Yes, quite possibly - but I think it's slightly ambiguous; you can also get a spline doing exactly what's described by using a Hermite and setting the tangents to:
 
m0 = (1/6)( 54q(1/3) - 27q(2/3) - 33p0 + 6p1 )
m1 = (1/4)( 7p0 + 2m0 + 20p1 - 27P(2/3)
 
(assuming I've worked it out right - quite possibly mistakes in there) :)
 
Cheers,
Andrew.


From: Simon Fenney [mailto:simon.fenney@powervr.com]
Sent: 04 November 2008 12:28
To: andrew.vidler@ninjatheory.com; Game Development Algorithms
Subject: RE: [Algorithms] spline name

No. AFAICS (and assuming my maths is correct) the cubic spline Jarkko has described has the following basis matrix:
                [ -9  27  -27  9]
  M_interp =1/2 [ 18 -45  36  -9]
                [-11 18   -9   2] 
                [  2 0    0    0]
 
Whereas for a Hermite spline we have (from Foley et al)
              [ 2  -2  1  1]
  M_hermite = [-3   3 -2 -1]
              [0    0  1  0] 
              [1    0  0  0]
 
Simon


From: Andrew Vidler [mailto:andrew.vidler@ninjatheory.com]
Sent: 04 November 2008 11:38
To: 'Game Development Algorithms'
Subject: Re: [Algorithms] spline name

I think you've just found a way of specifying the tangents for a cubic hermite curve?
http://en.wikipedia.org/wiki/Cubic_Hermite_spline
 
If you look at the formula for q(1/3) and q(2/3) then you'll get two equations in terms of the endpoints and the tangent at each endpoint - just rearranging for the tangents gives you two equations (one for each tangent) in terms of the endpoints and q(1/3), q(2/3) - which is what you've got.
 
Unless there's some other characteristic of the spline that means it's not a Hermite?
 
Cheers,
Andrew.


From: Jarkko Lempiainen [mailto:altairx@gmail.com]
Sent: 04 November 2008 11:10
To: 'Game Development Algorithms'
Subject: [Algorithms] spline name

Hi,

 

Does anyone know if there is a name for a cubic spline which goes through all the defined control points p0..p3 in the interval t=[0, 1], so that q(0)=p0, q(1/3)=p1, q(2/3)=p2 and q(1)=p3? I solved the basis matrix for it, but don’t know what’s the name of the wheel I just reinvented ;)

 

 

Cheers, Jarkko

 


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