Bezier interpolation

[#1]

Hi,

Is there a function to plot Bezier interpolation? 

[#2]

Multiple answers, depending what you need to do. Here are some. 

For 2D graphs you can use the path function. You can't find it by searching for Bezier because the scripting index spells it Bézier. Try searching for zier instead. @XanGregg @sheila_loring 

This example extends the scripting index example with dragable handles.

// initial data; points 3 and 5 are not initially constrained about point 4
points = [
10 10 1,// move-to
30 70 0,// control point
40 10 0,// control point
50 40 3,// bezier
60 60 0,// control point
70 20 0,// control point
90 10 3 // bezier
];

// moving a square handle will call this function to 
// copy the handle coordinates to the matrix
setpoint = Function( {index, x, y}, // parameters
	{angle, distance}, // local variables
	// copy to the matrix
	points[index, 1] = x; 
	points[index, 2] = y;
	// 3 and 5 are constrained about 4, comment out to see why...
	// if the angles don't match, it isn't smooth
	If( index == 3 | index == 5,
		other = If( index == 3, 5, 3 ); // if I'm 3, he must be 5, etc
		// trig in radians, 2*pi is 360
		// atan does all 4 quadrants with 2 arguments
		angle = ATan( points[index, 2] - points[4, 2], points[index, 1] - points[4, 1] );
		// maintain other's distance
		distance = Sqrt( (points[other, 2] - points[4, 2]) ^ 2 + (points[other, 1] - points[4, 1]) ^ 2 );
		// fix other's angle to match my angle, but opposite me
		points[other, 1] = points[4, 1] + distance * Cos( angle + Pi() );
		points[other, 2] = points[4, 2] + distance * Sin( angle + Pi() );
	);
	// redraw
	gb << inval;
);
New Window( "Example",
	gb = Graph Box(
		Fill Color( "blue" );
		// there is no good shortcut for creating handles; the JSL holds the instances
		Handle( points[1, 1], points[1, 2], setpoint( 1, x, y ) );
		Handle( points[2, 1], points[2, 2], setpoint( 2, x, y ) );
		Handle( points[3, 1], points[3, 2], setpoint( 3, x, y ) );
		Handle( points[4, 1], points[4, 2], setpoint( 4, x, y ) );
		Handle( points[5, 1], points[5, 2], setpoint( 5, x, y ) );
		Handle( points[6, 1], points[6, 2], setpoint( 6, x, y ) );
		Handle( points[7, 1], points[7, 2], setpoint( 7, x, y ) );
		// construction lines
		Pen Color( "red" );
		Line( {points[1, 1], points[1, 2]}, {points[2, 1], points[2, 2]} );
		Line( {points[3, 1], points[3, 2]}, {points[4, 1], points[4, 2]} );
		Line( {points[4, 1], points[4, 2]}, {points[5, 1], points[5, 2]} );
		Line( {points[6, 1], points[6, 2]}, {points[7, 1], points[7, 2]} );
		// the bezier
		Pen Color( "black" );
		Path( points );
	)
);

I realized I should also constrain moving point 4 to drag points 3 and 5 along as well. (Not done, should be easy.)

Bezier control points

Maybe you need a spline. I'm not sure what the distinction between spline, p-spline, b-spline in the scripting index is. I used spline HeatColor vs Spline vs Interpolate to make this curve

spline coef, spline eval

There is also support for Bezier splines in JMP's 3D OpenGL support: https://www.jmp.com/support/help/14-2/bezier-curves.shtml which might help draw a surface. A poor example is hidden in FFT Video 

Great post @Duane_Hayes Understanding cubic splines 

Something I ran into along the way: https://www.tinaja.com/glib/bezconn.pdf

[#3]

Hi Craige,

I am trying to understand path () function and move points, control points, bezier points.

I have these four points from the data I have 

point =
[ 0.93 77.04 1,
1.37 155.22 0,
1.96 390.46 0,
2.41 846.87 3];

When I put this in path function, I get a line between point 1 and point 4, but it doesn't go through point 2 and point 3. Am I missing something?

Thanks.

[#4]

What are you trying to do?

The wikipedia article is pretty good. The control points generally do not lie on the Bezier curve. 

JMP can fit a spline to a set of points and gives you a slider to set the stiffness of the spline. 

this spline is very stiff

Edit: bad caption! NOT very stiff!

also see How to see equation used for spline? and cubic spline regression   and How is the Smooth Line in the Graph Builder determined 

[#5]

This is what I am trying to do 

Replicate Excel's scatter plot with a smooth curve feature.

If I use JMP fit Y by X with a flexible spine, I have a different curve than what Excel is producing with scatter plot + smooth line. I believe Excel is using Bezier curve fit.

Datapoints:

Spin | X | Y

A | 2.41 | 846.8692
A | 0.93 | 77.043
A | 1.37 | 155.2193
A | 1.96 | 390.4639

I do not want to draw this line by using flexible spline, but I want to force bezier curve. I hope that makes sense.

Thanks

[#6]

@mitulshah,

It would help if you appended the Excel chart you expect.

If I use the data (series) you provided, without sorting it, Excel creates the first graph below. If I sort the series in Excel, the second graph is teh result.   

Note, that

• JMP platforms, by default fit a curve to the the data by value not by order in the table.  
• Fit Y by X, fit Flexible, Kernel Smoother Linear Tri-Cube, would be the option I would use instead a spline, assuming you want a plot like the second plot below.
• To produce a  plot like the first chart below where the curve accounts for the data order, you will need to create a script for a Path, as @Craige_Hales mentioned previously. 

From your repsonse, it was not clear to me what you are looking for.

 

[#7]

Thinking about @gzmorgan0  explanation, you can do it in graph builder:

Select Points, then SHIFT to add the line

[#8]

I apologize for the confusion here.

Data:

In JMP, I did Fit Y by X

If I input the same data in Excel 

I get this 

Notice in JMP green line there a bit of an oscillation/ kink in the curve around lower left (between X values 2-3), whereas in Excel that is not the case.

my end goal is to interpolate values from the line, so if Excel and JMP gives two different lines, then I would end up with different interpolation values.  

[#9]

The spline chooses a slope at the beginning and end that does not appear to be the same for JMP and Excel. The spline uses a different cubic between each pair of points and can smoothly join a large number of points, matching slopes and curvatures. If you only have four points, it might make sense to fit a cubic to the four points and use the formula to do the interpolation.

Maybe we'll get a statistician's answer.

Graph Builder can do it. So can Fit Y by X.

[#10]

Yes. I hoping for that.

Another way to do this -- > maybe we can calculate the control point based on the four known points I have on the curve and then draw bezier curve passing through all four points. is there a way to do that?