Determination of Enclosed Area

[jmas]

Is there any way to calculate area shown in the attached file, mathematically (not numerically and approximately). The restriction two lines are perpendicular and the arc is circular. Any feedback/reference will be highly appreciated.

 

[brandoncdg]

If you click the 3 lines, type mpedit, and join the lines together into 1 big polyline, you can then click and get the area. Or you can just hatch the area, click the hatch and get the area.

Civil Development Group, LLC
Los Angeles Civil Engineering specializing in Hillside Grading

http://civildevelopmentgroup.com

 

[jmas]

Thank brandoncdg for your reply. It seems there is a misunderstanding here. These lines are not existing ones in autocad. It is an area, on the ground, with following information:
1- two perpendicular lines with 210 and 260 m length.
2- a circular arc which connect end of these lines together.
The question: Is there any formula which give us enclosed area in terms of before mentioned data.
Regards,

 

[brandoncdg]

I would suggest you draw this in CAD and do what I said. Otherwise do this with basic math formulas.

Civil Development Group, LLC
Los Angeles Civil Engineering specializing in Hillside Grading

http://civildevelopmentgroup.com

http://civildevelopmentgroup.com/blog

 

[CarlB]

I haven't found a formula for a direct calculation, but some approximate formulas. The "problem" area is for a segment of a circle, given the arc length & chord but not the angle or radius. You know the chord length, it's the triangle's hypotenuse. Need to assume R & Delta that will satisfy equations for the chord & arc lengths, iterate.

 

[jmas]

Thank you CarlB. Would you please tell me how it can be drawn in autocad? after i create two lines, there isn't any option for draw of a circular arc with two "start" and "end" points and "length" of arc.

 

[RHTPE]

There is a device called a Planimeter that, when calibrated, can give you the area.

Knowing a bit more about the arc (like chord height & radius) can get you to a formula for the circular section. That, combined with the area for the triangle, will give you the total area.

Otherwise, it's an iterative process in CAD to draw an arc of matching length that connects the ends of the 2 perpendicular lines.


Ralph
Structures Consulting
Northeast USA

 

[CarlB]

I don't belive there is a way to directly draw the arc in AutoCAD, as you've found (not found).
You can put the 2 formulas in Excel & iterate.
I get a radius of about 181.36.

 

[chicopee]

Mathematically:
h=r-(r^2-l^2/4)^.5 If angle angle between radii at ends of arc is not more than 180 degrees.

h=r+(r^2-l^2/4)^.5 if angle is more than 180 degrees.

You are given the chord length "l" and the rise "h" value being the distance between midpoints of arc and chord.
Calculate radius "r".

Assuming the drawing is in ACAD, draw two tangents at end points or any where on the arc and their normals at point of tangencies. Intersection of the normals (or perpendiculars) will be the radius of the arc.

 

[jmas]

Let's try this approach.
Suppose length of circular arc is L and half of distance between end of lines is b (see attached file). then we will have:
RxAlpha=L/2
Rxsin(Alpha)=b
So we have:
Alpha-Cxsin(Alpha)=0
where C=L/2/b
This will give us a non-linear equation that i don't think whether it has a straight solution.

 

[CarlB]

You're finding what I found, no direct solution.

Although some responders above indicated otherwise, maybe they can enlighten us


Some formulas at-

http://mathworld.wolfram.com/CircularSegment.html

 

[RHTPE]

CarlB (& everyone else) - There IS a direct solution that involves some equation manipulation that eludes me right now. It would have to be in 3 forms, one for an arc angle less than 180°, one for a half circle (arc angle = 180°), and another for an arc angle greater than 180°.

See page 6-18 in the 8th Edition AISC Manual. I have been spoiled by 25 years of AutoCAD, so my geometric analysis skills are very rusty.



Ralph
Structures Consulting
Northeast USA

 

[CarlB]

RHTPE-

That's just not right, saying you know there is a direct solution then not providing it.

People much smarter than I have said there is not an exact solution, so if you can find one you would rock the mathematical world I think.

Here's a power series solution to thetranscendental equations, whatever that means

http://answers.google.com/answers/threadview/id/417212.html

 

[RHTPE]

Sorry CarlB. From the equations in the AISC Manual, and knowing these 3 things: Arc length, chord length, and that it is a circular arc, intuitively I know there is but one radius. Knowing the radius leads to the complete solution that the OP wants. Finding the solution intrigues me, and when I get some work completed that actually pays the bills, I'll return back to this challenge.


Ralph
Structures Consulting
Northeast USA

 

[brandoncdg]

Sorry my original assumption was that the drawing was a scaled drawing. Out of curiosity what is this problem trying to be solved? I am somewhat curious as to why and what this area needs to be solved.

Civil Development Group, LLC
Los Angeles Civil Engineering specializing in Hillside Grading

http://civildevelopmentgroup.com

http://civildevelopmentgroup.com/blog

 

[RHTPE]

brandoncdg -

Out of curiosity what is this problem trying to be solved? I am somewhat curious as to why and what this area needs to be solved.

Why does it matter? This is likely a lot area question. The OP wants input regarding a direct mathematical solution to the question. Those of us well-versed in AutoCAD know that we can enclosed the area with a polyline and find the area.

But there is no way in AutoCAD to draw an arc knowing just the chord & arc lengths, so it becomes an iterative process. Knowing these 3 things: Arc length, chord length, and that it is a circular arc can lead to only one radius - once that is solved, the rest is easy.



Ralph
Structures Consulting
Northeast USA

 

[chicopee]

CarlB, jmas did not attach an ACAD drawing (.dxf or .dwg) so there is no way to determine the rise "h" or to show the normals at points of tangency along the arc in his .pdf drawing.

 

[RHTPE]

chicopee - If you study the equations I referenced in AISC's Eigth Edition Steel Construction Manual, and if you carefully consider that given a circular arc and knowing its chord length and arc length, there is but one radius for the arc. Once the radius has been determined, the rest is easy.


Ralph
Structures Consulting
Northeast USA

 

[jmas]

Thanks everybody for responses. The drawing which i attached is actually drawn in Autocad but approximately. As CarlB said i also reached to the fact that there isn't any exact solution to the A/M broblem, until another Euler arises on the earth.

 

[btrueblood]

"As CarlB said i also reached to the fact that there isn't any exact solution to the A/M broblem, until another Euler arises on the earth."

??? Of course there is an exact solution. It may not be expressed in a single function, and may involve irrational and trancendtal numbers (square roots and pi), but the algorithms given above will find a single solution, to any degree of precision you wish to calculate.

The closest to a closed form solution, applicable to your diagram, where I replace your numbers by variables:

x (lateral distance=210 in your fig)
y (vertical distance= 260 in your fig)
s (arc length = 425 in your fig)
a = chord length = sqrt(x^2 + y^2)

unknowns:

A = angle of the arc "s" (in radians)
R = radius of the arc "s"

equations:

s = R*A
a = 2*R*sin(A/2)

substituting,

a = (2s/A)*sin(A/2)

this last equation must be solved, most readily by iteration, to find the angle A, knowing a and s. There is only one exact angle "A", between 0 and 2*pi, that will solve the equation, and it can be determined "exactly" for any given value of exactly. Once A is known, the radius R can be calculated, and the center of the arc "s" located. Calculation of the area follows by calculation of the triangles and the circular segment thus formed.