# Draw arc with specified angle difference

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Elysi Cochin 2021년 6월 9일
댓글: Star Strider 2021년 6월 10일
Having 2 datas as attached in Data.mat, how to draw arc with a specified angle difference (angle difference can vary say 15, 30 or any other value as given by user)
The data columns in order are angle, radius, depth
How can i find the center with the attached data, so that both the arcs pass through the center, and display it in x,y,z coordinate
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Elysi Cochin 2021년 6월 9일
편집: Elysi Cochin 2021년 6월 9일
Sir the given values are in degrees, we need to convert it into radians. Compute the center and both the curves must pass through the center. Is it possible sir, with this information I have only so much details
Elysi Cochin 2021년 6월 9일
Sir, can you show me how to plot, if the angles are radian values (descending spiral)

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### 채택된 답변

Star Strider 2021년 6월 9일
Try this —
Data1 = LD.Data1;
A1 = Data1(:,1);
R1 = Data1(:,2);
D1 = Data1(:,3);
Data2 = LD.Data2;
A2 = Data2(:,1);
R2 = Data2(:,2);
D2 = Data2(:,3);
ctrfcn = @(b,a,r,d) [sqrt((b(1)+r.*cosd(a)).^2 + (b(2)+r.*sind(a)).^2 + (b(3)-d).^2)];
[B1,fval] = fminsearch(@(b)norm(ctrfcn(b,A1,R1,D1)), -rand(3,1)*1E+4)
[B2,fval] = fminsearch(@(b)norm(ctrfcn(b,A2,R2,D2)), -rand(3,1)*1E+4)
figure
plot3(R1.*cosd(A1), R1.*sind(A1), D1, 'm')
hold on
plot3(R2.*cosd(A2), R2.*sind(A2), D2, 'c')
scatter3(B1(1), B1(2), B1(3), 30, 'm', 'p', 'filled')
scatter3(B2(1), B2(2), B2(3), 30, 'c', 'p', 'filled')
hold off
legend('Data_1','Data_2', 'Centre_1', 'Centre_2', 'Location','best')
grid on
The axes are not scaled to be equal, because it then appears to be a flat surface.
Data1 Center:
x = -740.85
y = -349.01
z = 3.83
Data2 Center:
x = -740.96
y = -348.77
z = 3.81
.
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Elysi Cochin 2021년 6월 10일
Sir, shall i ask you one thing, why does the plotted center point lie away from the curves?
What i meant by center is the center point of the curve (means if we take a line of 10cm the center point is 5cm). By center i meant that.
Star Strider 2021년 6월 10일
As a general rule when talking about arcs or circles, the center is the center of the circle. It can never be on any of the circumferences.
You are asking for the midpoint of the arc.
MP1 = median([R1.*cosd(A1)+B1(1), R1.*sind(A1)+B1(2), D1],1);
MP2 = median([R2.*cosd(A2)+B2(1), R2.*sind(A2)+B2(2), D2],1);
fprintf(1,'Arc Midpoint 1:\n\t\tx = %8.2f\n\t\ty = %8.2f\n\t\tz = %8.2f\n',MP1)
fprintf(1,'Arc Midpoint 2:\n\t\tx = %8.2f\n\t\ty = %8.2f\n\t\tz = %8.2f\n',MP2)
produces —
Arc Midpoint 1:
x = 745.33
y = 366.43
z = 3.83
Arc Midpoint 2:
x = 752.68
y = 364.83
z = 3.82
That is the best I can do.

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### 추가 답변 (2개)

Image Analyst 2021년 6월 9일
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Elysi Cochin 2021년 6월 9일
Sir, i saw that link, but the data I have, is different with the example shown in the link.

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darova 2021년 6월 9일
t = linspace(0,1,20)*pi/180; % angle array
[X,Y,Z] = deal( zeros(10,20) ); % preallocation matrices
for i = 1:10
[X(i,:),Y(i,:)] = pol2cart(t*s.Data1(i,1),s.Data1(i,2)); % create arc
Z(i,:) = s.Data1(i,3); % depth
end
surf(X,Y,Z)
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Elysi Cochin 2021년 6월 9일
편집: Elysi Cochin 2021년 6월 9일
Sir how to convert the radius value from degrees to radians and plot the figure
also, at a time i want to give only one angle

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