circle problem, circumscribed to another circle and the area, area per image detection

Question

David Alejandro Ramirez Cajigas 2021-7-23

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/884329-circle-problem-circumscribed-to-another-circle-and-the-area-area-per-image-detection

编辑： Andrew 2021-8-26

mini2.xlsx

I have been working on a transportation problem, I have a series of coordinates that always vary, but for this question let's suppose that I have 13 coordinates, each chord represents the center of a circle with a radius of 400m.

the circles are on an x, y plane. The idea is to find the total area that is inside each circle, the problem is that it is not simply multiplying the number of circles by (PI * R ^ 2), because many of them have 1 or more shared areas that do not they can be added more than once.

These circles may or may not intercept each other, through the program that I am designing, I even obtain the distance matrix between each one, at first I thought that the best way to solve the problem was mathematically, but today I realized that it is a problem of mathematical complexity .

So looking on the internet I found that more people have had a similar problem and that it can be solved analytically.

It turns out that there are people who say that the problem could be solved if each point is plotted on a 2d plane, then a 400 meter Buffet is placed at each point, each circle is painted in one color and then the background in another color.

Finally, an algorithm is required that is capable of finding the area that is colored, suppose that what is inside each circle is yellow and the outside is green.

could you please help me with this

attached document excel file

1)Once I have plotted the figure, how do I put the 400 meter buffer at each point, I know the x, y coordinate of each point)

2)Supposing that I manage to obtain a graph with the points and their respective buffer of 400 meters of radius, how I paint everything that is inside the buffer of the same color

3)how do i find the colored area inside each buffer

T=readtable('mini2.xlsx')
H = table(T.lon,T.lat);
plot(T.lon,T.lat,'ko*400') 

plot It has no buffer, I don't know how to implement it

Thanks for your time.

examples:

I have read about the problem here : https://qastack.mx/programming/1667310/combined-area-of-overlapping-circles

http://jsfiddle.net/broofa/WRpmW/5/

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David Alejandro Ramirez Cajigas 2021-7-23

the centers if I have them, in a coordinate file

Andrew 2021-8-26

编辑：Andrew 2021-8-26

Are approximate answers OK? Since the radii are always 400 m, I would consider finding out ahead of time how much two circles overlap as a function of distance. At x/r~=0, the second circle has basically zero area, and at x/r~=2, it's basically a whole circle. Then you can simply pairwise see which ones overlap and apply the function. The only serious issue I see is that if you have three or more overlapping, you'll overcount the overlap, so I suppose this would work best if it's a pretty sparse map or the little bit in the midst of three circles doesn't matter much.

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Answer 1

DGM 2021-7-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/884329-circle-problem-circumscribed-to-another-circle-and-the-area-area-per-image-detection#answer_752179

编辑：DGM 2021-7-23

在 MATLAB Online 中打开

This could be improved or elaborated, but it's a simple start:

% let's assume that 1px = 1 square meter

r = 400;

C = rand(10,2)*5000 + 1000; % test locations (x,y)

% get padded image size

p = 10; % extra padding

s = ceil([(max(C(:,2))-min(C(:,2))),(max(C(:,1))-min(C(:,1)))] + r*2 + p*2);

C = C - [min(C(:,1)) min(C(:,2))] + r + p; % shift image

% draw all circles

m = false(s);

for c = 1:size(C,1)

thisroi = images.roi.Circle('center',C(c,:),'radius',r);

m = m | createMask(thisroi,m);

end

totalarea = sum(m(:)) % total number of white pixels = area

totalarea = 4914749

imshow(m)

There is also bwarea(), but you might want to read the docs to make sure it's what you want. It's not the same as a simple sum.

Of course, if there are many such circles and they are very spread out, the image may be very large. It might be rather slow to do it by drawing circles like this. This is a different approach for the same thing, which should be faster than explicit circle drawing.

% same stuff

r = 400;

C = rand(10,2)*5000 + 1000; % test locations (x,y)

% same stuff

p = 10;

s = ceil([(max(C(:,2))-min(C(:,2))),(max(C(:,1))-min(C(:,1)))] + r*2 + p*2);

C = round(C - [min(C(:,1)) min(C(:,2))] + r + p);

% use distance xform

m = false(s);

m(sub2ind(s,C(:,2),C(:,1))) = 1;

m = bwdist(m)<=r;

totalarea = sum(m(:)) % total number of white pixels = area

totalarea = 4164733

imshow(m)

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DGM 2021-7-25

编辑：DGM 2021-7-25

在 MATLAB Online 中打开

I don't have the toolbox for latlon2local(). I don't know if that's the problem you're having. I used lla2flat(), but that's part of a different toolbox which you might also not have.

