---
title: "C0 and C1 continuity of Bezier sections"
subject: "Computer Graphics"
chapter: 5
key: "semester_5/computer_graphics:c0-c1-continuity-bezier-sections"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/c0-c1-continuity-bezier-sections
site: BCT Engineering
---

# C0 and C1 continuity of Bezier sections

Computer Graphics · Chapter 5: Curve Modeling

- Asked in: 2070 Asar, 6 marks
- Video: [2  Spline Terminology & C0,C1 and C2 continuity. (Hindi)](https://www.youtube.com/watch?v=wn-J6edEJX4) — SufiyanDesignLab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/c0-c1-continuity-bezier-sections
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-5

## Question

Explain how to satisfy C⁰ and C¹ continuity conditions when joining two Bezier curve sections together.

## Short answer

C0: P0' = Pn. C1: tangents equal. For equal degree n, P1' = 2Pn − Pn−1, making Pn−1, Pn, P1' collinear with Pn as midpoint. For unequal degrees, n(Pn−Pn−1)=m(P1'−P0').

## Answer

Let the first Bezier section have control points P0 … Pn and the second have P0' … Pm'. The join is at the end of the first (Pn) and the start of the second (P0').

**C⁰ (position) continuity:** the curves must meet at the join.

P0' = Pn

**C¹ (first-derivative) continuity:** the tangent vectors at the join must be equal. The end tangents are

- P'(1) of the first = n(Pn − Pn−1)
- P'(0) of the second = m(P1' − P0')

Setting them equal (for equal degree n = m): P1' − P0' = Pn − Pn−1, so

**P1' = Pn + (Pn − Pn−1) = 2Pn − Pn−1**

That is, Pn−1, Pn (= P0') and P1' must be collinear and Pn must be the midpoint of Pn−1 and P1' (same distances on both sides).

```text
 Pn-1        Pn = P0'        P1'
  o-----------o------------o
 (first section)  (second section)
```

If the sections have different degrees: n(Pn − Pn−1) = m(P1' − P0').

For geometric continuity G¹ only the direction needs to match, so P1' can be anywhere on the line through Pn−1 and Pn on the other side of Pn, at any distance (tangents proportional).

Example (cubic sections): first section P0(0,0), P1(1,2), P2(3,2), P3(4,0). Second section must start at P0' = (4,0); for C¹, P1' = 2(4,0) − (3,2) = (5,−2).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is C0 continuity achieved when joining two Bezier sections?
   A. P1' = Pn
   B. The first point of the second section equals the last point of the first: P0' = Pn
   C. P0' = P0
   D. The tangents are equal
   Answer: B. C0 only requires that the sections meet. (Key point: C0 condition)

2. For equal degree, C1 continuity requires:
   A. P1' = Pn-1
   B. P1' = Pn
   C. P1' = 2Pn - Pn-1
   D. P1' = 2Pn-1 - Pn
   Answer: C. Equal tangents n(Pn-Pn-1) = n(P1'-P0') give P1' = Pn + (Pn - Pn-1). (Key point: C1 condition)

3. In a C1 join, Pn is positioned:
   A. At the midpoint of Pn-1 and P1' (collinear, equal spacing)
   B. Anywhere on the line, any distance
   C. Away from the line through Pn-1 and P1'
   D. Equal to Pn-1
   Answer: A. Pn-1, Pn and P1' are collinear with equal distances. (Key point: Geometry of C1)

4. How does G1 relax C1 for the join?
   A. P1' can be anywhere in the plane
   B. P1' must equal Pn
   C. P1' can lie anywhere on the line beyond Pn, at any distance
   D. Pn-1 must equal Pn
   Answer: C. G1 needs only matching tangent direction. (Key point: G1)

## Related in Computer Graphics

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