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#X obj 3 927 doc_also;
#X obj 103 927 lop~;
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#X msg 228 97 load b001.jpg;
#X msg 234 116 load r001.jpg;
#X text 11 29 temporal low-pass filter;
#X text 232 305 indicates the amount of one image being blended with
the other. this is one divided by the fraction of the input that will
be blended with the previous output.;
#X text 232 414 for values that are not powers of two \, the computation
may be less precise.;
#X text 232 449 because this uses feedback \, you can compute the influence
of a certain input on a certain later output using a recursive reasoning
: for example \, for a [#fade 4] \, 3/4 of the previous output was
made from 3/4 of its own previous output \, which was in turn made
from 1/4 of the input of that time : thus the next-to-next-to-last
input counts for (1/4)*(3/4)*(3/4) = 9/64 of the current output.;
#X obj 195 179 #store;
#X obj 195 198 #fade 4;
#X obj 228 78 loadbang;
#X text 337 114 change picture here;
#X msg 240 135 load g001.jpg;
#X obj 329 91 doc_accolade;
#X obj 139 927 rpole~;
#X text 232 549 [#fade 4] is similar to [*~ 0.25]-[rpole~ 0.75] : that
is \, you do one divided by the #fade argument to make the *~ argument
\, and then 1 minus the *~ argument to make the rpole~ argument. This
in turn is equivalent to a certain [lop~] but that's depending on the
sampling rate.;
#X text 232 623 you can increase precision of [#fade] by using a certain
[# <<] and the opposite [# >>] \, around [#fade] \, to momentarily
widen the range of possible values used by [#fade]. This will lower
the amount of residue that may stay in the output for too long or not
long enough (compared to the theory of the fractions because the fractions
assume no rounding).;
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#X obj 16 492 # sq-;
#X obj 16 549 display;
#X obj 16 436 expr cos($f1*2*3.141593/$f2)*1000;
#X obj 16 530 # sqrt;
#X obj 16 511 #fade 64;
#X obj 16 473 #fade 8;
#X text 232 723 argument \$1 has a cutoff period of pi*\$1 frames \,
for values of \$1 that are neither too big nor too small (guess what
those are...).;
#X text 232 811 grid that gradually change the output.;
#X text 232 895 output being gradually changed by the input.;
#X text 232 353 for example \, an argument of 5 will output the sum
of 1/5th of the input and 4/5ths of the previous output. thus \, the
rate of change of the output is 1/5th of the difference between the
(previous) output and the input.;
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