How to get the area of round-like shape more correctly?

I have a set of radiuses with same interval angle of 0.1 degrees.

The radiuses are about the same but a little difference.

For example,[ 10.1 10.12 10.11 10.09 ...].

So the shape is round-like but not a really circle.

I use circular sector area formula to get sector area of each radiuses and sum all of them to get the round-like shape's area.

Are there any inetrpolation method or integral method in Matlab can help me to improve the area calculation?

[Scott MacKenzie]

@Lin Ming Che I'm just re-reading your question and WR's answer. Is your objective to get the exact area of the shape defined by the points?

[Walter Roberson]

You cannot use the sector area directly when the radii are different: you have to use the minimum of the two radii (underestimate) or the maximum (overestimate) or the mean (probably underestimate)... or use an interpolation model.

[Scott MacKenzie]

OK, noted. But why bother with the sector areas? What's wrong with just converting the polar coordinates to cartesian coordinates and working from there?

[Walter Roberson]

The user has a sampling of a system in which the system being modeled has a continuous curve boundary -- a boundary that might be near to 10 but fluctuates. For example it might be a circle that is being distorted by stress.

And the question is what is the most accurate prediction of the area of the enclosed curve ?

The polynomial area would match the curve area in the limit case of the number of sides approaching infinity, but in the finite case, the polynomial area will always underestimate the area of the enclosed curve.

[Walter Roberson]

If r were a continuous function of θ, then you could use

but we do not have a formula for , and a question becomes what a reasonable way is to model given a finite list of r values at known θ and the fact that it is a continuous curve being modeled.

[Scott MacKenzie]

Well, I don't see anything in the question about a continuous curve being modeled. Perhaps the data are empirical -- measurements taken radially of a property that fluctuates at random. Or perhaps the fluctuation is due to noise in the instrument. So, I saw the question in fairly simple terms. But, @Lin Ming Che accepted your answer, so you're probably right.

[Walter Roberson]

Poster said "So the shape is round-like but not a really circle." That implies a continuous curve.

Answer 1

[Walter Roberson]

Assume that the center of the near-circle is at (0,0) .

Now, for any two adjacent points, you can fit an ellipse through the points

Pi = sym(pi);
syms a b r1 r2 d_theta positive
syms x y theta1 real
theta1 = 2*Pi*randi([0 359])/360

theta1 = 

d_theta = 2*Pi/360;
theta2 = theta1 + d_theta

theta2 = 

eqn = x^2/a^2 + y^2/b^2 == 1

eqn = 

x1 = r1*sin(theta1); y1 = r1*cos(theta1);
x2 = r2*sin(theta2); y2 = r2*cos(theta2);
eqn1 = subs(eqn, [x, y], [x1, y1])

eqn1 = 

eqn2 = subs(eqn, [x, y], [x2, y2])

eqn2 = 

sol1 = solve([eqn1, eqn2], [a, b], 'returnconditions', true)

sol1 = struct with fields:
             a: [4×1 sym]
             b: [4×1 sym]
    parameters: [1×0 sym]
    conditions: [4×1 sym]

sa = simplify(sol1.a)

sa = 

sb = simplify(sol1.b)

sb = 

simplify(sol1.conditions)

ans = 

R = 10 + randi([-3 3], 1, 2)/10

R = 1×2

9.9000 9.8000

valid = simplify(subs(sol1.conditions, [theta1, r1, r2], [0,R]))

valid = 

sa = simplify(subs(sol1.a(valid), [theta1, r1, r2], [0,R]),5)

sa = 

vpa(sa)

ans = 

10.62005020475202152256927138888

sb = simplify(subs(sol1.b(valid), [theta1, r1, r2], [0,R]),5)

sb = 

vpa(sb)

ans = 

5.3239078115062981895990085871548

Then, with a and b in hand, you should be able to calculate the area of the sector of the ellipse.

However... the equation I used here is for a non-rotated ellipse. For non-rotated ellipses, at theta1 = 0, then the radius at the next location must be strictly less than the radius at theta == 0. Also, if the difference in radii is too large, the code will fail to find a solution, again reflecting the constraints of non-rotated ellipses.

