Tag Archives: materials

Transparent, super wood

Leave a reply

As mentioned in a previous post, wood is more among the strongest materials per unit weight, making it ideal for table tops and telephone polls. On a per pound basis, most species of wood are more than twice as strong as aluminum or mild steel. Wood’s structure is is the reason; it’s a natural composite of air-filled, aligned tubes of crystalline cellulose, held together by natural glue, lignin.

In terms of raw strength though, pounds/in2, wood is not particularly strong, only about 7000 psi (45MPa) both in tension and compression, about half the strength of aluminum. It is thus not well suited to supporting heavy structures, like skyscrapers. (I calculate the maximum height of a skyscraper here), but wood can be modified to make it stronger by removing most of the air, and replacing it with plastic. The result is a stronger, denser, flexible composite, that is typically transparent. The flower below is seen behind a sheet of transparent wood.

A picture of a flower taken through a piece of transparent super-wood.

To make a fairly strong, transparent wood, you take ordinary low-density wood (beech or balsa are good) and soak it in alkali (NaOH). This bleaches the wood, softens the cellulose, and dissolves most of the lignin. You next wash off the alkali and soak the wood in a low viscosity epoxy or acrylic. Now, put it in a vacuum chamber to remove the air — you’ll need a brick to hold the wood down in the liquid. You’ll see bubbles in the epoxy as the air leaves. Then, when the vacuum is released, the wood soaks up the epoxy or acrylic. On curing, you get a composite strong and transparent, but not super strong.

To make the wood really strong, super-strong, you need to compress the uncured, epoxy soaked wood. One method is to put it in a vice. This drives off more of the air and further aligns the cellulose fibers. You now cure it as before (you need a really slow cure epoxy or a UV-cure polymer). The resultant product have been found to have tensile strengths as high as 270 MPa in the direction of alignment, over 40,000 psi. This is three times stronger than regular aluminum, 90 MPa, (13,500 psi). It’s about the strength of the strongest normal aluminum alloy, 6061. It’s sort of expensive to make, but it’s flexible and transparent, making it suitable for space windows and solar cells. It’s the lightest flexible transparent material known. It’s biodegradable, and that’s very cool, IMHO. See here for a comparison with other, high strength, transparent composites.

Robert Buxbaum, November 10, 2022. I think further developments along this line would make an excellent high school science fair project, college thesis, or PhD research project. Compare different woods, or epoxies, different alkalis, and temperatures, or other processing ideas. How strong and transparent can you make this material, or look at other uses. Can you use it for roof solar cells, like Musk’s but lighter, or mold it for auto panels, it’s already lighter and stronger, or use it as bullet-proof glass or airplane windows.

A Nuclear-blast resistant paint: Starlite and co.

Leave a reply

About 20 years ago, an itinerate inventor named Maurice Ward demonstrated a super insulating paint that he claimed would protect most anything from intense heat. He called it Starlite, and at first no one believed the claims. Then he demonstrated it on TV, see below, by painting a paper-thin layer on a raw egg. He then blasting the egg with a blow torch for a minute till the outside glowed yellow-red. He then lifted the egg with his hand; it was barely warm! And then, on TV, he broke the shell to show that the insides were totally raw, not only uncooked but completely unchanged, a completely raw egg. The documentary below shows the demonstration and describes what happened next (as of 10 years ago) including an even more impressive series of tests.

Intrigued, but skeptical, researchers at the US White Sands National Laboratory, our nuclear bomb test lab, asked for samples. Ward provided pieces of wood painted as before with a “paper thin” layer of Starlite. They subjected these to burning with an oxyacetylene torch, and to a simulated nuclear bomb blast. The nuclear fireball radiation was simulated by an intense laser at the site. Amazing as it sounds, the paint and the wood beneath emerging barely scorched. The painted wood was not damaged by the laser, nor by an oxyacetylene torch that could burn through 8 inches of steel in seconds.

The famous egg, blow torch experiment.

The inventor wouldn’t say what the paint was made of, or what mechanism allowed it to do this, but clearly it had military and civilian uses. It seems it would have prevented the twin towers from collapsing, or would have greatly extended the time they stayed standing. Similarly, it would protect almost anything from a flame-thrower.

As for the ingredients, Ward said it was non-toxic, and that it contained mostly organic materials, plus borax and some silica or ceramic. According to his daughter, it was “edible”; they’d fed it to dogs and horses without adverse effects.

Starlite coasted wood. The simulated nuclear blast made the char mark at left.

