# magnetic separation of air

As some of you will know, oxygen is paramagnetic, attracted slightly by a magnet. Oxygen’s paramagnetism is due to the two unpaired electrons in every O2 molecule. Oxygen has a triple-bond structure as discussed here (much of the chemistry you were taught is wrong). Virtually every other common gas is diamagnetic, repelled by a magnet. These include nitrogen, water, CO2, and argon — all diamagnetic. As a result, you can do a reasonable job of extracting oxygen from air by the use of a magnet. This is awfully cool, and could make for a good science fair project, if anyone is of a mind.

But first some math, or physics, if you like. To a good approximation the magnetization of a material, M = CH/T where M is magnetization, H is magnetic field strength, C is the Curie constant for the material, and T is absolute temperature.

Ignoring for now, the difference between entropy and internal energy, but thinking only in terms of work derived by lowering a magnet towards a volume of gas, we can say that the work extracted, and thus the decrease in energy of the magnetic gas is ∫∫HdM  = MH/2. At constant temperature and pressure, we can say ∆G = -CH2/2T.

With a neodymium magnet, you should be able to get about 50 Tesla, or 40,000 ampere meters At 20°C, the per-mol, magnetic susceptibility of oxygen is 1.34×10−6  This suggests that the Curie constant is 1.34 x293 = 3.93 ×10−4  At 20°C, this energy difference is 1072 J/mole. = RT ln ß where ß is the concentration ratio between the O2 content of the magnetized and un-magnetized gas.

From the above, we find that, at room temperature, 298K ß = 1.6, and thus that the maximum oxygen concentration you’re likely to get is about 1.6 x 21% = 33%. It’s slightly more than this due to nitrogen’s diamagnetism, but this effect is too small the matter. What does matter is that 33% O2 is a good amount for a variety of medical uses.

I show below my simple design for a magnetic O2 concentrator. The dotted line is a permeable membrane of no selectivity – with a little O2 permeability the design will work better. All you need is a blower or pump. A coffee filter could serve as a membrane.

This design is as simple as the standard membrane-based O2 concentrator – those based on semi-permeable membranes, but this design should require less pressure differential — just enough to overcome the magnet. Less pressure means the blower should be smaller, and less noisy, with less energy use.  I figure this could be really convenient for people who need portable oxygen. With several stages and low temperature operation, this design could have commercial use.

On the theoretical end, an interesting thing I find concerns the effect on the entropy of the magnetic oxygen. (Please ignore this paragraph if you have not learned statistical thermodynamics.) While you might imagine that magnetization decreases entropy, other-things being equal because the molecules are somewhat aligned with the field, temperature and pressure being fixed, I’ve come to realize that entropy is likely higher. A sea of semi-aligned molecules will have a slightly higher heat capacity than nonaligned molecules because the vibrational Cp is higher, other things being equal. Thus, unless I’m wrong, the temperature of the gas will be slightly lower in the magnetic area than in the non-magnetic field area. Temperature and pressure are not the same within the separator as out, by the way; the blower is something of a compressor, though a much less-energy intense one than used for most air separators. Because of the blower, both the magnetic and the non magnetic air will be slightly warmer than in the surround (blower Work = ∆T/Cp). This heat will be mostly lost when the gas leaves the system, that is when it flows to lower pressure, both gas streams will be, essentially at room temperature. Again, this is not the case with the classic membrane-based oxygen concentrators — there the nitrogen-rich stream is notably warm.

Robert E. Buxbaum, October 11, 2017. I find thermodynamics wonderful, both as science and as an analog for society.

# A very clever hydrogen pump

I’d like to describe a most clever hydrogen pump. I didn’t invent it, but it’s awfully cool. I did try to buy one from “H2 Pump,” a company that is now defunct, and I tried to make one. Perhaps I’ll try again. Here is a diagram.

Electrolytic membrane H2 pump

This pump works as the reverse of of a PEM fuel cell. Hydrogen gas is on both sides of a platinum-coated, proton-conducting membrane — a fuel cell membrane. As in a PEM fuel cell, the platinum splits the hydrogen molecules into H atoms. An electrode removes electrons to form H+ ions on one side of the membrane; the electrons are on the other side of the membrane (the membrane itself is chosen to not conduct electricity). The difference from the fuel cell is that, for the pump you apply a energy (voltage) to drive hydrogen across the membrane, to a higher pressure side; in a fuel cell, the hydrogen goes on its own to form water, and you extract electric energy.

As shown, the design is amazingly simple and efficient. There are no moving parts except for the hydrogen itself. Not only do you pump hydrogen, but you can purify it as well, as most impurities (nitrogen, CO2) will not go through the membrane. Water does permeate the membrane, but for many applications, this isn’t a major impurity. The amount of hydrogen transferred per plate, per Amp-second of current is given by Faraday’s law, an equation that also shows up in my discussion of electrolysis, and of electroplating,

C= zFn.

