# Is acceleration sufficient?

1. Feb 20, 2010

### wirefree

Greetings.

If an object's x, y & z components of acceleration over time are available, could it be established whether it is moving in a plane or along a straight line? Or would a double integration to determine position coordinates be necessary?

Look forward to a prompt response.

Best regards,
wirefree

2. Feb 20, 2010

### torquil

I'll denote the components of acceleration ax, ay, az. Let us consider the movement starting at t=0.

Assume that the particle starts with an acceleration that is in the same direction as its initial nonzero velocity. This defines a line. The particle will stay on this line if and only if (ax,ay,az) is zero or tangent to this line at all later times.

If the particles starts with a velocity in a different direction that its initial acceleration, these two vectors already define a plane. The particle will stay in this plane if and only if any later acceleration is zero or tangent to this plane.

If the particle starts at rest, the initial acceleration defines a line. If the time derivative of the acceleration at t=0 points in a different direction, this defines the plane. It will stay in this plane if and only if the acceleration is tangent to this plane at all later times (or zero).

Torquil

3. Feb 20, 2010

### Staff: Mentor

You can tell if it is moving in a straight line. That only happens if a=0. You cannot tell if it is moving in a plane (think about the acceleration of uniform circular motion or helical motion).

4. Feb 20, 2010

### rcgldr

For the straight line case, as long as the ratios between the accelerations ax, ay, az remain the same (or zero), then the path is a straight line. The ratios would correspond to a directed vector in 3d space. You could "unitize" this into a vector, define h = sqrt(ax^2 + ay^2 + az^2), then the unit vector would be {ax/h, ay/h, az/h}.

The plane case is more complex. You could have a circular path. It's my guess that ratio of accelerations would need to comply with the formula for a plane in 3d space (ax + by + cz + d = 0), since any acceleration with a component perpendicular to the plane would result in a non-compliant path.

Last edited: Feb 20, 2010
5. Feb 20, 2010

### wirefree

Greatly appreciate all responses.

Please find my follow-up responses & queries below.

Would it be equally effective to say that as long as change in acceleration in only along one axis, the object is following a straight line?

In order to validate the guess above, I submit below a reading of XYZ-components of acceleration, as captured by the accelerometer in my mobile phone, while I take half a golf swing from rest position to top-of-the-swing position (see 1st three images in the sequence image below):

http://img534.imageshack.us/img534/4379/arodgolfswing2.th.jpg [Broken]

Code (Text):

X:36, Y:2, Z:50
X:35, Y:3, Z:51
X:36, Y:3, Z:51
X43:, Y:3, Z:48
X65:, Y:3, Z:34
X:85, Y:2, Z:29

NOTE: Since the phone's orientation changes as I move my hands up to the top-of-the-swing position, so does the orientation of the 3 axis.

Could we arrive at some conclusion about Jeff Reid's guess that ratio of accelerations would need to comply with the formula for a plane in 3D space?

Best regards,
wirefree

Last edited by a moderator: May 4, 2017
6. Feb 20, 2010

### rcgldr

It's clear that if an acceleration has no component perpendicular to a plane, than that acceleration results in the object remaining on the plane. You need the direction of the acceleration vector to be perpendicualr to the line perpendicular to the plane, which is how the equation for a plane is derived.

7. Feb 20, 2010

### Staff: Mentor

Only if the initial velocity is along the same line (or 0), not in general. My understanding of the OP's question is that he wants to know if the path is along a line or plane using only the acceleration and not any information about initial velocity or position.

8. Feb 20, 2010

### rcgldr

Then there would be insufficient information. You'd need to know the initial conditions in order to determine the initial position and/or direction. What's missing are the constants of integration.

9. Feb 20, 2010

### wirefree

As closure, could you please confirm if acceleration vectors may be calculated as standard position vectors are. For example, for the above provisioned data on accelerometer readings, would the acceleration vector obtained using the first two data points be <-1,1,1>, for the 3rd & 2nd data points, <1,0,0>, and so on?

Best regards,
wirefree

10. Feb 21, 2010

### Staff: Mentor

There is insufficient data in your accelerometer readings to reconstruct the swing. Your accelerometer is rotating and there is no measurement of how much it is rotating. However, if you start from rest at a known position then you could, in principle, reconstruct the swing if you had the rotation information also.

11. Feb 22, 2010

### wirefree

Suppose my purpose be simply to ascertain whether the accelerometer is moving parallel to a given plane, and not establishing its position vector. Hence, I adopt the following methodology:

a) For every reading of XYZ-components of acceleration (e.g. X:3,Y:31,Z:0), I dot product the vector with the normal to the given plane.
b) If dot product of the two vectors is zero, then I can establish whether the accelerometer is moving parallel to the plane.

Kindly confirm if the above suffices for the given purpose.

Best regards,
wirefree

12. Feb 23, 2010

### Staff: Mentor

No, it does not. In addition you must know the initial velocity. If the initial velocity is in the plane (or 0) then your method will work, but if the initial velocity is not strictly in plane then the path will be helical rather than in plane.

However, there is one further complication, your accelerometer XYZ axes are rotating so their relationship to a given plane at any moment is unknown. So you will not even be able to take the dot product with the given set up.