The White sands engineers speculate that the paint worked by combination of ablation and intumescence, controlled swelling. The surface, they surmised, formed a foam of char, pure carbon, that swelled to make tiny chambers. If these chambers are small enough, ≤10 nm or so, the mean free path of gas molecules will be severely reduced, reducing the potential for heat transfer. Even more insulting would be if the foam chambers were about 1 nm. Such chambers will be, essentially air free, and thus very insulating. For a more technical view of how molecule motion affects heat transfer rates, see my essay, here.

Sorry to say we don’t know how big the char chambers are, or if this is how the material works. Ward retained the samples and the formula, and didn’t allow close examination. Clearly, if it works by a char, the char layer is very thin, a few microns at most.

Because Maurice Ward never sold the formula or any of the paint in his lifetime, he made no money on the product. He kept closed muted about it, as he knew that, as soon as he patented, or sold, or let anyone know what was in the paint, there would be copycats, and patent violations, and leaks of any secret formula. Even in the US, many people and companies ignore patent rights, daring you to challenge them in court. And it’s worse in foreign countries where the government actively encourages violation. There are also legal ways around a patent: A copycat inventor looks for ways to get the same behavior from materials that are not covered in the patent. Ward could not get around these issues, so he never patented the formula or sold the rights. He revealed the formula only to some close family members, but that was it till May, 2020, when a US company, Thermashield, LLC, bought Ward’s lab equipment and notes. They now claim to make the original Starlite. Maybe they do. The product doesn’t seem quite as good. I’ve yet to see an item scorched as little as the sample above.

Many companies today are now selling versions of Starlite. The formulas are widely different, but all the paints are intumescent, and all the formulas are based on materials Ward would have had on hand, and on the recollections of the TV people and those at White Sands. I’ve bought one of these copycat products, not Thermashield, and tested it. It’s not half bad: thicker in consistency than the original, or as resistive.

There are home-made products too, with formulas on the internet and on YouTube. They are applied more like a spackle or a clay. Still, these products insulate remarkably well: a lot better than any normal insulator I’d seen.

If you’d like to try this as a science fair project, among the formulas you can try; a mix of glue, baking soda, borax, and sugar, with some water. Some versions use sodium silicate too. The Thermoshield folks say that this isn’t the formula, that there is no PVA glue or baking soda in their product. Still it works.

Robert Buxbaum, March 13, 2022. Despite my complaints about the US patent system, it’s far better than in any other country I’ve explored. In most countries, patents are granted only as an income stream for the government, and inventors are considered villains: folks who withhold the fruits of their brains for unearned money. Horrible.

Wood, the strongest material for some things, like table-tops

2 Replies

Natural wood has a lower critical strength than most modern materials, and a lower elastic constant, yet it is the strongest material for some applications because it is remarkably light and remarkably cheap on a per-volume or weight. In some important applications, high strength per volume is the important measure, and in virtually every case high strength per dollar is relevant. Consider the table top: it should support a person standing on it, as one might do to change a lightbulb, and it should not weigh too much, or cost too much.

A 250 lb man on a table. The table should not weight too much, nor cost too much, yet it should support the man.

I’ve drawn a 9 foot by 4 foot table at left, with a 250 lb person in the center. Assuming that the thickness of the table is t, the deflection in the center, ∂, is found by the formula ∂ =FL3/4Ewt3. Here, F is the downward force, 250 lbs (a bit higher if we include the weight of the table), L is the length between the supports, 6 feet = 72 inches, E is the elastic constant of the table top, 2,300,000 psi assuming ash wood, w is the width of the table, 48″, and t is the thickness, let’s say 1″.

Using the formula above, we fid that the deflection of this tabletop is 0.211″ for a force of 250 lbs. That’s not bad. The weight of the 9′ table top is 125 lbs, which is not too bad either, and the cost is likely going to be acceptable: ash is a fairly cheap, nice-looking wood.

By comparison, consider using a 1/4″ thick sheet of structural aluminum, alloy 6061. The cost will be much higher and the weight will be the same as for the 1′ thick piece of ash. That’s because the density of aluminum is 2.7 g/cc, more than three times that of ash. Aluminum 6061is four times stiffer than ash, with an elastic constant of 10,000,000 psi, but the resistance to bending is proportional to thickness cubed; and 1/4 cubed is 1/64. We thus find that the 125 lb tabletop of Al alloy will deflect 3.11 inches, about 16 times more than ash, far too much to be acceptable. We could switch to thicker aluminum, 3/8″ for example, but the weight would be 50% higher now, the cost would be yet 50% higher, and the deflection would still be too high, 0.92 inches. Things get even worse with steel since steel is yet-denser, a 1/4″ sheet of steel would deflect about as much as the 3/8″ aluminum, but would weigh about twice as muc. For this application, and many others like it, wood is likely the best choice; its light weight per strength and low cost can’t be beat.

