If I only know $x$- and $y$- coordinates of every point on a trajectory without knowledge of time information, is there any way to approximate Cartesian acceleration angle at each point? Time interval between every two points is very small, ~0.03 second.
[Physics] How to approximate acceleration from a trajectory’s coordinates
accelerationkinematicsvelocity
Related Solutions
OK... once more: let's assume we are looking at a map-coordinate system where the x-axis points East, the y-Axis points North and the z-axis points towards the zenith. In such a system there are three acceleration components $\{a_E, a_N, a_Z - g\}$. Since the surface of the Earth is not an inertial system, the gravity of Earth always acts on the Zenith component and there is no way to get rid of that.
IF our accelerometer would have its axes aligned perfectly with the North, East and Zenith directions, we would be done easily: subtract a constant g from the Zenith-component and voila... perfect acceleration data. THAT is exactly what a commercial inertial measurement unit does! It keeps its accelerometer inside a gimbaled mechanical unit that makes sure that all three axes are always pointing into the same directions! This is achieved with a gyroscopic stabilizer rotating at 10,000rpm and very expensive precision electro-mechanics. Good for your Boing 777 but not your pocket!
So what's in your pockets? It's a mediocre accelerometer next to a mediocre rotation sensor next to an unreliable compass which will pick up your car keys more often than it will pick up Earth's actual magnetic field. That's what you have to work with.
The rotation of the accelerometer against the $\{a_E, a_N, a_Z - g\}$ system is initially unknown. Let's assume the device rests on the table. The only acceleration acting on it is -g_Z. Since the device is rotated trough three Euler angles against our favorite map system, we will be seeing three acceleration readings $a_x, a_y, a_z$, which should add up vectorially to the gravitational acceleration. So we can naively hope that $a_x^2 + a_y^2 + a_z^2 = g^2$. That, however, will not be the case, because, as we said, we have a crappy accelerometer. Each channel will have an offset, a gain error and, worst of all, at least a first order crosstalk error plus a noise term. So the measured accelerations will be
$\vec{a}_{measured} = M_{gain/crosstalk} \vec{a}_{physical} + \vec{a}_{offset} + \vec{a}_{noise}$.
In the above equation $M_{gain/crosstalk}$ is almost a unity matrix with diagonal elements representing the channel gains and the off-diagonal elements representing the channel crosstalk. If left uncorrected, the channel offsets will lead to a slow drift of an integrated position. If left uncorrected the channel gain errors will make us over- or underestimate the actual distance and the crosstalk errors will make point us in the wrong direction. Most importantly, the zx- and zy-crosstalk will mix part of the constant acceleration g into x and y directions, making the x and y offsets even larger.
But wait! We still haven't taken care of the rotation of the device! This rotation is another Matrix $M_{rot}$, which rotates the physical accelerations in the North-East-Zenith system into the physical accelerations acting on our sensor. So in reality the mapping goes something like this:
$\vec{a}_{measured} = M_{gain/crosstalk} M_{rot} \vec{a}_{NEZ} + \vec{a}_{offset} + \vec{a}_{noise}$.
So how do we estimate M_{rot}? With the rotation sensor, of course. That sensor measures the angular velocities of the device's rotation, then it adds its own offset, gain and crosstalk errors to that and then it gives you a signal that, after proper one time integration can be transformed into the $M_{rot}$ matrix, assuming we knew, at some point in time the original orientation of the device.
Do I have to say it? The million dollar 777 inertial guidance system uses the GPS and a ground calibration routine to align itself properly before the plane takes off. And that's what you have to do, too, to get started turning rotation sensor data into $M_{rot}$. So how does one calibrate the rotation sensor? Using $g_{Zenith}$ to determine the direction of the Zenith and the compass to determine North and East, of course! Oh... wait... we don't know where that vector is pointing unless we have a properly calibrated (and rotated!) accelerometer signal!