Here, C is the current in Amp-seconds, z is the number or electrons transferred per molecule, in this case 2, F is Faraday’s constant, 96,800, n is the number of mols transferred.  If only one plate is used, you need 96,800 Amp-seconds per gram of hydrogen, 53.8 Amp hours per mol. Most membranes can operate at well at 1.5 Amp per cm2, suggesting that a 1.1 square-foot membrane (1000 cm2) will move about 1 mol per minute, 22.4 slpm. To reduce the current requirement, though not the membrane area requirement, one typically stacks the membranes. A 100 membrane stack would take 16.1 Amps to pump 22.4 slpm — a very manageable current.

The amount of energy needed per mol is related to the pressure difference via the difference in Gibbs energy, ∆G, at the relevant temperature.

Energy needed per mol is, ideally = ∆G = RT ln Pu/Pd.

where R is the gas constant, 8.34 Joules per mol, T is the absolute temperature, Kelvins (298 for a room temperature process), ln is the natural log, and Pu/Pd is the ratio of the upstream and downstream pressure. We find that, to compress 2 grams of hydrogen (one mol or 22.4 liters) to 100 atm (1500 psi) from 1 atm you need only 11400 Watt seconds of energy (8.34 x 298 x 4.61= 11,400). This is .00317 kW-hrs. This energy costs only 0.03¢ at current electric prices, by far the cheapest power requirement to pump this much hydrogen that I know of. The pump is surprisingly compact and simple, and you get purification of the hydrogen too. What could possibly go wrong? How could the H2 pump company fail?

One thing that I noticed went wrong when I tried building one of these was leakage at the seals. I found it uncommonly hard to make seals that held even 20 psi. I was using 4″ x 4″ membranes so 20 psi was the equivalent of 320 pounds of force. If I were to get 200 psi, there would have been 3200 lbs of force. I could never get the seals to stay put at anything more than 20 psi.

Another problem was the membranes themselves. The membranes I bought were not very strong. I used a wire-mesh backing, and a layer of steel behind that. I figured I could reach maybe 200 psi with this design, but didn’t get there. These low pressures limit the range of pump applications. For many applications,  you’d want 150-200 psi. Still, it’s an awfully cool pump,

Robert E. Buxbaum, February 17, 2017. My company, REB Research, makes hydrogen generators and purifiers. I’ve previously pointed out that hydrogen fuel cell cars have some dramatic advantages over pure battery cars.

# just water over the dam

Some months ago, I posted an engineering challenge: figure out the water rate over an non-standard V-weir dam during a high flow period (a storm) based on measurements made on the weir during a low flow period. My solution follows. Weir dams of this sort are erected mostly to prevent flooding. They provide secondary benefits, though by improving the water and providing a way to measure the flow.

A series of compound, rectangular weir dams in Maine.

The problem was stated as follows: You’ve got a classic V weir on a dam, but it is not a knife-edge weir, nor is it rectangular or compound as in the picture at right. Instead it is nearly 90°, not very tall, and both the dam and weir have rounded leads. Because the weir is of non-standard shape, thick and rounded, you can not use the flow equation found in standard tables or on the internet. Instead, you decide to use a bucket and stopwatch to determine the flow during a relatively dry period. You measure 0.8 gal/sec when the water height is 3″ in the weir. During the rain-storm some days later, you measure that there are 12″ of water in the weir. Give a good estimate of the storm-flow based on the information you have.

A V-notch weir, side view and end-on.

I also gave a hint, that the flow in a V weir is self-similar. That is, though you may not know what the pattern will be, you can expect it will be stretched the same for all heights.

The upshot of this hint is that, there is one, fairly constant flow coefficient, you just have to find it and the power dependence. First, notice that area of flow will increase with height squared. Now notice that the velocity will increase with the square root of hight, H because of an energy balance. Potential energy per unit volume is mgH, and kinetic energy per unit volume is 1/2 mv2 where m is the mass per unit volume and g is the gravitational constant. Flow in the weir is achieved by converting potential height energy into kinetic, velocity energy. From the power dependence, you can expect that the average v will be proportional to √H at all values of H.

Combining the two effects together, you can expect a power dependence of 2.5 (square root is a power of 0.5). Putting this another way, the storm height in the weir is four times the dry height, so the area of flow is 16 times what it was when you measured with the bucket. Also, since the average height is four times greater than before, you can expect that the average velocity will be twice what it was. Thus, we estimate that there was 32 times the flow during the storm than there was during the dry period; 32 x 0.8 = 25.6 gallons/sec., or 92,000 gal/hr, or 3.28 cfs.

The general equation I derive for flow over this, V-shaped weir is

Flow (gallons/sec) = Cv gal/hr x(feet)5/2.

where Cv = 3.28 cfs. This result is not much different to a standard one  in the tables — that for knife-edge, 90° weirs with large shoulders on either side and at least twice the weir height below the weir (P, in the diagram above). For this knife-edge weir, the Bureau of Reclamation Manual suggests Cv = 2.49 and a power value of 2.48. It is unlikely that you ever get this sort of knife-edge weir in a practical application. Be sure to measure Cv at low flow for any weir you build or find.

Robert Buxbaum, vote for me for water commissioner. Here are some thoughts on other problems with our drains.