Robert E. Buxbaum, January 11, 2022

Maximum height of an NYC skyscraper, including wind.

Leave a reply

Some months ago, I demonstrated that the maximum height of a concrete skyscraper was 45.8 miles, but there were many unrealistic assumptions. The size of the base was 100 mi2, about that of Sacramento, California; the shape was similar to that of the Eiffel tower, and there was no wind. This height is semi-reasonable; it’s about that of the mountains on Mars where there is a yellow sky and no wind, but it is 100 times taller than the tallest skyscraper on earth. the Burj Khalifa in Dubai, 2,426 ft., shown below. Now I’d like to include wind, and limit the skyscraper to a straight tower of a more normal size, a city-block square of manhattan, New York real-estate. That’s 198 feet on a side; this is three times the length of Gunther’s surveying chain, the standard for surveying in 1800.

Burj Khalifa, the world’s tallest building, Concrete + glass structure. Dubai tourism image.

As in our previous calculation, we can find the maximum height in the absence of windby balancing the skyscrapers likely strength agains its likely density. We’ll assume the structure is made from T1 steel, a low carbon, vanadium steel used in bridges, further assume that the structure occupies 1/10 of the floor area. Because the structure is only 1/10 of the area, the average yield strengthener the floor area is 1/10 that of T1 steel. This is 1/10 x 100,000 psi (pounds per square inch) = 10,000 psi. The density of T1 steel is 0.2833 pounds per cubic inch, but we’ll assume that the density of the skyscraper is about 1/4 this; (a skyscraper is mostly empty space). We find the average is 0.07 pounds per cubic inch. The height, is the strength divided by the density, thus

H’max-tower = 10,000psi / 0.07 p/in3 = 142, 857 inches = 11, 905 feet = 3629 m,

This is 4 1/4 times higher than the Burj Khalifa. The weight of this structure found from the volume of the structure times its average density, or 0.07 pounds per cubic inch x 123 x 1982x 11,905 = 56.45 billion pounds, or, in SI units, a weight of 251 GNt.

Lets compare this to the force of a steady wind. A steady wind can either either tip over the building by removing stress on the upwind side, or add so much extra stress to the down-wind side that the wall fails. The force of the wind is proportionals to the wind’s energy dissipation rate. I’ll assume a maximum wind speed of 120 mph, or 53.5 m/s. The force of the wind equals the area of the building, times a form factor, ƒ, times the rate of kinetic energy dissipation, 1/2ρv2. Thus,

F= (Area)*ƒ* 1/2ρv2, where ρ is the density of air, 1.29kg/m3.

The form factor, ƒ, is found to be 1.15 for a flat plane. I’ll presume that’s also the form factor for a skyscraper. I’ll take the wind area as

Area = W x H,

where W is the width of the tower, 60.35 m in SI, and the height, H, is what we wish to determine. It will be somewhat less than H’max-tower, =3629 m, the non-wind height. As an estimate for how much less, assume H = H’max-tower, =3629 m.
For this height tower, the force of the wind is found to be:

F = 3629 * 60.35* 2123 = 465 MNt.

This is 1/500 the weight of the building, but we still have to include the lever effect. The building is about 60.1 times taller than it is wide, and as a result the 465 MNt sideways force produces an additional 28.0 GNt force on the down-wind side, plus and a reduction of the same amount upwind. This is significant, but still only 1/9 the weight of the building. The effect of the wind therefore is to reduce the maximum height of this New York building by about 9 %, to a maximum height of 2.05 miles or 3300 m.

The tallest building of Europe is the Shard; it’s a cone. The Eiffel tower, built in the 1800s, is taller.

A cone is a better shape for a very tall tower, and it is the shape chosen for “the shard”, the second tallest building in Europe, but it’s not the ideal shape. The ideal, as before, is something like the Eiffel tower. You can show, though I will not, that even with wind, the maximum height of a conical building is three times as high as that of a straight building of the same base-area and construction. That is to say that the maximal height of a conical building is about 6 miles.

In the old days, one could say that a 2 or 6 mile building was inconceivable because of wind vibration, but we’ve found ways to deal with vibration, e.g. by using active damping. A somewhat bigger problem is elevators. A very tall building needs to have elevators in stages, perhaps 1/2 mile stages with exchanges (and shopping) in-between. Yet another problem is fire. To some extent you eliminate these problems by use of pre-mixed concrete, as was used in the Trump tower in New York, and later in the Burj Khalifa in Dubai.

