To build a pyramid

This was written by me as a freshman undergrad at Case Western Reserve University in 2006 for PHYS 123, Physics I Honors. It is reprinted in full, original form, with no attempt made to make the conclusion more realistic. Some of my assumptions were silly and made the problem trivial, but these were never graded for strict accuracy. 

The Great Pyramid was the tallest structure in the world from about 2570 BC to AD 1300 (when it was surpassed by the Lincoln Cathedral in England).  Its specifications are given below:

Length of one side of base (base is square) = 230.4 m

Height (original, estimated) = 146.6 m

Number of stones = 2.4 million

Total mass = 5.9 million tonnes

Average density = 2300 kg/m³

Egyptologists, from the Ancient Greeks who subjugated the old empire of the Pharaohs, to the British archaeologists who had such an interest in the Egyptian colony, to the present day scholars, have consistently marveled at the investment of labor and planning that must have gone into producing such a marvelous creation more than four and a half millennia past. 

How much work did it REALLY take to build the pyramid?  How many workers were involved?  What was the power of the labor machine that created it?  To put things in perspective, how much would this building cost today?

for W_x is the work involved in transporting the blocks horizontally across the desert, and W_y is the work involved in getting the bricks to their locations in height on the pyramid.

where the force is in opposition to dragging the blocks from their quarry (friction) and the displacement is how far from the pyramid building site the blocks must be dragged (the stones came from various far-away source; the average distance is probably around 500 miles).  The Egyptians had no means of locomotion for these stones except ropes and muscle.  Let us say that the force of friction was approximately equal to the normal force, due to the incredibly high friction generated by rocks on sand without lubrication.  Set F equal to the force of gravity and solve for W_x.

W_x = (2.6 million tonnes)*(9.81 m/s²)*(500 miles)

W_x = 2.1E+16 J

The pyramid has an angle with the normal provided by:

Take the average height of a block (we may be getting into rough territory here) to be 2.0m.  By the total height of the pyramid, we make the deduction that the pyramid is 73 layers high, and that for every layer the angle still holds true (that is to say, the slant height of the pyramid is a straight line).  The height and area of each level depends upon which numerical level it is, so our result is going to be a sum of works required for each level; work will be the volume of the layer multiplied by the density of the pyramid times the gravitational acceleration times the height.  To simplify that expression:

where h is the height at point n, d is one side of the base of the level at point n, ρ is the density of the rock, and g is the gravitational acceleration.  When values are given appropriately:

The sum of the two energies yields:

This amount of energy is approximately equal to what is released from a 5 megaton bomb.  If you wanted to fund a labor force of this size, consider that Egyptologists project that 30,000 workers on average were needed for 20 years, provided that they worked 10 hours a day every day.  If you think you can pay average wages of 10 dollars per day without mutiny, then you too can have your own pyramid for a mere 2.2 billion dollars.

Physical analysis of Planet of the Apes

This was written by me as a freshman undergrad at Case Western Reserve University in 2006 for PHYS 123, Physics I Honors. It is reprinted in full, original form, with no attempt made to make the conclusion more realistic. Some of my assumptions were silly and made the problem trivial, but these were never graded for strict accuracy. 

In the context of the book The Planet of the Apes by Pierre Boulle (also a big-budget 1968 movie by Franklin J. Schaffner and starring Charlton Heston, and a more recent film), we see an early popular understanding of time dilation serve as a major plot device.  The relativity of time was used as a convenient method for allowing travel to a distant star system.  It was imagined in 1963, amidst frenzied advancement in astronomy, and so placed its opening timeframe perhaps only a few decades into the future.

Professor Antelle, a genius scientist, has invented a special spacecraft that is able to move at such a high velocity (via unknown propulsion) that time itself is slowed significantly for the pilot.  This obviates the problem of impossibly low interstellar speed, and allows a huge amount of space to be traversed in even “less” time (from the pilot’s point of view) due to time dilation.  Ulysses, the main character, is part of the expedition, along with the professor, and Levain, a physician. 

Time dilation versus velocity

They intend to use this spaceship to travel to the nearest place they believe that extraterrestrial life may exist- a star system whereof the supergiant Betelgeuse is the local sun.  The time it would take their ship to reach there is 350 years, but for the individuals inside, the time will feel like a mere two years.  What velocity does this entail?  Rearranging the above equation for known values:

In order for time dilation to be that potent, one must get very close to the speed of light.  As we can see, earthly technology brings us nowhere close to even this velocity which would only shrink time by a factor of 175.  In order to reach space millions or billions of light-years away, the only possibility is to get even closer to the speed of light.

