How can I find the angle between a cubic smoothing spline and a line ?

Hello everyone,

As I started my internship in researching, I started using Matlab in some much more difficult way that I used to in my school, which explain my appearance on your forum.

Here is my problem:

I need to program something which allows me to detect the contact angle of a droplet on a surface. I have already programmed a part which allows me to detect the sides of the droplet, to mark it with some markers, and to get the coordinates of each markers, Then, I used the cubic smoothing spline function in order to get an approximated curve of my sides:

sz2 = size(tabl2);
Size2 = sz2(1);
for it = 1:Size2
  Xg(it) = tabl2{it}(1);
  Yg(it) = -tabl2{it}(2); % Minus added because of the axis direction
  end
Smooth = csaps (Xg,Yg,0.05); % Xg and Yg are basically the coordinates of the left side of my droplet.

I also used a basic polyfit in order to model the contact line.

Here is a picture of the curves I get :

The blue line is my cubic spline, which represent the side of my droplet.

The red line represents the surface, which isn't perfectly straight.

Now I need to get the contact angle between my cubic spline and my contact line, but I am stuck on it, Do you have any ideas?

Thank you very much for you help.

Answer 1

[John D'Errico]

You have the smoothing spline. Differentiate the spline. Compute the slope of the spline at the intersection point. You already know the slope of the straight line.

Now it is simple. You have two slopes. Use atan (or atand if you want the angle in degrees) to compute the angles of both curves, with respect to the coordinate axes. Subtract the resulting angles. The difference will be the angle between the curves at the point of intersection.

It is even easier if the straight line is a horizontal line, since then the corresponding angle you would compute is zero.

WTP?

[Clement Prunier]

Hello

Thank you for your answer, but I am still a bit confused. What do you mean by compute the slope of the spline at the intersection point? I know I should try to get the slope of my derivative at the intersection point, but i can't just do

    slope = deriv(x)

Because it doesn't return me the value of the derivative at this point, but instead

deriv(113)
Index exceeds matrix dimensions.

Do you know which syntax I should use to get this value?

Thank you

[Clement Prunier]

I finally found it, I'll put it for other peoples :

derivD = fnder(SmoothD,1); %Derivative of my spline

 ContactD = Xd(1); %Contact point between my spline and my line

 CAD = 180-(atand (fnval(derivD,ContactD))); %Contact angle between both, assuming that the contact line is a straight line // to the x axis

Thank you for your help.

Answer 2

[Konstantin Gulin]

I would recommend finding the contact vectors for your two curves.

For the spline, take two points; the point of contact [x2 y2] and the point immediately preceding that one [x1 y1].

Generate the vector (you need to add a third dimension, even if it's irrelevant):

v = [(x2 - x1) (y2-y1) 0]

Do the same thing for your line and generate a vector u.

To find the angle between the two use:

Theta = atan2d(norm(cross(u,v)),dot(u,v));

[Clement Prunier]

Hello, thank you for your answer.

The problem is that with your solution, I would use the points I used to make an approximation, which are not very precise, which is why I used a smoothing spline.

Do you know a way to get these values from the smoothing spline itself, and not from the initial points?

Thank you

[Konstantin Gulin]

You can use the Y values from your spline function. For example if your spline is:

s = spline(x,y,xq)

then s(x) will return all the y values in your spline function.

[Clement Prunier]

Unfortunatly, this syntax doesn't work for the kind of function I have. I used the csaps function, but I don't understand what kind of function it gives me and how to get values from it. Do you have any idea?

Thank you for your help