The compressive strength of high-silica, low aggregate, UHPC-3 concrete is 135 MPa (about 19,500 psi), and the density is 2400 kg/m3 or about 0.0866 lb/in3. I will assume that 60% of the volume is empty and that 20% of the weight is support structure (For the steel building, above, I’d assumed 3/4 and 10%). In the absence of wind,

H’max-cylinder-concrete = .2 x 19,500 psi/(0.4 x.0866  lb/in3) = 112,587″ = 9,382 ft = 1.77 miles. This building is 79% the height of the previous, steel building, but less than half the weight, about 22,000,000,000 pounds. The effect of the wind will be to reduce the above height by about 14%, to 1.52 miles. I’m not sure that’s a fire-safe height, but it is an ego-boost height.

Robert Buxbaum. December 29, 2019.

Thermal stress failure

1 Reply

Take a glass, preferably a cheap glass, and set it in a bowl of ice-cold water so that the water goes only half-way up the glass. Now pour boiling hot water into the glass. In a few seconds the glass will crack from thermal stress, the force caused by heat going from the inside of the glass outside to the bowl of cold water. This sort of failure is not mentioned in any of the engineering material books that I had in college, or had available for teaching engineering materials. To the extent that it is mentioned mentioned on the internet, e.g. here at wikipedia, the metric presented is not derived and (I think) wrong. Given this, I’d like to present a Buxbaum- derived metric for thermal stress-resistance and thermal stress failure. A key aspect: using a thinner glass does not help.

Before gong on to the general case of thermal stress failure, lets consider the glass, and try to compute the magnitude of the thermal stress. The glass is being torn apart and that suggests that quite a lot of stress is being generated by a ∆T of 100°C temeprarture gradient.

To calcule the thermal stress, consider the thermal expansivity of the material, α. Glass — normal cheap glass — has a thermal expansivity α = 8.5 x10-6 meters/meter °C (or 8.5 x10-6 foot/foot °C). For every degree Centigrade a meter of glass is heated, it will expand 8.5×10-6 meters, and for every degree it is cooled, it will shrink 8.5 x10-6 meters. If you consider the circumference of the glass to be L (measured in meters), then
∆L/L = α ∆T.

where ∆L is the change in length due to heating, and ∆L/L is sometimes called the “strain.”. Now, lets call the amount of stress caused by this expansion σ, sigma, measured in psi or GPa. It is proportional to the strain, ∆L/L, and to the elasticity constant, E (also called Young’s elastic constant).

σ = E ∆L/L.

For glass, Young’s elasticity constant, E = 75 GPa. Since strain was equal to α ∆T, we find that

σ =Eα ∆T 

Thus, for glass and a ∆T of 100 °C, σ =100°C x 75 GPa x 8.5 x10-6 /°C  = 0.064  GPa = 64MPa. This is about 640 atm, or 9500 psi.

As it happens, the ultimate tensile strength of ordinary glass is only about 40 MPa =  σu. This, the maximum force per area you can put on glass before it breaks, is less than the thermal stress. You can expect a break here, and wherever σu < Eα∆T. I thus create a characteristic temperature difference for thermal stress failure:

The Buxbaum failure temperature, ß = σu/Eα

If ∆T of more than ß is applied to any material, you can expect a thermal stress failure.

The Wikipedia article referenced above provides a ratio for thermal resistance. The usits are perhaps heat load per unit area and time. How you would use this ratio I don’t quite know, it includes k, the thermal conductivity and ν, the Poisson ratio. Including the thermal conductivity here only makes sense, to me, if you think you’ll have a defined thermal load, a defined amount of heat transfer per unit area and time. I don’t think this is a normal way to look at things.  As for including the Poisson ratio, this too seems misunderstanding. The assumption is that a high Poisson ratio decreases the effect of thermal stress. The thought behind this, as I understand it, is that heating one side of a curved (the inside for example) will decrease the thickness of that side, reducing the effective stress. This is a mistake, I think; heating never decreases the thickness of any part being heated, but only increases the thickness. The heated part will expand in all directions. Thus, I think my ratio is the correct one. Please find following a list of failure temperatures for various common materials. 

Stress strain properties of engineering materials including thermal expansion, ultimate stress, MPa, and Youngs elastic modulus, GPa.

You will notice that most materials are a lot more resistant to thermal stress than glass is and some are quite a lot less resistant. Based on the above, we can expect that ice will fracture at a temperature difference as small as 1°C. Similarly, cast iron will crack with relatively little effort, while steel is a lot more durable (I hope that so-called cast iron skillets are really steel skillets). Pyrex is a form of glass that is more resistant to thermal breakage; that’s mainly because for pyrex, α is a lot smaller than for ordinary, cheap glass. I find it interesting that diamond is the material most resistant to thermal failure, followed by invar, a low -expansion steel, and ordinary rubber.

