Extreme Geekery: Giant Sun Birthday Cake

sun_birthday_cakeHappy Birthday to me! Today I turn 41. Luckily, I’ll probably have a single candle on my cake (or two candles: a four shaped one and one shaped like the number one). I won’t have to deal with blowing out forty-one candles. Still, I began to wonder exactly how bright you could get with candles. Let’s suppose that the Sun was a giant birthday cake. How many candles would it need to keep outputting as much light as it currently does. (Since this is a thought experiment, we’ll ignore such mundane details as "How do the candles burn without an atmosphere" and "How do the candles not melt down with time".)

The Sun outputs about 6.84 x 1027 lumens. Written out, this is 6,840,000,000,000,000,000,000,000,000. That’s a LOT of lumens.  The best reference I could find for a candle’s output was 12.56 lumens. Of course, the birthday candles I use tend to be smaller and so likely generate less lumans than other candles. Let’s round down to 10 lumens for a birthday candle. (Plus, this makes the math easier.)

If each candle is providing 10 lumans of light, we would need 684,000,000,000,000,000,000,000,000 candles. That’s 684 trillion trillion candles!  I wonder if the warehouse stores carry ultra-mega packs of candles.

So we’ve got our candles and are ready to light them… Wait, first we need to put them in the cake.  So we make a HUGE spherical cake and place the candles all around it. How big of a cake do we need? Well, if each candle has about 1 square centimeter of space around it (we’re packing them in), the cake would need a surface area of 684,000,000,000,000,000,000,000,000 square centimeters or 68,400,000,000,000,000 square kilometers. (That’s over 26,000,000,000,000,000 square miles.)  The Sun itself has a surface area of 6.09 x 1012 square km.  That’s a big cake!

How much bigger, you ask? Stand back, I’m going to use Math!

Now the surface area of a sphere can be calculated by pi*d2.  Let’s say that the cake’s diameter is d1 and the Sun’s is d2. This gives us:

68,400,000,000,000,000 = pi*d12

6,090,000,000,000 = pi*d22

Obviously, the cake is some number, N, times bigger than the Sun so we can say:

d1 = N * d2

Plugging this into the first equation we get:

68,400,000,000,000,000 = pi*(N*d2)2


68,400,000,000,000,000 = pi*N2*d22

Now the second equation can also be written as:

d2 = square root(6,090,000,000,000/pi)

Plugging this into our calculations, we get:

68,400,000,000,000,000 = pi*N2*(square root(6,090,000,000,000/pi))2


68,400,000,000,000,000 = pi*N2*6,090,000,000,000/pi

The Pi’s cancel out and we can divide each side by 6,090,000,000,000 to get:

N2 = 1,123.15

This means that N is about 33.5.

Our birthday cake would need to be almost 34 times the size of the Sun just to be as bright as it is.

I think I’m going to need more frosting!

NOTE: The Sun Birthday Cake image above was made by combining Decorative Sun by ivak and Chocolate Birthday Cake(brown) by version2. Both images are available from OpenClipArt.org.

It’s Palindrome Week!

I’ve been a math geek for as long as I can remember.  When I was in school, I would devise complex mathematical equations and then solve them.  That was my idea of fun.  I couldn’t imagine why anyone would hate math.

You might think that, as I got older, my math geekery died out.  It didn’t.  If anything, it grew as it found more areas of my life to work into.  I’ll read an article, blog post, or comment, and suddenly be overcome with the urge to solve a related problem.  That’s how many of my Extreme Geekery posts begin.  (I have another one on tap.)

My math geekery isn’t just limited to solving equations, though.  I also love certain kinds of numbers.  If I have a blog post with 495 words, I feel compelled to bring that total up to 500.  A few years back, on December 12th, 2012, I loved seeing the date turn to 12/12/12.  Few numbers fill me with delight, though, as palindromes.

In case you’ve forgotten the grade school lessons, palindromes are strings that read the same backwards as forwards.  For example, "A man, a plan, a canal, Panama" reads the same both ways.  You can have palindromes using numbers as well.  If you read 10,401 backwards, you get 10,401.

This week is a palindrome week.  It began with 5/10/15 (i.e. 51015), continued on with 5/11/15 (51115), then 5/12/15 (51215), etc.  Palindrome week will actually continue through to this coming Tuesday with 5/19/15 (51915).  It will end on the 20th since 5/20/15 isn’t a palindrome.

I was even happier when, on Palindrome Week, my car’s odometer read:


That’s palindrome mileage on palindrome week!

By the way, this post is 313 words long so this post itself is a palindrome.  I’m also posting it at 8:38 AM.  The math geek in me is very happy!

