oUR iNITIAL vALUES aRE t AND r (r bEING THE rESOLUTION). wE tAKE THE cURVE, wHICH iS cURRENTLY A lINE fROM [XYZ=0] TO [XY=0 z=1] AND sUBDIVIDE iT uSING oUR vALUE r. iN aNOTHER iNSTANCE OF r wE sUBTRACT 1 AND dIVIDE oUR iNDEX BY iT, THE iNDEX bEING 0 AT THE oRIGIN OF THE cURVE ([XYZ=0]) AND 1 AT THE eND ([XY=0 z=1]). tHEN wE mULTIPLY tHAT BY oUR vALUE t, gIVING uS THE cURVE OF t wE wANT, wITH aNY vALUE FOR r. fOR THE sAKE OF sIMPLICITY, i wILLBE rEFERRING TO tHAT oUTCOME AS tA. nOW iM gONNA gET iNTO THE fORMULAS. fIRST oFF, wE mULTIPLY tA wITH t. tHEN wE sUBTRACT tA. aS wERE gONNA bE uSING tHIS oUTCOME mULTIPLE tIMES iM gONNA cALL iT tB. fIRST oFF, wE sUBTRACT 38.8 fROM tB, tHEN dIVIDE iT BY 54.2, gIVING oUR x vALUE. fOR y, wE tAKE THE sIN OF tB, mULTIPLY iT wITH 0.883 AND tHEN tAKE THE mAXIMUM OF tHAT AND THE vALUE wE gET fROM sUBTRACTING 145.41 fROM tB AND mULTIPLYING iT wITH 0.00558. tHEN wE tAKE THE mIN OF tHAT AND THE vALUE wE gET fROM sUBTRACTING 10.31 fROM tB AND mULTIPLYING iT wITH 0.0827. fINALLY, wE tAKE THE mIN OF tHAT AND THE vALUE wE gET fROM mULTIPLYING 2 BY THE pOWER OF THE vALUE wE gET fROM sUBTRACTING 73.39 fROM tB AND dIVIDING iT BY 12.71 wITH -0.9 AND aDDING 0.8. fOR z, wE jUST tAKE tB AND mULTIPLY iT wITH -0.0125 AND aDDING 0.05. tHEN wE jOIN tHAT cURVE AND THE cURVE wE gET fROM THE sAME fORMULAS, oNLY tHAT FOR tHIS oNE wE tAKE THE mAXIMUM OF oUR oTHER x vALUE AND -0.35 FOR x, wE tAKE THE vALUE wE uSE IN oUR y fORMULA TO dO THE lAST mINIMUM oPERATION wITH FOR y AND wE aDD 0.1 TO oUR oLD z vALUE FOR z. nOW wE tRANSFORM THE jOINED cURVE iNTO A mESH wITH THE vALUES 80.630 FOR t AND 200 FOR r. wE dO tHAT BY tAKING A cIRCLE cURVE pRIMITIVE wITH A rADIUS OF 0.1M AND mAKING iT fOLLOW THE cURVE wE cREATED AND sAVING THE pOINTS’ pOSITIONS eVERY fEW sTEPS AND tHEN cONNECTING eACH eDGE wITH iTS nEXT dUPLICATE. fINALLY, wE uSE A sUBDIVISION aLGORITHM fOLLOWED BY A rEMESHING aLGORITHM TO sMOOTH THE mESH oUT. nOW wE hAVE A pIZZA
[texturepreview='500']11798[/texturepreview]