Help Needed by Maths literate person

[Brian d marge]

Hi folk
I posted this under tuning as eventually thats what the above will be used for ......
What I need help with is help with intergration ....I was to busy looking up the skirt of the maths teacher ,,,,and that was a few years ago now .....

Anyway ... I need to find the area under the/ Binomial ( an upside down U just incase I got the name wrong ....
What I would like to do Is use intergration to arrive back at the equation of the graph ,,,,

Whats happening is I am getting the data ,,,constucting /plotting the graph ..then I can find the area ( Use acad and ask for the area ....!!!!) ,,,but it doesnt give me the equation ,,,which in some circumstances could be of use .....
Anyone Know of a method of working back to find the equation and hence the area ?????
Its worth a punt ,,,,,,( and before anyone asks I did a google search ,,,but me x s and Y s ,,,fell apart ,,,as it wasnt easy to understand ,,,,)

thanks in advance I hope some one can point me in the right direction ,,,,,

Stephen

who also hasnt access to any text books ,,,,,,

[erik]

you want the equation for the curve that fits the data?

Or you want teh equation for the area under the curve?

If you have excel, to get the equation for the line that fits the data, make an XY (scatter) plot graph, then go to the chart menu and "add trendline" and you can choose what type you want (polynomial, I guess.) and make sure you click the next tab on the window and select "show equation on graph"

If you haven't got excel, I dunno how you'd do it :p

[ZorsT]

the curve you speak of sounds like a parabola. (a -ve one at that)

If I had my books here I might be able to tell you.

I'm not sure if 6th form is enough though

[sAsLEX]

what eric said is the goer for the equation of the line, then its integration of that to find the area, also a pic always helps! post it up here there are enough of us engine-neers here to try sort it out

[marty]

it's 42. everyone knows that.

[marty]

oops sorry gareth - i've gone off track. my apologies.

[Drum]

Sounds like a polynomial equation (-ax^2-bx-c) picture needed.
Area doesnt matter as that area could be bounded by any number of equations.More helpful to provide a few sets of co-ordinates. Then it can be solved.

Or as previously mentioned. Plot it in excel. Add a polynomial 'trend line', then right click the trendline, go to options, and click display equation on graph.

[R6_kid]

oops sorry gareth - i've gone off track. my apologies.

how did you know i was watching!!! i was thinking the exact some thing, 42 is the answer to the universe, life, and everything...

its not off track, tis good humour

now what im doing is.... :argh:

stephen - put a pic up and i'll see if my D- in Maths 108 (calculus) can help you at all :niceone:

[Brian d marge]

you want the equation for the curve that fits the data?

If you haven't got excel, I dunno how you'd do it :p

dunno about having Excel....I havent a clue ....that doesnt help matters !! but yes I will post a pict when I get home

Stephen

[T.W.R]

have worked with excel quite a bit and don't really think you'll be able to calculate a specific arethmetic progression or volume through excel unless you have the complete data range of the parabolic curve( i presume thats what your working with) and know the set points/area of the curve you want unless you use a isological or isref calculation in excel.
you may have to use a CAD program to identify the points for calculation
ie. excel can't equate volumetric sequences or measurements into a true figure

[FzerozeroT]

Hmm, could help if you were a bit more specific about the data gathered, i.e. if it is the resonant value of an exhaust etc, because you need to make sure that the equation has a distingushable variable for each physical thing. Sampling errors could give you a polynomial that is missing a factor such as airbox resonance/volume, and you'll end up using a bollox equation. Or ignition mapping??

[Brian d marge]

Ok..hope this works ..BTW thanks for the response terriffic ...Mongo ere aint to good wiv da numbers ,,,I just hit em wiv big ammer till they look good ...Eric can vouch for that !!!

hope the attached file helps

I use acad at the moment ,,,but one day Im not going to have access to acad ,,so really must learn the method ..( again ..was to busy looking up teachers skirt ,,,,can remember her clearly ...black stockings ,,little black dress,,,must of dropped me pencil a thousand time a class.....)

So yeah both the area under the graph ..( ralf half pissed as a newton method ..adding squares ...
and the equation of the line would be great .....

