<![CDATA[Bezier curvatures as matrix operations and without a slider]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider
Mon, 28 Dec 2015 22:13:16 +0000Thu, 24 Dec 2015 11:12:14 +0000Zend_Feed<![CDATA[Hi, Curve[(1 - t)³ P_0 + 3(1 - t)² t P_1 + 3 (1 - t) t² P_2 + t³ P_3,t,0,1] should produce the Bezier curve as a curve object (which has some slight advantages compared to locus object, eg. you should be able to find intersections with a line) Cheers, Zbynek]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186181
Thu, 24 Dec 2015 12:06:20 +0000<![CDATA[ Curve[(1 - t)³ P_0 + 3(1 - t)² t P_1 + 3 (1 - t) t² P_2 + t³ P_3,t,0,1] Okay, I didn't know that and it's really nice. I'll try to figure out matrices myself...]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186183
Fri, 25 Dec 2015 17:39:46 +0000<![CDATA[As far as the matrices are concerned, I have 4 points P0 to P4 and 3 matrices: - m_1 = {{1, t, t², t³}} - m_2 = {{1, 0, 0, 0}, {-3, 3, 0, 0}, {3, -6, 3, 0}, {-1, 3, -3, 1}} - m_3 = {{P_0}, {P_1}, {P_2}, {P_3}} But: T = m_1 * m_2 * m_3 = {{}, {}, {}, {}} I guess I am getting something wrong, but I am not seeing it...]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186185
Sat, 26 Dec 2015 20:40:09 +0000<![CDATA[Hi, for now the input bar supports matrix multiplication only for matrices of numbers, not variables or points. To get more symbolic results you can use the CAS view: enter the matrices into the first three rows of CAS then f(t):=ToPoint[Element[$1 $2 $3,1,1]] gives you the Bezier curve (also note that product of the three matrices is a 1x1 matrix, hence Element[] command is needed to extract the matrix entry. ToPoint is used to make sure the result is a cartesian point, not a complex one -- workaround for a minor bug.). Cheers, Zbynek]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186187
Sun, 27 Dec 2015 01:15:18 +0000<![CDATA[The results are right, but I am not seeing a point/curve. Am I supposed to one? Or is a step missing? https://ggbm.at/2348709]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186189
Sun, 27 Dec 2015 18:52:09 +0000<![CDATA[Just for the record: Using "L:=ToPoint[Element[$1 $2 $3,1,1]]" and turning on visibility there is a point. Although the locus tool doesn't seem to work...]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186191
Mon, 28 Dec 2015 18:26:00 +0000<![CDATA[Hi, your screenshots show that t is a number, therefore you only get one particular point. In the first CAS cell please replace t by s and in the last one f(t) by f(s), then delete t. Cheers, Zbynek]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186193
Mon, 28 Dec 2015 20:00:07 +0000<![CDATA[ your screenshots show that t is a number, therefore you only get one particular point. In the first CAS cell please replace t by s and in the last one f(t) by f(s), then delete t. This way it didn't work for me. But starting all over again not using a slider at all it worked. Thank you, Zbynek.]]>
https://help.geogebra.org/topic/bezier-curvatures-as-matrix-operations-and-without-a-slider#comment-186195
Mon, 28 Dec 2015 22:13:16 +0000