The practical difficulty is not so much in what a person would do for years on a spaceship (although this is a bit mind-boggling) but the quantity of energy it would take to transport anything at speeds close to that of light.  The kinetic energy of an object traveling at the speed of light is phenomenal.  The craft described in the book is not miniscule, either.  Let us say, for example, that using miraculous miniaturization technologies that the spacecraft can be able to carry its engines, three passengers, and enough supplies for two years forward, two years back- in a mass no greater than that of the Space Shuttle.

This figure is a bit large, to say the least.  If this spaceship spread out its acceleration over the ridiculously long interval of 20 days, then the power required would be:

which is equal to 3.8 billion horsepower.  If it were to accelerate to that speed in the same time that it took for the Shuttle to clear the atmosphere, then over one trillion horsepower would be required. 

            The conclusion I draw from this is that, in order for humans to attain speeds close to that of light, the mass involved must be infinitesimal enough so that the energy can be produced to power it, or else new methods of power (e.g. not derived from chemical or electrical propulsion) must be found. 

            But audiences would never have suspected this in the optimistic year of 1963, and as The Planet of the Apes shows, it was not a picnic when the light-speed travelers arrived at their destination.  Society involved the subjugation of humans by their primate overlords.  When our hero Ulysse finally fled, and returned to Earth, 700 years had passed and the same fate of human enslavement had befallen his planet.

            The moral of the story, if one can be said to exist, was stated aptly by a student in Physics 123 on the day of the relativity lecture: “Stay the hell away from the speed of light.”

The sacrifice of the HMS Thunder Child

This was written by me as a freshman undergrad at Case Western Reserve University in 2006 for PHYS 123, Physics I Honors. It is reprinted in full, original form, with no attempt made to make the conclusion more realistic. Some of my assumptions were silly and made the problem trivial, but these were never graded for strict accuracy.

“About a couple of miles out lay an ironclad, very low in the water, almost, to my brother's perception, like a water-logged ship. […]It was the torpedo ram, Thunder Child, steaming headlong, coming to the rescue of the threatened shipping."

~H.G. Wells, The War of the Worlds

The prelude:

The year is approximately 1900.  The Dreadnought is not yet conceived, and in the late 1890s, ironclad rammers still represent the pinnacle of naval technology.  Fighting desperately for survival against the Martian war machines, the Royal Navy selects the finest ramming ship they had in their arsenal, and the one with the greatest nimbleness and speed. 

She was the Thunder Child, blessed of agility and formidable guns and armor, yet of size small enough to make a tactical naval battle with the Martians on its own terms.  Indeed, her skirmish would be the single bare victory had by the humans of the Victorian era Earth that attempted to fight for their lives against extraterrestrial invaders.  In the Thunder Child they found a symbol- she was built with amazing care and represented the pinnacle of the technology of the world in 1900.

 

Now was the time.  There would be no other.  Thousands of refugees were fleeing after London fell, and the entire British merchant marine could be destroyed by the horrendous Martian war machines.  Three of these devices were dispatched to the seas around England to intercept any and all human vessels, killing them with black smoke.  The lives of thousands were at stake.

The engagement:

Thunder Child was at full steam when she sighted the Martian war machines.  Not used to water, the Martians were not quite sure what to make of the ramming warship.  They had seen no mechanical device at the humans’ disposal that was as large as a warship.  They made the assumption that the device was organic, and deployed the sinister black smoke against it.  Thunder Child’s crew retreated into the ship and they did not inhale any of the poison.  The smoke clouds gave cover to the Thunder Child, and she steamed on a direct collision course with the first war machine, at full 20 knots:

The Martians finally wised up and attempted to strike it with their Heat Ray.  One hit was successful, and the Thunder Child was extremely damaged; still she steamed on.  The pointed bow, with an edge merely an inch thick and twelve feet high, struck hard and pierced the extraterrestrial metal.  The impact was devastating and Thunder Child cleaved the war machine in half very jarringly, losing half of its momentum within a second. (Assumptions made regarding the dimensions and characteristics of the ramming action are all guesses by me.)

(As we find, the armor of the Martian war machines had a tensile strength of greater than 280MPa- superior to modern rolled homogenous steel.  Steel of this quality was nonexistent in 1900, and may have seemed alien.)

