with length 800 km/h that starts at the tip of the wind velocity and ends somewhere on the horizontal line you just made. The total velocity is a vector from the base of the wind velocity arrow to the tip of the airspeed velocity arrow. Do you get a right triangle? Which velocity is the hypotenuse in it?

    The alternative is adhering to the other suggestion you got. Total velocity is the aircraft velocity plus wind velocity and must have north-south component zero. This gives a requirement on the north-south component of the aircraft velocity. Since you know one component and the absolute value, you can solve for the second (westward) component. Since the wind has no east/west component, the total velocity is just the westward component of the aircraft velocity.
     
    Last edited: Sep 17, 2014
  7. Sep 17, 2014 #6

    gneill

    User Avatar

    Staff: Mentor

    Yup.

    Fig1.gif Fig1.gif
     
    Last edited: Sep 20, 2014
  8. Sep 20, 2014 #7
    gneill, so VAB is the effective velocity (speed) vector?

    Something like this?
    diagram.png
    It's drawn in Inkscape, not some strictly math or physics oriented program, so expect minor inaccuracies. Assuming that I've followed your instructions properly ('tis all new to me, so sorry if I failed somewhere), the hypotenuse's magnitude is equal to airplane's speed relative to wind (hope I've used the physics/maths terminology properly here :p).
    Well, okay. I believe I've expressed such a concept in the concluding part of my first post, just not by these words. However, regardless of perception and the way you interpret the task, it all, as far as I can see, boils down to setting v_aw as the hypotenuse, v_wg for the right cathetus and calculating the v_ag out of that. All I'm missing are the whys and hows. Correct me if I'm wrong.

    EDIT: XenForo messed up my first post in this thread, so there are just bullets instead of the a., b., c. lists. I don't see the Edit button, so it's doomed to stay this way forever :p.

    OT: Regardless of noticeable, yet negligible effects on my first post here, congrats on moving to XenForo! PF.com looks more sleek and modern now!
     
    Last edited: Sep 20, 2014
  9. Sep 20, 2014 #8

    gneill

    User Avatar

    Staff: Mentor

    Yes, VAB is the plane's net velocity vector with respect to the the ground.

    The whys and hows flow from the understanding that, as far as the airplane is concerned, wind doesn't exist; there's just a big sea of air that the plane moves through at its rated speed with respect to that "still" air. Wind is a concept of the sea-of-air moving as a whole with respect to some observer on the ground. Because the sea-of-air is moving with respect to the ground, an observer there has to add the plane's velocity vector with respect to the air to his own wind velocity vector in order to determine the net velocity of the plane in his frame of reference (the ground).

    I'll discard that post so you won't have to look at it forever ;)
     
  10. Sep 20, 2014 #9

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, not something like that - exactly that ;)

    It even inspired me to actually get an avatar ... :p
     
  11. Sep 21, 2014 #10
    Yup, saw the avatar. I like it, and it's definitely relevant to the PF name and topic (which, according to Mr. Google, looks like Earth's gravity field anomalies), but I fear it's a little bit too detailed for such a low resolution profile picture, making the textual part of the image illegible. Perhaps something like this would work better, but would eradicate the factual aspect, so feel free to act according to your will, this is just my opinion as a free-time designer.

    Regarding the problem, whose name shall not be spoken, it seems that this is the one and only way to perceive this problem (CMIIW), so I just need to chew and digest the scientific (physics) standpoint and put it to use properly.
     

    Attached Files:

    Last edited: Sep 21, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Go home airplane, you are drunk! (Another Airplane Problem Thread)
  1. Airplane Problem (Replies: 1)

  2. Airplane problem (Replies: 3)

Loading...