Quote of the Day
You must accept 1 of 2 basic premises: Either we are alone in the universe or we are not alone. Either way, the implications are staggering!
— Wernher von Braun
Figure 1: Photograph of 2017 Eclipse. (Source)
Yesterday, a reader asked me how to compute the totality path width for the eclipse that will cross the US on 21 Aug 2017. I wrote a post on how to perform this calculation years ago. NASA has published a path width value of 114.7 km. This width will actually vary a bit as the shadow moves across the Earth because the distance change slightly between all the bodies involved. Also, the Earth and Moon are not perfectly round, which I assume. NASA has very detailed models that even include the Moon's shape variations due to mountains and valleys.
In today's post, I will show how to compute a good approximation to NASA's result. I provide Figures 2 and 3 to show how the various parameters are defined. For the details on the analysis, please see my original post. Figure 1 shows my model for the Sun-Earth-Moon system during the eclipse.
Figure 2: Solar Eclipse Geometry
Figure 3 shows the details on my approximation for the umbra width.
Figure 3: Illustration of Umbra on the Earth.
I used Equation 1 to compute the totality path width. I grabbed the data specific to 21-Aug-2017 from this web site. The rest of the information I obtained from Google searches.
| Eq. 1 |  |
where
- sumbra is the totality path width on the Earth
- dsun is the distance between the Earth and Sun.
- dmoon is the distance between the Earth and Moon.
- rmoon is the radius of the Moon.
- rsun is the radius of the Sun.
- α is the vertex angle of the moon's shadow cone (see Figure 1). We compute α using
- dumbra is the length of the moon's shadow cone (see Figure 1). We compute dumbra using
Figure 4 shows my calculations. I obtained 116 km, which compares favorably with NASA's 114.7 km.
Figure 4: My Totality Path Width Calculation.
I enjoy your posts because they investigate the "deeper story", so hopefully I can offer this little nitpick without risk of offense. When I read your Mark Twain quote for this article, something didn't quite ring true, so I checked into it and came across this gem of applying math to questions of attribution. It's behind a paywall, so the link is to the Google cache version.
https://webcache.googleusercontent.com/search?q=cache:eIbRR6jiRDMJ:https://www.bloomberg.com/view/articles/2015-05-18/a-guide-to-fake-quotes-on-the-internet+&cd=14&hl=en&ct=clnk&gl=us
Regarding the eclipse, any thoughts on how to compute how much light energy it takes to damage the retina? I've always wondered if simple lack of pain in viewing the sun on, say, a strongly overcast day is a reliable indicator of safety or not.
Thank you for the link on the quote. I have changed the Quote of the Day for this post.
I have a physicist on my team who does all of our eye safety calculations for our lasers. There are standards for this sort of thing. I will talk to him tomorrow – he has gone home for the day. I would not depend on simple pain.
mark
I talked to our safety guy, and we went through the calculations. However, I really am not comfortable providing the details of the calculations because they are too easy to screw up. There is an app for viewing the eclipse that you may want to download (it cost $2). This does all the calculations for you. It also provides you LOTS of other stuff.
This article goes into the history of eye damage after staring at the sun. The history is not good and the damage can occur without any pain. You can still buy glasses here.
A coworker of mine has a great web page with all sorts of info on the eclipse – check it out.
mark
I had suspected pain was not a reliable indicator. Great information, thanks. That app looks terrific.
Being near the path of the totality, I bought glasses for my family that are supposedly ISO certified, but they are made in China and did not come with a certificate. They are nearly mirror opaque and seem to work correctly, but it would be interesting if there were a simple way to test them using a photodiode or something.
? where did you get the formula for the 0.2635 deg calculation? I just did a scaled test and the shadow of the moon is for all intents and purposes the same size of the moon (approx. 2,159 miles). As far as I can determine your calculation indicates the amount by which the radius is reduced, not the size of the diameter of the totality. at 93 million miles light from the sun is for all intents and purposes parallel, and given the relative sizes of the moon and sun, since the moon just happens at that distance to completely cover the sun, it is impossible for the moons shadow to effectively reduced by 2,150/70 = 30 times reduced or for the shadow to be 0.3% of its actual size.
exllent article to explain the phomena of eclipse path,