# More French engineering, the Blitzkrieg motorcycle.

There’s something fascinating that I find in French engineering. I wrote a previous essay about French cars, bridges, and the Eiffel tower. Here’s a picture or two more. Things I wanted to include but didn’t. First here’s a Blitzkrieg Vespa motorcycle; the French built some 800 of these from 1947 to 1962 and used them in Vietnam and Algeria. What’s remarkable is how bizarrely light and unprotected it is. It’s a design aesthetic that follows no one, and that American engineers would not follow.

French Blitzkrieg Vespa used in Vietnam; cannon range is 4.5 miles.

The key engineering insight that allows this vehicle to make sense is that recoil-less rifles are really recoil-less if you design them right. Thus, one can (in theory) mount them on something really light, like a Vespa. Another key (French) insight is that a larger vehicle may make the soldier more vulnerable rather than less by slowing him down and by requiring more gasoline and commissariat services.

Americans do understand the idea of light and mobile, but an American engineers idea of this is a jeep or an armored truck; not a Vespa. From my US engineering perspective, the French went way overboard here. The French copy no one, and no one copies the French, as they say. Still, these things must have worked reasonably well or they would not have made 800 of them over 15 years. A Vespa is certainly cheaper than a Jeep, and easier to transport to the battle zone….

Robert Buxbaum, February 18, 2016. The Italians have a somewhat similar design aesthetic to the French: they like light and cheap, but also like maneuverable and favor new technology. See my blog about a favorite Fiat engine.

# Weird flow calculation

Here is a nice flow problem, suitable for those planning to take the professional engineers test. The problem concerns weir dams. These are dams with a notch in them, somethings rectangular, as below, but in this case a V-shaped notch. Weir dams with either sort of notch can be used to prevent flooding and improve the water, but they also provide a way to measure the flow of water during a flood. That’s the point of the problem below.

A series of weir dams with rectangular weirs in Maine.

You’ve got a classic V weir on a dam, but it is not a knife-edge weir, nor is it rectangular or compound as in the picture at right. Instead it is nearly 90°, not very tall, and both the dam and weir have rounded leads. Because the weir is of non-standard shape, thick and rounded, you can not use the flow equation found in standard tables or on the internet. Instead, you decide to use a bucket and stopwatch to determine that the flow during a relatively dry period. You measure 0.8 gal/sec when the water height is 3″ in the weir. During the rain-storm some days later, you measure that there are 12″ of water in the weir. The flow is too great for you to measure with a bucket and stopwatch, but you still want to know what the flow is. Give a good estimate of the flow based on the information you have.

As a hint, notice that the flow in the V weir is self-similar. That is, though you may not know what the pattern of flow will be, you can expect it will be stretched the same for all heights.

As to why anyone would use this type of weir: they are easier to build and maintain than the research-standard, knife edge; they look nicer, and they are sturdier. Here’s my essay in praise of the use of dams. How dams on drains and rivers could help oxygenate the water, and to help increase the retention time to provide for natural bio-remediation.

If you’ve missed the previous problem, here it is: If you have a U-shaped drain or river-bed, and you use a small dam or weir to double the water height, what is the effect on water speed and average retention time. Work it out yourself, or go here to see my solution.

Robert Buxbaum. May 20-Sept 20, 2016. I’m running for drain commissioner. send me your answers to this problem, or money for my campaign, and win a campaign button. Currently, as best I can tell, there are no calibrated weirs or other flow meters on any of the rivers in the county, or on any of the sewers. We need to know because every engineering decision is based on the flow. Another thought: I’d like to separate our combined sewers and daylight some of our hidden drains.

# Advanced windmills + 20 years = field of junk

Everything wears out. This can be a comforting or a depressing thought, but it’s a truth. No old mistake, however egregious, lasts forever, and no bold advance avoids decay. At best, last year’s advance will pay for itself with interest, will wear out gracefully, and will be recalled fondly by aficionados after it’s replaced by something better. Water wheels, and early steamships are examples of this type of bold advance. Unfortunately, it is often the case that last years innovation turns out to be no advance at all: a technological dead end that never pays for itself, and becomes a dangerous, rotting eyesore or worse, a laughing-stock blot or a blot on the ecology. Our first two generations of advanced windmill farms seem to match this description; perhaps the next generation will be better, but here are some thoughts on lessons learned from the existing fields of rotting windmills.

The ancient design windmills of Don Quixote’s Spain (1300?) were boons. Farmers used them to grind grain or cut wood, and to to pump drinking water. Holland used similar early windmills to drain their land. So several American presidents came to believe advanced design windmills would be similar boons if used for continuous electric power generation. It didn’t work, and many of the problems could have been seen at the start. While the farmer didn’t care when his water was pumped, or when his wood is cut. When you’re generating electricity, there is a need to match the power demand exactly. Whenever the customer turns on the switch, electricity is expected to flow at the appropriate amount of Wattage; at other times any power generated is a waste or a nuisance. But electric generator-windmills do not produce power on demand, they produce power when the wind blows. The mismatch of wind and electric demand has bedeviled windmill reliability and economic return. It will likely continue to do so until we find a good way to store electric power cheaply. Until then windmills will not be able to produce electricity at competitive prices to compete with cheap coal and nuclear power.

