Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have an Accelerometer connected to a device that feeds the instant values of the acceleration in the 3 directions. I've tried to calculate the distance for a vertical movement using these values with excel (applying two times an integration), but somehow it doesn't seem to work properly.

How would be a good way to calculate the traveled distance from the measured acceleration using excel tables?

share|improve this question
    
If you are only interested in vertical displacement, you only need vertical acceleration as well. Can you show us what you did? –  Raskolnikov Sep 13 '12 at 11:51
    
In fact I'm interested in displacement in all directions, but for starters I've tryed so simple as possible... –  Francisco Sep 13 '12 at 11:56
    
I have a long list of values, and since I know the time interval for the measures, I've tried calculating first the mean of the acceleration, and then x = a * 0.5 * t^2 ... the test result is far different to the expected distance –  Francisco Sep 13 '12 at 12:01
    
Look at Eurequa ( creativemachines.cornell.edu/eureqa ) where you can take measured data, smooth, fit, and apply calculus functions. –  ja72 Sep 13 '12 at 14:40

1 Answer 1

up vote 2 down vote accepted

Starting at ${position}_z$ = $z$ = 0 and $v(z) = 0$ and by tracking multiple acceleration values either with a time interval or at fixed intervals, $t$, then you can get the position.... somewhat. It will drift over time. Also, your device cannot rotate whatsoever, or else you need a gyroscope to track that and then use trigonometry to properly orient the x y and z values from the accelerometer. Assuming it's always oriented such that the $a(z)$ is always perfect vertical acceleration (if you're in a vehicle that's always flat, in which case z doesn't matter, or you're on a vertical guide rail),

$$p(z) = \int_0^t v(z) ~dt = \iint_0^t a(z) ~dt $$

Also, from here:

Short answer: Forget about it.

Longer answer: Unless you're on a perfectly straight rail, you will not achieve what you want to do without (a) a set of gyros; and (b) Far more accurate sensors than what you have.

Accelerometers measure acceleration in the body fixed reference frame, whereas you need some displacement in an earth-fixed frame.

Therefore, you need not only to integrate the accelerometers, but rotate them into the earth-fixed frame before doing the integration.

This is assuming perfect sensors. MEMS sensors are far from perfect - I have written up a post on some of the errors here.

Consider two errors: 1. A bias on the accelerometer. 2. An initial attitude (tilt) error.

In addition to whatever acceleration signal there is, integrate a bias and you get a ramp error with time. Integrate the ramp and you get a quadratically increasing error with time. This will add up really, really quickly.

Consider a tilt error. You'll now be measuring some of the gravity vector in the forward (or whatever) direction. Integrate this error twice and you'll have the same problem as the bias.

So, my advice again is DON'T! Find another method.

Also, check this book out for more detailed designs, or use whatever sensors and algorithm these guys are on:

http://www.youtube.com/watch?v=6ijArKE8vKU

If you still want to give this a shot, use the Trapezoidal method in Excel, it's pretty easy. There's an explanation page here with a sample, but here's a more complete way:

1  Time [A]  Acceleration [B]   Velocity [C]                Distance [D]
2  0         a(z)               0                           0
3  1         a(z)               =C2+(A3-A2)*(B2+B3)/2       =D2+(B3-B2)*(C2+C3)/2
4  2         a(z)               =C3+(A4-A3)*(B3+B4)/2       =D3+(B4-B3)*(C3+C4)/2
5  3         a(z)               =C4+(A5-A4)*(B4+B5)/2       =D4+(B5-B4)*(C4+C5)/2
6  4         a(z)               =C5+(A6-A5)*(B5+B6)/2       =D5+(B6-B5)*(C5+C6)/2
7  5         a(z)               =C6+(A7-A6)*(B6+B7)/2       =D6+(B7-B6)*(C6+C7)/2
8  6         a(z)               =C7+(A8-A7)*(B7+B8)/2       =D7+(B8-B7)*(C7+C8)/2
9  7         a(z)               =C8+(A9-A8)*(B8+B9)/2       =D8+(B9-B8)*(C8+C9)/2
10 8         a(z)               =C9+(A10-A9)*(B9+B10)/2     =D9+(B10-B9)*(C9+C10)/2
11 9         a(z)               =C10+(A11-A10)*(B10+B11)/2  =D10+(B11-B10)*(C10+C11)/2
12 10        a(z)               =C11+(A12-A11)*(B11+B12)/2  =D11+(B12-B11)*(C11+C12)/2
   ...       ...                ...                         ...
share|improve this answer
    
While you are totally correct that this is not a reliable method it really depends on the application. For something like a model rocket with a large acceleration in one direction this can work nicely, where as to track the position of a person using a smartphone without GPS this is hopeless. –  Alexander Sep 13 '12 at 12:57
    
The model rocket will experience some amount of 'tipping' and may even reverse direction on descent, which will skew your measurements by providing smaller z accelerations than accurate, and then by adding to the position when falling, respectively. On a ground based vehicle x and y might be useful, but once you introduce suspension it really needs a gyroscope to track the roll, pitch, and yaw of the accelerometer to have much accuracy. –  Ehryk Sep 13 '12 at 17:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.