Little background first:
I take part in preparing environmental reports for wind turbines/farms and now I'm exploring shadow flickering effect of moving blades.

From some sources (in Polish) I got to know that the shadow casted by turbine blades is solid, black only to some distance and above this distance the shadow is blurred. This distance was described as approx. $108\ * \ width$ of blade. As an explanation of this arguments the umbra-penumbra-antumbra diagram was presented.

I wanted to find where from this $108$ value comes from.
I found that for spherical bodies, when the occulting object is smaller than the star, the length (L) of the umbra's cone-shaped shadow is given by:
where $R_s$ is the radius of the star, $R_o$ is the occulting object's radius, and $r$ is the distance from the star to the occulting object.
I've also found that this equation comes from simple triangles similarity in Sun-Earth arrangement.

When I use the radius of the Sun and the distance from the sun to the Earth, the equation actually simplifies ($R_s >> R_o=3.5m$) to: $L=215R_o \approxeq108D_o$ ($D_o$ - object diameter)

My question is:
Can this equation be used to approximate the length of solid part (umbra part) of a shadow of objects on the Earth?


You seem to have answered your own question. The only requirement seems to be that the light source behave as you have assumed in your derivation. When there is a lot of scattering, as in a cloudy day, there are many secondary sources that will wash out this arrangement, and obviously, somewhere in between there is a point in which the approximation breaks down.

  • $\begingroup$ The behaviour of light is actually of my concern. But not in local scale (scattering). Rather between Sun and object. That's why I ask if I can use the same relations as between Sun and Earth to Sun and object. $\endgroup$ – Marcin Nov 9 '13 at 15:32
  • $\begingroup$ Yes, I think you can use the same relationship. $\endgroup$ – lionelbrits Nov 10 '13 at 11:12

Under the assumption that reflection or scattering from other sources are neglected the maximum length of the shadow is achieved when your object covers the full solid angle of the sun which is $9.35\cdot10^{−3}$ rad. Therefore the tangent of this angle gives you the length of the triangle to $L = r_{\text {object}}\cdot 107$.

  • $\begingroup$ I made an edit and fixed simplified equation. There's L=2*108*Ro. Sorry. Btw, can you explain the term "solid angle of the sun"? $\endgroup$ – Marcin Nov 9 '13 at 14:35
  • $\begingroup$ Imagine the sky as an imaginary sphere of radius $R_{sky}$. It has area $4\pi R_{sky}^2$. Solid angle is just this area divided by $R^2$, just like angle on a circle is the distance along the perimeter divided by radius. $\endgroup$ – lionelbrits Nov 9 '13 at 15:14

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