Bitcoin

Why do the declines in the mempool have a slope? shouldn’t it be entirely vertical/instantaneous?



https://jochen-hoenicke.de/queue/#2h

when a block is found, it may take time to propogate through the network, but that’s on the order of a few seconds right?
so why does jochen’s mempool show a slope, rather than a totally vertical dropoff?



View Reddit by TheTrillionthApeView Source


7 Comments on Why do the declines in the mempool have a slope? shouldn’t it be entirely vertical/instantaneous?

  1. Zudafrica

    The mempool is completely dynamic-changing second by second—–Go to Blockchain.info to see the activity. The other thing is the mempool is a basket of unconfirmed transactions second by second——Not blocks —blocks have nothing to do with the mempool times.

  2. nowitsalllgone

    I wonder if his node drops transactions from the mempool as it scans a new block and finds them included in it. If that was happening, it could result in a slight angle because it takes a few seconds to scan the block. But some of those slopes are two minutes in width, so idk.

  3. igadjeed

    You’re probably right, but is anything instantaneous, and how would calculus work if graphs had infinite gradients (vertical lines)?
    To have a vertical line – because when a block is mined, all its transactions disappear instantly from the mempool – there would have to be 2 mempool counts at exactly the same time with 2 different numbers
    The graph represents the fact that data points are sampled periodically. The slope measures the time between the last data point before block mining and the first data point after

    Same issue: https://www.blockchain.com/charts/difficulty

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