Draw arc with specified angle difference

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

[Star Strider]

If the angles are radian values, the data each describe a descending spiral with angles that span nearly 10 radians, or about 1.6 times the circumference of the circle. If they are in degrees, it is almost a straight line, spanning only about 10°.

Please provide details.

[Elysi Cochin]

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]

Sir, can you show me how to plot, if the angles are radian values (descending spiral)

Answer 1

[Star Strider]

Try this —

LD = load('Data.mat');

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

.

[Elysi Cochin]

Sir why is it line?

When the center values are round, both will become same, with the method you have shown above.

Cant i get an arc or curve structure, with the values given? Can we draw both the curves through any one center or the rounded center value?

If the above is not possible, using the data attached, can i get 2 curve which intersect almost in the center?

If it is not possible with the attached data only, what additional data is needed to plot so

[Star Strider]

Try this —

LD = load('Data.mat');
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) (b(1)+cosd(a)).^2 + (b(2)+sind(a)).^2 - (b(3).*(r+d)).^2;
[B1,fval] = fminsearch(@(b)norm(ctrfcn(b,A1,R1,D1)), rand(3,1)*1E+1)
[B2,fval] = fminsearch(@(b)norm(ctrfcn(b,A2,R2,D2)), rand(3,1)*1E+1)

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), 50, 'm', 'p', 'filled') 
scatter3(B2(1), B2(2), B2(3), 50, 'c', 'p', 'filled') 
plot3(R1.*cosd(A1)+B1(1), R1.*sind(A1)+B1(2), D1, '--c')
plot3(R2.*cosd(A2)+B2(1), R2.*sind(A2)+B2(2), D2, '--m')
hold off
legend('Data_1','Data_2', 'Centre_1', 'Centre_2', 'Location','best')
grid on
view(-20,30)
% axis('equal')
fprintf(1,'Data1 Center:\n\t\tx = %8.2f\n\t\ty = %8.2f\n\t\tz = %8.2f\n',B1)
fprintf(1,'Data2 Center:\n\t\tx = %8.2f\n\t\ty = %8.2f\n\t\tz = %8.2f\n',B2)

There was a slight error in the previous code, now fixed. This also draws the arcs.

Data1 Center:

x = 3.32

y = 6.34

z = 0.01

Data2 Center:

x = 3.02

y = 9.67

z = 0.01

.

[Star Strider]

First, the curves do not intersect. They are a reasonably constant distance from each other. I did not test them to see if they intersected beyond the arcs provided.

Second, the centres are approximately equal, as are the other parameters. That is simply the nature of the data.

[Elysi Cochin]

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]

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.

Answer 2

[Image Analyst]

See the FAQ again:

https://matlab.fandom.com/wiki/FAQ#How_do_I_create_an_arc.3F

[Elysi Cochin]

Sir, i saw that link, but the data I have, is different with the example shown in the link.

Answer 3

[darova]

Wha about this representation?

s = load('data.mat');
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)

[Elysi Cochin]

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