6 ms·
A drop with a radius of 5.7 km... http://www.wolframalpha.com/input/?i=10MW+times+a+year+divided+by+143.851+PWh+times+1.3+billion+cubic+kilometres http://www.w
by pepve 13y ago
A drop with a radius of 5.7 km...
http://www.wolframalpha.com/input/?i=10MW+times+a+year+divided+by+143.851+PWh+times+1.3+billion+cubic+kilometres http://www.wolframalpha.com/input/?i=10MW+times+a+year+divid...
- 6d0debc071 13y agoI'm not sure whether you're agreeing with me or not, and in a way that's why people shouldn't just use large numbers and then act like they've said something meaningful. Ratios matter, large numbers without a relevant basis for comparison on the other hand are just misleading. There are roughly 116 million households in the US, saving the energy of 7,500 is not a big change. You're solving roughly 1 - 15,466 th of the problem. And that's assuming that all the savings could even be applied to the US, which they most certainly couldn't. That's not a big impact. Chances are no-one - even if they were looking - could notice the figurative needle move on a change that small at all the power-stations serving the aggregated demand.