% recreate example table

lon = [-76.498 -76.502 -76.501 -76.5 -76.502 -76.505 -76.506 -76.503 -76.501 -76.503 -76.501 -76.499 -76.498];

lat = [3.4309 3.4321 3.4356 3.4355 3.432 3.4336 3.4336 3.434 3.4328 3.4337 3.4328 3.4318 3.4314];

lonlat = table(lon.',lat.');

% process table and convert to meters

lla = [fliplr(table2array(lonlat)) zeros(size(lonlat,1),1)];

position = lla2flat(lla,[min(lla(:,1)) min(lla(:,2))],90,0);

% same as before

r = 400;

C = position(:,1:2); % locations (x,y)

p = 10; % padding

s = ceil([(max(C(:,2))-min(C(:,2))),(max(C(:,1))-min(C(:,1)))] + r*2 + p*2);

C = round(C - [min(C(:,1)) min(C(:,2))] + r + p);

m = false(s);

m(sub2ind(s,C(:,2),C(:,1))) = 1;

m = bwdist(m)<=r;

totalarea = sum(m(:)) % total number of white pixels = area

totalarea = 1555013

imshow(m)

The points are a lot closer than I expected, but that seems about right.

If you don't have lla2flat(), then there are also user-submitted tools. I just did a quick search and found this:

https://www.mathworks.com/matlabcentral/fileexchange/45699-ll2utm-and-utm2ll

which gives similar results.

% process table and convert to meters
lonlat = fliplr(table2array(lonlat));
position = ll2utm(lonlat(:,1),lonlat(:,2)); % [xpos, ypos] in meters
% and so on...

David Alejandro Ramirez Cajigas 2021-8-17

It is posted in this same forum below

David Alejandro Ramirez Cajigas 2021-8-17

I have posted the question right here at the end, it will be very kind if you could help me

Thank you

down in this forum, is the question

Thank you

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Answer 2

David Alejandro Ramirez Cajigas 2021-7-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/884329-circle-problem-circumscribed-to-another-circle-and-the-area-area-per-image-detection#answer_752609

在 MATLAB Online 中打开

for Java I found this on the web could matlab be adapted?

author : http://jsfiddle.net/broofa/WRpmW/5/

// Get HTML canvas element (and context) to draw into
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
// Lil' circle drawing utility
function circle(x,y,r) {
  ctx.beginPath();
  ctx.arc(x, y, r, 0, Math.PI*2);
  ctx.fill();
}
// Clear canvas (to black)
ctx.fillStyle = 'black';
ctx.fillRect(0, 0, canvas.width, canvas.height);
// Fill shape (in white)
ctx.fillStyle = 'white';
circle(40, 50, 40);
circle(40, 10, 10);
circle(25, 15, 12);
circle(35, 90, 10);
// Get bitmap data
var id = ctx.getImageData(0, 0, canvas.width, canvas.height);
var pixels = id.data; // Flat array of RGBA bytes
// Determine area by counting the white pixels
for (var i = 0, area = 0; i < pixels.length; i += 4) {
  area += pixels[i]; // Red channel (same as green and blue channels)
}
// Normalize by the max white value of 255
area /= 255;
// Output result
document.getElementById('result').innerHTML = area.toFixed(2);

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DGM 2021-7-23

编辑：DGM 2021-7-23

This is JS, and is conceptually equivalent to the first example.

David Alejandro Ramirez Cajigas 2021-7-24

Thank you for all

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Answer 3

David Alejandro Ramirez Cajigas 2021-8-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/884329-circle-problem-circumscribed-to-another-circle-and-the-area-area-per-image-detection#answer_767697

编辑：David Alejandro Ramirez Cajigas 2021-8-17

在 MATLAB Online 中打开

@DGM @Image Analyst

Will it be possible to make the plotted image of black and white circles to measure the area, be drawn within specific margins? that is to say that it is drawn, for example, inside a straight line and where a circle touches the "wall" of the rectangle, is that area not counted?

I could have the coordinates of the rectangle and from there I would already have the "wall" where the circuit cannot advance.I made a small drawing and I will post it within the same link as the question already solved. with an exampleThank you.

