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If earth were completely smooth, excluding space for current water bodies, what weather should i expect?

I'm under the assumption that wind would be a lot stronger, but i have nothing to back that up.

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    $\begingroup$ How would the change between water and dry land go? Gentle slope onto a sandy beach? Also is there a moon? $\endgroup$
    – Mormacil
    Commented Mar 19, 2017 at 16:37
  • $\begingroup$ Is there shrubbery? Trees? Wildlife? Vegetation? $\endgroup$ Commented Mar 19, 2017 at 16:43
  • $\begingroup$ Yes, the accepted answer of the older question specifically includes weather detail. $\endgroup$
    – JDługosz
    Commented Mar 19, 2017 at 20:20

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Down below the tip of Cape Horn, there is an open expanse of water which circumnavigates the globe with almost no obstructing land masses to disrupt the winds.
Referred to as The Roaring Forties, The Furious Fifties and the Screaming Sixties, the winds in this region are among the strongest and most consistent on the planet.

From that, I think we can assume that your version of earth, occupied by non-obtrusive, wind-friendly land masses would be a little more windy than our current world.

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  • $\begingroup$ I believe that your higher winds are caused by air currents falling from elevated regions and into your flat unobstructed region. Given you've described a path of least resistance in an otherwise resisting world I would expect its behavior to be rather different than a world where there is no unique path like this. I don't believe it can be generalised. $\endgroup$ Commented Mar 19, 2017 at 17:43
  • $\begingroup$ @LioElbammalf, I hadn't thought of it that way.. Thanks for the clarification. Does that leave us with no method of figuring out the wind speed on a perfectly smooth planet? $\endgroup$ Commented Mar 19, 2017 at 19:32
  • $\begingroup$ I think it does make it difficult to figure out. Wind is really just air of different densities (temperatures) mixing such that they would tend to a uniform density. On a uniform world I think it would be dominated by the difference in temperature between the poles and equator. I'm not quite sure how you would determine the strength of that wind though. $\endgroup$ Commented Mar 19, 2017 at 20:13

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