Those fellows at Rolling Rock, not being content with merely brewing
skunky beer, have created a clearly hoaxish ad campaign to make people
believe that they're going to shine a laser on the moon with their logo.

They've apparently hired the great engineers of the fabled
spaceship that visited the sun (but did so safely because they did it at
night,) in that they plan on shining their message on the full moon.
Yes, the full moon that's illuminated by the full force of the sun,
which shines every bit as bright on the moon as it does on earth. (A
bit brighter on the moon, actually.) I understand that Rolling Rock
treats their employees to drive-in movies during the day, too.

Just for argument's sake, let's see how much power they'd have to
expend to make this happen. The following calculations are Frink
notation ( http://futureboy.us/frinkdocs/ ):

The moon (and the earth) both receive about 1372 watts per square
meter of energy from the sun. (The derivation of that is in the Frink
documentation here: http://futureboy.us/frinkdocs/#Superman )

Of course, it's hard to project an image in direct sunlight. Go try
it. The sun is bloody bright. Let's give them the benefit of the doubt
and pretend that we could actually detect their lasers on the moon if
they were only 10% as bright as the light from the sun, though. That's
generous. The calculations are:

sunpower = 1372 W/m^2
logopower = sunpower * 10 percent

Now, the moon presents a face to us that's:

moonarea = pi moonradius^2

which gives about 3.66 million square miles. This is very close to
the area of the contiguous 48 United States. And they're saying they'll
project an image that lights that whole thing up, which is visible ...
in direct sunlight? Think about the sun beating down on the entire
country and think about some projector from the sky trying to project an
image on top of that. Across the whole country. Is this starting to
sound feasible?

Let's say that they're only going to light up 5% of the surface of
the moon with their logo at any time. Thus, the lit area is only:

litarea = moonarea * 5 percent

So, the total amount of power their laser will have to deliver is:

laserpower = litarea * logopower

So, the amount of power they'd have to deliver is 65 TRILLION watts.

(This is equal to the power of detonating one Hiroshima-sized nuclear
weapon every 0.8 seconds constantly for as long as you're going to run
the laser. If your laser's not perfectly efficient, it's going to dump
much more than that amount of power as heat. You don't want to be
standing around it.)

65 trillion watts is about 80 times as much electrical power as every
electrical power plant in the U.S. can generate, put together. I'm not
sure where they're going to plug in their laser. Maybe they have
awesome batteries. This may get expensive.

How expensive would it be? Well, Pennsylvania electricity rates in
December 2007 were about 10.6 cents/kWh.

electricity = 10.6 cents/kWh

So, if they kept the display on for just 5 minutes, it would cost:

electricity * laserpower * 5 minutes

Which would give a cost of 575 million dollars. Well, we'd have to
expect that prices would have to go up, considering they'd have to build
79 times more electric plants than the U.S. already *has*, just to power
that laser.

Oh, and consider that Rolling Rock was bought by Anheuser-Busch for
only 82 million dollars. So, for the value of the whole worth of the
company, they could only run the display for:

82 million dollars / (laserpower electricity) -> sec

Which gives 42.7 seconds.

So, in a nutshell, don't bother looking up.

Perhaps if they used some of that wonderful optical technology to
keep ultraviolet light out of their beer bottles (hint: try brown glass,
geniuses) then it wouldn't taste like wet dead skunk we found under our
porch once. Actually, the skunk was less skunky. And he'd been dead a
while.

--
Alan Eliasen | "Furious activity is no substitute
eliasen@... | for understanding."
http://futureboy.us/ | --H.H. Williams