# Did the experiment, now more confused

1. Jun 5, 2015

### houlahound

some time ago I posted a question re acceleration data. I bought an accelerometer and logged some data.

when stationary the accelerometer (can mention the brand if that is not spamming) read 1g, when thrown in the air got what was expected with 0g ie weightless when the accelerometer switched from upward motion to downward motion due to gravity as expected.

the problem is this when the accelerometer was stationary ie velocity = 0 you can integrate the acceleration curve as a function of time and you clearly get a non zero value because acc = 1g for this time.

the integral gives the acceleration in theory but the data suggests the device had a non zero motion.

how do I rationalise this??

2. Jun 5, 2015

### A.T.

It had non zero motion, relative to a free falling frame. If you want another frame, you have to subtract their relative acceleration from the measurement.

Last edited: Jun 5, 2015
3. Jun 5, 2015

### houlahound

if you were just given the data and did not know how it was obtained like I do and asked to calculate the velocity the only option would be to integrate wrt time and get a false description of the motion would you not??

but yah I can see how subtracting 1g would fix the interpretation and calculation but would also displace the free fall bit ie 0g in original frame to -1g which again leads to a nonsense???

4. Jun 5, 2015

### A.T.

Why is that nonsense?

5. Jun 5, 2015

### houlahound

because I should get velocity = 0 at the peak which is now -1g which on integration will not be 0 it non-zero.

altho the integral is now zero at the place where the device was actually stationary.

will collect some data again and try post it, trying to describe graphs and integrals in text form is just distracting.

6. Jun 5, 2015

### A.T.

From which time point on did you integrate it?

7. Jun 5, 2015

### Staff: Mentor

Integrating the acceleration equation tells you nothing about the velocity expect how it is changing over time.

If F(t) = A, where A is a constant representing constant acceleration, then integrating gives you F(t) = At + C, where C is the velocity at t=0. But, C must either be provided to you, or you have to find it out some other way, if possible. Without having the value for C, you cannot say what the velocity of an object is, you can only say that it is changing according to the equation

If you throw your accelerometer into the air at an initial velocity of 10 m/s, then its velocity changes according to the equation: V = 10 - 9.81t. Setting V equal to zero gives: 0 = 10-9.81t. Solving for t: -10 = -9.91t, or t = 1.02. So the velocity after 1.02 seconds in the air is zero m/s.

If you were to start with the acceleration equation A = -9.81, then integrating with respect to time gives you V = C - 9.98t. As I said before, if you don't know C you cannot find when the velocity is zero because you don't know the initial velocity. But if I give you C = 10 m/s, then the equation becomes identical to the one before and everything works fine.
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8. Jun 5, 2015

### houlahound

can't really do this textually, but why should it matter vel = ∫(accel)dv for any time.

9. Jun 5, 2015

### houlahound

can't edit, my last reply was to A.T.

10. Jun 5, 2015

### Staff: Mentor

Isn't it V = ∫ A dt, not dv?

11. Jun 5, 2015

### houlahound

Drakkith, that moved a few gears in my head, will try apply that to the data, I think I am heading to a resolution with that post.

thanks for replies, going to think about it all.

12. Jun 5, 2015

### A.T.

Right, and only if you start the integration at V = 0. That's why I asked.

13. Jun 5, 2015

### houlahound

thanks for the correction.

14. Jun 5, 2015

### houlahound

the constant of integration does not seem to come into it when doing a numerical integration from a graph, not using mathematical functions to integrate where there would be a "C". doing it dirty straight off the area under a curve.

15. Jun 5, 2015

### A.T.

These methods, applied to acceleration, will give you the change in velocity during the integrated period. You need the initial velocity to compute the actual velocity and displacement.

16. Jun 5, 2015

### rumborak

I figured I'd mention this, but integrating over sampled data will increase your error margin over time. That's why autonomous robots can only go so far by integrating over the acceleration/velocity data before they need to resort to absolute position measurements again.

17. Jun 5, 2015

### A.T.

Throwing the acceleromter up is a short period of time. The biggest error will likely come from the uncontrolled orientation of the accelerometer, which makes it difficult to interpret the measured acceleration given in a local coordinate system. Unless it has gyroscopes too, to track the orientation.

18. Jun 5, 2015

### rumborak

Well, "short" is all a question of sampling rate and inherent noise in the measurement. If you sample a 1MHz, one second is an eternity. Keep in mind that by integrating (i.e. adding) over noisy values, you are convolving the noise Gaussians iteratively, which corresponds to adding the variances together. So, if you integrate over 100 values, your error variance has grown by a factor of 100.

19. Jun 5, 2015

### Staff: Mentor

Of course. That's why I said it needs to be supplied to you or measured. You cannot integrate to find it. Have you taken a calculus class or calculus based physics class yet?