Extreme Geekery: Give Me A Lever Long Enough…

LeverArchimedes once said "Give me a lever long enough and a fulcrum on which to place it, and I shall move the world."  He was waxing poetic about the power of the lever.  The lever is one of several simple machines along with the wheel and axle, the pulley, the inclined plane, the wedge, and the screw.  These machines can help to perform feats that a man acting alone using only his own strength couldn’t hope to achieve.  They can also be combined to increase the machines’ abilities.  Of course, I don’t think Archimedes really meant that one could set up a lever and move the entire Earth.

But could you do this?  How long would the lever need to be?

Obviously, there are some stumbling blocks to our plan.  First of all, there’s no ground in space to position the lever’s fulcrum.  Even if we made a giant lever and somehow put it against the Earth , pushing down on one side would cause the entire contraption to drift through space.  There’s also the problems that gravity would cause – the gravity on the lever close to the Earth would be more than the gravity further away.  What’s more, unlike a lever on Earth, we’ll be battling against the gravity of the Earth orbiting the Sun, not an object resting on the Earth.


All of this could, at least, seriously mess up the calculations for how long our lever would need to be or, at worst, make the whole affair pointless.  Since there’s no such thing as "pointless" when it comes to Extreme Geekery, .  Let’s simplify things, by making two assumptions.  First will be some "ground" and Earthlike gravity.  Our fulcrum will rest on the "ground" and will push up against the Earth attempting to lift it as if it were just an extremely large rock.  The second assumption is that there isn’t any other source of gravity to mess things up.  We’re ignoring the Sun, other planets, the Moon.  Everything.

So, how long would a lever need to be to move the world?  The formula for this is quite simple:

F1 * D1 = F2 * D2

In other words, the force applied down (F1) times the length (distance) of the "pushed down" portion of the lever (D1) is equal to the force upward on the other side of the lever (F2) times the length of that side of the lever (D2).  We’ll position the lever and fulcrum so that the "Earth side" is under half of the planet.  Initially, you might think that this means that the Earth side of the lever is the radius of the Earth.  Don’t worry.  I did too.  I figured out all of my calculations before realizing the truth.  (The good news is that I had to do more math.  This is always good news to a math geek.)


The radius of the Earth at the equator is 6,378 km.  The radius from center to pole is 6,360 km.  The rest depends on how inclined the lever is, but let’s say it’s raised so that the it is up 25% of the center-to-pole radius.  We need to figure out what side C is.  Easy enough.  Using the formula A2 + B2 = C2 gives us a result of about 6,573 km.

The weight of the Earth is easily Googled: 5.972 x 1024 kg.  This only leaves us the force pushing down on the other side.  Given that Archimedes said he could do it, I’m going to say that only one person should attempt it.  No cheating and gathering an army or pushing on it with rocket thrusters.  We’ll do this see-saw style and sit the person on the other end of the lever.  We’ll have the person be 100kg – perhaps slightly overweight but not unrealistically so.

Now how long is the "pushed down" side of the lever?  Well, we have:

D1 * 100 kg * 9.8 m/s2 = 5.972 * 1024 * 9.8 m/s2 * 6,573 km

(Side note: In case you’re wondering where those 9.8s came from, force equals mass times acceleration.  F = ma.  This means that the forces we need for each side of our lever equation are really the weight of our objects times the acceleration – our faked Earth gravity down.  Also, yes I know they cancel each other out, but I’m including them in there for completeness.)

Simplified, this gives us a lever distance of 3.93 * 1026 km.  That seems pretty long, but how long is it really?

Light is the fastest substance known.  The speed it travels at can’t be matched by anything we know of.  That’s why we measure distances in light years – or the distance that light travels in a year.  One light year is 9.46 * 1012 km.  This means our side of the lever would need to be about 4.15 * 1013 (or 41.5 trillion) light years long.  How long is this?  Well, it’s certainly longer than the Milky Way.  That’s 100,000 light years across.  It’s even bigger than the diameter of our local supercluster of galaxies.  That’s 110 million light years.  In fact, our observable Universe is only 45.7 billion light years so our lever would need to be over 908 times the length of the known Universe.

That’s one big lever.

Ok, so one person probably couldn’t do it alone – even by Extreme Geekery standards.  But what about the power of teamwork?  If every human on Earth got together and sat down on the lever, how long would it need to be to move the Earth?

All humans together weigh 632 billion pounds or 286.67 billion kg.  Plugging this into our formula above means that our lever would need to be 1.37 * 1017 km or about 14,474 light years.  This distance is much more reasonable.  Yay teamwork!  Still, maybe we could improve the results.

All of the biomass on Earth (except bacteria – those guys claim to have lost their invitations) hopped onto our galactic see-saw now.


Totaled up, Earth’s biomass weighs 560 billion tonnes or 560 trillion kg.  This translates into a lever of 7 * 1015 km or about 741 light years.  Even better still.

