What  are the variables, parameters and constraints of data

jjackson2004
jjackson2004 used Ask the Experts™
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given gravitational data points

Time (min)                   gforce
.01                                35
.03                                 28
.1                                   20
.3                                   15
1                                    11
3                                     9
10                                  6
30                                  4.5

I understand what the variables are,  what are the parameters(it is 11?)  and what constraints are there.  figured out it was an exponential function similar to f(x)=11* x^-0.250

Also need to figure the correlation coefficient for this function.  Think it is .35 or similar (trying to do linear regression on ti-84).

Might be a couple of additional questions once I better understand

thanks
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"figured out it was an exponential function similar to f(x)=11* x^-0.250"

I have some difficulty with your notation.
Using your equation I would expect something like
gforce (t) = 11* t^(-0.250)
which is not an exponential equation and does not come close to fitting the data.
The curve is obviously non-linear so you cannot use a linear regression to check for goodness of fit

Author

Commented:
it is very close to the function that graph 4.3 calculated for the data points

if you take f(1)=11 you have the starting point.
from there I had to figure out how to match to the rest of the data.

back original question:  what are the parameters and what are the constraints?

What is the correlation coefficient?
Might be a couple of additional comments once I better understand.
-
"it is very close to the function that graph 4.3 calculated for the data points"
I do not understand this at all.
If you understand what the variables are, what are they?
-
I am sure that the parameters are not 11  (you might be able to find a parameter using that data point, but any other would be as good.
-
What do you mean by "do linear regression on ti-84"?
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Author

Commented:
I think now that it is a power function, not exponential.

Apologies, but I am not sure how to explain it further.  plugged data point into graph 4.3 and had it plot the best fit using different graph types.  found one that fit the data points.

Do you know what my parameters are?

to answer the question about "do linear regression on ti-84" please refer to http://mathbits.com/mathbits/tisection/statistics2/correlation.htm
"graph 4.3"
and what is graph 4.3. It does not seem to be available to me.
I looked at your program. It is a little much to learn a new program but it is an interesting and good one. After spending about two hours on it I can give you some additional comments.
It is NOT a power function.
I would suggest
f(x)=10/(log(10+1))+1
in this case the variables are on the x axis x (really time) and acceleration (g but not gravitational).
the parameters (to be adjusted for best fit are 10 and the two 1s.
The constraint is that x must be <0. I leave it to you to get the correlation coefficient.
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Actually your function f(x)=11*x^0.25 is probably better than mine. In which case the parameters are 11 and 0.250 and there are no constraints although because of the data points it would not be out of place to consider only the x>0.

Author

Commented:
My focus has moved on.  Only so much time to spend on mental exercises.  Thanks.  

On to the next pursuit.

I will give you the points since you are the only one who took the time to help out.

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