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Physics distance question

  1. Jul 3, 2005 #1
    i got this in one of my exam. (simmilar)

    a plane going from city A to B. the distance is S and bearing B1, the speed the plane is V1 and there are wind blowing from bearing B2 with speed V2. find B1 in order so the plane arrive at the city B.

    just answer is expression if you insist but, the problem is that is this question make any sense, to answer B1?
    Last edited: Jul 3, 2005
  2. jcsd
  3. Jul 3, 2005 #2
    The question is how to allow for the drift. If the wind is a pure headwind or tailwind, then the plane must be pointed straight at the destination as if it were calm. But if the wind has a crosswind component, then the nose of the pane must be tilted towards the wind so that it makes good the track it wants over the ground.

    The maths is just a vector addition/subtraction question.
  4. Jul 7, 2005 #3
    Again... I'm not really a math person; but is this correct?

    http://img43.imageshack.us/img43/4344/trajectory5gs.png [Broken]

    The trajectory V3 would be what the plane uses.

    Oh, wait... you're asking what B1 is, so the plane apparently left from city A; but not on a straight path and now we have to know where it must start... In that case just add the V2 to where he was from and move the plane over that much and that's where B1 would need to be.
    Last edited by a moderator: May 2, 2017
  5. Jul 21, 2005 #4
    what i am saying is
    that they give us the speed
    not the force
  6. Jul 21, 2005 #5
    You didn't even like my picture??? :frown:

    :cry: :cry: :cry:
  7. Jul 22, 2005 #6
    no, it's not that, i like the picture and that's the one i used in my exam, but your picture ( and what i had drawn as well) is for the vector as FORCE not SPEED.

    just think that the air is made up of very heavy material, even not of the variable above changed, the B1 obviously change right ?
  8. Jul 22, 2005 #7
    You don't have enough variables defined:
    1)B1 is fixed as the direct path to B from A over a distance S.
    2)you have a wind at B2 distance function of time V2
    3)Airplane covering distance as function of time V1

    You don't need to find B1, you cann't that's given in your first line.
    you need to find the bearing the airplane will use = call it B3
    so that when B3 @ V1 vectored w/ B2 @ V2
    Will produce a new vector running on the desired bearing of B1
  9. Jul 23, 2005 #8
    no, what i meant is that this problem cannot be solved by vector.

    you see, we use vector when we are dealing with force, not speed.

    and no, RandallB, i'm sorry not to state that B1 is a variable.

    my point is that if B1 is the variable while the other is constant, then it is still missing variable to convert the speed into force which then can be solve ussing vector.
  10. Jul 24, 2005 #9
    Your first sentance defined B1 as fixed
    To solve this problem you do not need the Distance S at all.
    You do not need "Force"
    Pilots deal with vectors and speed everyday the wind blows.
    You only need two speeds Airplane and wind.
    But you cannot solve this if your not using THREE bearings.
    1)Real direction "bearing" to city B (you defined that as B1)
    2)Current bearing of wind.
    3)And then figure the bearing the plane will "aim" itself with.
    Not a big deal - pilots figure this vector problem everyday in less than a couple minutes.
  11. Jul 27, 2005 #10
    Originally Posted by ArielGenesis
    B1 is a variable.

    Your first sentance defined B1 as fixed

    yes, I am sorry RandallB, i meant it to be a variable.

    And just think that the weight of the plane that we use is folded up to 10000000 times and somehow someway still able to fly with that same speed and size and shape. Bad example, sorry can't find better. Will the wind of the same speed and density and fludity and material and ect, will give the same effect to the plane? Obviously no, right?
  12. Jul 28, 2005 #11
    I'm not sure your grasping the point.
    Your just drawing a triangle here,
    Three sides, three angles, three bearings
    You cannot build a "formula" for this with accounting for and using three bearings call them what ever you like.
    One of them will be the real direction to B from A.
    The cities don't move, this bearing is fixed.
    you only need two "Lengths" on your triangle.
    (finding time to reach B dosn't seem to be part of the question)
    Speed works, or distance for a common unit of time, fine.
    As I understand your test problem or howework problem is that all values are given as unknown fixed values - Except the direction or the bearing of the airplane, that is to be solved for.
    Put in some real numbers, draw some of the triangles, include some extreame values where the plane could "fly backwards" and get there (wind faster than plane but headed toward B) and you should see how to write your vector solution.

    I assume you intended this as a homework help question rather than a "brain teaser".
    So unless your instuctor somehow limited you to only two bearings - yes the problem makes sense and you'll find it to be a basic vector problem.

  13. Jul 28, 2005 #12


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    The problem does not make sense as written because B1 is used to identify both the heading from city A to city B, and the initial heading that the airplane takes in order to account for the wind.
  14. Jul 30, 2005 #13
    RandallB, I had understood what had you are trying to explain, quite well. And i had used it even before i started the thread.

    Take my example, there are other factor other then the speed of the wind and other that i had state that determine the B1.
  15. Jul 30, 2005 #14
    I don't understand - you know the answer and this is a riddle?
    We need to find the "other factor" needed to find bearing used by plane in flying to B?

