cancel
Showing results for
Did you mean:
cancel
Showing results for
Did you mean:

Community Tip - Did you get an answer that solved your problem? Please mark it as an Accepted Solution so others with the same problem can find the answer easily. X

1-Newbie

## How to calculate area?!

Hello! I have a 2 graphs. I don't know how to calculate the area under the graph. Graph with red is half of a circle, and graph with black is defined by 3 functions. Normally, the graph area with red is higher than the graph area with black. Thank you for help!

1 ACCEPTED SOLUTION

Accepted Solutions
24-Ruby V
(To:studor)

As you know where the different segments join and as the curve seems to be symmetrical wrt x071 it should be easy to calculate the area under the black curve using three different integrals.

I am still concerned as of your unit mismatch, but here is a way whoch uses the "units" you provide:

BTW, you talk about the red "half-circle" but is height is 7.5 (whatever unit) and its width is 2*71 (whatever unit).

EDIT: My result differs from Freds because he sticked to the range variables you used for plotting, but I think that those ranges overlap and so I tried to compensate for that. I may be wrong, though, as i am still a bit confused about the complicated and clumsy way you defined and plotted your functions.

Also I assumed symmetry which upon closer inspection seems not to apply. So you would have to a get rid of the front factor 2 and add three more integrals.

7 REPLIES 7
24-Ruby IV
(To:studor)

Monte Carlo Method!

24-Ruby V
(To:studor)

I have a problem with the units in your application

You label the abscissa of your plot as being an angle and obviously (despite of the fact you write °RAD) you mean the angle be measured in degree.

But the absicissa of the red curve (X(...) is of dimensions length, probably unit mm. Thats a discrepancy!

Furthermore, what should the unit of that area you are looking for be? mm^2, degree times mm or simply mm (=radians times mm).

1-Newbie
(To:Werner_E)

I forget to say this. Consider Y axis simmilary with X axis. In mm units. Maximum is 7.5 mm like on X Axis. The half of circle will have area 3,14/2 * (7.5)^2. This means 88.312 mm^2. But I don't know how to calculate area for black graph.

24-Ruby V
(To:studor)

As you know where the different segments join and as the curve seems to be symmetrical wrt x071 it should be easy to calculate the area under the black curve using three different integrals.

I am still concerned as of your unit mismatch, but here is a way whoch uses the "units" you provide:

BTW, you talk about the red "half-circle" but is height is 7.5 (whatever unit) and its width is 2*71 (whatever unit).

EDIT: My result differs from Freds because he sticked to the range variables you used for plotting, but I think that those ranges overlap and so I tried to compensate for that. I may be wrong, though, as i am still a bit confused about the complicated and clumsy way you defined and plotted your functions.

Also I assumed symmetry which upon closer inspection seems not to apply. So you would have to a get rid of the front factor 2 and add three more integrals.

1-Newbie
(To:Werner_E)

Thanks! This is graph of lifting height for a cam. 7.5 mm is maximum height and 142 degrees is lifting time measured in camshaft rotation. And elipse is for a ideal cam which have maximum randament.

23-Emerald I
(To:studor)

You can integrate to find the areas.  The curve that's only two functions is easiest:

For the other, you need to either make a single function or do three piecewise integrations.

But I agree with Werner, you're not plotting apples and apples.

• The first (black curve) is plotting length (h) against angle,
• The second curve (red) is plotting Y against X length against length.

Thanks!

Announcements
Top Tags