There is also the problem of repair. The old windmills of Holland still turn a century later because they were relatively robust, and relatively easy to maintain. The modern windmills of the US stand much taller and move much faster. They are often hit, and damaged by lightning strikes, and their fast-turning gears tend to wear out fast, Once damaged, modern windmills are not readily fix, They are made of advanced fiberglass materials spun on special molds. Worse yet, they are constructed in mountainous, remote locations. Such blades can not be replaces by amateurs, and even the gears are not readily accessed to repair. More than half of the great power-windmills built in the last 35 years have worn out and are unlikely to ever get repair. Driving past, you see fields of them sitting idle; the ones still turning look like they will wear out soon. The companies that made and installed these behemoth are mostly out of the business, so there is no-one there to take them down even if there were an economic incentive to do so. Even where a company is found to fix the old windmills, no one would as there is not sufficient economic return — the electricity is worth less than the repair.

Komoa Wind Farm in Kona, Hawaii, June 2010; A field of modern design wind-turbines already ruined by wear, wind, and lightning. — Friends of Grand Ronde Valley.

A single rusting windmill would be bad enough, but modern wind turbines were put up as wind farms with nominal power production targeted to match the output of small coal-fired generators. These wind farms require a lot of area,  covering many square miles along some of the most beautiful mountain ranges and ridges — places chosen because the wind was strong

Putting up these massive farms of windmills lead to a situation where the government had pay for construction of the project, and often where the government provided the land. This, generous spending gives the taxpayer the risk, and often a political gain — generally to a contributor. But there is very little political gain in paying for the repair or removal of the windmills. And since the electricity value is less than the repair cost, the owners (friends of the politician) generally leave the broken hulks to sit and rot. Politicians don’t like to pay to fix their past mistakes as it undermines their next boondoggle, suggesting it will someday rust apart without ever paying for itself.

So what can be done. I wish I could suggest less arrogance and political corruption, but I see no way to achieve that, as the poet wrote about Ozymandias (Ramses II) and his disastrous building projects, the leader inevitably believes: “I am Ozymandias, king of kings; look on my works ye mighty and despair.” So I’ll propose some other, less ambitious ideas. For one, smaller demonstration projects closer to the customer. First see if a single windmill pays for itself, and only then build a second. Also, electricity storage is absolutely key. I think it is worthwhile to store excess wind power as hydrogen (hydrogen storage is far cheaper than batteries), and the thermodynamics are not bad

Robert E. Buxbaum, January 3, 2016. These comments are not entirely altruistic. I own a company that makes hydrogen generators and hydrogen purifiers. If the government were to take my suggestions I would benefit.

# The french engineering

There is something wonderful about French Engineering. It is good, but different from US or German engineering. The French don’t seem to copy others, and very few others seem to copy them. Nonetheless French engineering managed to build an atom bomb, is a core of the Airbus consortium, and both builds and runs the fastest passenger trains on earth, the TGF, record speed 357 mph on the line between Paris and Luxembourg.

JULY 14, 2015 Female engineering students of the Ecole Polytechnique, march in the Paris Bastille Day military parade. (Photo by Thierry Chesnot/Getty Images).

France was almost the only country to sell Israel weapons for the first 20 years of its existence, and as odd as the weapons they sold were, they worked. The Mirage jet was noted for short-range and maneuverability; in 1967, they handily defeated Egypt and Syria’s much larger force of Russian Migs. More recently, Argentina used French Exocet missiles to sink 3 British warships in the Argentine war, and last week, Turkey used a french missile to down a Su24, the new main Russian fighter-bomber. not bad for a country whose main engineering school marches in Napoleonic garb.

The classic of French Engineering, of course is the Eiffel Tower. It is generally unappreciated that this is not the only Eiffel structure designed this way. Eiffel designed railroad bridges, aqueducts. Here’s an Eiffel railroad bridge.

Eiffel railroad bridge, still in use. American, German, or British bridges of the era look nothing like this.

To get a sense of the engineering artistry of the Eifflel tower, consider that when the tower was built, in 1871, self-financed by Eiffel, it was more than twice as tall as the next-tallest building on earth. ff one weighed the air in a cylinder the height of the tower with a circle about its base, the air would weigh more than the steel of the tower. But here are some other random observations, while first level of the tower houses a restaurant, a normal American space-use choice,the second level housed, when the tower opened the print shop and offices of the International Herald Tribune; not a normal tenant. And, on the third level, near the very top, you will find Mr Eiffel’s apartment. The builder lived there; he owned the place. It’s still there today, but now there are now only mannequins in residence. It’s weird, but genius, like so much that is French engineering.

Eiffel’s apartment atop the tower, now occupied by mannequins of Eiffel and Edison, a one-time guest.

Returning to the French airplane, The french were the first to make mono-planes. But having succeeded there, they made a decent-enough plane-like automobile, the 1932 Helicon car. It’s a three-man car with a propeller out front and rear-wheel steering. At first, you’d think this is a slow, unmanageable, deathtrap, like Buckminster Fuller’s Dymaxion,.  But you’d be wrong, the Helicon (apparently) is both speedy and safe it moves at 100 mph or more once it gets going, still passed French safety standards in 2000, and gets taken out for (semi-normal) jaunts. Don’t stand in front of the propeller (there’s a bicycle version too).

