Suppose an aircraft is moving at a certain fixed altitude above the ground. It follows a path defined by latitude and longitude. Now if we want to define the position of an aircraft at any point in the air, three variables are required for example X for latitude, Y for longitude and Z for the altitude. Suppose an aircraft flies and reach a certain fixed altitude Z, it then follows a route defined by X and Y. Now suppose that at any stage during the flight the aircraft decides to take a turn.The aircraft does a maneuvre in such a way that it turns either to the left or right and increases or decreases it's altitude in the same time. Assuming knowing the speed and the current three Dimensional position of the aircraft, how can the future position of the aircraft after a known time t can be predicted? Also assume that other aircraft related constant parameters are also known like radius of the turn, Bank angle etc
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$\begingroup$ A set of coordinates defines a position, not a course! $\endgroup$– GeorgCommented Jun 8, 2011 at 11:55
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1$\begingroup$ Two comments: 1.) Not really sure what the intent of the question is; 2.) You use x,y and z, which implies a Cartesian coordinate system; but latitude and longitude are parts of a spherical coordinate system. $\endgroup$– VintageCommented Jun 8, 2011 at 17:00
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2$\begingroup$ In keeping with the comment by @Vintage, it would be helpful if you could explain why you have asked the question - I.e. the circumstances that let you to it and what you want an answer to achieve. As it stands it quite difficult to interpret and therefore answer. $\endgroup$– qftmeCommented Jun 8, 2011 at 19:55
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$\begingroup$ A bit of clarification to my question. Suppose an object is moving on a circular path in three dimensional space in cartesian coordinate system. If the initial position of the object is known, how can the position of the object can be located after time t in terms of X,Y,Z coordinates. Assuming we know the speed of the object and the radius of the circular path. $\endgroup$– user3963Commented Jun 10, 2011 at 13:55
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$\begingroup$ @Abdul: now that's a much easier question to answer. In fact, consider just editing the question and replacing what you had written with the text of your comment. But also please take a look at physics.stackexchange.com/questions/10777/… and see whether it tells you what you want to know. $\endgroup$– David ZCommented Jun 10, 2011 at 18:47
2 Answers
I assume you are going to program this. First you need to write the vector differential equations of motion. The derivative of position is velocity, and the derivative of velocity is acceleration. Acceleration is force divided by mass. Force is the sum of a number of components, mainly gravity, lift, drag, and thrust.
Then you just integrate those with any ODE solver you like. The simplest is Euler. If you want more accuracy, you can use Runge-Kutta. You probably don't need a stiff solver such as Gear or Adams-Moulton.
So if the motion of the aircraft is along a straight line with wings level at constant speed, then the lift vector is equal and opposite to the gravity force vector. The way you turn it is to bank the wings to an angle. For example, if you bank the wings 45 degrees to the left, that tilts the lift vector 45 degrees to the left, so it's vertical component is .707 of what it was, and .707 of it is directed to the left. The sideways force causes the path to be a circular arc. The reduced vertical force causes the aircraft to start descending.
To compensate for that tendency to descend, the pilot increases the magnitude of the lift vector by adding back pressure on the control yoke. You'll notice that you feel a little heavier in the turn. The increased lift results in increased drag, so the pilot increases engine power to maintain speed.
Then when the aircraft has travelled far enough around the circular arc to be headed in the desired direction, the pilot levels the wings, releases the back pressue, and reduces the engine power. If he doesn't, the aircraft will start to climb.
You'll notice this the next time you fly.
That's how turning works. Now I'll tell you how straight flight works. The lift and drag you get from the wing depends on two things, speed and angle of attack. If you reduce speed, but want to stay in level flight, you need more angle of attack to get the same lift. So the way a pilot slows down a plane is to reduce power and then gradually pull the nose up by applying back pressure on the yoke.
Since it would be tiring to continue to apply back pressure when flying slowly, the pilot has a "third hand", the trim wheel. Rolling that wheel backward applies back pressure on the yoke so the pilot doesn't have to hold it. In fact, the primary speed control of the airplane is the trim wheel. The power doesn't actually control speed, it controls whether the aircraft ascends or descends at the speed it is travelling.
One last point is balance. If you take a plane on the ground, and place a jack under each wing at the center of lift and hoist it up, its nose will fall toward the ground. It's center of gravity is forward of the center of lift. So when it's flying, as the main wing is pushing up, the tail is pushing down. The moment between the two is what keeps the nose from dropping. This is very important for safety, because it stabilizes the speed. If the speed decreases, that moment decreases, and the nose drops, causing the aircraft to start to go "downhill". Since it is going downhill, it picks up speed. Conversely if it's speed increases, it starts to go uphill, causing its speed to decrease. If the aircraft is loaded too far aft, its speed and its up/down directional stability is lost. In fact, it is possible to get nosed up, and then go into a backslide all the way to the ground!
This problem cannot be solved, because the airplane has all sorts of flaps that control the lift, so even if you know X(t) and Y(t), so you know the horizontal speed of all the turns, you don't know the lift or the engine thrust used, so you can't find the height.
It is an ill posed problem, in either formulation.
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$\begingroup$ It's reasonable to assume no change of flaps in the turn & negligible aileron drag. Then for a given bank angle, you still need to know 1) how much additional lift is pulled (which affects drag), and 2) how much additional thrust is applied. Then it's straightforward to determine the path. $\endgroup$ Commented May 16, 2012 at 16:14
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$\begingroup$ @MikeDunlavey: I see, this does work, but I can't imagine it's a sensible approximation. But then the OP should say so, and explain what this problem is about. $\endgroup$ Commented May 16, 2012 at 17:55