Comments (2)
dcaoyuan
commented on May 24, 2024
1
Bloom filter is 256 bytes, that is 256 * 6,000,000blocks * (tx per block) = 1.43G * (tx per block).
Assume tx per block is (600,000 tx per day) / (5760 blocks per day) = 100
The size in bytes of 6,000,000 blocks would be 1.43G * 100 = 143G.
Of course, the (tx per block) is less than 100 before i.e. 4,000,000 block.
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dcaoyuan
commented on May 24, 2024
After (at block 6285787)
23G .khipu.eth/leveldb
457M .khipu.eth/kesque.logs/receipts_idx-0
654M .khipu.eth/kesque.logs/evmcode-0
655M .khipu.eth/kesque.logs/td-0
50G .khipu.eth/kesque.logs/body-0
3.8G .khipu.eth/kesque.logs/header-0
13G .khipu.eth/kesque.logs/account-0
457M .khipu.eth/kesque.logs/td_idx-0
25G .khipu.eth/kesque.logs/storage-0
38G .khipu.eth/kesque.logs/receipts-0
4.4G .khipu.eth/kesque.logs/account_idx-0
12G .khipu.eth/kesque.logs/storage_idx-0
457M .khipu.eth/kesque.logs/body_idx-0
262M .khipu.eth/kesque.logs/evmcode_idx-0
457M .khipu.eth/kesque.logs/header_idx-0
147G .khipu.eth/kesque.logs
4.0K .khipu.eth/keystore
169G .khipu.eth/
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