In PoS how is Net Stake Weight calculated In PoS coins there is a common 'time to stake' estimate, this is based on the users own accumulated stake amount as a proportion of the network wide stake weight.

However there is no way for any node to reliably estimate the network stake weight, in theory it's just all nodes individual weights added together, but as no single node is guaranteed to see all others without duplication, I wanted to understand how this is calculated.
While I don't know how other PoS blockchains operate, I can give some insight about the PoSv2/3 version based on Peercoin's implementation of proof of stake; one such example would be Novacoin.

The total network weight is ~computed by this function:

double GetPoSKernelPS()
{
    int nPoSInterval = 72;
    double dStakeKernelsTriedAvg = 0;
    int nStakesHandled = 0, nStakesTime = 0;

    CBlockIndex* pindex = pindexBest;;
    CBlockIndex* pindexPrevStake = NULL;

    while (pindex &amp;&amp; nStakesHandled &lt; nPoSInterval)
    {
        if (pindex-&gt;IsProofOfStake())
        {
            dStakeKernelsTriedAvg += GetDifficulty(pindex) * 4294967296.0;
            nStakesTime += pindexPrevStake ? (pindexPrevStake-&gt;nTime - pindex-&gt;nTime) : 0;
            pindexPrevStake = pindex;
            nStakesHandled++;
        }

        pindex = pindex-&gt;pprev;
    }

    if (!nStakesHandled)
        return 0;

    return dStakeKernelsTriedAvg / nStakesTime;
}


How the weight estimation algorithm works is pretty interesting: it iterates backwards through the blockchain, from the latest known best block until it finds 72 blocks that were produced using a proof of stake (basically all the past 72 blocks).

Using these 72 PoS blocks, it calculates the total amount of time that has passed between each consecutive block pair (added up to a total -- nStakesTime) and the total amount of "difficulty" that was satisfied during the production of those blocks (dStakeKernelsTriedAvg).

The final network weight estimation is given by dividing the "amount of work" by the "amount of time". 

This algorithm does not attempt to calculate the exact amount of coins that are being "at stake" at any given time, because as you noticed that wouldn't be possible; yet the estimations it gives are pretty accurate.