Everyone’s heard of the mysterious [cue echo-y scary voice] Bermuda Triangle. If the Triangle were as dangerous as the stories would have you believe, there would be piles of shipwrecks jutting out of the water. How many sunken ships would it take to fill the Bermuda Triangle?
According to various websites, the Bermuda Triangle covers 500,000 square miles with depths ranging anywhere from a few hundred feet to about 30,000 ft. Assuming an average depth of 15,000 ft, we can compute the total volume we’ll need to fill with sunken ships
volume = (depth) · (area)
= (15,000 ft) · (500,000 square miles)
= 5.9×1015 m3.
Ships come in all different sizes. Since it’s the sunken ship everyone knows, I’ll assume we’re dealing with Titanic-sized ships1. According to Wikipedia, the Titanic had a length, beam, and height given by 269.1 m, 28.0 m, and 53.3 m respectively. This gives a volume of approximately 4.0×105 m3. We can then easily compute the number of Titanic-sized ships that would fit in the Bermuda Triangle,
# of ships = (volume of ocean) / (volume per ship)
= (5.9×1015 m3) / (4.0×105 m3per ship)
= 1.5×1010 ships.
That’s 15 billion ships. If you sunk one million ships per day, it would take 40 years to fill the Bermuda Triangle.
[1] Yes, I know there are no icebergs in Bermuda. Work with me, people. Work with me!
We’ve all heard the idiom “As far as the eye can see,” but just how far is this? If you’re standing on the Earth, how far can you see?
There are many different ways to interpret my original question, so I should have been more specific. As a friend wrote,
“I frequently see M31 (Andromeda galaxy) which is about 2 million light years. I did NOT see the GRB that was bright enough to be seen -5 billion light years- (visible only for a few seconds).”
I admit, I wasn’t thinking astronomically when I first posted the question. Since the age of the universe is thought to be 14 billion years, it stands to reason that the absolute farthest the eye could see is 14 billion light years1. I was thinking more terrestrially. If we confine ourselves to the quasi-two-dimensional world that is the surface of the Earth, how far can the eye see?
Consider the Earth to be a sphere of radius RE = 6400 km. The eyes of a standing person might be some distance H = 1.7 m off the ground. By constructing a triangle (see diagram), the Pythagorean theorem predicts the equation
(RE + H)2 = RE2 + X2,
which can be rearranged to solve for the farthest terrestrial distance X one can see
X = [ (RE + H)2 - RE2 ]1/2
= [ (6400 km + 1.7 m)2 – (6400 km)2 ]1/2
= 4.7 km.
This is the distance at which ships start to sink on the horizon. Note that the higher up you go (i.e. larger H), the farther you can see. At the top of Mount Everest, you can see 336 km in each direction.
[1] A light year is the distance light travels in one year, so any object further away than this could not be seen since the light would not have had time to reach you.
Author, journalist, and TV producer Simon Singh1 has a great question for us today. Dr. Singh has written numerous popular science books including Fermat's Enigma, The Code Book, and Big Bang: The Origin of the Universe. His question is this:
“How many razor strokes does the average clean-shaven grown up man make each morning? People typically say 20 or 30, but when they count them they see that it is about ______.”
I’ve left Dr. Singh’s estimate blank. Without counting, how many razor blade strokes does a man make each day?
How many strokes you use clearly depends on how much facial hair you have, your shaving technique, and what kind of razor you use. You almost certainly make more strokes with a dull single blade then you will with a brand new 5-blade razor2. It’s difficult to measure the area of ones face since it’s not a flat surface, but a reasonable estimate3 is 7 in. by 8 in = 56 in2. Some times you use strokes that cover several inches, other times your razor moves only a few millimeters. Let’s assume the average stroke covers one square inch of area. If you’re like me, you usually like to shave each area a few—let’s say 3—times to make sure it’s smooth. From this we can estimate the total number of strokes,
# strokes = (total area) · (strokes per unit area)
= (56 in2) · (3 strokes per in2)
= 170 strokes.
This is within an order of magnitude of Dr. Singh’s estimate of 100 strokes. For the record, I just counted and got 257 strokes with a fairly dull Gillette Fusion.
Dr. Singh, thank you for both for the great question and for your heroic work fighting for free speech.
[1] For those of you who read this earlier and are wondering why the attribution changed, it's because I screwed up. I had written Dr. Singh and he was kind enough to reply with the above question. Somehow I mixed up his email address with Simon Tofield's who had submitted a question earlier. I'd like to give my sincerest apologies to both Simons for my stupid and very embarrassing mix up.
[3] I actually measured this to be sure. It may seem large, but I’ve got one of those necks where you can’t quite tell where the facial hair stops and the chest hair begins.
