 Aug 17th 2014, 03:10 AM #1 Junior Member   Join Date: Aug 2014 Posts: 29 Please help me with this airplane problem I have to write a C# program for an assignment for school. Programming is not a problem, but physics is :P So I am kindly asking you guys for helping me out with this.. Sorry for my bad english, I'm not a native speaker. A glider pilot starts a back-loop with a plane, flying horizontally at a speed of 60 m/s. Maneuver is carried out so that he pulls the stick back and holds it in the same position througout the whole loop. The initial centripetal acceleration is 30 m/s2 in the direction vertically upwards. I must calculate the position, velocity and acceleration of the aircraft as a function of time during the interval from the beginning implementation of acrobatics to the time in which the aircraft at a constant acceleration is outlining a full circle. Does the aircraft come back to the original/initial position? The origin is placed at the point at which the airplane moves from the horizontal position of the ribbon/loop. I'm not really good at physics so I don't even know how to start :/ Acceleration due to gravity is 9.8 m/s2 Air resistance can be neglected, so that the only force acting on the plane is the force of gravity and the lift force, which is perpendicular to the direction of the current speed of the airplane. Surrounding air is stationary relative to Earth and a plane (throughout the maneuver) doesn't rotate around its horizontal or vertical axis. I really hope I translated it so you can understand. Thank you!   Aug 17th 2014, 06:58 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 This is going to be an interesting project. First you need to clarify a few assumptions: does the plane fly with a constant amount of thrust? If so, we will need to know the amount of thrust and the plane's weight, and will have to estimate the drag due to air resistance as being proportonal to the square of the speed. And do you assume that the center of rotation is constant with speed? For example when traveling level at 60 m/s and having 30 m/s^2 centripetal acceleration implies a radius of the loop as 120 m. When flying vertically upward the plane's speed is somewhat slower - do we assume the pilot manipulates controls to maintain that 120 m radius, or does the plane continue to have 30 m/s^2 centripetal acceleration, or is centripetal acceleration proportional to air speed?   Aug 18th 2014, 01:48 AM #3 Junior Member   Join Date: Aug 2014 Posts: 29 Here is the text, if it helps: The glider pilot launch ribbon (England. Looping) in level flight at a speed of 60 m / s. Maneuver carried out so that the moment you pull the stick to set as back and then stick throughout the maneuver, retained in the same position. The initial centripetal acceleration maneuver at 30 m / s2 and in the direction of vertically upwards. Calculate the position, velocity and acceleration of the aircraft as a function of time during the interval from the beginning implementation of acrobatics to the time in which the aircraft at a constant acceleration outlining full circle. or this time the aircraft comes back to the original course of the year? The origin placed at the point at which moves from the horizontal plane of the ribbon. In addition to the specified data using the following assumptions:  acceleration due to gravity is 9.8 m / s2  air resistance can be neglected, so that the only force acting on the plane, the force of gravity and lift force, which is perpendicular to the direction of the current speed of the airplane  lift force is proportional to the square of the speed of the airplane  the maneuver the airplane is rotated about its horizontal or vertical axis  surrounding air is stationary relative to Earth Some explanations about the task: Mechanics of Flight is fairly complicated, but it can be for a basic understanding simplified to a few basic concepts. On the airplane operating force of gravity and the aerodynamic force as a result of the movement of aircraft in the ambient air. The aerodynamic forces contribute mainly skirts and the fuselage and control surfaces. Due to the hull and krill operates on a plane drag force, which is due to lower losses are trying to minimize the aerodynamic shape of the aircraft. The wings give the aircraft buoyancy, with which in the normal regime in particular, balance the gravitational force. Buoyancy depends mainly on the the speed of the airplane with respect to the surrounding air, and the shape of the profile and the angle of attack (the angle between the center line of the profile and the relative circumferential velocity of air in relation to the plane). By tilting controls to create Having these areas additional buoyancy, which make war distance from the center of gravity of the aircraft creates a torque which causes rotation letla in different axes. Deflection flaps (on the outer strain wings) causes rotation about a horizontal axis deviation rudder (vertical control surfaces on the tail) vrtnje about a vertical axis and deviation elevator (horizontal control surfaces on the tail) deviation around the transverse axis. When the airplane is not inclined to dismantle the resultant aerodynamic forces on the aerodynamic drag, which operates in the opposite direction from the direction of flight, and the buoyancy, which acts in the perpendicular direction with respect to the speed of aircraft, which coincides with the direction of the vertical axis. At speeds of the tasks of the air resistance is certainly noteworthy, however, are modern gliders so well designed that the sliding relationship 1:45 and more. In level flight the lift force acting in the vertical direction and exactly balances the gravitational force. when pilot tilts the elevator back to the airplane to rotate around a transverse axis, causing an increase in the incident as air in relation to the wings of the airplane and therefore increase buoyancy. Excess buoyancy causes Numerical methods in physics - a selection of finished tasks 4 cenripetalni acceleration and the plane starts to move around in an arc. from the initial centripetal acceleration reduced by the gravitational acceleration and velocity can be calculated from proportionality coefficient between the lift force and the square of the speed of the aircraft, which was assumed to be a constant. Due to the centripetal acceleration starts to rise and the airplane due to the force of gravity to lose speed (because the weight is no longer working at right angles to the direction of velocity). Due to the reduced speed decreases buoyancy (as it is proportional to the square of the speed), which affects the smaller centripetal acceleration. If the speed does not decrease, to the movement of the arc upward centripetal acceleration grow as the force of gravity is no longer directed opposite to the force of buoyancy. The situations described are valid until the aircraft reaches the upper point of tangles, then the situation turn. Useful links: http://en.wikipedia.org/wiki/File:Loop1.gif http://en.wikipedia.org/wiki/Circular_motion http://en.wikipedia.org/wiki/Flight_dynamics ---------------------------------------------------------------- Can this be calculated? Damn, I'm such a dummy when it comes to physics   Aug 18th 2014, 05:00 AM #4 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 I missed the fact that this is a glider, so my questions abut thrust are not relevant. And ignoring air drag really simplifies things. With no drag the principle of conservation of energy says that the glider will complete a closed loop, returning to its starting position and speed at the end of the loop. For your computer program I suggest that you divide time into small increments - say 1/10 second, and determine the change in acceleration, velocity, and position for each increment of time. The equations you will need are: Since lift force is proportional to the square of the velocity, so is centripetal acceleration. From a = v^2/r, this means that Kv^2 = v^2/r, and r is therefore a constant. Not what I would have expected. 3. For each increment of time you have a starting position, velocity, and acceleration, in both x and y coordinates. You need to treat velocity and acceleration as vectors (i.e. they have magnitudes in both the x and y directions) The ending acceleration position can be calculated from p_2 = p_1 + v_1t + (1/2)a_1(delta_t)^2 where v_1 and a_1 are the values of initial velocity and acceleration. Other equations: v_2 = v_1 - a_1 delta t a_2 = a_1 + a_lift - a_gravity where a_lift is v^2 x (30/60^2) in the direction perpendicular to the velocity vector, and a_gravity is -g in the vertical direction. Hope this helps. Post back if you have further questions. Last edited by ChipB; Aug 18th 2014 at 05:02 AM.   Aug 18th 2014, 05:10 AM #5 Junior Member   Join Date: Aug 2014 Posts: 29 OK, thank you very much for your reply. Ive got another theory too, so we can compare: If the pilot is, indeed flying in a perfect circle then his acceleration must be constant and always pointing toward the center of the circle. You say "initial centripetal acceleration is 30 m/s2 in the direction vertically upwards" so his acceleration is always 30 m/s^2 pointing toward the center of the circle. Any circle can be given in parametric equations as x=−Rsin(θ), y=Rcos(θ) where, here, x is the horizontal distance right or left of the center of the circle and y is the vertical distance above or below the center of the circle. R is the radius of the circle and (x, y), when θ=0, is the starting position at the bottom of the loop. Assuming a constant angular speed, we can write θ=ωt where ω is the angular velocity and t the time. Differentiating, dx/dt=−Rωcos(ωt) and dy/dt=−Rωsin(ωt). We are told that, at the beginning of the loop, when t= 0, the airplane has a horizontal speed of 60 m/ so we have dx/dt(0)=−Rω=44. The net acceleration is the second derivative, d2x/dt=Rω2sin(ωt) and dy/d2=Rω2cos(ωt). We are told that, at the beginning of the loop, when t= 0, the airplane has an initial vertical acceleration of 30 m/s^2 so we must have Rω2=30. From those two equations, Rω=60 and Rω2=30, you can solve for both R and ω and so find the parametric equations for the airplane's trajectory. Add R to the y component to move the origin to the bottom of the loop. ----------------- What do you say about that written above? Now I'll study both and try to understand it..   