{"id":424,"date":"2007-09-01T09:23:41","date_gmt":"2007-08-31T23:23:41","guid":{"rendered":"http:\/\/www.somethinkodd.com\/oddthinking\/2007\/09\/01\/the-sum-of-one-million-frickin-laser-beams\/"},"modified":"2012-09-25T21:43:55","modified_gmt":"2012-09-25T11:43:55","slug":"the-sum-of-one-million-frickin-laser-beams","status":"publish","type":"post","link":"https:\/\/www.somethinkodd.com\/oddthinking\/2007\/09\/01\/the-sum-of-one-million-frickin-laser-beams\/","title":{"rendered":"The Sum of… One Million Frickin’ Laser Beams"},"content":{"rendered":"

Something has been bugging me since I posted the red laser pointers plus moon = red eclipse<\/a> idea.<\/p>\n

It’s not that it is morally wrong<\/a>.<\/p>\n

It’s not that it is astronomically wrong.<\/p>\n

Well, actually, it is. I wrongly suggested that this experiment should occur during a full moon. That is precisely the wrong<\/em> time to do it. Instead, you should do it during a new moon, when you are not competing with the Sun’s light.<\/div>\n

What’s been bugging me is the need for a back-of-a-really-big-envelope calculation…. just how many laser pointers would it take<\/strong> to fake a red eclipse?<\/p>\n

Let me fill that gap now.<\/p>\n

Part 1: What’s the cross-section of the moon?<\/h4>\n

Approximation: The moon is a perfect sphere.<\/p>\n

The cross-section of the moon is a circle of area π×R2<\/sup>, where R=radius of the moon.<\/p>\n

Fact: The moon has a diameter of 3474 kilometres<\/a>, which is 3.47 × 106<\/sup> metres.<\/p>\n

Thus, the cross section of the moon is π × ((3.47 × 106<\/sup>)\/2)2<\/sup>) = 9.4 × 1012<\/sup> m2<\/sup>.<\/p>\n

Part 2: How much sunlight falls per square metre on the moon?<\/h4>\n

Approximation: The moon is equidistant from the Sun as the Earth is, and thus the Sun’s radiation on the top of the Earth’s atmosphere is the same as the sun’s radiation on the top of the moon’s atmosphere.<\/p>\n

Approximation: The moon has no atmosphere.<\/p>\n

Fact: The Solar Constant<\/a> (total energy in all solar electromagnetic radiation at the top of the Earth’s atmosphere) averages 1366 W.m-2<\/sup>.<\/p>\n

Part 3: How much sunlight falls on the moon?<\/h4>\n

Total energy due to sunlight = Energy per square metre × cross-section area
\n= 1366 W.m-2<\/sup> × 9.4 × 1012<\/sup> m2<\/sup>
\n= 1.3 × 1016<\/sup> watts.<\/p>\n

We don’t need all of this energy, though, to light up the moon. The red eclipse wasn’t as bright as a full moon.<\/p>\n

Part 4: How much of that energy fell on the moon during the red eclipse?<\/p>\n

To start with, the red eclipse only had the red wavelengths. <\/p>\n

Approximation: The Earth’s atmosphere absorbed all of the visible spectrum except the red range (between about 625 nm and 740 nm), and it attenuated the red range by 50%. <\/p>\n

Warning: This is a very coarse<\/em> approximation. It is based only on my perception of the relative brightness of the moon during the eclipse. Can anyone find a better approximation? I would prefer approximations with a tendency to underestimate (rather than over-estimate) the total number of lasers, so I can’t be accused of exaggerating.<\/div>\n

How much of the sunlight’s energy is in the red wavelengths? Well, this is described in the following chart (from Robert A. Rohde, via Wikipedia<\/a>).<\/p>\n

Solar Spectrum diagram (courtesy of Wikipedia)<\/a><\/p>\n

So the red light is the area under the curve in the red part of the spectrum.<\/p>\n

Oh dear, how do you find the area under a curve?<\/p>\n

There are two ways: <\/p>\n

One is to find the integral of the function. That only requires two items that I don’t have: a description of the function, and any recollection of how to integrate<\/a>.<\/p>\n

The other is the old-fashioned way – draw the curve on a grid and count the squares underneath it.<\/p>\n

Here we go then:<\/p>\n

First, we colour the graph into red and non-red areas. <\/p>\n

Solar Spectrum (approximation) divided into red and non-red energy<\/a><\/p>\n

Approximation: The red area of the graph reasonably represents the wavelengths associated with the red hue of the eclipse.<\/p>\n

Then we run a simple Python based on the Python Imaging Library (PIL) script:<\/p>\n

Approximation: That the pixels in the mock-up curve accurately represent the area under the real curve.<\/p>\n

>> import Image<\/strong>
\n>> Image.open(\"red_sunlight_graph.png\").getcolors()<\/strong>
\n[(522179, (255, 255, 255)), (6516, (238, 3, 17)), (57505, (3, 64, 238))]<\/code><\/p>\n

This means the total energy arriving at the moon is (6516+57505) pixel units, while the total red energy is 6516 pixel units, or just over 10%. Remembering to attenuate by 50%, we get only 5% of the solar energy making it past the Earth’s atmosphere.<\/p>\n

Total red light energy we need to transmit to the moon: 1.3 × 1016<\/sup> watts × 5% = 6.6 × 1014<\/sup><\/p>\n

Part 5: How much power does a laser pointer emit?<\/h4>\n

According to the UK Health Protection Authority’s advice on laser pointers<\/a>, laser pointers should have less than 1 milliwatt of power. Wikipedia claims a five milliwatt maximum in the USA, but a citation is needed.<\/p>\n

Approximation: Laser pointers are 10-3<\/sup> watts.<\/p>\n

Part 6: So, how many laser pointers does it take to change a moon?<\/h4>\n

First, we need a few more rather coarse approximations that under-estimate the real number of lasers required :<\/p>\n

Approximation: The Earth’s atmosphere does not attenuate red laser light – there is no Raleigh scattering<\/a> or dust in the air.<\/p>\n

Approximation: The laser beam is completely parallel, and does not diverge, even over the average distance of 384,000 kilometres between the Earth and moon.<\/p>\n

Total power required divided by power per laser = 6.6 × 1014<\/sup> watts \/ 1 × 10-3<\/sup> watts = 6.6 × 1017<\/sup>.<\/p>\n

We have an answer! Two-thirds of a quintillion lasers: 660,000,000,000,000,000.<\/p>\n

Everybody on Earth (except the astronomers, of course) will need to carry 100 million laser pointers each.<\/p>\n

That’s almost as many lasers as kernels of wheat that have ever produced<\/a>.<\/p>\n

I guess what I am saying is, we’ve got a lot of work ahead of us in the next few weeks, if we want to pull off this little prank – we’d better get started soon!<\/p>\n","protected":false},"excerpt":{"rendered":"

Just how many laser pointers would it take to fake a red eclipse?<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_s2mail":"yes","footnotes":""},"categories":[31,27],"tags":[61,59,60,55,56,54],"class_list":["post-424","post","type-post","status-publish","format-standard","hentry","category-geek","category-thoughts-from-the-shower","tag-maths","tag-moon","tag-physics","tag-puzzle","tag-real-life-puzzle","tag-solution"],"_links":{"self":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts\/424","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/comments?post=424"}],"version-history":[{"count":1,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts\/424\/revisions"}],"predecessor-version":[{"id":1684,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/posts\/424\/revisions\/1684"}],"wp:attachment":[{"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/media?parent=424"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/categories?post=424"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.somethinkodd.com\/oddthinking\/wp-json\/wp\/v2\/tags?post=424"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}