1) the point cloud in longitude and latidut is

lat	lon
4310571	-76.5249196
4330517	-76.5281573
4329578	-76.5278992
4259733	-76.5169571
4141012	-76.5322045
4132041	-76.5212801
4147463	-76.5212291
4200984	-76.5090573
4193326	-76.5304345
4191612	-76.5226856
4195618	-76.5180319
4194799	-76.5200061
4197733	-76.514478
4199018	-76.5121552
4191972	-76.5076382
4186178	-76.5074345
4172711	-76.5061631
4174853	-76.5059834
4131446	-76.5021961
413158	-76.5019253
410971	-76.5001202
4120227	-76.4916586
4146176	-76.4889863
4187462	-76.5249977
4121893	-76.5258832
4266255	-76.5285806
4291025	-76.5167687
4347968	-76.5141764
4321028	-76.5101736
4271533	-76.5053988
4185984	-76.4869869
4218167	-76.4909459
4248689	-76.4945159
4282086	-76.5327531
4354939	-76.5187478
4334187	-76.5101116
4352097	-76.5118256
4277076	-76.5368692
4309797	-76.5358927
4333837	-76.5351849
4349393	-76.5347021
4327974	-76.5364751
4182723	-76.5367191
4090406	-76.5355765
4077261	-76.5315237
4109503	-76.4893841
4086604	-76.5217632
4107276	-76.5215164
4166607	-76.5211731
4197919	-76.520905
4226876	-76.5205857
4247333	-76.5203818
4252902	-76.5180966
4295177	-76.5165489
4295177	-76.5165489
431641	-76.5163424
434254	-76.5147331
4347146	-76.5147384
4324815	-76.5163367
4213276	-76.517367
4175283	-76.5222486
412915	-76.5227287
410417	-76.523005
4071183	-76.5233778
4253297	-76.5015155
4339325	-76.4964618
4309045	-76.4980119
4244061	-76.5032478
4351374	-76.5232265
4346636	-76.5230613
4352634	-76.5206634
4330464	-76.5264623
4329822	-76.525545
432503	-76.5272398
431776	-76.528092
4305833	-76.5299112
4284118	-76.5320919
42642	-76.5337575
4241897	-76.533956
4235712	-76.5346507
4203582	-76.5356673
4178333	-76.5365309
4186983	-76.5357638
4215123	-76.5348599
4260317	-76.5333472
431708	-76.53415
4312579	-76.5281195
4321469	-76.5268189
4104115	-76.5293914
4129524	-76.5282836
4116382	-76.5231909
4225161	-76.5052968
4223848	-76.5046797
4201063	-76.5077996
4198948	-76.5072551
4282383	-76.4985591
4307088	-76.4971941
4295397	-76.498683
4341627	-76.4954614
4186526	-76.5283958
420709	-76.5345703
4148539	-76.5343228
4143741	-76.5182786
4143097	-76.5204408
4127084	-76.5180348
4121921	-76.5152421
4101545	-76.5174233
4144032	-76.5129681
4171883	-76.5100628
4339407	-76.53169
4313357	-76.5324383
435523	-76.5321513
4296035	-76.527203
4293248	-76.5276801
4261009	-76.5289884
4230866	-76.5299733
4235895	-76.5295651
4198576	-76.5310217
4198307	-76.5307561
4168025	-76.5317512
416891	-76.5319739
4140367	-76.5326686
4140957	-76.5328912
4119804	-76.5336127
4118654	-76.5333713
4093833	-76.5341652
4096271	-76.534353
4072306	-76.5351255
4085317	-76.5191425
4089332	-76.5194188
4076618	-76.5207116
4112626	-76.5170021
4133751	-76.5148108
4176403	-76.510361
4163313	-76.5116723
4153969	-76.5043551
4084892	-76.4974891
4101436	-76.4990257
4144384	-76.503151
4351054	-76.5284981
4261281	-76.5275993
4261523	-76.527849
4229822	-76.5274654
4228698	-76.5277042
4193409	-76.5272482
4195142	-76.5274917
4174159	-76.5276934
4175039	-76.5274625
4152952	-76.5282755
4153355	-76.5280341
4136218	-76.5285115
4128266	-76.5289863
4105909	-76.5295951
4122988	-76.4918565
4172202	-76.4868609
414781	-76.4891394
4114073	-76.4938627
4161733	-76.4946323
413493	-76.4958315
4153673	-76.4960917
409121	-76.5029153
4107622	-76.5010752
4078625	-76.5035831
4223742	-76.5071477
4225534	-76.5073462
4296915	-76.5034007
4300717	-76.5034999
427595	-76.504774
4321119	-76.5024404
4273729	-76.5045942
4354533	-76.5004663
4320209	-76.5022017
4252149	-76.5060265
4250836	-76.5057395
4355899	-76.5261781
4306315	-76.5255236
4283743	-76.5244554
4276114	-76.5261325
4235978	-76.5259367
4216031	-76.5252152
4217371	-76.5227609
4199565	-76.5219831
4077286	-76.5265107
4108104	-76.525537
42046	-76.5240967
423772	-76.5242335
4260048	-76.5243488
4275149	-76.5116888
4254534	-76.5118604
42231	-76.5121072
4119459	-76.5144917
4120689	-76.5143116
410232	-76.5128663
4103927	-76.5127
4080554	-76.5104764
4079321	-76.5106856
410584	-76.5180612
4136912	-76.5177206
4146069	-76.519405
4281973	-76.5199898
4324517	-76.5213684
4319432	-76.5237341
4256217	-76.5249679
4257289	-76.5229053
4234424	-76.5215964
4102992	-76.5182481
4148131	-76.5178014
4211585	-76.517155
414896	-76.5176163
4173218	-76.5175815
4173192	-76.517375
4198733	-76.5172382
4197957	-76.5174501
4226928	-76.517037
4080927	-76.5165086
4072227	-76.5159668
4076885	-76.5163987
4144512	-76.4930341
4129628	-76.4915964
4176748	-76.4959094
419292	-76.4974544
4224996	-76.5005174
4207702	-76.4988491
4227033	-76.5040283
4205988	-76.5019228
4180153	-76.4994632
4162721	-76.4980081
4352152	-76.5141913
4335874	-76.5092426
4330037	-76.5074616
433582	-76.5052032
4336247	-76.5059327
4339756	-76.5031808
4327895	-76.501204
4337159	-76.5030869
4328028	-76.5008097
4318442	-76.4993533
4314239	-76.4980551
4344601	-76.5118336
4192084	-76.5207168
4193369	-76.5181929
4194627	-76.5159237
4196073	-76.5133407
4197599	-76.5103903
428547	-76.5330845
4121579	-76.4873705
4297281	-76.5109285
4261112	-76.5168337
4229298	-76.5171669