    Other than the four already given, I'd say there is no other factor.
    So I give up - tell us.
  16. Aug 8, 2005 #15
    I know the expected answer of this and so any one of us, ut us what you and other had give before. Why i thought this is a riddle is just because that we acutally cannot give the answer that way, because as what i had said before, that vector should be presented in force instead of speed. I suspect that most of the poeple would not realisze that and that is how it is brainteassing.
  17. Aug 8, 2005 #16
    Well no it's more of just a homework problem, because --
    No, as in somewhere else in this thread it said the FORCE has nothing to do with using a vector, presenting vectors only in FORCE is NOT true, speed is fine and in this case required.

    As in a pilot wanting to Fly straight north to his destination (city B)
    but he has a 50 knot wind blowing from the West directly to the East.
    IF he flies at a speed of 100 knots he would want to fly a bearing of Northwest (45 degrees off due North).
    Then his flying would put him “off course" to the city at a rate of 50 knots as he flys to far to the west,
    but the wind would be pushing him back on course at the same rate of 50 knots.
    No ‘force’ needed, just vector speeds.

    Converting this ‘triangle’ into vector formulas should use speeds not forces,
    hope that’s what you used on the test or quiz.
  18. Aug 9, 2005 #17
    I am sorry that you seems not to understand me as well as i dont understand you. You are trying to say that we should use vector formulas with the triangle using speed in which i disagree. I said that we should use force instead of vector. My opinion is that we don't know how much the plane move. If there are wind blowing from east to west 50 knots, will the plane also move from east to west as much as 50 knots too (assuming that the plane suppose to go from north to south)? No, we dont know. It could move 50 knots, 40 or even 10. Depending on the weight of the plane and many other factor. That is my opinion. I write this way due to the fact that we are nearly discussing nothing on the previous page. And i would be happy to see this end with you being satisfy with my view as well as having me satisfy with yours.
  19. Aug 9, 2005 #18
    This really is a homework issue. You are not understanding vectors. You need to sit down with who ever gave you the exam and be sure you understand vectors – they will not agree that “Force” is required.

    AS to the example - Of course you know where you are going. Cities do not randomly move around. I picked a city due NORTH not south, SO to fly North I choose to fly northwest at 100 knots. Now going northwest “where” am I going in relation to N & W? – using square of sin and cosine of 45 degrees I see I’m going 50 knots NORTH and 50 knots WEST. Thus IF I was driving on flat ground I would be heading north towards the city I want to go to, but I’d also be moving to the west OFF COURSE at 50 knots.
    BUT, I’m not driving - I’m in the air and the air is moving – moving to the EAST blowing me off course to the east! SO since I chose to fly “off course” on a VECTOR that took me to far west and the wind is moving me to far east by the same SPEED the east-west components cancel each other out and the “ground track” I travel will be due NORTH.
    Although slower than my air speed of 100 knots, I’ll see the city coming to me from my view in the plane at 50 knots not over the nose of the plane but from a point between the nose and the right wing.

    Can you write a vector formula to input any city at any bearing – YES
    Can the formula include variables to put in any wind speed blowing on any bearing – YES
    Can it include a variable flying speed – YES but pilots usually pick the most efficient speed for the plane.
    Can the formula solve for the “compass reading”, “bearing”, “vector direction” the pilot must fly to go to that city. Sure once all the above info is included – NO FORCE VALUES required.

    Again - Pilots due this everyday with what amounts to a small graphical slide rule – stop in a pilot shop sometime.
    After review with your instructor you might want to start a new thread under homework or maybe this can be moved there.
  20. Aug 15, 2005 #19
    Could you please take a look at my analogy. I could not have anything else to say unless you agree or disagree(which you certainly do) with an argument againts my point.

    Can you write a vector formula to input any city at any bearing – YES
    Can the formula include variables to put in any wind speed blowing on any bearing – YES (but not only wind speed)
    Can it include a variable flying speed – YES but pilots usually pick the most efficient speed for the plane.
    Can the formula solve for the “compass reading”, “bearing”, “vector direction” the pilot must fly to go to that city. Sure once all the above info is included – (that is needed but still need more)

    let me repeat my analogy:
    1. suppose the wind blow west 10 km/hr
    2. the plane is suppose (w/o wind) to move perpendicular to the wind (north or south)
    3. will the plane move to the west as fast as 10km/hr (and still moving north or south)?4. no, definitely less then 10 km/hr
    5. Depend on the weight of the plane, shape and size, fluidity, air density and others.

    I might stop in a pilot shop sometimes.
  21. Aug 15, 2005 #20
    Nothing here to justify saying no.
    Where is the suggested direction of flight?
    I'd recommend "FLYING" North by North-northwest in order to "GO" North.
    None of #5 is useful info at all.
    Have them show you a Manual - Flight Computer. If you don't understand vectors and the old style Manual version to work wind drift you'll never be able to use an Electronic Flight Computer.

    Did you actually sit down with you instructor yet?
    Have them help you draw the speed vectors.
    Convert them to distance vectors.
    NO force vectors.
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