Defined matrix in Catmull Spline Curve
I am trying to use Catmull spline curve in my program , I am trying to understand it but why we only use below given Matrix , because the examples I saw I only found the below one
In Catmull spline curve , why its speciality ? and if we can use other matrix then how and what change it effects ?
geometry algebraic-curves interpolation spline
asked May 12 '14 at 9:48
ARG
158111
add a comment |
I am trying to use Catmull spline curve in my program , I am trying to understand it but why we only use below given Matrix , because the examples I saw I only found the below one
In Catmull spline curve , why its speciality ? and if we can use other matrix then how and what change it effects ?
geometry algebraic-curves interpolation spline
asked May 12 '14 at 9:48
ARG
158111
add a comment |
I am trying to use Catmull spline curve in my program , I am trying to understand it but why we only use below given Matrix , because the examples I saw I only found the below one
In Catmull spline curve , why its speciality ? and if we can use other matrix then how and what change it effects ?
geometry algebraic-curves interpolation spline
asked May 12 '14 at 9:48
ARG
158111
I am trying to use Catmull spline curve in my program , I am trying to understand it but why we only use below given Matrix , because the examples I saw I only found the below one
In Catmull spline curve , why its speciality ? and if we can use other matrix then how and what change it effects ?
geometry algebraic-curves interpolation spline
geometry algebraic-curves interpolation spline
asked May 12 '14 at 9:48
ARG
158111
asked May 12 '14 at 9:48
ARG
158111
edited May 12 '14 at 13:58
asked May 12 '14 at 9:48
ARG
158111
asked May 12 '14 at 9:48
ARG
158111
asked May 12 '14 at 9:48
ARG
158111
158111
add a comment |
add a comment |
1 Answer
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A Catmull-Rom spline is just a sequence of cubic segments, strung together so that their tangents match at the junctions.
You can think of each cubic segment in "Hermite" form. This means the segment is defined in terms of its end-points $mathbf{p}_0$ and $mathbf{p}_1$ and its end tangents $mathbf{m}_0$ and $mathbf{m}_1$. Because two adjacent cubic segments share the same point and tangent, they join smoothly.
Of course, you do this same construction with any set of points $mathbf{p}_i$ and tangent vectors $mathbf{m}_i$. Usually the points $mathbf{p}_i$ are given, and the splining function somehow makes up a set of tangent vectors $mathbf{m}_i$. Different splining techniques make up the tangent vectors in different ways. The Catmull-Rom spline calculates the tangent vector $mathbf{m}_i$ just by scaling the vector $mathbf{p}_{i+1} - mathbf{p}_{i-1}$. That's where the matrix comes from. This page shows the derivation.
You can invent some other way to fabricate the tangents $mathbf{m}_i$, and this would give rise to a different matrix. There is nothing really special about the Catmull-Rom approach or its associated matrix.
answered May 16 '14 at 2:49
bubba
30k32986
add a comment |
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1 Answer
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1 Answer
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active
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votes
A Catmull-Rom spline is just a sequence of cubic segments, strung together so that their tangents match at the junctions.
You can think of each cubic segment in "Hermite" form. This means the segment is defined in terms of its end-points $mathbf{p}_0$ and $mathbf{p}_1$ and its end tangents $mathbf{m}_0$ and $mathbf{m}_1$. Because two adjacent cubic segments share the same point and tangent, they join smoothly.
Of course, you do this same construction with any set of points $mathbf{p}_i$ and tangent vectors $mathbf{m}_i$. Usually the points $mathbf{p}_i$ are given, and the splining function somehow makes up a set of tangent vectors $mathbf{m}_i$. Different splining techniques make up the tangent vectors in different ways. The Catmull-Rom spline calculates the tangent vector $mathbf{m}_i$ just by scaling the vector $mathbf{p}_{i+1} - mathbf{p}_{i-1}$. That's where the matrix comes from. This page shows the derivation.
