1

For the purpose of this question, let's say that I have an accelerometer that is on a foot of a human. ( red square on by the heel in the below photo )

enter image description here

I'd calculate the angle the following way, as the leg goes from step 1 to step 2.

$\theta = tan^{-1}(\frac{x}{y})$

At step 1, the angle would be $0$, as $y = -1$ and $x = 0$. As the leg moves towards step 2, it would come closer and closer to $90^{\circ}$. If the leg could be raised enough to be parallel to the waist, it would be a full $90^{\circ}$, given the assumption that if $y == 0$, the result equates to $90^{\circ}$.

Now that I have illustrated what I am doing, I am curious to monitor what happens on the z axis. The base assumption thus far has been that the person will go from Step 1, to Step 2, and not swing their leg back and forth. So what I want to do is, as the person is moving their leg, I want to measure the angular change on the Z axis. i.e Is the person moving their leg back and forth while going from Step 1 to Step 2.

This has me a bit stumped at the moment, because I do not know which of the other two axis I should use to determine the change in angle on the Z axis. I'd expect to see the following:

As the person moves the leg back over time, but continues from step 1 to step 2, that the angle goes from $0^{\circ}$ to $-90^{\circ}$. As the person moves the leg forwards, but continues from step 1 to step 2, that the angle on the z axis goes from $0^{\circ}$ to $90^{\circ}$.

angryip
  • 147

1 Answers1

1

In the plane $y-z$ the rotation would be

$$\alpha = \tan^{-1}\left(\frac{z}{y}\right)$$

with positive values for $\alpha$ when moving back.

A general expression for the absolute angle between the leg and y axis is

$$\phi = \tan^{-1}\left(\frac{\sqrt{x^2+z^2}}{|y|}\right)$$

For a general description you could use spherical coordinates system using the standard axis system

enter image description here

$$\begin{align} r&=\sqrt{x^2 + y^2 + z^2} \\ \theta &= \arccos\frac{z}{\sqrt{x^2 + y^2 + z^2}} = \arccos\frac{z}{r} \\ \varphi &= \arctan \frac{y}{x} \end{align}$$

Note that the inverse tangent must be suitably defined, taking into account the correct quadrant.

Setting, with your reference system

  • $z=-Y$
  • $y= Z$
  • $x=-X$
user
  • 154,566
  • can you tell me why you are using the y axis? i.e how come you are not using the x axis? For some reason, it would make sense to me more if you combined x and y, as gravity is going over form one axis to the other, no? – angryip Mar 25 '18 at 20:42
  • @angryip I'm assuming that the forward movements for the leg is in y-z plane, thus the coordinates are z in horizontal direction and y in vertical. I just noticed that y is negative thus I need to correct the part for the sign of the angle. – user Mar 25 '18 at 20:48
  • @angryip Are you interested also to the sign for the angle or only to the absolute value? – user Mar 25 '18 at 20:48
  • the leg can move back and forth while going from step 1 to step 2. ( eventually, when the leg reaches waist height, gravity will fully be on the x axis ) hence, i want to capture the angular change on the z axis in respect to the entire movement, regardless if the x or y axis has gravity on them. && I just want the angle. so i am sure i'll have to use a piecewise function, but that should be easy – angryip Mar 25 '18 at 20:53
  • @angryip thus you are looking for the angle between the leg and y axis? – user Mar 25 '18 at 20:57
  • i must admit that I am not too sure i am following. so let's say that the person moves their leg high enough that it is parallel to their waist. ( gravity is now on -x axis, and y is zero ) When they move their leg back and forth, ( like they are doing a karate kick ) but keeping it at the waist level, i'm not too sure that th y axis would capture anything. right? – angryip Mar 25 '18 at 21:03
  • @angryip ah ok now is clear, to decribe the general movement maybe you should refer to spherical coordinates – user Mar 25 '18 at 21:14
  • i see. what i’m curious about now is how you knew where to put the x and y acis into your equation for sperical coordinates? – angryip Mar 25 '18 at 21:30
  • @angryip It depends what are you givens (I suppose cartesian coordinates) and if you are free to choose other direction for the axes – user Mar 25 '18 at 21:41
  • hmm not sure i am following. So I will know all of the values. ( i.e i can ready the x, the y, and the z coordinates of the accelerometer ) as the leg moves up and side to side. i know that gravity at one point will move from being on the y axis to being on the x axis. So the spherical equation you shared, does this show me the angle on the z axis as the leg performs the back and forth action ( like a karate kick ) as it is moved horizontally from point 1 to point 2? – angryip Mar 25 '18 at 22:05
  • @angryip then you know and have as given the coordinates x,y,z? – user Mar 25 '18 at 22:09
  • right, i’ll always have all the coordinates. the accelerometer has these readily available – angryip Mar 25 '18 at 22:45