1932 Helicon car; 100 mph, seats 3, propeller-driven. Photo by Yalon.

The Helicon never quite took off, as it were, but an odd design motorcycle did quite well, at least in France, the Solex, front wheel motorcycle.Unlike US motorcycles, it’s just a bicycle with an engine above the front wheels. The engine runs “backwards” and drives the front wheel via a friction-cam. The only clutch action involves engaging the cam. Simple, elegant, and unlikely to be duplicated elsewhere.

A Solex motorcycle and an e-Solex, the battery-powered version. A Citroen and a Peugeot sport are in the background. Popular in France.

The reason I’m writing about French Engineering is perhaps because of the recent attacks. Or perhaps because of aesthetic. It’s important to have an engineering aesthetic — an idea you’re after — and to have pride in one’s craft too. The French stand out in how much they have of both. Some months ago I wrote about a more American engineering aesthetic, It’s a good article, but interestingly, I now note that some main examples I used were semi-French: the gunpowder factory of E. I. Dupont, the main productions facility of a Frenchman’s company in the US.

Robert Buxbaum, December 13, 2015. Some months ago, I wrote about a favorite car engine, finally being used on the Fiat 500 and Alfa Romeo. Fast, energy-efficient, light, maneuverable, and (I suspect) unreliable; the engine embodies a particularly Italian engineering aesthetic.

# It’s rocket science

Here are six or so rocket science insights, some simple, some advanced. It’s a fun area of engineering that touches many areas of science and politics. Besides, some people seem to think I’m a rocket scientist.

A basic question I get asked by kids is how a rocket goes up. My answer is it does not go up. That’s mostly an illusion. The majority of the rocket — the fuel — goes down, and only the light shell goes up. People imagine they are seeing the rocket go up. Taken as a whole, fuel and shell, they both go down at 1 G: 9.8 m/s2, 32 ft/sec2.

Because 1 G ofupward acceleration is always lost to gravity, you need high thrust from the rocket engine, especially at the beginning when the rocket is heaviest. If your engine provides less thrust than the weight of your rocket, your rocket sits on the launch pad, and if your thrust is merely twice the weight of the rocket your waste half of your fuel doing nothing useful. Effectively, the upward acceleration of the shell, a = F/m -1G where F is the force of the engine, and m is the mass of the rocket and whatever fuel is in it, and the 1 G is the upward acceleration lost to gravity.  My guess is that you want to design a rocket engine so that the upward acceleration, a, is in the range 8-10 G. This range avoids wasting lots of fuel without requiring you to build the rocket too sturdy. At a = 9G, the rocket engine force, F, has to be about 10 times the rocket weight; it also means the rocket structure must be sturdy enough to support a force of ten times the rocket weight. This can be tricky because the rocket will be the size of a small skyscraper, and the structure must be light so that the majority is fuel. It’s also tricky that this 9-11 times the rocket weight must sit on an engine that runs really hot, about 3000°C. Most engineering projects have fewer constraints than this, and are thus “not rocket science.”

Basic force balance on a rocket going up.

A space rocket has to reach very high speeds; most things that go up, come down almost immediately. You can calculate the minimum orbital speed by balancing the acceleration of gravity, 9.8 m/s2, against the orbital acceleration of going around the earth, a sphere of 40,000 km in circumference (that’s how the meter was defined). Orbital acceleration, a = v2/r, and r = 40,000,000 m/2π = 6,366,000m. Thus, the speed you need to stay up indefinitely is v=√(6,366,000 x 9.8) = 7900 m/s = 17,800 mph. That’s roughly Mach 35, or 35 times the speed of sound. You need some altitude too, just to keep air friction from killing you, but for most missions, the main thing you need is velocity, kinetic energy, not potential energy, as i’ll show below. If you achieve more speed than 17,800 m/s, you circle the earth higher up; this makes docking space-ships tricky, as I’ll explain also.

It turns out that kinetic energy is quite a lot more important than potential energy for sending an object into orbit, and that rockets are the only way practical to reach orbital speed; no current cannon or gun can reach Mach 35. To get a sense of the energy balance involved in rocketry, consider a one kg mass at orbital speed, 7900 m/s, and 200 km altitude. You can calculate that the kinetic energy is 31,205 kJ, while the potential energy, mgh, is only 1,960 kJ. For this orbital height, 200 km, the kinetic energy is about 16 times the potential energy. Not that it’s easy to reach 200 miles altitude, but you can do it with a sophisticated cannon, or a “simple”, one stage, V2-style rocket, you need multiple stages to reach 7,900 m/s. As a way to see this, consider that the energy content of gasoline + oxygen is about 10.5 MJ/kg (10,500 kJ/kg); this is only 1/3 of the kinetic energy of the orbital rocket, but it’s 5 times the potential energy. A fairly efficient gasoline + oxygen powered cannon could not provide orbital kinetic energy since the bullet can move no faster than the explosive vapor. In a rocket this is not a constraint since most of the mass is ejected. I’ll explain further below.

