LabVIEW
Acquire coordinates of an arc
If you have two points and a radius, you will have zero (2*radius<distance between points), one (2*radius=distance), or two circles that match. Since there is an infinite number of coordinates on the arc, you must limit yourself to a finite number of coordinates (e.g. 100 equally spaced points).
Typically you'll have two solutions for the circle, and two possible arcs for each solution, so which one of the four do you pick? (see picture for possible arcs given two points and a radius)
Can you provide more details on the problem? Do you have some example data?
Message Edited by altenbach on 12-27-2005 01:09 PM
Hi Giridhar,
Your question is more a solving issue than a LabVIEW specific question.You can use the expression for a circle to find the coordinates for the centers of the two circles that Altenbach mentions. The expression for a circle is
R^2 = (x-a)^2+(y-b)^2
where R is the radius of the circle and (a,b) is the center of the circle.
After finding the center of the circle that the arc is part of, you should be able to calculate the arc as the coordinates between the two points.
What heve you done so far? Are you getting stuck with anything specific to LabVIEW?
Your question is more a solving issue than a LabVIEW specific question.You can use the expression for a circle to find the coordinates for the centers of the two circles that Altenbach mentions. The expression for a circle is
R^2 = (x-a)^2+(y-b)^2
where R is the radius of the circle and (a,b) is the center of the circle.
After finding the center of the circle that the arc is part of, you should be able to calculate the arc as the coordinates between the two points.
What heve you done so far? Are you getting stuck with anything specific to LabVIEW?
To avoid all the ambiguity of multiple solutions, it might be easier to define the start and end points PLUS one point lying on the arc. Three points uniquely define a circle and the solution to the problem is trivial. 🙂 There will always be exactly one solution!
Attached is a draft of a simple demo (LabVIEW 7.0) that does just that.
I am sure you could easily generate "point three" from your desired "radius + endpoints" and re-use most of my code, with the problem that there will be zero, two or four solutions depending on the inputs.
Note that I use mostly complex math to avoid explicit trigonometric functions. Let me know if you have questions. 😄
Message Edited by altenbach on 12-30-2005 11:12 AM
Just on a side note:
If you have LabVIEW 8.0, things get quite a bit nicer, for example:
- We have events for "cursor move", thus we don't need to constantly poll the cursors.
- xy graphs draw complex data directly in the complex plane (real=x, imaginary=y).
Message Edited by altenbach on 12-30-2005 01:08 PM
Your modification assumes that the center of the circle is a (0,0). This does not agree with the figure in your first attachment.
If you assume the center at the origin, you'll most likely won't find a solution. Notice that in your modification your arc will only go through one of your two points, you're throwing away the "r" part of the other one and only retaining the angle. Your system os overdetermined.
In this case you might as well throw both away and turn the R indicator into a control. 😮
Why don't you define the arc from e.g. "center+radius+direction1+direction2"?