You can invent some other way to fabricate the tangents $mathbf{m}_i$, and this would give rise to a different matrix. There is nothing really special about the Catmull-Rom approach or its associated matrix.
answered May 16 '14 at 2:49
bubba
30k32986
add a comment |
A Catmull-Rom spline is just a sequence of cubic segments, strung together so that their tangents match at the junctions.
You can think of each cubic segment in "Hermite" form. This means the segment is defined in terms of its end-points $mathbf{p}_0$ and $mathbf{p}_1$ and its end tangents $mathbf{m}_0$ and $mathbf{m}_1$. Because two adjacent cubic segments share the same point and tangent, they join smoothly.
Of course, you do this same construction with any set of points $mathbf{p}_i$ and tangent vectors $mathbf{m}_i$. Usually the points $mathbf{p}_i$ are given, and the splining function somehow makes up a set of tangent vectors $mathbf{m}_i$. Different splining techniques make up the tangent vectors in different ways. The Catmull-Rom spline calculates the tangent vector $mathbf{m}_i$ just by scaling the vector $mathbf{p}_{i+1} - mathbf{p}_{i-1}$. That's where the matrix comes from. This page shows the derivation.
You can invent some other way to fabricate the tangents $mathbf{m}_i$, and this would give rise to a different matrix. There is nothing really special about the Catmull-Rom approach or its associated matrix.
answered May 16 '14 at 2:49
bubba
30k32986
add a comment |
A Catmull-Rom spline is just a sequence of cubic segments, strung together so that their tangents match at the junctions.
You can think of each cubic segment in "Hermite" form. This means the segment is defined in terms of its end-points $mathbf{p}_0$ and $mathbf{p}_1$ and its end tangents $mathbf{m}_0$ and $mathbf{m}_1$. Because two adjacent cubic segments share the same point and tangent, they join smoothly.
Of course, you do this same construction with any set of points $mathbf{p}_i$ and tangent vectors $mathbf{m}_i$. Usually the points $mathbf{p}_i$ are given, and the splining function somehow makes up a set of tangent vectors $mathbf{m}_i$. Different splining techniques make up the tangent vectors in different ways. The Catmull-Rom spline calculates the tangent vector $mathbf{m}_i$ just by scaling the vector $mathbf{p}_{i+1} - mathbf{p}_{i-1}$. That's where the matrix comes from. This page shows the derivation.
You can invent some other way to fabricate the tangents $mathbf{m}_i$, and this would give rise to a different matrix. There is nothing really special about the Catmull-Rom approach or its associated matrix.
answered May 16 '14 at 2:49
bubba
30k32986
A Catmull-Rom spline is just a sequence of cubic segments, strung together so that their tangents match at the junctions.
You can think of each cubic segment in "Hermite" form. This means the segment is defined in terms of its end-points $mathbf{p}_0$ and $mathbf{p}_1$ and its end tangents $mathbf{m}_0$ and $mathbf{m}_1$. Because two adjacent cubic segments share the same point and tangent, they join smoothly.
Of course, you do this same construction with any set of points $mathbf{p}_i$ and tangent vectors $mathbf{m}_i$. Usually the points $mathbf{p}_i$ are given, and the splining function somehow makes up a set of tangent vectors $mathbf{m}_i$. Different splining techniques make up the tangent vectors in different ways. The Catmull-Rom spline calculates the tangent vector $mathbf{m}_i$ just by scaling the vector $mathbf{p}_{i+1} - mathbf{p}_{i-1}$. That's where the matrix comes from. This page shows the derivation.
You can invent some other way to fabricate the tangents $mathbf{m}_i$, and this would give rise to a different matrix. There is nothing really special about the Catmull-Rom approach or its associated matrix.
answered May 16 '14 at 2:49
bubba
30k32986
edited Nov 27 at 6:17
answered May 16 '14 at 2:49
bubba
30k32986
answered May 16 '14 at 2:49
bubba
30k32986
answered May 16 '14 at 2:49
bubba
30k32986
30k32986
add a comment |
add a comment |
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