2) the result image is

3) the idea is that the points are plotted in a "rectangular" range demarcated by coordinates

for example:

lat         lon
3.4356   -76.5369  %lan and lot 1
3.4071    -76.4868 %lan and lot 2

How do I get it to plot and measure me within the geometric limits of the rectangle, formed by the given coordinates?

Thank you

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David Alejandro Ramirez Cajigas 2021-8-16

编辑：David Alejandro Ramirez Cajigas 2021-8-16

在 MATLAB Online 中打开

another simpler graphic example

lat          lon
3.3622      -76.5243
3.3498      -76.5121

z

%points to evaluate
lat	            lon
3.3534876	-76.523249
3.3531877	-76.52344
3.3597285	-76.5240052
3.3591999	-76.5241002

The idea is that I paint and piss within the parameters of the rectangle given by the initial coordinates, currently plotting and measuring within the maximum limit of the area of the outermost circuit, generating the error that it has more area than necessary, because in some cases the boundary of the rectangular wall can cut a circle and for other cases the circles are in the center of the image and external space is lost.

For example now I measure the% are of area adding and subtracting the black and white area, if however it is an error because the black area should be delimited by the coordinates of the walls of the rectangle

            %E = J{:,:}
            lla = [fliplr(table2array(J)) zeros(size(J,1),1)]
            position = lla2flat(lla,[min(lla(:,1)) min(lla(:,2))],90,0)
            % same as before
            %r = ;
            v5= 300
            r = v5 ;
            C = position(:,1:2); % locations (x,y)
            p = 10; % padding
            s = ceil([(max(C(:,2))-min(C(:,2))),(max(C(:,1))-min(C(:,1)))] + r*2 + p*2);
            C = round(C - [min(C(:,1)) min(C(:,2))] + r + p);
            m = false(s);
            m(sub2ind(s,C(:,2),C(:,1))) = 1;
            m22 = bwdist(m)<=r;
            totalarea = sum(m22(:)) % total number of white pixels = area
            m23 = bwdist(m)>=r;
            totalarea2 = sum(m23(:))
            imshow(m22)
            
            
            m24=(totalarea*100)/(totalarea+totalarea2)  %shows area percentage in white
            

you need to put limits with the coordinates of the rectangle so that it would give me the real percentag

                        lat          lon
3.3622      -76.5243
3.3498      -76.5121
%with this he formed the limits of the rectangle
%vertices are (lon1, lat1) (lon2, lat2) (lon2, lat1) (lon1, lat 2)

In example 1 of the stars, there is a reference frame (rectangular), the idea is to measure the area of all the internal stars and the part cut by the reference frame

z

DGM 2021-8-19

在 MATLAB Online 中打开

Oof. Sorry about that. Sometimes site notifications get buried, and I almost never check email anymore. Anyway...

