calculate the angle between a line (between two points) and the perpendicular line
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if point1 is a(x1,y1) and point 2 is b(x2,y2). I want to draw a vertical from point b in order to calculate the angle between the line ab and the verticle line.
I wrote this code but I think i can't get the verticle line correctly as I got the same value of the angle and i'm sure the angle value is not the same.
for i=1:size(a,1)
angle = atand(b(1,i) - a(1,i))-(b(2,i) - a(2,i)))
end
any one can help me please?
3 件のコメント
Geoff Hayes
2019 年 3 月 20 日
Amneh's answer moved here
I'm using a loop because points a and b are moving and changing their positions. I want to caluctae the angle for every fram. I assumed that the vericle line is between (point a (x1,y1) and Point (x1,y2))
Geoff Hayes
2019 年 3 月 20 日
I assumed that the vericle line is between (point a (x1,y1) and Point (x1,y2)) Can you provide a picture? I don't understand how the vertical line is between (x1,y1) and (x1,y2) (or do you mean (x2,y2) for the second one?). Also I thought I want to draw a vertical from point b which suggests that the vertical line is drawn from point b and is not in between the two points. Please clarify.
採用された回答
Jan
2019 年 3 月 20 日
Maybe you mean:
angle = atand(b(1,:) - a(1,:)) - (b(2,:) - a(2,:)))
% ^???
without a loop. But here the number of parentheses does not match. I'm not sure which problem you want to solve.
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その他の回答 (2 件)
Geoff Hayes
2019 年 3 月 20 日
編集済み: Geoff Hayes
2019 年 3 月 20 日
Amneh - if a is (x1,y1) and b is (x2,y2) then let c be (x2,y1). With these three points, you should have a right-angled triangle. We can determine the opposite and adjancent as
opposite = abs(a(1) - c(1));
adjacent = abs(c(2) - b(2));
and so the angle is
myAngleDegrees = atand(opposite / adjacent);
(I think!)
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