Quote of the Day

You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals.

— Booker T. Washington

For more on the up-coming eclipse, see this post.

## Introduction

A couple of weeks ago, I was watching the Wonders of the Solar System with Brian Cox on the Science channel.In this episode, he was talking about the Moon and solar eclipses. He made a comment that the region of totality (i.e. complete darkness) during a solar eclipse is only a few hundred kilometers across. To illustrate this point, I found a great picture (Figure 1) taken from the Mir space station of the moon's shadow on the Earth.

I have never been through a total eclipse, but I must admit that I have always found them interesting. Maybe I can talk my lovely bride into taking a solar eclipse cruise? She still has not responded to my request for a vacation touring World War II Pacific battefields. Aren't I romantic? Anyway, I started to wonder whether I could verify the statement that Brian Cox made during the program.

I began my little exercise by looking for an exact number for the size of the region of totality and some details on the geometry of the situation. It did not take long. In the book, Historical Eclipses and Earth's Rotation by Stephenson, I found the following statement.

Under a vertical sun the umbra can never exceed about 270 km in diameter. However, at lower solar altitudes the elongated shadow of the Moon may be much wider than this, occasionally exceeding 500 km.

This quote is completely consistent with Brian Cox's statement and gives me a bit of information on the geometry of the situation. Let's see if a little geometry can verify this statement.

## First, A Few Definitions

Let's work with the definitions from dictionary.com.

- Totality
- The quality or state of being total; as, the totality of an eclipse.For a solar eclipse, totality is the state of the sun being completely obscured to an observer by the moon. Totality depends on the position of the observer.
- Umbra
- A region of complete shadow resulting from the total obstruction of light by an opaque object, especially the shadow cast by the moon onto the earth during a solar eclipse.

## Analysis

### Umbra Geometry

Figure 2 illustrates the basic solar eclipse geometry.

The situation in Figure 2 produces the smallest region of totality on the earth. A little high-school trigonometry gives me the following equation.

Next, I will use this angle to compute the length of the umbra (*d _{Umbra}*).

### Umbra Spot Size on the Earth

Figure 3 illustrates how I estimated the umbra spot size on the Earth. I am ignoring the curvature of the Earth because the spot is so small and the Earth is so big.

Again, a little bit of trigonometry gives me the following equation.

is roughly the diameter of the umbra spot on the Earth. This completes my derivation.

### Solar Eclipse Data

Symbol |
Description |
Value |
Units |

r_{Sun} |
Radius of the Sun | 696,000 | km |

r_{Moon} |
Radius of the Moon | 1738 | km |

d_{Smax} |
Maximum distance from the Sun to the Earth | 152,100,000 | km |

d_{Mmax} |
Maximum distance from the Moon to the Earth | 356,400 | km |

r_{Earth} |
Radius of the Earth | 6378 | km |

### Results

Simply substitute the eclipse data into the equations and you get 273 km for the maximum diameter region of totality when the sun is vertical with respect to the observer. As far as I am concerned, this analysis confirms both Cox's and Stephenson's statements. This was a nice little problem.

Here is a screen capture of my calculations.

Here is a copy of these calculations in Excel.

You don't have to wait that long for a total solar eclipse. There is one going across the US in 2017. Monday, 21 Aug 2017 to be exact. You can find out where it will be at this website, http://www.eclipse2017.org.

I saw a partial one around 1999 in Southern California. I could have gone to Mexico to seen the total one. It is eerie to have the sunlight dim in the middle of the day with no clouds. I could see why early people were freaked out by them.

I had no idea there would be one that close!

Somehow I think I would have a harder time convincing my wife to take a vacation to Nebraska to see an eclipse than I would to get her to take a cruise to see an eclipse. Maybe I could convince her to do both!

I am not getting 273, help? -_-

I had some typos in my table. I am so sorry. This is the reason that I now use screen shots from my Mathcad worksheets. I have corrected my table of parameters and I now include my calculations as the bottom of the post.

Again, sorry for the trouble.

Mathscinotes

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Nice l like it

This is a very nicely presentations of the calculation. Thanks for enligthenening me 🙂

any idea how big the penumbra is?

...i make it about 6700 km in diameter. Seems pretty big, but I guess it would really fade off to nothing in that space.

how do you calculate the area of the umbra.

Very useful to Research scholar , But not full & Complete calculation is given

Wait, wouldn't the combined radius over the combined diameter be 1/2?

I am not getting the answer???

Could you show me some intermediate results? I would love to help.

mark

I have now included an Excel workbook with the example calculations. Hopefully that will help.

mark

I found a possible error in your analysis... I believe that it should be:

tan θ = opposite / adjacent --- not sin θ

asin/atan are very close for small angles but asin is slightly too large.

I also assumed distances are from Earth Surface to make the last step easier, so I get:

m = moon radius

M = moon distance (from Earth's surface)

s = sun radius

S = sun distance (from Earth's surface)

θ = arctan((s-m)/(S-M)) -- the angle of the shadow

adjacent = opposite / tan(θ) -- trigonometric identity, to find shadow length

tan(arctan()) cancels out, and opposite is moon radius, giving:

adjacent = m / ((s-m)/(S-M))

Umbral Diameter ~ 2 * (adjacent - M) * tan(θ)

substitute

Umbral Diameter ~ 2 * ((m / ((s-m)/(S-M))) - M) * ((s-m)/(S-M))

simplify

Umbral Diameter ~ 2 * (m S - M s) / (S - M)

You can try this out on wolfram|alpha:

http://www.wolframalpha.com/input/?i=2*(M+s+-+m+S)%2F(M+-+S),+where+m%3D1737.5,M%3D353622,s%3D695700,S%3D149600000

I've used slightly different values than you but you can change them easily.

