# Rate of change of acceleration - basic!

1. Feb 4, 2004

### Micky

I'm plotting rate of change of acceleration against time. Acceleration is measured as "g". Time is plotted on the x axis, rate of change of acceleration is plotted on the y axis. Is "dg" a valid label for the "y" axis?

Thanks

2. Feb 4, 2004

### Staff: Mentor

I'm not clear as to what you are actually plotting. Are you plotting acceleration versus time in order to find the rate of change of the acceleration?

Or are you plotting the rate of change of acceleration versus time, as you state? If the later, you can call the y-axis $\frac{da}{dt}$ or $\frac{dg}{dt}$.

Or are you plotting change of g versus t?

3. Feb 4, 2004

### Micky

Hi Doc

I already have rate of change of acceleration (in the vertical direction) measured as rate of change of "g" which I can plot against time, I can then show a certain rate of change of acceleration during a unit of time (in this case 0.01s)

I've assumed that rate of change of acceleration is the same as rate of change of "g", is this correct?

I'm uncertain if the "y" axis should be "dg" or "dg/dt"

Thanks

4. Feb 4, 2004

### Staff: Mentor

Are you measuring your acceleration in terms of g the acceleration due to gravity? Or is it just coincidence that you use the letter "g" to refer to acceleration?

It sounds like you are actually measuring changes in g during fixed intervals (0.01 sec) at various times. Is this right? If so then whether you plot &Delta;g or &Delta;g/dt doesn't matter, since they will be proportional. But you better know which you are plotting, since that will determine your label.
You tell me. If you are calling the acceleration "g" (why not call it "a"?) then the rate of change of g is the same as the rate of change of a.
What you call the y-axis depends on what you are plotting: are you plotting &Delta;g or &Delta;g/(0.01) ?

5. Feb 4, 2004

### Micky

Hi Doc
Acceleration measured in terms of gravity.
Yes.
Absolutely.
I can certainly call it "a".
Right, I think I see it know The Y axis is dg, but the graph is dg/t [?]

Thanks Doc

How do you post the delta triangle?

6. Feb 4, 2004

### Staff: Mentor

Ah... You are measuring acceleration "a" in units of "g". Do not call the y-axis "g" or "&Delta;g", call it "a" or "&Delta;a" or "&Delta;a/&Delta;t", depending on what you are measuring.

I still don't know what you are actually measuring. Here's my guess at what your data might look like:
T= 1 sec; &Delta;a = .12 g (measured over a 0.01 sec interval)
T= 2 sec; &Delta;a = .04 g (measured over a 0.01 sec interval)

Etc.... Am I even close? What is it that you are measuring the &Delta;a of? How are you making the measurements?

(Note that if you were to list the rate of change of a, it would be measured in units of g/sec not g.)

To find out to make the &Delta;, just quote this post and look.

7. Feb 4, 2004

### Micky

I'm measuring the acceleration of a body in the vertical direction using an accelerometer and datalogger. The raw data is downloaded to a laptop and saved as a text file. The data is presented in two columns, time (measured in units of 0.01s) and acceleration (measured as g). So this means that at any given point in time I know the acceleration of the body.

I load the text file into Excel and calculate how acceleration changes from one measurement to the next, therefore these units must still be g, not &Delta;g [?] .... confused again!

I can then plot a graph of the results of the calculation against time to show the &Delta;g/t [?] [?]

M

8. Feb 4, 2004

### Staff: Mentor

I think I see what you are doing now. Your spreadsheet calculates &Delta;a at each time t (I'll bet that you say &Delta;a(at t = n) = a(at t = n+1) - a(at t = n)... or something like that). Good.

Note that "g" is a constant (equals about 9.8 m/s2) so &Delta;g makes no sense. You are going to graph &Delta;a as a function of time. Also note that both a and &Delta;a will have units of "g".
Assuming your spreadsheet finds &Delta;a, your graph would show &Delta;a plotted against time. You will be able to see how the acceleration varies over time.

9. Feb 4, 2004

### Micky

Thanks for that Doc