Robert E. Buxbaum, July 3, 2019. I should note that, for several of these materials, those with very high thermal conductivities, you’d want to use a very thick sample of materials to produce a temperature difference of 100*C.

How tall could you make a skyscraper?

3 Replies

Built in 1931, the highest usable floor space of the Empire State building is 1250 feet (381m) above the ground. In 1973, that record was beaten by the World Trade Center building 1, 1,368 feet (417 m, building 2 was eight feet shorter). The Willis Tower followed 1974, and by 2004, the tallest building was the Taipei Tower, 1471 feet. Building heights had grown by 221 feet since 1931, and then the Burj Khalifa in Dubai, 2,426 ft ( 739.44m):. This is over 1000 feet taller than the new freedom tower, and nearly as much taller than the previous record holder. With the Saudi’s beginning work on a building even taller, it’s worthwhile asking how tall you could go, if your only  limitations were ego and materials’ strength.

Burj Khalifa, the world’s tallest building, Concrete + glass structure. Dubai tourism image.

Having written about how long you could make a (steel) suspension bridge, the maximum height of a skyscraper seems like a logical next step. At first glance this would seem like a ridiculously easy calculation based on the math used to calculate the maximum length of a suspension bridge. As with the bridge, we’d make the structure from the strongest normal material: T1, low carbon, vanadium steel, and we’d determine the height by balancing this material’s  yield strength, 100,000 psi (pounds per square inch), against its density, .2833 pounds per cubic inch.

If you balance these numbers, you calculate a height: 353,000 inches, 5.57 miles, but this is the maximum only for a certain structure, a wide flag-pole of T1 steel in the absent of wind. A more realistic height assumes a building where half the volume is empty space, used for living and otherwise, where 40% of the interior space contains vertical columns of T1 steel, and where there’s a significant amount of dead-weight from floors, windows, people, furniture, etc. Assume the dead weight is the equivalent of filling 10% of the volume with T1 steel that provides no structural support. The resulting building has an average density = (1/2 x 0.2833 pound/in3), and the average strength= (0.4 x 100,000 pound/in2). Dividing these numbers we get a maximum height, but only for a cylindrical building with no safety margin, and no allowance for wind.

H’max-cylinder = 0.4 x 100,000 pound/in2/ (.5 x 0.2833 pound/in3) = 282,400 inches = 23,532 ft = 4.46 miles.

This is more than ten times the Burj Khalifa, but it likely underestimates the maximum for a steel building, or even a concrete building because a cylinder is not the optimum shape for maximum height. If the towers were constructed conical or pyramidal, the height could be much greater: three times greater because the volume of a cone and thus its weight is 1/3 that of a cylinder for the same base and height. Using the same materials and assumptions,

The tallest building of Europe is the Shard; it’s a cone. The Eiffel tower, built in the 1800s, is taller.

H’max-cone = 3 H’max-cylinder =  13.37 miles.

A cone is a better shape for a very tall tower, and it is the shape chosen for “the shard”, the second tallest building in Europe, but it’s not the ideal shape. The ideal, as we’ll see, is something like the Eiffel tower.

Before speaking about this shape, I’d like to speak about building materials. At the heights we’re discussing, it becomes fairly ridiculous to talk about a steel and glass building. Tall steel buildings have serious vibration problems. Even at heights far before they are destroyed by wind and vibration , the people at the top will begin to feel quite sea-sick. Because of this, the tallest buildings have been constructed out of concrete and glass. Concrete is not practical for bridges since concrete is poor in tension, but concrete can be quite strong in compression, as I discussed here.  And concrete is fire resistant, sound-deadening, and vibration dampening. It is also far cheaper than steel when you consider the ease of construction. The Trump Tower in New York and Chicago was the first major building here to be made this way. It, and it’s brother building in Chicago were considered aesthetic marvels until Trump became president. Since then, everything he’s done is ridiculed. Like the Trump tower, the Burj Khalifa is concrete and glass, and I’ll assume this construction from here on.

let’s choose to build out of high-silica, low aggregate, UHPC-3, the strongest concrete in normal construction use. It has a compressive strength of 135 MPa (about 19,500 psi). and a density of 2400 kg/m3 or about 0.0866 lb/in3. Its cost is around $600/m3 today (2019); this is about 4 times the cost of normal highway concrete, but it provides about 8 times the compressive strength. As with the steel building above, I will assume that, at every floor, half of the volume is living space; that 40% is support structure, UHPC-3, and that the other 10% is other dead weight, plumbing, glass, stairs, furniture, and people. Calculating in SI units,

H’max-cylinder-concrete = .4 x 135,000,000 Pa/(.5 x 2400 kg/m3 x 9.8 m/s2) = 4591 m = 2.85 miles.

