Two recent fractals by Sebastien Helie; Par files below: http://www.emarketingiseasy.com/TESTS/FOTD/STORMY.GIF http://www.emarketingiseasy.com/TESTS/FOTD/UNCLE.GIF These were calculated with my default: float=yes Uncle's formula (at the bottom of this post) specifies two bailout criteria: z<|10| and potential=256/200/4 Am I correct in assuming that specifying two bailout criteria doesn't make sense? Anyone know which one gets used? An excerpt from the Fractint docs re: potential= is below: - Hal Lane ######################## # hallane@earthlink.net ######################## ----------------------------------------------------------- 3.9 Continuous Potential Fractint Version 20.04 . . . The parameters for continuous potential are: potential=maxcolor[/slope[/modulus[/16bit]]] These parameters are present on the <Y> options screen. "Maxcolor" is the color corresponding to zero potential, which plots as the TOP of the mountain. Generally this should be set to one less than the number of colors, i.e. usually 255. Remember that the last few colors of the default IBM VGA palette are BLACK, so you won't see what you are really getting unless you change to a different palette. "Slope" affects how rapidly the colors change -- the slope of the "mountains" created in 3D. If this is too low, the palette will not cover all the potential values and large areas will be black. If it is too high, the range of colors in the picture will be much less than those available. There is no easy way to predict in advance what this value should be. "Modulus" is the // bailout // value used to determine when an orbit has "escaped". Larger values give more accurate and smoother potential. A value of 500 gives excellent results. As noted, this value must be <128 for the integer fractal types (if you select a higher number, they will use 127). "16bit": If you transform a continuous potential image to 3D, the illumination modes 5 and 6 will work fine, but the colors will look a bit granular. This is because even with 256 colors, the continuous potential is being truncated to integers. The "16bit" option can be used to add an extra 8 bits of goodness to each stored pixel, for a much smoother result when transforming to 3D. . . . --------------------------------------------------------------------------- ================== Stormy { ; a fractal noise area Sebastien Helie 2011-22-20 ; IN # a good to develope maybae ; with google reset=2000 type=formula formulafile=2012frm.frm formulaname=distant2 corners=-20/20.00001/-14.99999/15 potential=256/200/4 colors=00000A0e00eee00e0eeL0eeeLLLLLzLzLLzzzLLzLzzzLzzz000555<3>HHHKKKOO\ O<3>ccchhhmmmssszzz00z<3>z0z<3>z00<3>zz0<3>0z0<3>0zz<2>0GzVVz<3>zVz<3>zV\ V<3>zzV<3>VzV<3>Vzz<2>Vbzhhz<3>zhz<3>zhh<3>zzh<3>hzh<3>hzz<2>hlz00S<3>S0\ S<3>S00<3>SS0<3>0S0<3>0SS<2>07SEES<3>SES<3>SEE<3>SSE<3>ESE<3>ESS<2>EHSKK\ S<2>QKSSKSSKQSKOSKMSKK<2>SQKSSKQSKOSKMSKKSK<2>KSQKSSKQSKOSKMS00G<3>G0G<3\

G00<3>GG0<3>0G0<3>0GG<2>04G88G<2>E8GG8GG8EG8CG8AG88<2>GE8GG8EG8CG8AG88G\ 8<2>8GE8GG8EG8CG8AGBBG<2>FBGGBGGBFGBDGBCGBB<2>GFBGGBFGBDGBCGBBGB<2>BGFBG\ GBFGBDGBCG000<6>000 }

frm:distant2{;sinpix z=pixel: x=real(sin(z)*0.66) y=imag(z*0.66) a=sin(x) b=tan(y) c=2+z^a z=z+a+b-c+pixel z<|10| } ================== ================== uncle { ; baby uncle Sebastien Helie 2011-11-20 ; the guardian ; with google reset=2000 type=formula formulafile=2012frm.frm formulaname=forsake corners=-0.5199512/0.6000251/-0.20963693/0.63035184 params=0/1 potential=256/200/4 colors=CCC<15>gCgiCikCk<3>sCs<16>MCs<4>CCs<16>Cis<4>Css<16>CsM<4>CsC<16>\ isi<4>sss<16>ssM<4>ssC<16>sMC<4>sCC<17>KCC<4>ACCCCC<15>gCgiCikCk<3>sCs<1\ 6>MCs<4>CCs<16>Cis<4>Css<10>CsY } frm:forsake{; z=pixel: x=real(1.4142) y=imag(1.4142) q=x/y/z z=z^(-2)-q^(p1*z)+pixel z<|10| } ==================