Bézier Curves: where the 3 (or 1/3) constant comes from when moving from Hermite curves?
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I'm learning Bézier Curves, and I'm stuck at the reason why they use 3 (or 1/3) constant when moving from Hermite curves?
Like in this, this, and this source.
e.g. why t0 = 3(q1 - q0) ?
or why v1 = v0 + 1/3 d0 in the first link:
x0 → v0
→ v1 = v0 + ⅓d0
→ v2 = v3 – ⅓d1
x1 → v3
bezier-curve
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
$endgroup$
add a comment |
$begingroup$
I'm learning Bézier Curves, and I'm stuck at the reason why they use 3 (or 1/3) constant when moving from Hermite curves?
Like in this, this, and this source.
e.g. why t0 = 3(q1 - q0) ?
or why v1 = v0 + 1/3 d0 in the first link:
x0 → v0
→ v1 = v0 + ⅓d0
→ v2 = v3 – ⅓d1
x1 → v3
bezier-curve
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
$endgroup$
add a comment |
$begingroup$
I'm learning Bézier Curves, and I'm stuck at the reason why they use 3 (or 1/3) constant when moving from Hermite curves?
Like in this, this, and this source.
e.g. why t0 = 3(q1 - q0) ?
or why v1 = v0 + 1/3 d0 in the first link:
x0 → v0
→ v1 = v0 + ⅓d0
→ v2 = v3 – ⅓d1
x1 → v3
bezier-curve
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
$endgroup$
I'm learning Bézier Curves, and I'm stuck at the reason why they use 3 (or 1/3) constant when moving from Hermite curves?
Like in this, this, and this source.
e.g. why t0 = 3(q1 - q0) ?
or why v1 = v0 + 1/3 d0 in the first link:
x0 → v0
→ v1 = v0 + ⅓d0
→ v2 = v3 – ⅓d1
x1 → v3
bezier-curve
bezier-curve
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
asked Dec 19 '18 at 15:30
David RefaeliDavid Refaeli
1216
1216
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A cubic Bezier curve is represented as
$$B(t)=(1-t)^3Q_0+3(1-t)^2tQ_1+3(1-t)t^2Q_2+t^3Q_3$$
and its first derivative is
$$B'(t)=3(1-t)^2(Q_1-Q_0)+6t(1-t)(Q_2-Q_1)+3t^2(Q_3-Q_2)$$
For cubic Hermite curve, $P_0,T_0, P_1$ and $T_1$ are the position and first derivative vectors at curve's start ($t=0$) and end ($t=1$). So, we can evaluate the cubic Bezier curve's position vector and first derivative at $t=0$ and $t=1$ to obtain
$P_0=B(0)=Q_0$
$T_0=B'(0)=3(Q_1-Q_0)$
$P_1=B(1)=Q_3$
$T_1=B'(1)=3(Q_3-Q_2)$.
answered Dec 19 '18 at 17:45
fangfang
2,472166
$endgroup$
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
add a comment |
$begingroup$
As the answer from @fang shows, the $tfrac13$ arises from the use of the Bernstein polynomials (the blending functions you use to define Bezier curves).
But there are other forms of the same curve that use different constants. The Timmer form of the curve uses a constant of $tfrac14$, and the Ball form uses $tfrac12$. Different blending functions will give you different constants. There's nothing magic about $tfrac13$; in some ways $tfrac14$ (the Timmer form) is a better choice.
There's a picture showing the Timmer control points in my answer to this question.
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A cubic Bezier curve is represented as
$$B(t)=(1-t)^3Q_0+3(1-t)^2tQ_1+3(1-t)t^2Q_2+t^3Q_3$$
and its first derivative is
$$B'(t)=3(1-t)^2(Q_1-Q_0)+6t(1-t)(Q_2-Q_1)+3t^2(Q_3-Q_2)$$
For cubic Hermite curve, $P_0,T_0, P_1$ and $T_1$ are the position and first derivative vectors at curve's start ($t=0$) and end ($t=1$). So, we can evaluate the cubic Bezier curve's position vector and first derivative at $t=0$ and $t=1$ to obtain
$P_0=B(0)=Q_0$
$T_0=B'(0)=3(Q_1-Q_0)$
$P_1=B(1)=Q_3$
$T_1=B'(1)=3(Q_3-Q_2)$.
answered Dec 19 '18 at 17:45
fangfang
2,472166
$endgroup$
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
add a comment |
$begingroup$
A cubic Bezier curve is represented as
$$B(t)=(1-t)^3Q_0+3(1-t)^2tQ_1+3(1-t)t^2Q_2+t^3Q_3$$
and its first derivative is
$$B'(t)=3(1-t)^2(Q_1-Q_0)+6t(1-t)(Q_2-Q_1)+3t^2(Q_3-Q_2)$$
For cubic Hermite curve, $P_0,T_0, P_1$ and $T_1$ are the position and first derivative vectors at curve's start ($t=0$) and end ($t=1$). So, we can evaluate the cubic Bezier curve's position vector and first derivative at $t=0$ and $t=1$ to obtain
$P_0=B(0)=Q_0$
$T_0=B'(0)=3(Q_1-Q_0)$
$P_1=B(1)=Q_3$
$T_1=B'(1)=3(Q_3-Q_2)$.