The factor 9.8 m/s2 is necessary when using SI units to account for the acceleration of gravity; it converts convert kg-weights to Newtons. Pascals, by the way, are Newtons divided by square meters, as in this joke. We get the same answer with less difficulty using inches.

H’max-cylinder-concrete = .4 x 19,500 psi/(.5 x.0866  lb/in3) = 180,138″ = 15,012 ft = 2.84 miles

These maximum heights are not as great as for a steel construction, but there are a few advantages; the price per square foot is generally less. Also, you have fewer problems with noise, sway, and fire: all very important for a large building. The maximum height for a conical concrete building is three times that of a cylindrical building of the same design:

H’max–cone-concrete = 3 x H’max-cylinder-concrete = 3 x 2.84 miles = 8.53 miles.

Mount Everest, picture from the Encyclopedia Britannica, a stone cone, 5.5 miles high.

That this is a reasonable number can be seen from the height of Mount Everest. Everest is rough cone , 5.498 miles high. This is not much less than what we calculate above. To reach this height with a building that withstands winds, you have to make the base quite wide, as with Everest. In the absence of wind the base of the cone could be much narrower, but the maximum height would be the same, 8.53 miles, but a cone is not the optimal shape for a very tall building.

I will now calculate the optimal shape for a tall building in the absence of wind. I will start at the top, but I will aim for high rent space. I thus choose to make the top section 31 feet on a side, 1,000 ft2, or 100 m2. As before, I’ll make 50% of this area living space. Thus, each apartment provides 500 ft2 of living space. My reason for choosing this size is the sense that this is the smallest apartment you could sell for a high premium price. Assuming no wind, I can make this part of the building a rectangular cylinder, 2.84 miles tall, but this is just the upper tower. Below this, the building must widen at every floor to withstand the weight of the tower and the floors above. The necessary area increases for every increase in height as follows:

dA/dΗ = 1/σ dW/dH.

Here, A is the cross-sectional area of the building (square inches), H is height (inches), σ is the strength of the building material per area of building (0.4 x 19,500 as above), and dW/dH is the weight of building per inch of height. dW/dH equals  A x (.5 x.0866  lb/in3), and

dA/dΗ = 1/ ( .4 x 19,500 psi) x A x (.5 x.0866  lb/in3).

dA/A = 5.55 x 10-6 dH,

∫dA/A = ∫5.55 x 10-6 dH,

ln (Abase/Atop) = 5.55 x 10-6 ∆H,

Here, (Abase/Atop) = Abase sq feet /1000, and ∆H is the height of the curvy part of the tower, the part between the ground and the 2.84 mile-tall, rectangular tower at the top.

Since there is no real limit to how big the base can be, there is hardly a limit to how tall the tower can be. Still, aesthetics place a limit, even in the absence of wind. It can be shown from the last equation above that stability requires that the area of the curved part of the tower has to double for every 1.98 miles of height: 1.98 miles = ln(2) /5.55 x 10-6 inches, but the rate of area expansion also keeps getting bigger as the tower gets heavier.  I’m going to speculate that, because of artistic ego, no builder will want a tower that slants more than 45° at the ground level (the Eiffel tower slants at 51°). For the building above, it can be shown that this occurs when:

dA/dH = 4√Abase.  But since

dA/dH = A 5.55 x 10-6 , we find that, at the base,

5.55 x 10-6 √Abase = 4.

At the base, the length of a building side is Lbase = √Abase=  4 /5.55 x 10-6 inches = 60060 ft = 11.4  miles. Artistic ego thus limits the area of the building to slightly over 11 miles wide of 129.4 square miles. This is about the area of Detroit. From the above, we calculate the additional height of the tower as

∆H = ln (Abase/Atop)/ 5.55 x 10-6 inches =  15.1/ 5.55 x 10-6 inches = 2,720,400 inches = 226,700 feet = 42.94 miles.

Hmax-concrete =  2.84 miles + ∆H = 45.78 miles. This is eight times the height of Everest, and while air pressure is pretty low at this altitude, it’s not so low that wind could be ignored. One of these days, I plan to show how you redo this calculation without the need for calculus, but with the inclusion of wind. I did the former here, for a bridge, and treated wind here. Anyone wishing to do this calculation for a basic maximum wind speed (100 mph?) will get a mention here.