The Thunder Child tried then to open up with her six inch guns, but the range was too short for them to be effective.  Instead she, with foundering keel but usable rudder and engines, accelerates to full speed again, to attack the second war machine.  Persistent and desperate salvos destroy the Thunder Child before she can ram the second ship, but the ships boiler and ammunition explode into a massive hailstorm of steel that crushes the second war machine with thousands of tons of molten iron and wounds the third.

The outcome:

A marginal victory for humanity… the destruction of two Martian war machines.  This raid saved the lives of thousands, but there was to be no respite in the struggle to survive against the extraterrestrial invaders.

Structural failure of a CD

This was written by me as a freshman undergrad at Case Western Reserve University in 2006 for PHYS 123, Physics I Honors. It is reprinted in full, original form, with no attempt made to make the conclusion more realistic. Some of my assumptions were silly and made the problem trivial, but these were never graded for strict accuracy.

Professor Starkman once asked us to use rotational mechanics to find out the properties of a CD spinning at 7200 RPM; this gave us some appreciation of the stress on a CD as it is spun. The Mythbusters once tackled the objective of trying to cause structural failure to a CD by creating unusually high rotational speeds to the CD to investigate the myth that a standard CD drive can under certain circumstances spin fast enough to cause a CD to break apart and turn into a lethal disc of shrapnel.

Let’s mesh these worlds, and see what it would physically take to destroy a standard CD.  The figure we were given in our physics class was 7200 RPM.  In certain disc drives, the regular speed may be more. 

Physical Information of CD-

Thickness (X) = 1.20 mm

Material = 100% Polycarbonate (tensile strength, σ_t, of polycarbonate is about 75 MPa)

Radius (R) = 12.0 cm

Density (d) = 1.20 g/cm^3

Let us make the assumption that since the hole is filled in, we have a complete volume of disc.  Plugging that in to our density:

m/V = d

m =dV

m = dπR²X

Mass (m) = 0.0650 kg

We have a radius and a thickness, which corresponds to a cross-sectional area of a CD on one side.  Recall that it only requires a break at one of these cross-sectional areas to fail.  This material is very brittle; do not expect much strain as a result of stress.  It ought to shatter.  This should simplify things.

I plan to evaluate the centripetal force caused by the spinning of the disc as a function of ω.  This disc must respond to a centripetal force with a normal force.  This normal force is dictated by its structural integrity.  Given that we have a specific area of interest, this force may be divided by area, leaving us with units of N/m^2… the same units as Pa, which is proportional to our tensile strength.  The units of tensile strength and pressure are identical.  Evaluate for the maximum possible ω which will cause a force that exceeds our tensile strength.

σ_t = F/A           (Force required to break divided by area equals tensile strength)

A = XR            (Cross-sectional area is equal to radius times height)

XRσ_t = F         (The force that is required)

F = mv^2 / r

v = Rω

F = mRω²

XRσ_t = mRω²

(XRσ_t / mR) ^½ = ω

(Xσ_t / m) ^½ = ω

This is the maximum possible angular velocity that we can achieve.  After plugging in the appropriate values:

((0.0012m)*(75E+6Pa)/(0.065kg))^ ^½ = 1177 rad/s

We now have a figure in radians per second, but disc drives are never advertised in such figures.  What does this translate to in terms of revolutions per minute, the preferred angular velocity measurement of the West?

1177 rad/s*(radian / second)*(1 revolution / 2π radian)*(60 second / 1 minute) =

11200 RPM

Our ceiling figure for angular speed of a CD is 11200! Um, wasn't it way faster on Mythbusters? =/

In all probability, as we estimate for error in this problem, our estimate is extremely liberal with its notion of structural failure.  In actuality, the polycarbonate material may be higher or lower than the one we listed; but every CD has a bottom and top layer which would likely enhance structural integrity.  Additionally, we did not account for the removed section of the disc (the hole in the center, into which an electric motor pushes a rotor that spins the disc.  

My point in this experiment is to reflect on the magnitude of stress on the CD in your disc drive as it whizzes around at 120 to 170 revolutions per second.  A modern engine will be on the redline when a CD drive is operating properly.

The GAU-8 Avenger Autocannon

This was written by me as a freshman undergrad at Case Western Reserve University in 2006 for PHYS 123, Physics I Honors. It is reprinted in full, original form, with no attempt made to make the conclusion more realistic. Some of my assumptions were silly and made the problem trivial, but these were never graded for strict accuracy.

