• @TropicalDingdong
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    101 year ago

    The numbers are so mind-bogglingly out there, its just hard to fathom, let alone believe it can actually happen.

    I was working on some back of the napkins this weekend. I came up with the number 70 trillion megagrams of CO2 in the atmosphere from burning fossil fuels. If you’ve got a better number, would love to know but its what I got.

    For reference, Mauna Loa is about 70 trillion cubic meters from ocean floor. So just round numbers, hand-wavey estimates, thats how much carbon we have to get out of the atmosphere. The largest mountain on earth.

      • @TropicalDingdong
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        11 year ago

        However it is propably smarter to store just the carbon in the form of coal. That would only be 0.475 trillion t. Thats like 1200 times the global paper production in mass.

        So I tried doing my calculations with better data.

        My estimate is that there are 2041 gigatonnes of co2 in the atmosphere remaining post industrial revolution. This equivalent to about 5% of the mass of Mauna Loa, in the form of wood equivalent carbon. Still a bewilderingly large amount, but much smaller.

        We would need to sequester approximately 1,133,900,000,000 megagrams of wood from trees to offset these emissions.

        Would also love a sanity check. I’m using this dataset and some conversion factors for my math: https://ourworldindata.org/grapher/cumulative-co-emissions?country=~OWID_WRL

    • @Feathercrown
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      1 year ago

      Wouldn’t it be a 1m square base tower the same height as the mountain? Nowhere near the mountain’s volume.

      • Skua
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        fedilink
        41 year ago

        A 1m square tower that size would only be 4,169 m ^3 . Mauna Loa is mostly basalt, which weighs about 2.9 Mg per m ^3 . This tower would therefore weigh about 12,000 tonnes. You’d need almost 6 billion of those towers to get to the 70 trillion Mg figure.

        Mauna Loa’s actual mass is probably about three times that number because of the density of the rocks in it, but in terms of orders-of-magnitude estimation it was about right.