It is a very fast aprx. 1 kb (928 b)FILT RAM, which saves data on filter by the basis of filter's spectrum tech.It is 29 bit wide 8 bit address RAM.
fram
harddisk
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device
computer
data
binary
fast
memory
ssdmemory
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To operate this RAM just input address and value to save at that address (by sparking the INST's) and spark write wire.To read input address and spark read. For clearing just spark clear.
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Didnt suppose we would write a post at the same time :D, so to clarify, my comment below is referred to @TehBoxtrot94
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Input Data and Address (this means put some data on some place, you need to specify what data and place by this). Every cell (place) contains 29 bits worth of data. So yeah, then write or read depending on what you want to do. Keep in mind the data is in binary.
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After fast reading few hounders lines of tpt code I did'n saw any special use for this last bits int filt, last one is a sign of stored number(sometimes elem id), but this don't matter in getting color. Maybe beacouse ignoring 2 give us 30 bits and we have 10 bits for every r,g,b channel? But now it's using diffrend method of drawing. Back compatibiliti? Maybe I missed somenthing?
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How do i operate this? can someone help me a little?
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"I suppose"
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However, NoVIcE, there are 181 elements as of TPT 93.3, AND filt's ID is 125, which is 001111101, is MUCH greater than 3, or 11 in binary. You implied that you can compress 9 bits LOSSLESSLY into 2, which is impossible.
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Actually it uses 30 out of 32 bits. However people use the last bit as in case in some operation we get zero. If zero is outputted, then 0 bits are activated and the bray will "die" (it will disappear). To prevent this from happening, people turn on the last bit and dont use it in any calculations to make sure the bray will stay on, hence 29 bits. I suppose the 2 out of 32 bits are occupied for the element itself.
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mario3154: The material FILT can only store a maximum of 29 bits per pixel, and this FRAM uses all of them.
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Why is it only 29 bit, and not 32? Can someone explain?