% This is only part of the dataset!
lon = [-76.5249196 -76.5281573 -76.5278992 -76.5169571 -76.5322045 -76.5212801 -76.5212291 -76.5090573 -76.5304345 -76.5226856 -76.5180319 -76.5200061 -76.514478 -76.5121552 -76.5076382 -76.5074345 -76.5061631 -76.5059834 -76.5021961 -76.5019253 -76.5001202 -76.4916586 -76.4889863 -76.5249977 -76.5258832 -76.5285806 -76.5167687 -76.5141764 -76.5101736 -76.5053988 -76.4869869 -76.4909459 -76.4945159 -76.5327531 -76.5187478 -76.5101116 -76.5118256 -76.5368692 -76.5358927 -76.5351849 -76.5347021 -76.5364751 -76.5367191 -76.5355765 -76.5315237 -76.4893841 -76.5217632 -76.5215164 -76.5211731 -76.520905 -76.5205857 -76.5203818 -76.5180966 -76.5165489 -76.5165489 -76.5163424 -76.5147331 -76.5147384 -76.5163367 -76.517367 -76.5222486 -76.5227287 -76.523005 -76.5233778 -76.5015155 -76.4964618 -76.4980119 -76.5032478 -76.5232265 -76.5230613 -76.5206634 -76.5264623 -76.525545 -76.5272398 -76.528092 -76.5299112 -76.5320919 -76.5337575 -76.533956 -76.5346507 -76.5356673 -76.5365309 -76.5357638 -76.5348599 -76.5333472 -76.53415 -76.5281195 -76.5268189 -76.5293914 -76.5282836];
lat = [3.4310571 3.4330517 3.4329578 3.4259733 3.4141012 3.4132041 3.4147463 3.4200984 3.4193326 3.4191612 3.4195618 3.4194799 3.4197733 3.4199018 3.4191972 3.4186178 3.4172711 3.4174853 3.4131446 3.413158 3.410971 3.4120227 3.4146176 3.4187462 3.4121893 3.4266255 3.4291025 3.4347968 3.4321028 3.4271533 3.4185984 3.4218167 3.4248689 3.4282086 3.4354939 3.4334187 3.4352097 3.4277076 3.4309797 3.4333837 3.4349393 3.4327974 3.4182723 3.4090406 3.4077261 3.4109503 3.4086604 3.4107276 3.4166607 3.4197919 3.4226876 3.4247333 3.4252902 3.4295177 3.4295177 3.431641 3.434254 3.4347146 3.4324815 3.4213276 3.4175283 3.412915 3.410417 3.4071183 3.4253297 3.4339325 3.4309045 3.4244061 3.4351374 3.4346636 3.4352634 3.4330464 3.4329822 3.432503 3.431776 3.4305833 3.4284118 3.42642 3.4241897 3.4235712 3.4203582 3.4178333 3.4186983 3.4215123 3.4260317 3.431708 3.4312579 3.4321469 3.4104115 3.4129524];
lonlat = table(lon.',lat.');
% boundary
box = [-76.5369 3.4356; -76.4868 3.4071]; % lon/lat
% process table & box; convert to meters
lla = [fliplr(table2array(lonlat)) zeros(size(lonlat,1),1)];
position = lla2flat(lla,[min(lla(:,1)) min(lla(:,2))],90,0);
boxpos = lla2flat([fliplr(box) [0; 0]],[min(lla(:,1)) min(lla(:,2))],90,0);
boxpos = boxpos(:,1:2);
% same as before
r = 400;
C = position(:,1:2); % locations (x,y)
p = 10; % padding
s = ceil([(max(C(:,2))-min(C(:,2))),(max(C(:,1))-min(C(:,1)))] + r*2 + p*2);
refpoint = [min(C(:,1)) min(C(:,2))];
C = round(C - refpoint + r + p);
corners = round(boxpos - refpoint + r + p);
m = false(s);
m(sub2ind(s,C(:,2),C(:,1))) = 1;
m = bwdist(m)<=r;
m(:,[1:corners(1,1) corners(2,1):end]) = 0;
m([1:corners(1,2) corners(2,2):end],:) = 0;
size(C,1)*round(pi*r^2)
totalarea = sum(m(:)) % total number of white pixels = area
imshow(m); hold on
plot(corners([1 2 2 1 1],1),corners([1 1 2 2 1],2),':')

David Alejandro Ramirez Cajigas 2021-8-20

GRACIAS!!!

Thanks a lot.

I hope one day I can help someone like that. Thanks again

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