# Given a radius and velocity calculate position of an aircraft banking to make a turn

I have a radius, R, for an aircraft traveling at velocity, V. If we start at point, (X,Y), what is the position of the point at time, t.

For example:
The aircraft is at point (0,0) and traveling at 250 knots and initiates a turn with a bank angle, phi, of 5 degrees. Assume that the aircraft can instantaneously rotate to the five degree bank. The equation for the turn radius, R where g is the acceleration due to gravity (9.81) is: $$\text{R} = \frac{V^2}{\text{g} \tan{\phi}}$$

For this example, R = 10.4 nautical miles. Where is the aircraft at t = 2 if the aircraft is traveling at a heading of 90 degrees (straight along the y axis)?

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1) I assume the movement is circular. To define the position of a point moving along a circular path with constant speed it is necessary to know the centre of the path, besides its radius. 2) In the example the starting point is $(x,y)=(0,0)$. What is the movement centre? 3) Is $V$ the angular speed or the magnitude of the tangential speed? 4) And what is $\theta$? –  Américo Tavares Nov 29 '10 at 21:33
@Americo, question updated –  Elpezmuerto Nov 29 '10 at 21:47
What is the direction of the turn? In my answer I assumed it was to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis. –  Américo Tavares Nov 30 '10 at 12:32

In a motion along a circular path centered at $(0,0)$ (see figure below) with constant angular velocity $\omega$ the position of a point $P(x,y)$ (in Cartesian coordinates and $P(r,\varphi)$ in cylindrical coordinates) starting at $(r,0)$ at $t=0$ is given by the law of the uniform circular motion:

$$x=r\cos(\omega t)$$

$$y=r\sin(\omega t),$$

where $r$ is the radius of the circumference and $\omega=\dfrac{d\varphi}{dt}$. The magnitude of the tangential velocity is $v=\omega r$ and the radial velocity is zero. Thus $\omega=\dfrac{v}{r}$.

$$x=r\cos\left(\frac{vt}{r}\right)$$

$$y=r\sin\left(\frac{vt}{r}\right)$$

In your case $r=R$, $v=V$ (it is assumed that the aircraft mantains the speed $V$) and the centre of the circumference is $(X,Y)=(-R,0)$. If you make the change of coordinates $X=x-R$ and $Y=y$, you get the following $X,Y$ coordinates as a function of time $t$:

$$X=R\left(\cos\left(\frac{V}{R}t\right)-1\right)$$

$$Y=R\sin\left(\frac{V}{R}t\right)$$

In this $XY$-coordinate system the motion starts at $(X,Y)=(0,0)$ ($t=0$).

I assumed that the direction of the turn is to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis:

$$X=R\left(-\cos\left(\frac{V}{R}t\right)+1\right)$$

$$Y=R\sin\left(\frac{V}{R}t\right).$$

The numerical result is obtained for $t=2$ and the other given data.

Added: For a different starting position $\left( X_{0},Y_{0}\right)$ at $t=0$ we have to incorporate these values in the motion equations. Integrating $% \omega =\dfrac{d\varphi }{dt}$, we obtain $\varphi =\omega t+C$, where $C$ is a constant. Then

$$x=r\cos \left( \omega t+C\right) ,y=r\sin \left( \omega t+C\right)$$

$$x_{0}=r\cos \left( C\right) ,y_{0}=r\sin \left( C\right)$$

and

$$\dfrac{y_{0}}{x_{0}}=\dfrac{\sin \left( C\right) }{\cos \left( C\right) }% =\tan \left( C\right) ,C=\arctan \left( \dfrac{y_{0}}{x_{0}}\right)$$

$$x=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) ,y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) .$$

Since

$$X=x-r=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) -r,Y=y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}% \right) \right)$$

we get

$$X_{0}=x_{0}-r,Y_{0}=y_{0},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+r}$$

$$r=R,\omega =\dfrac{v}{r}=\dfrac{V}{R},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+R}.$$

Finally, we obtain the equations of the motion of the aircraft in the $X,Y$-plane:

$$X=x-R=R\left( \cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) -1\right)$$

$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right),$$

where $\left( X_{0},Y_{0}\right)$ is the position of the aircraft at $t=0$.

If the direction of the turn is to the right, then the motion is symmetric with respect to the $y-$axis:

$$X=-x+R=R\left( -\cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) +1\right)$$

$$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right).$$

Note: I changed the notation; in the question the position of the aircraft at $t=0$ is $\left( X,Y\right)$.

The velocity vector is $\overrightarrow{v}=v\overrightarrow{e}_{\varphi }$ and the acceleration vector $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$

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Note: I changed the notation; in the question the position of the aircraft at $t=0$ is $\left( X,Y\right)$, while here is $(X_0,Y_0)$. –  Américo Tavares Nov 30 '10 at 13:03
+1 This is exactly what I was looking for! –  Elpezmuerto Nov 30 '10 at 15:41

There is still not enough information to answer the question. The initial speed is specified, but not the direction. Given the direction of travel, you can find the center of the circle. Then (as in my last comment) the angular velocity is V/R and you can apply the same trig functions.

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direction added –  Elpezmuerto Nov 29 '10 at 22:21
Now you should be able to follow my last comment in the previous answer. –  Ross Millikan Nov 29 '10 at 22:37
@Elpezmuerto: The direction (left/right) of the turn has to be known. –  Américo Tavares Nov 30 '10 at 12:34

I don't know what you mean by R. If you start at the origin, $v_x=v\cos(\theta)$ and $v_y=v\sin(\theta)$, $x=v_xt$ and $y=v_yt$.

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R is the radius –  Elpezmuerto Nov 29 '10 at 20:52
But if you start at the origin, the radius is zero. At any time, including t=2, the radius is $\sqrt{x^2+y^2}$. So I don't see what to do with R=52. –  Ross Millikan Nov 29 '10 at 20:55
The point is traveling along a curve specified by radius, R = 52. Consider the first point along that curve (0,0) –  Elpezmuerto Nov 29 '10 at 20:57
Then I don't know what to do with theta. It is more convenient to put the center at the origin, but if you want the starting point at the origin we can translate later. The angular rate is V/R, so the angle around the circle is 2V/R at time t=2. If we start at (52,0) then we have x=52cos(2V/R) and y=52sin(2V/R). Translating x becomes 52cos(2V/R)-52. –  Ross Millikan Nov 29 '10 at 21:14

Use the following:

• $L$ = lift
• $R$ = radius
• $\theta$ = bank angle
• $M$ = mass
• $C$ = centrifugal force
• $G$ = gravity

Then

$$L\cos(\theta)=MG\\L\sin(\theta)=C=\frac{MV^2}{R}$$

and so

$$\tan(\theta)=\cfrac{\frac{MV^2}{R}}{MG}=\frac{V^2}{RG}$$

Ergo

$$R=\frac{V^2}{\tan(\theta)G}$$

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It would be helpful if you wrote your solutions using MathJax and proper formatting. –  Ian Coley Oct 10 '13 at 15:28