A shell fired at a 45° angle that reaches 200 km altitude would go about 800 km — the distance between North Korea and Japan, or between Iran and Israel. That would require twice as much energy as a shell fired straight up, about 4000 kJ/kg. This is a value still within the range for a (very large) cannon or a single-stage rocket. For Russia or China to hit the US would take much more velocity, and orbital, or near orbital rocketry. To reach the moon, you need more total energy, but less kinetic energy. Moon rockets have taken the approach of first going into orbit, and only later going on. While most of the kinetic energy isn’t lost, I’m still not sure it’s the best trajectory.

The force produced by a rocket is equal to the rate of mass shot out times its velocity. F = ∆(mv). To get a lot of force for each bit of fuel, you want the gas exit velocity to be as fast as possible. A typical maximum is about 2,500 m/s. Mach 10, for a gasoline – oxygen engine. The acceleration of the rocket itself is this ∆mv force divided by the total remaining mass in the rocket (rocket shell plus remaining fuel) minus 1 (gravity). Thus, if the exhaust from a rocket leaves at 2,500 m/s, and you want the rocket to accelerate upward at 9 G, you must exhaust fast enough to develop 10 G, 98 m/s2. The rate of mass exhaust is the mass of the rocket times 98/2500 = .0392/second. That is, about 3.9% of the rocket mass must be ejected each second. Assuming that the fuel for your first stage engine is less than 80% of the total mass, the first stage will flare-out in about 20 seconds at this rate. Your acceleration at the end of the 20 seconds will be greater than 9G, by the way, since the rocket gets lighter as fuel is burnt. When half the weight is gone, it will be accelerating at 19 G.

If you have a good math background, you can develop a differential equation for the relation between fuel consumption and altitude or final speed. This is readily done if you know calculous, or reasonably done if you use differential methods. By either method, it turns out that, for no air friction or gravity resistance, you will reach the same speed as the exhaust when 64% of the rocket mass is exhausted. In the real world, your rocket will have to exhaust 75 or 80% of its mass as first stage fuel to reach a final speed of 2,500 m/s. This is less than 1/3 orbital speed, and reaching it requires that the rest of your rocket mass: the engine, 2nd stage, payload, and any spare fuel to handle descent (Elon Musk’s approach) must weigh less than 20-25% of the original weight of the rocket on the launch pad. This gasoline and oxygen is expensive, but not horribly so if you can reuse the rocket; that’s the motivation for NASA’s and SpaceX’s work on reusable rockets. Most orbital rocket designs require at least three stages to accelerate to the 7900 m/s calculated above, and the second stage is almost invariably lost. If you can set-up and solve the differential equation above, a career in science may be for you.

Now, you might wonder about the exhaust speed I’ve been using, 2500 m/s. If you can achieve higher speeds, the rocket design will be a lot easier, but doing so is not easy for a gasoline/ oxygen engine like Russia and the US uses currently. The heat of combustion of gasoline is 42 MJ/kg, but burning a kg of gasoline requires roughly 2.5 kg of oxygen. Thus, for a rocket fueled by gasoline + oxygen, the heat of combustion is about 10.5 MJ/kg. Now assume that the rocket engine is 30% efficient. Per unit of fuel+ oxygen mass, 1/2 v2 = .3 x 10,500,000; v =√6,300,000  = 2500 m/s. Higher exhaust speeds have been achieved, e.g. with hydrogen-fueled rockets. The sources of inefficiency are many including incomplete combustion in the engine, gas flow off the center-line, and friction flow in the engine and between the atmosphere and gases leaving the rocket nozzle. If you can make a reliable, higher efficiency engine, a career in engineering may be for you.

At an average acceleration of 10 G = 98 m/s2 and a first stage that reaches 2500 m/s you find that the first stage will burn out after 25.5 seconds. If the rocket were going straight up (bad idea), you’d find you are at an altitude of about 28.7 km. A better idea would be an average trajectory of 30°, leaving you at an altitude of 14 km or so. At that altitude you can expect to have far less air friction, and you can expect the second stage engine to be more efficient. It seems to me, you may want to wait 15 seconds or so before firing the second stage: you’ll be another few km up and it seems to me that the benefit of this altitude will be worthwhile. I guess that’s why most space launches wait a few seconds before firing the second stage.

As a final bit, I’d mentioned that docking a rocket with a space station is difficult, in part, because docking requires an increase in angular speed, w, but this generally goes along with a decrease in altitude; a counter-intuitive behavior. Setting the acceleration due to gravity equal to the angular acceleration, we find GM/r2 = w2r, where G is the gravitational constant, and M is the mass or the earth. Rearranging, we find that w2  = GM/r3. For high angular speed, you need small r: a low altitude. When we first went to dock a space-ship, in the early 60s, we had not realized this. When the astronauts fired the engines to dock, they found that they’d accelerate in velocity, but not in angular speed: v = wr. The faster they went, the higher up they went, but the lower the angular speed got: the fewer the orbits per day. Eventually they realized that, to dock with another ship or a space-station that is in front of you, you do not accelerate, but decelerate. When you decelerate you lose altitude and gain angular speed: you catch up with the station, but at a lower altitude. Your next step is to angle your ship near-radially to the earth, and accelerate by firing engines to the side till you dock. Like much of orbital rocketry, it’s simple, but not intuitive or easy.

