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A cubic is more general than a quadratic, so clearly a quadratic is going to be a special case of a cubic. We can demonstrate this by looking at the equation f
by nanofortnight 12y ago
A cubic is more general than a quadratic, so clearly a quadratic is going to be a special case of a cubic.
We can demonstrate this by looking at the equation for a quadratic Bézier:
(1 - t)^2 P_0 + 2(1 - t)t P_1 + t^2 P_2
Now, let's increase the degree.
= [ (1 - t)^2 P_0 + 2(1 - t)t P_1 + t^2 P_2 ] * [ 1 - t + t ]
clearly this equation is equal to the above one because 1-t+t == 1 and multiplying by one does nothing.
= [ (1 - t)^2 P_0 + 2(1 - t)t P_1 + t^2 P_2 ] * [ (1 - t) + t ]
= (1 - t)^3 P_0 + t(1 - t)^2 P_0 + 2(1 - t)^2t P_1 + 2(1 - t)t^2 P_1 + (1 - t)t^2 P_2 + t^3 P_2
Group by powers of t:
= (1 - t)^3 P_0 + t(1 - t)^2 [P_0 + 2*P_1] + (1 - t)t^2 [2P_1 + P_2] + t^3 P_2
Which is exactly a cubic Bezier curve.
Thus, all quadratic Beziers can represented cubic ones, but the reverse does not hold, thus quadratics are specializations of cubics.
QED.
This is known as degree elevation.
- scannerlicker 12y agoOh, I see my mistake! Thank you very much! I'm not very math savvy and I was trying to get this right. Anyway, thank you very much for making this clear! Cheers!