MATLAB Answers

0

nearest tangent point from Ginput point to line ?

Kaleesh Bala さんによって質問されました 2018 年 10 月 22 日
最新アクティビティ Image Analyst
さんによって コメントされました 2018 年 10 月 30 日
I'm trying to get the nearest by creating a ginput and then ginput point finds the nearest point to the line.
x = [0,20]
y= [20,50]
plot(x,y)
[x,y] = ginput(1);
h1 = text(x,y,'o', ...
'HorizontalAlignment','center', ...
'Color', [1 0 0], ...
'FontSize',8);

  1 件のコメント

Kaleesh Bala 2018 年 10 月 22 日

サインイン to comment.

タグ

製品

2 件の回答

Rik
回答者: Rik
2018 年 10 月 22 日
 採用された回答

My FEX submission should help here. Unlike what the documented behavior should be, pt is actually not extended to 3D automatically, so you'll have to do that yourself until I update the file. The code below should work as intended.
x = [0,20];
y= [20,50];
v1=[x(1) y(1) 0];
v2=[x(2) y(2) 0];
plot(x,y)
[x,y] = ginput(1);
pt=[x(:),y(:),zeros(numel(y),1)];
h1 = text(x,y,'o', ...
'HorizontalAlignment','center', ...
'Color', [1 0 0], ...
'FontSize',8);
distance=point_to_line_distance(pt, v1, v2)

  11 件のコメント

Rik
2018 年 10 月 23 日
Actually my function is equivalent to the line below (as the description states as well):
distance=norm(cross(v1-v2,pt-v2))/norm(v1-v2)
The distance you get will depend on which point you randomly select. Look at your axes, do you believe the distance output? The few I tested I believe the output to be correct.
The code below should project your point to the line
x = [0,20];
y= [20,50];
figure(1),clf(1)
plot(x,y),hold on
P=ginput(1);%'random' point
%assign shorthand
AP=P-A;AB=A-B;A=[x(1) y(1)];B=[x(2) y(2)];
%calculate projection
Pp=A+(AB/norm(AB))*dot(AP,AB)/norm(AB);
%plot P and P'
plot(P(1),P(2),'o')
plot(Pp(1),Pp(2),'*')
Rik
2018 年 10 月 30 日
Did this suggestion solve your problem? If so, please consider marking it as accepted answer. It will make it easier for other people with the same question to find an answer. If this didn't solve your question, please comment with what problems you are still having.
Image Analyst
2018 年 10 月 30 日
Perhaps have your code draw a line from the point perpendicularly to the closest point on the line, then run your code, and save a screenshot of the figure and upload it. Maybe if he sees that he will accept it.

サインイン to comment.


Image Analyst
回答者: Image Analyst
2018 年 10 月 22 日

Use sqrt():
uiwait(helpdlg('Click one point.'));
[xUser, yUser] = ginput(1);
distances = sqrt((xUser - xLine).^2 + (yUser - yLine) .^ 2);
[minDistance, indexOfMin] = min(distances);
hold on;
% Put a marker on the line.
plot(xLine(indexOfMin), yLine(indexOfMin), 'r*');
% Draw a line from the user-clicked point to the point on the line.
line([xUser, xLine(indexOfMin)], [yUser, yLine(indexOfMin)], 'LineWIdth', 2, 'Color', 'r');

  3 件のコメント

Kaleesh Bala 2018 年 10 月 23 日
Thanks Image analyst I tried to add it to the code ,but it provides different result as it's taking to 0,0 or even lower axis value,I'm bascially trying to project the point to the line @ shorter distance location
x = [0,20]
y= [20,50]
plot(x,y)
uiwait(helpdlg('Click one point.'));
[xUser, yUser] = ginput(1);
distances = sqrt((xUser - x).^2 + (yUser - y) .^ 2); [minDistance, indexOfMin] = min(distances);
hold on;
% Put a marker on the line.
plot(x(indexOfMin), y(indexOfMin), 'r*');
% Draw a line from the user-clicked point to the point on the line.
line([xUser, x(indexOfMin)], [yUser, y(indexOfMin)], 'LineWIdth', 2, 'Color', 'r');
Kaleesh Bala 2018 年 10 月 23 日
It's like the point getting projected to the line @ shortest travel distance.
Image Analyst
2018 年 10 月 23 日
OK, I thought you had a bunch of points along a line, not just two. In that case you'll have to use the point-to-line distance formula, which are readily avalable all over the web.

サインイン to comment.



Translated by