Thanks Stephen

Stephen

[Drum]

Brian, do you have the raw data in some sort of file? If so, post that instead.

[avgas]

How far apart are the data points brian? cos if they are close enought to be accurate then you can use the simple summation rule.
ie add all the totals together to give you the total area. There looks to be sufficent resolution of data points to do that and will get you to within .01mm of the actuall integration.
If you give me the raw data i can probably punch it through MATLAB if i still remember how to use damn program....cursed black magic.

[Jeremy]

As nice as the data plot is the raw data would be far more useful.

Your best bet is probably to use Simpson's rule even though chances are your function isn't a polynomial with a degree of cubic or less. [Looks like a sine function with a simple transform to me]

[sAsLEX]

punch it through MATLAB if i still remember how to use damn program.


in a strong cockney accent * MATLAB its DEAD EASY* for all the AU enginners!

[Brian d marge]

As nice as the data plot is the raw data would be far more useful.



I am on to that will post the raw data as soon as I get my hands on it ....

The simpsons rule ,now that would give me the area easy enough

( something like ...hieght/3 { Yo +4Y1+Y2.......}=area

Thats easy ,,,cool ...but I should be able to work backwards to find the equation ..

Lets say for exaple I use simpsons rule and find the area ..is 16.7 whamajamas ..

Using an old copied example I found

Area = integral between 3 and 1 ( X^2 +4)dx
= { X^3/3+4X}
= (3^3/3+4 x 3)-( 1^3/3+4 x 1 )=16.7

I am answering my own question here ,,,I can see it ..:doh:

Anyway I will carry on ...so yes From the area I should be able to go back to the equation ,,,,,

and with that equation I should then be able to add and subtract mutiply equations together ,,,

Anyway I will post the data as soon as I get my mitts on it ,,,,,,,
see how we go

Stephen

[Brian d marge]

Ok
when the raw data turns up Ill post it ...till then ..try this see if I am in the ball park

Take a parabola ...X^2 cutting the X axis at -2 and 2 and the Y axis at 4

When Y = 0
x = -2 and 2

( X-2) (X+2)
= 4-X^2
Intergrate between -2 and 2 to get the area I get 10 2/3

checked with simpsons Rule get the same ....which is cool ..and exactly what I wanted great ..

Now the original graph as was pointed out was a sine function SOOOOOOO all I have to do is expand the above to include ,,,the sine function ????


So like a plan??????


Stephen

BTW thanks for all the help

[zooter]

There are common integrals listed in the back of your logarithm tables which should see you right.

[erik]

I don't actually know the proper way to do this kinda stuff.

But, from plotting the data in excel and messing around with sine equations, the closest I've got to an equation that gives a similar curve is:

=0.316*SIN(RADIANS(A1-11))-0.316/2*SIN(RADIANS(A1-11))+0.158

Take a look at the graph in the excel file. There is the original data curve, then the approximation based on the above equation. And I've plotted the difference between the two. If you could figure out an equation for the difference between the two and add it to the other approximation, you'd get the equation you wanted. 'course, that's just as hard or harder than finding the equation for the original data...

Dunno if this helps or not.

I don't think you'll have much luck with a polynomial equation or parabola. I think it has to be based on trig functions like sine and cos etc.

[erik]

Actually, the trick might be combining a parabola plus a sinewave.

I spent some time with excel again, the parabola+sinewave curve matches more closelly than the previous sinewave only.
Maybe with some more fiddling with the figures, you could get it an even closer match.
Or maybe there's some other easier way to do it...

[parsley]

I suppose the proper way of getting the equation of a curve is to use Fourier analysis, but I've always preferred Bezier curves. I've used them quite a lot for curve editing functions in GUIs, but they're also a simple way of defining a curve.

[k14]

Nah just fit a 6 order poly line of best fit to teh line in excel, gets a pretty damn close fit then. That gives a formula of y = 1E-10x6 - 2E-08x5 + 2E-06x4 - 0.0001x3 + 0.0025x2 - 0.0095x + 0.0082

Then you integrate it and fine the finite integral from the lower to the upper limit (what were they again?). Not too hard to do. I'll do it later if you want.