From the above, it’s clear that our present buildings are nowhere near the maximum achievable, even for construction with normal materials. We should be able to make buildings several times the height of Everest. Such Buildings are worthy of Nimrod (Gen 10:10, etc.) for several reasons. Not only because of the lack of a safety factor, but because the height far exceeds that of the highest mountain. Also, as with Nimrod’s construction, there is a likely social problem. Let’s assume that floors are 16.5 feet apart (1 rod). The first 1.98 miles of tower will have 634 floors with each being about the size of Detroit. Lets then assume the population per floor will be about 1 million; the population of Detroit was about 2 million in 1950 (it’s 0.65 million today, a result of bad government). At this density, the first 1.98 miles will have a population of 634 million, about double that of the United States, and the rest of the tower will have the same population because the tower area contracts by half every 1.98 miles, and 1/2 + 1/4 + 1/8 + 1/16 … = 1.

Nimrod examining the tower, Peter Breugel

We thus expect the tower to hold 1.28 Billion people. With a population this size, the tower will develop different cultures, and will begin to speak different languages. They may well go to war too — a real problem in a confined space. I assume there is a moral in there somewhere, like that too much unity is not good. For what it’s worth, I even doubt the sanity of having a single government for 1.28 billion, even when spread out (e.g. China).

Robert Buxbaum, June 3, 2019.

The hydrogen jerrycan

1 Reply

Here’s a simple invention, one I’ve worked on off-and-on for years, but never quite built. I plan to work on it more this summer, and may finally build a prototype: it’s a hydrogen Jerry can. The need to me is terrifically obvious, but the product does not exist yet.

To get a view of the need, imagine that it’s 5-10 years in the future and you own a hydrogen, fuel cell car. You’ve run out of gas on a road somewhere, per haps a mile or two from the nearest filling station, perhaps more. You make a call to the AAA road-side service and they show up with enough hydrogen to get you to the next filling station. Tell me, how much hydrogen did they bring? 1 kg, 2 kg, 5 kg? What did the container look like? Is there one like it in your garage?

The original, German "Jerry" can. It was designed at the beginning of WWII to help the Germans to overrun Europe.

The original, German “Jerry” can. It was designed at the beginning of WWII to help the Germans to overrun Europe. I imagine the hydrogen version will be red and roughly these dimensions, though not quite this shape.

I figure that, in 5-10 years these hydrogen containers will be so common that everyone with a fuel cell car will have one, somewhere. I’m pretty confident too that hydrogen cars are coming soon. Hydrogen is not a total replacement for gasoline, but hydrogen energy provides big advantages in combination with batteries. It really adds to automotive range at minimal cost. Perhaps, of course this is wishful thinking as my company makes hydrogen generators. Still it seems worthwhile to design this important component of the hydrogen economy.

I have a mental picture of what the hydrogen delivery container might look like based on the “Jerry can” that the Germans (Jerrys) developed to hold gasoline –part of their planning for WWII. The story of our reverse engineering of it is worth reading. While the original can was green for camouflage, modern versions are red to indicate flammable, and I imagine the hydrogen Jerry will be red too. It must be reasonably cheap, but not too cheap, as safety will be a key issue. A can that costs $100 or so does not seem excessive. I imagine the hydrogen Jerry can will be roughly rectangular like the original so it doesn’t roll about in the trunk of a car, and so you can stack a few in your garage, or carry them conveniently. Some folks will want to carry an extra supply if they go on a long camping trip. As high-pressure tanks are cylindrical, I imagine the hydrogen-jerry to be composed of two cylinders, 6 1/2″ in diameter about. To make the rectangular shape, I imagine the cylinders attached like the double pack of a scuba diver. To match the dimensions of the original, the cylinders will be 14″ to 20″ tall.

I imagine that the hydrogen Jerry can will have at least two spouts. One spout so it can be filled from a standard hydrogen dispenser, and one so it can be used to fill your car. I suspect there may be an over-pressure relief port as well, for safety. The can can’t be too heavy, no more than 33 lbs, 15 kg when full so one person can handle it. To keep the cost and weight down, I imagine the product will be made of marangeing steel wrapped in kevlar or carbon fiber. A 20 kg container made of these materials will hold 1.5 to 2 kg of hydrogen, the equivalent of 2 gallons of gasoline.

I imagine that the can will have at least one handle, likely two. The original can had three handles, but this seems excessive to me. The connection tube between two short cylinders could be designed to serve as one of the handles. For safety, the Jerrycan should have a secure over-seal on both of the fill-ports, ideally with a safety pin latch minimize trouble in a crash. All the parts, including the over- seal and pin, should be attached to the can so that they are not easily lost. Do you agree? What else, if anything, do you imagine?

Robert Buxbaum, February 26, 2017. My company, REB Research, makes hydrogen generators and purifiers.