Of course, even a "mere" 741 light years is a long distance for a lever.  As much of a genius that Archimedes was, he might have overestimated himself when it came to levers.  Unless…


NOTE: The Earth image I used is by stevepetmonkey and is available via OpenClipArt.org.

Common Core and Fourth Dimensional Math

I couple of weeks ago, I saw a video on Twitter.  In it, a girl tried to solve 1,568 + 1,423 + 680.  First, she tried to solve it using a Common Core tactic.  Next, she tried to solve it using the normal "stacking" method that kids have been learning for years.  Take a look:

Not only did the Common Core method take her 10 times as long, but the complexity resulted in an incorrect answer.  Meanwhile, the old-school stacking method was faster AND more accurate.  Even if she had made an error using the stacking method, it would have been easier to check and would have taken less space to write out.

However, while watching this video, I saw another problem.


Doing the common core math, 1’s were represented by dots, 10’s were lines, 100’s were squares, and 1,000’s were cubes.  Suppose the problem had been 15,680 instead of 1,568, though.  How would the child have been able to represent this?  Sure, she could draw 15 cubes, but the natural progression seems to be to add a dimension to the figure for each additional place value.  Going by this logic, we’ll have grade school children drawing hypercubes to solve math equations.


Yes, a simple math problem like 15,682 + 23,624 would require tapping into the fourth dimension to solve.  Clearly, this is yet another non-scalable Common Core method.

Why do we continue to confuse our kids when traditional math teaching methods work so much better?

In New York, sadly, the answer is that Governor Cuomo has decided that the public schools must go to clear the way for charter schools.  He attacks teachers – even going so far as to insinuate that teachers oppose high stakes test scores being tied to their jobs to protect teachers having inappropriate relations with students – and calls public schools "monopolies" while he backs the charter schools and pushes for more of them.  If Governor Cuomo has his way, all public schools would be charter schools run by private businesses.  After all, privatization fixes everything, right?

The only bright spot is his claim to support "anti-creaming" legislation which would force charter schools to not only accept ESL, low income, and special needs students, but to report on how many they have at the beginning and end of every school year.  Still, having one good idea doesn’t make me support all of the bad ideas he wants to implement.

Maybe it’s time to stop seeking overly complicated answers in the fourth dimension and when a real world simple solution is present.  Fund public schools fairly and let our teachers teach instead of forcing them to focus on how well the students can perform on multiple high stakes tests.  No hypercube is needed to solve this problem.

NOTE: The hypercube image above is by mate2code, was released into the public domain, and is available from Wikimedia Commons.

The Return of Common Core Math

Last year, I wrote about how NHL went into a full blown panic over his math homework.  I featured one of his problems and compared the Common Core method of solving it against the "Old School" method.  I concluded at the time that not only was the Common Core method more work than was necessary (drawing pictures instead of actually dealing with the numbers), but its methodology wouldn’t scale to larger problems.  I haven’t seen much of NHL’s math work this year, mainly because he finishes it all in school.  I managed to steal a glance at his homework tonight and it actually seems sane.

JSL, on the other hand, actually shed tears thanks to his homework recently.  After helping him with it for a few days, I figured I’d do another Common Core vs. Old School Math comparison.

Let’s start with a random problem from JSL’s most recent homework.  What is 52 – 7?

We’ll start with the Old School method.  First, we display the problem like so.


Next, we try to subtract 7 from 2.  Seeing this isn’t possible, we take one off of the 5 and add 10 to the 2.


Now, we can subtract 7 from 12 and get 5.


Finally, we subtract in the tens place – easy since it is 0 from 4.


And we have our answer: 45.  Not hard at all, right?


Now, let’s see how the Common Core method does it.  First you write out the problem.


So far, so good.  Now you break 10 off of the 52 like so:


Taking 10 off of 52 leaves 42 left so we’ll fill that in too.


Now we subtract the 7 from the 10 and get 3.


Finally, we add that 3 to the 42 to get 45.

So we have our answer: 45.


Looking at both methods, I understand what the Common Core authors were shooting for.  They were trying to take the "borrow 10" step from Old School math and recreate it in a more visual fashion.  The problem I have with this is that it needlessly complicates the math problem.  52 – 7 turns into three separate problems: 52 – 10 = 42, 10 – 7 = 3, and 42 + 3 = 45.  There’s also the unwritten math problem of what tens break out do you need to do. After all, if the problem was 52 – 16, breaking 10 out wouldn’t help.  Instead, you would need to pull 20 out of 52 and adjust the rest of the problem accordingly.

In the end, the Common Core method isn’t horrible.  Certainly, not as bad as last year’s math.  Still, it is needlessly complex and not developmentally appropriate for a second grader.  I’m all for progress, but everything new isn’t necessarily better.  In this case, the Old School method is easier and more direct while the Common Core method is more confusing.  The good news is that I think my son has finally figured out how to do this math.  The bad news is that he shouldn’t have had to.

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