[Walter Roberson]

An alternate approach might be to take each group of three adjacent points, and find a circle that passes through the three points, and calculate the sector area of that; https://www.futurelearn.com/info/courses/maths-linear-quadratic/0/steps/12130 for the circle fitting.

This would have the difficulty that each sector is counted twice, with slightly different models. As a first approximation, you could divide the total by 2... but you can probably do better.

Answer 2

[Image Analyst]

What about using polyarea to compute the area of each (nearly circular) polygon?

% I have a set of radiuses with same interval angle of 0.1 degrees.
angles = 0 : 0.1 : 360;
% The radiuses are about the same, but a little different.
% For example,
radii = [10.1 10.12 10.11 10.09]
% Compute the area for each "circle" (actually a polygon) using polyarea().
for k = 1 : length(radii)
	x = radii(k) * cosd(angles);
	y = radii(k) * sind(angles);
	% Plot circle.
	plot(x, y, '-');
	grid on;
	hold on;
	% Compute area.
	thisArea(k) = polyarea(x, y);
	fprintf('The area of the polygon for radius = %.2f = %f.\n', radii(k), thisArea(k));
end

You get

The area of the polygon for radius = 10.10 = 320.473704.

The area of the polygon for radius = 10.12 = 321.744163.

The area of the polygon for radius = 10.11 = 321.108619.

The area of the polygon for radius = 10.09 = 319.839417.

Is that what you mean?

[Walter Roberson]

No, it is not what the user means.

The user has an object that is not quite circular, but is circular-like. For example,

N = 20;
r = 10 + randn(1,N)/3; r(end+1) = r(1);
theta = linspace(0,2*pi,N+1);
polarplot(theta,r)

except with N = 360.

This is a sampling of system in which the system being modeled has a continuous curve boundary -- a boundary that might be near to 10 but fluctuates. For example it might be a circle that is being distorted by stress.

And the question is what is the most accurate prediction of the area of the enclosed curve ?

[Scott MacKenzie]

My read on this question is that there is only one shape, something like this but with more points:

[Image Analyst]

When he said:

I have a set of radiuses with same interval angle of 0.1 degrees.

The radiuses are about the same but a little difference.

For example,[ 10.1 10.12 10.11 10.09 ...].

So the shape is round-like but not a really circle.

I took it to mean that there was a set of angles for every shape, that went from 0 to 360 in steps of 0.1 degrees, and that one radius would apply to every vertex in the shape. So you get a polygon with 3600 vertices. Then he wants to investigate how quantization affects the area. Like if it was real small, like an octogon of course it would not match the perfect circle very well. As the circle gets bigger and with more vertices, the shape gets rounder and rounder and closer and closer to a perfect circle. At least that's how I took it.

Regardless, if it's a irregular shape like have, or a "perfect" polygon like I assumed, either way can have it's area calculated if you put the 3600 vertices into polyarea().

[Walter Roberson]

If it is intended to be a continuous curve, then polyarea() will typically underestimate.

Consider the case of an approximation of a circle, with no variation in the radius. Then polyarea() would correspond to inscribing a polygon with 3600 vertices inside the circle, and that is certain to have a smaller enclosed area than the circle itself.

[Scott MacKenzie]

The words curve or continuous do not appear in the question. It's a round-like shape defined by a colleciton of points and the questionner wishes to compute the area. That's how I read the question. However, I suspect WR's inference is correct since the questionner accepted his answer.

I'd be curious to see, for a given set of points, the difference in areas calculated using WR's approach and the use of polyarea. Not much difference, I suspect. For a third comparion, how about this. Since the round-like shape has 3600 radii at 0.1 degree spacing, and the radii are the same except for a little difference ...

area = pi * mean(radii)^2;

[Walter Roberson]

I think you would be better taking sqrt(mean(radii.^2)) as the mean radius. That would lead to pi*mean(radii.^2) as the area

[Scott MacKenzie]

Ah, yes. Noted. Thanks.

[Bjorn Gustavsson]

@Walter Roberson: It will only underestimate a sector-area in the case that the curve is convex, which it might be but that is not necessarily so since you will have sectors where the radius is smaler than their neighbours. For such sectors it seems to be a bit peculiar to assume that they are circle-sectors...