An Aesthetic of Mechanical Strength

6 Replies

Back when I taught materials science to chemical engineers, I used the following poem to teach my aesthetic for the strength target for product design:

The secret to design, as the parson explained, is that the weakest part must withstand the strain. And if that part is to withstand the test, then it must be made as strong as all the rest. (by R.E. Buxbaum, based on “The Wonderful, One-hoss Shay, by Oliver Wendell Holmes, 1858).

My thought was, if my students had no idea what good mechanical design looked like, they’d never  be able to it well. I wanted them to realize that there is always a weakest part of any device or process for every type of failure. Good design accepts this and designs everything else around it. You make sure that the device will fail at a part of your choosing, when it fails, preferably one that you can repair easily and cheaply (a fuse, or a door hinge), and which doesn’t cause too much mayhem when it fails. Once this failure part is chosen and in place, I taught that the rest should be stronger, but there is no point in making any other part of that failure chain significantly stronger than the weakest link. Thus for example, once you’ve decided to use a fuse of a certain amperage, there is no point in making the rest of the wiring take more than 2-3 times the amperage of the fuse.

This is an aesthetic argument, of course, but it’s important for a person to know what good work looks like (to me, and perhaps to the student) — beyond just by compliments from the boss or grades from me. Some day, I’ll be gone, and the boss won’t be looking. There are other design issues too: If you don’t know what the failure point is, make a prototype and test it to failure, and if you don’t like what you see, remodel accordingly. If you like the point of failure but decide you really want to make the device stronger or more robust, be aware that this may involve strengthening that part only, or strengthening the entire chain of parts so they are as failure resistant as this part (the former is cheaper).

I also wanted to teach that there are many failure chains to look out for: many ways that things can wrong beyond breaking. Check for failure by fire, melting, explosion, smell, shock, rust, and even color change. Color change should not be ignored, BTW; there are many products that people won’t use as soon as they look bad (cars, for example). Make sure that each failure chain has it’s own known, chosen weak link. In a car, the paint on a car should fade, chip, or peel some (small) time before the metal underneath starts rusting or sagging (at least that’s my aesthetic). And in the DuPont gun-powder mill below, one wall should be weaker so that the walls should blow outward the right way (away from traffic).Be aware that human error is the most common failure mode: design to make things acceptably idiot-proof.

Dupont powder mills had a thinner wall and a stronger wall so that, if there were an explosion it would blow out towards the river. This mill has a second wall to protect workers. The thinner wall should be barely strong enough to stand up to wind and rain; the stronger walls should stand up to explosions that blow out the other wall.

Dupont powder mills had a thinner wall and a stronger wall so that, if there were an explosion, it would blow out ‘safely.’ This mill has a second wall to protect workers. The thinner wall must be strong enough to stand up to wind and rain; the stronger walls should stand up to all likely explosions.

Related to my aesthetic of mechanical strength, I tried to teach an aesthetic of cost, weight, appearance, and green: Choose materials that are cheaper, rather than more expensive; use less weight rather than more if both ways worked equally well. Use materials that look better if you’ve got the choice, and use recyclable materials. These all derive from the well-known axiom, omit needless stuff. Or, as William of Occam put it, “Entia non sunt multiplicanda sine necessitate.” As an aside, I’ve found that, when engineers use Latin, we look smart: “lingua bona lingua motua est.” (a good language is a dead language) — it’s the same with quoting 19th century poets, BTW: dead 19th century poets are far better than undead ones, but I digress.

Use of recyclable materials gets you out of lots of problems relative to materials that must be disposed of. E.g. if you use aluminum insulation (recyclable) instead of ceramic fiber, you will have an easier time getting rid of the scrap. As a result, you are not as likely to expose your workers (or you) to mesothelioma, or similar disease. You should not have to pay someone to haul away excess or damaged product; a scraper will oblige, and he may even pay you for it if you have enough. Recycling helps cash flow with decommissioning too, when money is tight. It’s better to find your $1 worth of scrap is now worth $2 instead of discovering that your $1 worth of garbage now costs $2 to haul away. By the way, most heat loss is from black body radiation, so aluminum foil may actually work better than ceramics of the same thermal conductivity.

Buildings can be recycled too. Buy them and sell them as needed. Shipping containers make for great lab buildings because they are cheap, strong, and movable. You can sell them off-site when you’re done. We have a shipping container lab building, and a shipping container storage building — both worth more now than when I bought them. They are also rather attractive with our advertising on them — attractive according to my design aesthetic. Here’s an insight into why chemical engineers earn more than chemists; and insight into the difference between mechanical engineering and civil engineering. Here’s an architecture aesthetic. Here’s one about the scientific method.

Robert E. Buxbaum, October 31, 2013