Robert Buxbaum, August 12, 2015. A cannon that could reach from North Korea to Japan, say, would have to be on the order of 10 km long, running along the slope of a mountain. Even at that length, the shell would have to fire at 450 G, or so, and reach a speed about 3000 m/s, or 1/3 orbital.

# Gatling guns and the Spanish American War

I rather like inventions and engineering history, and I regularly go to the SME, a fair of 18th to 19th century innovation. I am generally impressed with how these machines work, but what really brings things out is when talented people use the innovation to do something radical. Case in point, the Gatling gun; invented by Richard J. Gatling in 1861 for use in the Civil war, it was never used there, or in any major war until 1898 when Lieut. John H. Parker (Gatling Gun Parker) showed how to deploy them successfully, and helped take over Cuba. Until then, they were considered another species of short-range, grape-shot cannon, and ignored.

A Gatling gun of the late 1800s. Similar, but not identical to the ones Parker brought along.

Parker had sent his thoughts on how to deploy a Gatling gun in a letter to West Point, but they were ignored, as most new thoughts are. For the Spanish-American War, Parker got 4 of the guns, trained his small detachment to use them, and registered as a quartermaster corp in order to sneak them aboard ship to Cuba. Here follows Theodore Roosevelt’s account of their use.

“On the morning of July 1st, the dismounted cavalry, including my regiment, stormed Kettle Hill, driving the Spaniards from their trenches. After taking the crest, I made the men under me turn and begin volley-firing at the San Juan Blockhouse and entrenchment’s against which Hawkins’ and Kent’s Infantry were advancing. While thus firing, there suddenly smote on our ears a peculiar drumming sound. One or two of the men cried out, “The Spanish machine guns!” but, after listening a moment, I leaped to my feet and called, “It’s the Gatlings, men! It’s our Gatlings!” Immediately the troopers began to cheer lustily, for the sound was most inspiring. Whenever the drumming stopped, it was only to open again a little nearer the front. Our artillery, using black powder, had not been able to stand within range of the Spanish rifles, but it was perfectly evident that the Gatlings were troubled by no such consideration, for they were advancing all the while.

Roosevelt, his volunteers, and the Buffalo soldiers charge up Kettle Hill, Frederick Remington.

Soon the infantry took San Juan Hill, and, after one false start, we in turn rushed the next line of block-houses and intrenchments, and then swung to the left and took the chain of hills immediately fronting Santiago. Here I found myself on the extreme front, in command of the fragments of all six regiments of the cavalry division. I received orders to halt where I was, but to hold the hill at all hazards. The Spaniards were heavily reinforced and they opened a tremendous fire upon us from their batteries and trenches. We laid down just behind the gentle crest of the hill, firing as we got the chance, but, for the most part, taking the fire without responding. As the afternoon wore on, however, the Spaniards became bolder, and made an attack upon the position. They did not push it home, but they did advance, their firing being redoubled. We at once ran forward to the crest and opened on them, and, as we did so, the unmistakable drumming of the Gatlings opened abreast of us, to our right, and the men cheered again. As soon as the attack was definitely repulsed, I strolled over to find out about the Gatlings, and there I found Lieut. Parker with two of his guns right on our left, abreast of our men, who at that time were closer to the Spaniards than any others.

From thence on, Parker’s Gatlings were our inseparable companion throughout the siege. They were right up at the front. When we dug our trenches, he took off the wheels of his guns and put them in the trenches. His men and ours slept in the same bomb-proofs and shared with one another whenever either side got a supply of beans or coffee and sugar. At no hour of the day or night was Parker anywhere but where we wished him to be, in the event of an attack. If a troop of my regiment was sent off to guard some road or some break in the lines, we were almost certain to get Parker to send a Gatling along, and, whether the change was made by day or by night, the Gatling went. Sometimes we took the initiative and started to quell the fire of the Spanish trenches; sometimes they opened upon us; but, at whatever hour of the twenty-four the fighting began, the drumming of the Gatlings was soon heard through the cracking of our own carbines.

Map of the Attack on Kettle Hill and San Juan Hill in the Spanish-American War, July 1, 1898 The Spanish had 760 troops n the in fortified positions defending the crests of the two hills, and 10,000 more defending Santiago. As Americans were being killed in “hells pocket” near the foot of San Juan Hill, from crossfire, Roosevelt, on the right, charged his men, the “Rough Riders” [1st volunteers] and the “Buffalo Soldiers [10th cavalry], up Kettle Hill in hopes of ending the crossfire and of helping to protect troops that would charge further up San Juan Hill. Parker’s Gatlings were about 600 yards from the Spanish and fired some 700 rounds per minute into the Spanish lines. Theyy were then repositioned on the hill to beat back the counter attack. Without the Parker’s Gatling guns, the chances of success would have been small.