Those with even a cursory knowledge of national air forces will know that there are several roles to fill- air superiority fighter, heavy bomber, stealth bomber, and close-in ground support.  The A-10, ugly and ungainly as it is, is a magnificent piece of work that fills the role of the last category.  With a low flying speed, “titanium bathtub” armor, immense endurance, and a phenomenally powerful main gun, the A-10 is a wonder of military technology and of physics.

Notice that autocannon in the nose?  This is the GAU-8 Avenger, a scaled-up version of the more familiar 7.62mm Minigun and 20mm Vulcan cannon.  It fires depleted uranium shells of 30 mm diameter at a rate of 4200 rounds per minute.  Here our cannon is shown to scale.

One of the most persistent claims by enthusiastic armchair strategists is that the Avenger is so powerful that its recoil is at least as powerful as the engines of the plane- thus, it is capable of slowing down and even stopping our A-10 in the air.  Is this true?  Let’s consider momentum and force.

(momentum) = (force)*(time)                            p = Ft

What is the momentum of our stream of fire?  Since we have a rate of fire and a mass, we can use dimensional analysis to determine momentum, and hence force.

momentum = (mass of shell)*(velocity)*(rate of fire)*(time)                   p = mrt

(mass of shell)*(rate of fire)*(velocity)*(time)/(time) = force                  (mrt)/t = F

(mass of shell)*(rate of fire)*(velocity) = force                                       F = mrv

Think of r as a frequency instead of rate of fire.  We are firing 4200rpm, which equates to 70 rounds per second or 70Hz.  Doing some additional research, we find that the mass of a round is 0.425kg and our muzzle velocity (highest velocity ever attained by the bullet) is 1036m/s.  In comparison, the makers of the A-10 claim that their two engines will produce 80kN of thrust.  Let’s see if we really have the power to stop a plane.

F = mrv

80.kN < (0.425kg)*(70.Hz)*(1036m/s)

80.kN </ 31kN

We can produce “only” thirty-one thousand newtons of force by our Avenger cannon.  This means that the plane will experience a possible deceleration if it is only using partial power, but this can be overcome by employing a steady, large amount of thrust, which is in practice what the pilots tend to do as they are well aware (and possibly fearful) of the myth of planes stopping in midair.

But the GAU-8 Avenger is still incredibly powerful.  Let’s say we mounted all 281kg of the gun on the Terminator’s back, gave him 1000 rounds of ammo and convinced him that he could fire the weapon from a standing position.  His hydraulic joints lock up (he will not drop the weapon and there will be no force lost to excess motion) and he prepares to fire at the full 4200rpm, perhaps too confident of his abilities after playing around with a 7.62mm Minigun in T2.  The Minigun is well beyond the range of a single soldier to carry and operate, but he manages it without difficulty.  However, this weapon can be mounted on a helicopter door, while the Avenger, as we have seen, generates nearly as much recoil as one the jet engines on the A-10.

The Terminator weighs 200.kg.  Of course, he manages to shoulder the behemothic weapon with a bit of effort, and starts firing at T-1000 who has found himself a nifty T-72 tank.  Arnold digs his heels into the muddy ground and achieves a coefficient of friction of 1.0.  How fast will the Terminator accelerate backwards, or can he actually stand in place?

frictional force = (coefficient of friction)(mass)(gravity)               Ff = μmg

Remember to add up all the weight that is now being held on the Terminator’s hyper-alloy legs.

Ff = (1.0)*(281kg + 425kg + 200.kg)*(9.81m/s^2)

Ff = 8.9kN

The friction force is massive, but it is less than a third of the GAU-8 recoil.  There is a net force and Arnold starts to slide immediately.  The net force, accounting for friction, on the Terminator and his gun is now 22kN.

F = ma

22kN = (281kg + 425kg + 200.kg)a

a = 2.4 m/s^2

Buh-bye Arnie. You'll be all over the place. Good luck aiming the thing.  Was the T-1000 terminated?

At 500m, the Avenger can pierce about 70mm of modern composite armor.  The armor of a T-72 is thicker than 70mm- but its armor is old-fashioned steel, meaning that it would take approximately twice as much armor to achieve the same strength.  Narrowing the range to near point-blank, the effect would be catastrophic to this pensionable commie tank.

At negligible range, the 30mm rounds would turn the tank into a hailstorm of steel confetti, and the pyrophoric depleted uranium rounds, upon piercing the frontal armor, would also set fire to the shrapnel within the tank.  They would have so much kinetic energy left that some of the rounds would indeed pierce the rear armor of the tank as well.