In principle, an ice bullet should be able to kill a person without leaving a trace since the ice would melt away destroying any evidence. In practice, this is not the case as MythBusters demonstrated in a 2003 episode. Will this always be the case? What if the bullets started off really cold? How cold would a bullet have to be to make it out of a gun barrel before evaporating?
Wow! I’m impressed. I like ntm’s analysis in the comments. My only concern would be that not all of the energy goes into heating the bullet since some must be used to propel it forward. I’ll consider the 45 g bullet as a test case. Real bullets have speeds ranging anywhere from 180-1500 m/s. We can use the formula for kinetic energy to determine how much of the 18,942 J goes into linear motion. Plugging in the slowest bullet speed, we get
kinetic energy = (mass) · (velocity)2 / 2
= (45 g) · (180 m/s)2 / 2
= 730 J,
while the fastest bullet speed gives a kinetic energy of
= (45 g) · (1500 m/s)2 / 2
= 51,000 J.
In the case of the former, 730 J is barely a significant figure compared to the muzzle energy, which means that most of the energy will likely go into heating the bullet1. In the latter case, there’s not even enough energy to propel the bullet forward never mind heating it. If the muzzle energy is 18,942 J, then the absolute fastest a 45 g lead bullet can move is about 920 m/s, and that would mean all the energy goes into kinetic motion rather than heat. For a 3.97 g ice bullet, the fastest speed achievable for a bullet absorbing all that energy is 3100 m/s.
Even if all the energy is transferred directly to kinetic energy, there is still air friction that will heat the bullet. At high speeds, the frictional drag force is proportional to the velocity squared:
drag force = - (drag coefficient) · (air density) · (area of the bullet) · (velocity)2 / 2.
The air density is 1.2 kg/m3. I’ll assume the cross-sectional area of the bullet is about 0.1 cm2 (i.e. about 0.3 cm by 0.3 cm.) The drag coefficient for a bullet2 is about 0.295. Plugging these numbers in, we find the total drag force is,
We can compute the total work done by air friction on the bullet by multiplying this force by the distance traveled. Some guns have barrels that are as small as 10 cm, so the total work done on the bullet can be estimated by3
work = (force) · (distance)
= (-17 N) · (10 cm)
= -1.7 J.
If this is the only work that goes into heating the bullet, then its temperature change can computed as
temperature change = (energy) / [(mass) · (specific heat)]
= (1.7 J) / [(3.97 g) · (2.05 J/g·K)]
= 0.21 K.
This means the bullet could be 1 K (or equivalently 1°C) below the melting temperature and it would still fire!
What gives? Didn’t MythBusters bust this myth? It turns out our assumption that all the energy went directly into the kinetic energy of the bullet is bupkis. The truth lies somewhere between ntm’s result and my own. If all the energy goes into translational motion, then the bullet will make it out of the barrel with no problems. If all the energy goes into heating the bullet, it will melt almost instantaneously. As ntm very accurately points out, you can only lower the temperature of the bullet to absolute zero, so there’s only so much we can freeze it. Whether or not an ice bullet will ever work depends on how efficient your rifle is. If we cool the bullet to zero temperature, it will be able to absorb 2200 J before it melts4. Whether or not the bullet melts depends on how efficient the rifle is at transferring energy to the bullet. So long as less than 12% of the muzzle energy is converted to heat, the bullet should make it out of the barrel fine.
At any rate, very good work, ntm. I am suitably impressed.
[1] There is also some energy in the recoil of the rifle. This should have much less kinetic energy than the bullet. You can see this if you solve for the conservation of both energy and momentum.
[2] Since they’re made of different materials, it’s unclear whether or not ice bullets and lead bullets will had the same drag coefficient, but this is a reasonable order of magnitude estimate.
[3] Because of friction, the bullet will slow down so technically we can’t assume the velocity is constant over all 10 cm. Because of this, our estimate here should be considered an upper bound for the work done on the bullet.
[4] We can also include the latent heat necessary to undergo a phase change, but this will be small.
In a 1991 episode of the sitcom Cheers, Boston Celtics forward Kevin McHale becomes obsessed with knowing how many bolts are in the Boston Garden basketball court, and the obsession ruins his game. If only Kevin had a copy of How Many Licks?How many bolts are in the basketball court at the old Garden?
An NBA basketball court is 94 ft by 50 ft. As you can see, there are about 10 floorboards running across the width of the court and 18 running down its length giving a total of 180 floorboards. If each floorboard is held down by 4 bolts (one at each corner), there will be 720 total bolts.
A quick Google search shows that there are actually 988 bolts. Presumably, my estimate is lower because I only counted the playing surface and not the several meters floorboards that make up the sidelines.