Aug 18th 2014, 05:16 AM #6 Junior Member   Join Date: Aug 2014 Posts: 29 What is p_2 and p_1? position? path? In meters? Thank you for helping me!   Aug 18th 2014, 06:35 AM #7 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 I should have realized that he doesn't fly in a perfect circle - that would be true only if we ignore gravity. As the angle of the plane relative to ground increases the force due to gravity is added to the lift of the wings to increases the acceleration towards the center of the loop. So the loop gets tighter near the apex, and hence attempts to apply equations appropriate for a circle like a=v^2/r don't work. P_2 is "position 2" and P_1 is position 1. Since this is supposed to be a computer programming project, there's not much point in trying to explicitly determine the path in mathematical form. By the way, I did a quick simulation in Excel, and got a loop that looks like the attached. Last edited by ChipB; Aug 18th 2014 at 06:51 AM.   Aug 18th 2014, 08:35 AM #8 Junior Member   Join Date: Aug 2014 Posts: 29 ChipB, thank you! This looping looks right. So 'r' is not constant. Did you change equations? Can you send me the Excel file maybe? How did you do that? Programming will not be difficult to do, after I get all the theory and equations right. So what do you suggest the mathematical path should be? Still struggling to perfectly understand this. What do you mean by p_2 (the ending acceleration position) ? Is that a position of a glider plane in a certain time? So if I want to know where would a plane be in 20 sec, this equation tells me a length of a path made by a plane within 20s on this curve you made in Excel? Path in mathematical form .. What do you mean by that? How should I determine a position of a plane? An assignment is to calculate position, speed and acceleration of a plane in certain time (in interval from start to finish of a full loop). So do you think that I should program this so the program would ask you to type in time, and then it should write you a position, speed and acceleration? I initially looked at it that way. Can this be done? Which acceleration is changing? a = 9,8 m/s2 is constant for sure. Than we have centripetal acceleration (initial 30 m/s2) and acceleration in movement? Thank you!!   Aug 18th 2014, 08:51 AM #9 Junior Member   Join Date: Aug 2014 Posts: 29 Ok, program does not have to ask you about time. Time flies and when we'll lunch the program it will write every single dot on this curve you just pasted from Excel (so every single dot must me defined with position, speed and acceleration - let's say for every 1/10 of a second). Program will stop when a plane finishes a loop. Which equation did you use for that curve of a loop in Excel? Thank you Last edited by StudentTM; Aug 18th 2014 at 08:54 AM.   Aug 18th 2014, 09:38 AM #10 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 At time 0 you have position P_1=(0,0), velocity V_1 = (60, 0) and acceleration A_1 = (0, 30). After time T (say, 1/10 second) using standard equations of motion for constant acceleration you have: P_2 = (P_1x+V_1xT + (1/2)A_1xT^2, P_1y+V_1yT + (1/2)A1_yT^2) V_2 = (V_1x+A_1xT, V_1y+A_1yT) Here P_1x means the x-component of P1, V_1x means the x-component of V_1, etc. We know that acceleration is not actually constant, so these equations will introduce some error. So the trick is to make T small enough that the error is insignificant. I think you'll find setting T to 0.05 seconds gives reasonable good results, though T = 0.01 is better. The acceleration of the plane in the x- and y-directions is equal to the x- and y-components of the lifting force minus the vertical force of gravity: A_2 = (A_Lx, A_Ly-g) The equation for the magnitude of A_L (meaning acceleration due to lift) is A_L = (30+g)(V_1/60)^2. Note that it includes g - this is because the amount of lift at the start must provide enough force to overcome gravity plus accelerate upwards at 30 m/s^2. The direction of A_L is perpendicular to direction of travel, so: A_Lx = A_L (-V_1y/V1) A_Ly = A_L(V_1x/V1) where V1 is the magnitude of velocity, which equals sqrt(V_1x^2 + V_1y^2). Once I calculate the values for P_2, V_2 and A_2, I copy them to a second line of the spreadsheet to be the new values of P_1, V_1 and A_1, and redo the calculation again. Rinse and repeat, over and over again. Last edited by ChipB; Aug 18th 2014 at 12:43 PM.  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