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是否可以检测使用风玫瑰科里奥利效应?

我曾经在高中学到的,我以前住的地方经历了城市上空的空气流动的科里奥利效应。我从来没有注意到过,但我最近听说过一遍,我不知道这是真的还是假的。我绝对没有办法验证这个理论利用科里奥利效应的更广泛的定义,但是我有访问风向玫瑰图看起来像下面的图:

有什么办法可以理解如果有任何Coriolis-like风通过使用这个图或类似的吗?网上我发现一个漂亮的素描的效果,但我不明白如何关联这两个。

有谁能帮我理解如果一个从风向玫瑰图可以得到这种效果?

回答

像许多其他问题我回答的地球科学,答案是复杂的。江南体育网页版基于[地转状态原理][1],科里奥利力将指向右边的风矢量在北半球在南半球(左)。* *能*验证基于风玫瑰*科里奥利力?* *号你需要多个同时观察在一个广泛的观察空间。你可以看到这个天气地图显示风向。变得更容易看到科里奥利效应在高空地图上,他们没有影响的表面。它也更容易看到进一步从赤道,在科里奥利是零。[地球NullSchool漂亮的地图显示全球风][2],可以帮助你看到科里奥利效应(颜色是平均海平面压力):[!(在这里输入图像描述)[3]][3][![在这里输入图像描述][4]][4]在你的风向玫瑰图,可以近似科里奥利效应,将其横盘整理。科里奥利效应可以经验的气象学领域之外的(我不是指的是一种“马桶式”的谬误)。 Long before Gaspard Coriolis lived, [it was recognized that cannonballs were deflected by the Earth's rotation][5], though it did have some inaccuracies. One question I anticipate that I anticipate you might have is "How far would a something (wind or a cannonball) have to travel before the Coriolis effect would have a meaningful effect?" Projectiles are, in my opinion, the easiest way to show that this is a real thing. This "acceleration" is approximately $a_{cor}=fv$ where $f$ is the [latitudinally dependent Coriolis parameter][6]. and $v$ is the speed. [You can see how this impacts ballistics through this Wikipedia entry][7]. Given various assumptions, like a constant latitude, ignoring drag, and ignoring the loss of forward momentum from the perpendicular acceleration, the distance one would be off after firing a projectile would be: $$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u_0^{-1} $$, where $\Omega$ is the angular velocity of the earth, $\phi$ is the latitude, and $u$ is the speed, and $\Delta x$ is the distance to the target. So at say 45 degrees north and going 5 meters per second over a 100 meter distance, you would be off by 0.2 meters, [but if you are the Paris gun with a horizontal velocity of 1009.7 m/s, trying to bombard the city 120 km away at a latitude of 49 degrees north, you'll be 1.3 km off if you don't account for the Coriolis effect][8]. And because I know I will get someone who wants me to show my work for the previous equation here it is: $$\frac{d\vec{v}}{dt}=-f\hat{k}\times\vec{v}$$. If we consider the initial conditions $\vec{v}(t=0)=u(t=0)\hat{i}+v(t=0)\hat{j}=u_0\hat{i}+0\hat{j}$, then we have the system $$\frac{du}{dt}=fv$$ $$\frac{dv}{dt}=-fu$$. Since I made the assumption that momentum along the trajectory is not altered by the Coriolis force, the system reduces to $$\frac{du}{dt}=0$$ $$\frac{dv}{dt}=-fu$$. If we think that wind is the changing coordinates ($u=\frac{dx}{dt}$, $v=\frac{dy}{dt}$), then the system can be represented as $$\frac{d^2x}{dt^2}=0$$ $$\frac{d^2y}{dt^2}=-fu=-fu_0$$. Therefore $$\Delta y= fu_0t^2$$. And since the projectile must travel distance $\Delta x$ at velocity $u_0$, then $$t=\frac{\Delta x}{u_0}$$ which makes $$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u_0^{-1} $$ [1]: http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/geos.rxml [2]: https://earth.nullschool.net/#current/wind/isobaric/500hPa/orthographic=-58.35,45.51,734 [3]: https://i.stack.imgur.com/dOVxg.jpg [4]: https://i.stack.imgur.com/ENs84.jpg [5]: https://en.wikipedia.org/wiki/Giovanni_Battista_Riccioli#The_%22coriolis_effect%22_argument [6]: https://en.wikipedia.org/wiki/Coriolis_frequency [7]: https://en.wikipedia.org/wiki/Coriolis_force#Ballistic_trajectories [8]: http://academickids.com/encyclopedia/index.php/Paris_Gun

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