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A Total Solar Eclipse is Coming to the United States!

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(You're going to need them!)
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"...And we'll see YOU... in the shadow!!"

We are out of the office watching the eclipse! No emails will be able to be answered until we return on August 22 wishes everyone CLEAR SKIES on Monday!

If you don't have SAFE eclipse glasses, your students don't have to miss out!
You can still view the partial phases using Pinhole Projection!

Check out the safe viewing techniques from the American Astronomical Society!

And also see our viewing instructions!

Did you order from us? Are you worried about the big recall?
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Also, please read the blog post we wrote about the safety of glasses ordered from!


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2017 Eclipse Blog / FAQ

How fast is the shadow moving across the US during the eclipse?

A great question! The Moon does indeed “carry” its shadow along with it, as it moves in its orbit around the Earth.


And whenever that shadow happens to scrape across the Earth, we have a total eclipse!  (Of course, you have to be in the Path of Totality to see the total eclipse, and you have to use eye protection whenever the eclipse is not total – regardless of whether you’re in the path or not!)


Because of the geometry of the Earth’s shape, the shadow will travel faster across its surface and the ends of the eclipse path, and slowest right in the middle.  By using one of the best eclipse calculators out there (thanks to Xavier Jubier!), we can see that the Moon’s shadow (also called the “Umbra”) is moving:

  • 2410mph in Western Oregon
  • 1747mph in central Nebraska
  • 1462mph in Western Kentucky
  • 1502mph near Charleston SC

Make sure you are in the path on eclipse day!  And we’ll see you… in the shadow!

22 Responses to “How fast is the shadow moving across the US during the eclipse?”

  1. Thank you for the info on the speed of the moon’s shadow during the eclipse. I live in southeast Missouri, and will probably watch the eclipse from Farmington.

  2. Ron Hadfield says:

    I will be in Driggs, ID on August 21st and I believe it is in the centerline. Is there a way for me to determine the speed of the shadow in Driggs?

  3. Jim Boyle says:

    What day and time will the eclipse appear over the various Cities, ie Casper Wy?

  4. Randy P. says:

    Thank you for the cool info on the eclipse!

  5. Andrea L. says:

    So here is a great math problem: If I am in a vehicle going 65 mph under the solar eclipse, how long will I be in the dark? As a point of reference… Driving south from Columbia, SC to Charleston, SC on US I 26, on August 21st… Barring any traffic impasses.

  6. Tom says:

    This page says the umbra moves at 2955 mph in Western Oregon. I don’t think that’s right. Xavier’s eclipse calculator for Lincoln City, OR (on the coast) says 2410 mph. Should this be corrected or am I missing something? I love your website! Thanks.

  7. Harry J MacNeil says:

    I’ll be in Beatrice Nebraska.

  8. Tom Gordon says:

    I will be in Beatrice NE. Taking the day off work to travel an hour south from Omaha so I can be in the path.

  9. Ben says:

    Hi I’ll be in garden of the gods illinois.the cliffs there are quite a bit higher elevation than the adjacent valley. Is the shadow of the moon something tjat could be visibly discerned approaching me. Thinking that would he really neat video footage if so!

    • Admin says:

      If you have a clear view to the western horizon, the sight is spectacular. It’s like a big black curtain rushing toward you … from the West! 😉

  10. Gene Bahnsmath, not calculus. says:

    No worry, your Learjet can fly in the shadow. It is simple math, not calculus!

    Google for GPS coordinates:
    Madras, OR =44.6N, 121.1W Time Eclipse begins = 10:19 AM
    Paducah, KY=39.0N, 84.7W Time Eclipse begins = 1:22 PM

    Length of an arc = radius x angle in radians (1.0 radian = 57.3 deg approx)
    Radius of Earth = 3959 (somewhere). I used 3960.

    N-S displacement = 3960 x (44.6-39)/57.3 = 390 mi.
    Radius for E-W displacement = distance to polar axis (take an average elevation)
    So Avg elevation = (44.6+37.1)/2=40.85 deg N
    EW radius = 3960 x cosine(40.85 deg) = 3960 x 0.756 = 2994 mi.
    E-W displacement= 2994 x (121.1-84.7)/57.3 = 1902 mi.
    Approx Total Displacement = sqrt( 390^2 + 1902^2) = 1942 mi.

    Speed of Shadow = 1942/(13h 22m – 10h 19m) = 1942/(3+3/60)= 637 mph
    Speed of sound at 20,000 feet = 707 mph (Google again)
    637/707 = 0.9 Mach Number
    So the Learjet might fall behind. Better take an F-16.

    • Gene Bahnsmath, not calculus. says:

      Correction: Paducah KY is at 37.1 n, 88.6 W (I confused it with Covington)

      So, not bothering to redo the math right now, I believe it would correct to about 680 mph. Supersonic now, so definitely need the F-16.

      Gene B

      • Gene Bahnsmath, not calculus. says:

        Correction 2: I was too hasty. Forgot that 88 deg W Lat ia WEST of 84 deg W Lat. So it would be 45 mph less, not more.

        So rough estimate is about 592 mph, or 0.83 Mach at 20,000 ft, The Learjet might come close.

        Gene B

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