As an update here I think I realized why you used arcsin() -- it's because you correctly calculated the apparent size from the tangent to the moon sphere -- not the absolute size. I had assumed absolute size.

Hi Dark Star,

Thanks for writing me. I was going to look at your initial response this week. I appreciate you helping me keep the site accurate.

mark

Great problem. I came up with the same solution before I found your analysis.

I was confused as to why NASA states the diameter of the umbra in the august 2017 eclipse is only 70 miles in diameter.

Wouldn't an umbra only 70 miles in diameter make the moon much farther away?

Just putting it out there.

NASA also says the moon is travelling at twice the angular rotation as the earth and that's why this upcoming eclipse will track from WEST to EAST in the sky.

I have to laugh at NASA. They think we are stupid.

I know the moon rotates in the same direction as earth but the earth rotates much faster than the moon.

So from a perspective on earth the moon rises in the east and sets in the west.

Please someone correct if I'm wrong but wouldn't simple logic tell you that the umbra cast by the moon would travel EAST to WEST not WEST to EAST.

Am I stupid? Or is NASA and the MSM trying to hide some other event?

Like some other celestial body that everyone should be discussing but nobody is.

I hope to see something during the eclipse but I bet the chemtrails and fabricated rain will hide everything.

skiier

OK I am stupid. Stupid for not looking at the West to East thing from a different perspective.

If one considers for a very short duration of the eclipse that the moon is rotating the sun and not the earth one can easily understand the movement of the umbra on earth.

Because the earth and moon are 150 million km away from the sun we can make the assumption that for a short duration ( like derivatives as one approaches zero) that the moon is a planet in retrograde orbit to the earth. So then the moon produces an eclipse that makes the umbra move from West to East on the surface of the earth.

Using this model one has to only consider the angular velocity of the earth in relation to its point of rotation and the angular velocity of the moon in relation to it's point of rotation. The point of rotation of the moon being the sun. The point of rotation of the earth being the center of the earth itself.

Its becomes very easy then to calculate the velocity of the umbra in comparison the velocity of the earth at the point the umbra hits the earth.

And the velocity of the umbra is much much faster than the velocity of the point on the surface of the earth where the umbra hits.

Basically it s like a flashlight sweeping by a point very quickly but the point is moving much slower in the same direction as the light.

So viewed on earth the umbra will sweep West To East.

It would be very easy to make a visual representation in some form of intuitive animation to explain this phenomenom. I just don't know how.

A lot of people are confused on this point.

j r,

I had the same "logical" belief you did, about west to east being wrong, until I found the same explanation you also discovered. It seems to make sense, but what I don't get is this: Based on the "the moon is rotating the sun and not the earth" scenario... all eclipse shadows would move "generally" from west to east. If that proves out in history, then what's the big deal about this one all of a sudden?

X-man

Thanks for deleting my comments. I wouldn't want to have anyone question why the umbra in the upcoming eclipse is only 112km in diameter when it should be at least 273km wide according to your calculations.

Any body with any intuitive math skills would look at your model and come to the conclusion that a moon that creates a 112km wide umbra on the earth must be farther than 384000 km away. Simple geometry.

If the sun and the earth are fixed in position then the moon must be closer to the sun than we think for some reason and conversely farther away from the earth than normal. I don't think you will provide an analysis for this 70 mile wide umbra since all your interested is in showing off your high school math abilities. Who wants to show the truth? It doesn't suit your model.

Thanks again loser.

There is much to unpack here:

ve

Have a nice day.

mark

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OK,

I've read some of the other comments and realize I'm in way over my head. However, I thought the light rays from the sun hit the earth, for the most part, parallel to each other and at a 90 degree angle to the earth... And that shadows must always be the same size or larger than the subject. Will there be different size shadows of varying density/darkness based on the rays from all parts of the sun hitting the moon from different angles...or what? A simple, talk slow, man-on-the-street answer would be appreciated if you can get down on ground level with me.

Respectfully,

X-man

Sorry for being a dick.

I thought u deleted my comments.

Sorry for being a dick.

I am just rude sometimes.

Great info here. I'm putting everyone on this link.

wow great info Sir, geometry calculations seems to be simple but until someone like you enlighten it i will never figure out. Keep up the good work. Thanks for this.

Also if possible please provide info. and width about Penumbra width over earth or how much curvature of earth will have penumbra . There is video on youtube by Prof. Walter Lewin giving problem regarding same, but somehow i didnt get geometry calculations, but this info. was too great. Thank you.

* this could be very easy compare to your formula..... use similar triangle method and u will find that distance of umbra by moon is (distance between the sun and the moon)/399

* tan(x/2)= radius of moon / lenth of umbra

* distance of umbra - distance of moon and earth + earth radius will give u radius of the umbral cone from the surface of the earth..

and the last steps from the angle x and radius at the earth of the cone... find lenth of the arch...