Differential Equations Aircraft guidance in a crosswind Differnetial Equations q

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Differential Equations Aircraft guidance in a crosswind

Differnetial Equations question Group Projects for chapter

3 D Aircraft Guidance in a crosswind

An aircraft flying under the guidence of nondirectional beacon (a fixed radio transmitter, abbreviated NDB) moves so that its longitudinal axis always points torward the beacon (see figure 3.19) A pilot sets torward an NBD from a point at which the wind is at right angles to the initial direction af the aircraft; the wind maintains this direction. Assume that the wind speed and the speed of the aircraft through the air (its "airspeed") remain constant. (Keep in mind that the latter is different from the aircraft's speed with respect to the ground.)

(a) Locate the flight in the xy-plane, placing the start of the trip at (2,0) and the destination at (0,0). Set up the differential equation describing the aircrafts path over the ground. [Hint: dy/dx = (dy/dt)/(dx/dt).]

(b) Make an appropriate substitution and solve this equation. [Hint: See Section 2.6.]

(c) Use the fact that x = 2 and y = 0 at t = 0 to determine the appropriate value of the arbitrary constant in the solution set.

(d) Solve to get y explicitly in terms of x. Write your solution in terms of a hyperbolic function.

below is what I have so far....

Place the NDP at the origin and the airplane initially at location {2,0}. Since the plane is always flying directly toward the origin we can make an angle q at the origin which points at the plane. Let v be the plane's velocity through the air. The wind has no effect on the x component of the plane's velocity which is -v cos(q), the minus being necessary because the plane is flying in the direction of decreasing x as the time t advances. Ifw is the speed of the wind, then the plane's y component of velocity is w-vsin(q).Then the following equations must be true due to definitions of cosine and sine: dx/dt = -vcos(theta) = -(vy)/(sqrt(x^2+y^2)) dy/dt = w-vsin(theta) = w-(vy)/(sqrt(x^2+y^2)) Dividing the second equation by the first we obtain: dy/dx = (vy-w(sqrt(x^2+y^2)))/(vx) You can now solve this differential equation to obtain the following solution with the starting condition that y[2] = 0. I believe the portion above is correct but we are stuck on the rest.

Your differential equation is going to be based on the fact that the aircraft's velocity over the ground at any moment is the sum of two velocities:

the wind speed, which has a constant direction (as well as speed), and

the aircraft's velocity relative to the air, which has a constant speed but a direction pointing radially toward the origin .

So you'll need to work out how to express the radial direction. It will probably be easier to work in Cartesian coordinates. You will also probably want to start with two differential equations, one for dx/dt and another for dy/dt, and then construct a single equation from those. It will also make the later questions easier to deal with by calling the aircraft's speed v and the windspeed (gamma)

Explanation / Answer

follow the link.....http://www.scribd.com/doc/91448995/Wiley-Military-Avionics-Systems

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