I have had too little experience to make my judgment final; but certainly, if I were to command either a regiment or a brigade, whether of cavalry or infantry, I would try to get a Gatling battery–under a good man–with me. I feel sure that the greatest possible assistance would be rendered, under almost all circumstances, by such a Gatling battery, if well handled; for I believe that it could be pushed fairly to the front of the firing-line. At any rate, this is the way that Lieut. Parker used his battery when he went into action at San Juan, and when he kept it in the trenches beside the Rough Riders before Santiago.”

Here is how the Gatling gun works; it’s rather like 5 or more rotating zip guns; a pall pulls and releases the firing pins. Gravity feeds the bullets at the top and drops the shells out the bottom. Lt’ Parker’s deployment innovation was to have them hand-carried to protected positions, near-enough to the front that they could be aimed. The swivel and rapid fire of the guns allowed the shooter to aim them to correct for the drop in the bullets over fairly great distances. This provided rapid-fire accurate protection from positions that could not be readily hit. Shortly after the victory on San Juan HIll, July 1 1898, the Spanish Caribbean fleet was destroyed July 3, Santiago surrendered July 17, and all of Cuba surrendered 4 days later, July 21 (my birthday) — a remarkably short war. While TR may not have figured out how to use the Gatling guns effectively, he at least recognized that Lt. John Parker had.

Roosevelt gave two of these, more modern, Colt-Browning repeating rifles to Parker’s detachment the day after the battle. They were not particularly effective. By WWI, “Gatling Gun” Parker would be a general; by 1901 Roosevelt would be president.

The day after the battle, Col. Roosevelt gifted Parker’s group with two Colt-Browning machine guns that he and his family had bought, but had not used. According to Roosevelt, but these rifles, proved to be “more delicate than the Gatlings, and very readily got out-of-order.” The Brownings are the predecessor of the modern machine gun used in the Boxer Rebellion and for wholesale deaths in WWI and WWII.

Dr. Robert E. Buxbaum, June 9, 2015. The Spanish-American War was a war of misunderstanding and colonialism, but its effects, by and large, were good. The cause, the sinking of the USS Maine, February 15, 1898, was likely a mistake. Spain, a decaying colonial power, was a conservative monarchy under Alfonso XIII; the loss of Cuba seems to have lead to liberalization. The US, a republic, became a colonial power. There is an inherent friction, I think between conservatism and liberal republicanism, Generally, republics have out-gunned and out-produced other countries, perhaps because they reward individual initiative.

# Brass monkey cold

In case it should ever come up in conversation, only the picture at left shows a brass monkey. The other is a bronze statue of some sort of a primate. A brass monkey is a rack used to stack cannon balls into a face centered pyramid. A cannon crew could fire about once per minute, and an engagement could last 5 hours, so you could hope to go through a lot of cannon balls during an engagement (assuming you survived).

Small brass monkey. The classic monkey might have 9 x 9 or 10×10 cannon balls on the lower level.

Bronze sculpture of a primate playing with balls — but look what the balls are sitting on: it’s a dada art joke.

But brass monkeys typically show up in conversation in terms of it being cold enough to freeze the balls off of a brass monkey, and if you imagine an ornamental statue, you’d never guess how cold could that be. Well, for a cannonball holder, the answer has to do with the thermal expansion of metals. Cannon balls were made of iron and the classic brass monkey was made of brass, an alloy with a much-greater thermal expansion than iron. As the temperature drops, the brass monkey contracts more than the iron balls. When the drop is enough the balls will fall off and roll around.

The thermal expansion coefficient of brass is 18.9 x 10-4/°C while the thermal expansion coefficient of iron is 11.7 x10-4/°C. The difference is 7.2×10-4/°C; this will determine the key temperature. Now consider a large brass monkey, one with 10 x 10 holes on the lower level, 81 at the second, and so on. Though it doesn’t affect the result, we’ll consider a monkey that holds 12 lb cannon balls, a typical size of 1750 -1830. Each 12 lb ball is 4.4″ in diameter at room temperature, 20°C in those days. At 20°C, this monkey is about 44″ wide. The balls will fall off when the monkey shrinks more than the balls by about 1/3 of a diameter, 1.5″.

We can calculate ∆T, the temperature change, °C, that is required to lower the width-difference by 1.5″ as follows:

-1.5″ = ∆T x 44″ x 7.2 x10-4

We find that ∆T = -47°C. The temperature where this happens is 47 degrees cooler than 20°C, or -27°C. That’s 3.2°F, not a very unusual temperature on land, e.g. in Detroit, but on sea, the temperature is rarely much colder than 0°C or 32°F, the temperature where water freezes. If it gets to 3.2°F on the sea, something is surely amiss. To avoid this problem, land-based army cannon-crew uses a smaller brass monkey — e.g. the 5×5 shown. This stack holds 1/7 as many balls, but holds them to -74°F, a really cold temperature.

Robert E. Buxbaum, February 21, 2015. Some fun thoughts: Convince yourself that the key temperature is independent of the size of the cannon balls. That is, that I didn’t need to choose 12 pounders. A bit more advanced, what is the equation for the number of balls on any particular base-size monkey. Show that the packing density is no more efficient if the bottom lawyer were an equilateral triangle, and not a square. If you liked this, you might want to know how much wood a woodchuck chucks if a woodchuck could chuck wood, or on the relationship between mustaches and WWII diplomacy.