answered Dec 19 '18 at 17:45
fangfang
2,472166
$endgroup$
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
add a comment |
$begingroup$
A cubic Bezier curve is represented as
$$B(t)=(1-t)^3Q_0+3(1-t)^2tQ_1+3(1-t)t^2Q_2+t^3Q_3$$
and its first derivative is
$$B'(t)=3(1-t)^2(Q_1-Q_0)+6t(1-t)(Q_2-Q_1)+3t^2(Q_3-Q_2)$$
For cubic Hermite curve, $P_0,T_0, P_1$ and $T_1$ are the position and first derivative vectors at curve's start ($t=0$) and end ($t=1$). So, we can evaluate the cubic Bezier curve's position vector and first derivative at $t=0$ and $t=1$ to obtain
$P_0=B(0)=Q_0$
$T_0=B'(0)=3(Q_1-Q_0)$
$P_1=B(1)=Q_3$
$T_1=B'(1)=3(Q_3-Q_2)$.
answered Dec 19 '18 at 17:45
fangfang
2,472166
$endgroup$
A cubic Bezier curve is represented as
$$B(t)=(1-t)^3Q_0+3(1-t)^2tQ_1+3(1-t)t^2Q_2+t^3Q_3$$
and its first derivative is
$$B'(t)=3(1-t)^2(Q_1-Q_0)+6t(1-t)(Q_2-Q_1)+3t^2(Q_3-Q_2)$$
For cubic Hermite curve, $P_0,T_0, P_1$ and $T_1$ are the position and first derivative vectors at curve's start ($t=0$) and end ($t=1$). So, we can evaluate the cubic Bezier curve's position vector and first derivative at $t=0$ and $t=1$ to obtain
$P_0=B(0)=Q_0$
$T_0=B'(0)=3(Q_1-Q_0)$
$P_1=B(1)=Q_3$
$T_1=B'(1)=3(Q_3-Q_2)$.
answered Dec 19 '18 at 17:45
fangfang
2,472166
answered Dec 19 '18 at 17:45
fangfang
2,472166
answered Dec 19 '18 at 17:45
fangfang
2,472166
answered Dec 19 '18 at 17:45
fangfang
2,472166
2,472166
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
add a comment |
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
$begingroup$
ok, this makes perfect sense if you take a Bezier and check its Hermite constraints. But in all these lectures they start from Hermite and then "evolve" it to a Bezier... Maybe it's just cheating ("heuristics") as they know what is needed to get a Bezier. I tried to understand if there's a stronger intuition of why the 3 constant is used
$endgroup$
– David Refaeli
Dec 20 '18 at 6:26
add a comment |
$begingroup$
As the answer from @fang shows, the $tfrac13$ arises from the use of the Bernstein polynomials (the blending functions you use to define Bezier curves).
But there are other forms of the same curve that use different constants. The Timmer form of the curve uses a constant of $tfrac14$, and the Ball form uses $tfrac12$. Different blending functions will give you different constants. There's nothing magic about $tfrac13$; in some ways $tfrac14$ (the Timmer form) is a better choice.
There's a picture showing the Timmer control points in my answer to this question.
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
$endgroup$
add a comment |
$begingroup$
As the answer from @fang shows, the $tfrac13$ arises from the use of the Bernstein polynomials (the blending functions you use to define Bezier curves).
But there are other forms of the same curve that use different constants. The Timmer form of the curve uses a constant of $tfrac14$, and the Ball form uses $tfrac12$. Different blending functions will give you different constants. There's nothing magic about $tfrac13$; in some ways $tfrac14$ (the Timmer form) is a better choice.
There's a picture showing the Timmer control points in my answer to this question.
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
$endgroup$
add a comment |
$begingroup$
As the answer from @fang shows, the $tfrac13$ arises from the use of the Bernstein polynomials (the blending functions you use to define Bezier curves).
But there are other forms of the same curve that use different constants. The Timmer form of the curve uses a constant of $tfrac14$, and the Ball form uses $tfrac12$. Different blending functions will give you different constants. There's nothing magic about $tfrac13$; in some ways $tfrac14$ (the Timmer form) is a better choice.
There's a picture showing the Timmer control points in my answer to this question.
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
$endgroup$
As the answer from @fang shows, the $tfrac13$ arises from the use of the Bernstein polynomials (the blending functions you use to define Bezier curves).
But there are other forms of the same curve that use different constants. The Timmer form of the curve uses a constant of $tfrac14$, and the Ball form uses $tfrac12$. Different blending functions will give you different constants. There's nothing magic about $tfrac13$; in some ways $tfrac14$ (the Timmer form) is a better choice.
There's a picture showing the Timmer control points in my answer to this question.
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
answered Dec 30 '18 at 3:59
bubbabubba
30.8k33188
30.8k33188
add a comment |
add a comment |
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