And there begins your adventure of combining the signals from three unreliable sensors into one hopefully reliable signal. Part of the problem is to quantify all the error terms and to reconstruct and invert the coordinate transformation matrix.
I don't know how much of this Apple has done already. That should be documented somewhere... in the API or maybe just in an internal document that they won't show to you unless you are under an NDA with the Apple legal department.
Let's say you have a prescribed path parametrized as $\vec{r}(\alpha)$ for the position vector.
- The velocity vector is decomposed as $$ \vec{v}(\alpha,\dot{\alpha}) = \dfrac{\partial \vec{r}(\alpha)}{\partial \alpha} \dot{\alpha}$$
- The speed along the tangential direction is $$ v(\alpha,\dot{\alpha}) = \lvert d\dfrac{\partial \vec{r}(\alpha)}{\partial \alpha}\rvert \dot{\alpha} $$
- The tangential direction is thus $$\hat{e} = \dfrac{\vec{v}}{v} = \dfrac{ \dfrac{\partial \vec{r}(\alpha)}{\partial \alpha} }{\lvert\dfrac{\partial \vec{r}(\alpha)}{\partial \alpha}\rvert } $$
- The acceleration vector is decomposed as $$ \vec{a}(\alpha,\dot{\alpha},\ddot{\alpha}) = \dfrac{\partial \vec{v}}{\partial \alpha} \dot{\alpha}+\dfrac{\partial \vec{v}}{\partial \dot{\alpha}} \ddot{\alpha}$$
- The curvature is $$ {\kappa} = \dfrac{1}{\rho} = \dfrac{ \lvert \vec{v} \times \vec{a} \rvert}{ \lvert \vec{v} \rvert^3} $$
- The acceleration along the tangential direction is $$a_T = \hat{e} \cdot \vec{a} $$
- The acceleration towards the center of curvature is $$a_C = \dfrac{v^2}{\rho} = \vec{a} - a_T\; \hat{e}$$
I think this should be sufficient to deduce all the components you need.
Example
A particle is moving with constant speed along a helix with radius $R$ and pitch $h$.
- Position $\vec{r}(\alpha) = \begin{pmatrix} R \cos \alpha & R \sin \alpha & h \alpha \end{pmatrix}$
- Velocity $\vec{v}(\alpha,\dot{\alpha}) = \begin{pmatrix}-\dot{\alpha} R \sin \alpha & \dot{\alpha} R \cos\alpha & \dot{\alpha}\,h \end{pmatrix}$
- If the speed is $v$ then $\dot{\alpha} = \dfrac{v}{\sqrt{R^2+h^2}}$
- Similarly $\ddot{\alpha} = \dfrac{\dot{v}}{\sqrt{R^2+h^2}}$
- Acceleration $\vec{a} = \begin{pmatrix} -\dfrac{R v^2 \cos \alpha}{R^2+h^2}-\dfrac{R \dot{v} \sin \alpha}{\sqrt{R^2+h^2}} & -\dfrac{R v^2 \sin \alpha}{R^2+h^2}+\dfrac{R \dot{v} \cos \alpha}{\sqrt{R^2+h^2}} & \dfrac{\dot{v} h}{\sqrt{R^2+h^2}}\end{pmatrix}$
- Curvature $\kappa = \dfrac{R}{R^2+h^2}$
Best Answer
Yes. Provided you are only interested in the direction of the acceleration, and not it's magnitude. And further assuming your time samples are equally spaced, you can take the second derivative of the path and this will be proportional to the acceleration.
A decent method in practice would be to use a second order central finite difference scheme wherein you say that:
$$ a_x(t) = x(t-1) - 2x(t) + x(t+1) $$ and $$ a_y(t) = y(t-1) - 2y(t) + y(t+1) $$ this will give you decent estimates for the cartesian components of acceleration at every time, caveat to an overall scaling in magnitude that you won't know without knowing the actual timing, but the direction should be alright.