how to make rectangle in circle?

조회 수: 6 (최근 30일)
Sierra
Sierra 2022년 11월 19일
댓글: Matt J 2022년 11월 20일
I want to make rectangles in circle.
I know the radius, rectangle's width and length.
please let me know how to do this
thanks
  댓글 수: 3
Sierra
Sierra 2022년 11월 20일
I answerd to Matt J's!
could you guys look at that?

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채택된 답변

Matt J
Matt J 2022년 11월 19일
편집: Matt J 2022년 11월 19일
[X,Y]=ndgrid(-10:10);
r=arrayfun( @(x,y) nsidedpoly(4,'Center',[x,y],'Radius',1/sqrt(2)) , X(:),Y(:));
c=intersect(nsidedpoly(1000,'Radius',5), scale(r,[1,1.5]));
plot(c,'FaceColor','none')
  댓글 수: 4
Sierra
Sierra 2022년 11월 20일
Thanks again.
but My circle's origin point is not zero. so I added my circle's origin point.
width=0.0073;
length=0.0097;
circleRadius=0.0833;
[X,Y]=ndgrid(-circleRadius:width:circleRadius+width);
X(:) = X(:) + ARP_lon % origin x 126.7975
Y(:) = Y(:) + ARP_lat % origin y 37.5569
r=arrayfun( @(x,y) nsidedpoly(4,'Center',[x,y],'Radius',width/sqrt(2)) , X(:),Y(:));
c=intersect(nsidedpoly(1000,'Radius',circleRadius), scale(r,[1,length/width]));
plot(c,'FaceColor','none'); %axis equal
but it didn't work well.
could you explain more?
Matt J
Matt J 2022년 11월 20일
width=1;
length=1.3;
circleRadius=10;
[X,Y]=ndgrid(-circleRadius:width:circleRadius+width);
r=arrayfun( @(x,y) nsidedpoly(4,'Center',[x,y],'Radius',width/sqrt(2)) , X(:),Y(:));
c=intersect(nsidedpoly(1000,'Radius',circleRadius), scale(r,[1,length/width]));
c=translate(c,ARP_lon,ARP_lat);
plot(c,'FaceColor','none'); axis equal

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추가 답변 (1개)

Image Analyst
Image Analyst 2022년 11월 19일
See the FAQ:
Then plot it with plot and then for each x and y value you want find the end points of a line segment and use line() or plot() to draw the line from one side of the circle to the other. It's easy but if you really can't figure it out then write back.
  댓글 수: 3
Image Analyst
Image Analyst 2022년 11월 20일
You do not ned to know the crossing/intersection coordinates of the vertical and horizontal lines inside the circle. All you need to know for the horizontal lines is the y value and the two points on the circle closest to that y value. Similar for the lines in the other direction. But it looks like @Matt J suggested a different approach and you accepted that so I won't proceed with my approach.
Sierra
Sierra 2022년 11월 20일
Thanks for your answer!

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