Continuity of the Bezier Curve, Question

  • Thread starter Thread starter mymachine
  • Start date Start date
  • Tags Tags
    Continuity Curve
AI Thread Summary
The discussion centers on the continuity and smoothness of two cubic Bezier curves, A and B, particularly at the junction where A3 meets B0. For the curves to be smooth at this point, the segments A2-A3 and B0-B1 must be collinear, ensuring that their slopes match. While A3 and B0 can be the same point, they do not need to be midpoints of the segments. Both curves are described by 3rd-degree Bernstein polynomials, confirming their classification as cubic Bezier curves. The overall conclusion is that while the curves can be continuous, they may not be smooth without the specified conditions.
mymachine
Messages
42
Reaction score
0
Hi everyone,

I would like to ask about the continuity of the cubic Bezier curve.

There are two cubic Bezier curves, A and B, shown as below two images:

image.png

image.png


The coordinates of the A curve are:

A0 = (x0,y0) = (0,0)
A1 = (x1,y1) = (2,3)
A2 = (x2,Y2) = (5,4)
A3 = (x3,y3) = (7,0)

The coordinates of the B curve are:

B0 = (x0,y0) = (0,4)
B1 = (x1,y1) = (3,1)
B2 = (x2,y2) = (7,0)
B3 = (x3,y3) = (9,8)

If I join these two curves together by connecting the point A3 and B0, it looks such as below image:

A_B.png


However, the curve doesn't looks smooth at point A3 = B0.

The question is, what is the equation of the Bezier curve start from
point A0 > A1 > A2 > A3=B0 > B1 > B2 > B3
where the endpoint is A0, A3 = B0, and B3
and so that the curve is continue and looks smooth?

Also, does this A+B curve is 7th degree of the Bernstein polynomial?

Thank you
 
Mathematics news on Phys.org
Is it possible if

A2, A3 = B2, and B1

are not colinear

and

A3 = B2

is not the midpoint of A2 and B1?
 
Hi, mymachine,
yes (more or less): the points A2, A3 = B0, and B1 should be collinear. This is because the curve "A" is tangent at A3 to the straight segment A2-A3, and the curve "B" is tangent at B0 to the segment B0-B1. If these two slopes on the curves are to be the same, the segments A2-A3 and B0-B1 must have the same slope too. And since A3=B0, this puts these points on the same straight line.

The two segments A2-A3 and B0-B1 do not need to have the same length (that is, A3 = B0 does not need to be a midpoint of A2-B1).

Your last question, I couldn't understand. The Bernstein polynomials that constitute both the "A" and "B" curves are 3rd-degree polynomials (which is why these are called "cubic" Béziers).

(Just as a side comment, mathematicians use the word "continuity" to refer to the fact that, in your example, A3 = B0; that is, that you didn't need to lift the "pencil" to continue drawing, that the curve does not have a "hole" because of A3 and B0 not coinciding.) Your curve, as described, is continuous; it's just not smooth.

Hope this helps!
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top