Question

An aircraft flying towards a transmitter (NDB) moves so that it always points towards it (see diagram will post afterwards). The pilot starts flying towards the NDB directly from the cast, and the wind is blowing due north. Assume that the wind speed and the speed of the aircraft through the air(airspeed) are constant. (this is different from ground speed). If the trip starts at (2,0) and ends at (0,0), set up the ODE describing the aircraft's path over the ground given dy/dx =(dy/dt)/(dx/dt).

Answer 1

The answer is this http://s3.amazonaws.com/answer-board-image/cramster-equation-20089201527116335752123113180668984.gif but how?

Answer 2

The diagram is this: http://s3.amazonaws.com/answer-board-image/20089201425406335751754007822667224.jpg

Answer 3

So I'm just going to define some variables from the answer: v= constant velocity of the aircraft and w = constant velocity of the wind

Answer 4

So the equation obtained should have been homogeneous.

Answer 5

http://www.chegg.com/homework-help/questions-and-answers/differnetial-equations-question-group-projects-chapter-3-d-aircraft-guidance-crosswind-air-q3808571?cp=CHEGGFREESHIP

Answer 6

I cannot see the answer because I don't have membership there, Can you please screenshot it for me?

Answer 7

Refer to the attachment.

Answer 8

Thanks. I see the question but not the answer. Can you screenshot the answer?

Answer 9

The answer is not given that I can tell. I am not a member there.
I can provide the solution to the DF using Mathematica if you wish.

Answer 10

Sure! Anything will help right now. LOL

Answer 11

You probably have this already, but in case not:

The speed in the y direction is the sum of the wind speed and the plane's y speed. The wind speed is just v_w, since the wind is always due north. The plane's y speed is just the y component of its total speed, so -(y/sqrt(x^2+y^2))v_p (the negative sign since the plane is always pointing down towards the origin). The x speed is just the other component of the plane speed, -(x/sqrt(x^2+y^2))v_p. So, the two DE's are:

\[\begin{align*}\frac{dy}{dt} &= v_w-\frac{y}{\sqrt{x^2+y^2}}v_p \\ \frac{dx}{dt} &= -\frac{x}{\sqrt{x^2+y^2}}v_p\end{align*}\]

You can easily divide those to obtain dy/dx.

Answer 12

Tanks so much but can you tell me why you have thsoe for dy/dt and dx/dt?

Answer 13

Oh sorry you did explained it.

Answer 14

Hold on.. why is it sqrt(x^2+y^2) again?

Answer 15

Can someone tell me how \[\cos(\Theta)\] = \[\frac{ y}{ \sqrt{x^{2}+y^2} }\]