Why Don’t We All Get A Balloon, And Then We Can Fly Into The Sky?

A couple of weeks ago, I was at a birthday party for one of my son’s friends. It was a great day, at a little park a half hour drive north and east of where I live, situated on a tributary of the Ohio River. The kids all had squirt guns and the like, and got each other soaked down while the adults sat back and watched and took pictures and were grateful that they brought extra clothes and towels. There were balloons as well, because there were kids.

One child was super excited about the balloons. They were your ordinary latex kind, that you blow up with your own lungs, but he was bouncing them around and laughing. “Why don’t we all get a balloon?” he asked excitedly. “And then we can fly into the sky!”

So, yeah. It wasn’t my son that asked it. But it’s the kind of question he could have asked, so I’ll answer it.

How does a balloon float?

The same way a boat does.

Care to elaborate?

Of course.

It’s tempting to say that things float because they’re light, but that’s not quite accurate. For example, an oil tanker floats but it is not light – they can carry anywhere from 1,500 to 550,000 deadweight tons, depending on size. No. Floating has everything to do with the mass of the object, and the fluid that surrounds it. See, all objects placed into a fluid displace some of the fluid (put a rock in a cup of water to see for yourself). If the mass of the fluid you displace is greater than your mass, you float. And air, for these purposes, can be considered a fluid.

But let’s look at some math, since the University of Chicago was kind enough to put together a document (Lighter Than Air: Why Do Balloons Float?) that explains all of this in some detail. There are two forces in play, the downward force (which is the pull of gravity) and the upward force (which is how much the fluid resists the downward force). The downward force (Fg) is the mass of the object (M) x gravitational strength (g), which is also how you calculate “weight” in physics. Weight, after all, is mass times gravity (which is why you weigh less on the moon, even though you retain the same mass). Upward force (Fb) is the mass of the fluid displaced (m) x gravitational strength (g).

Once you have Fg and Fb, you can calculate life=t. All that is is Fb – Fg. If the result is positive (meaning Fb is larger than Fg), you are sinking. If the result is negative (meaning Fb is larger than Fg), you are rising. And if the result is 0 (meaning the two forces are equal), you are floating immobile in midfluid.

Uhm. Okay.

Let’s do an actual example, shall we?

Yeah. Lets.

After consulting Google, I found an estimate that the average-sized party balloon masses 1.7 grams, and several notes that they can weigh more depending on the actual size, thickness, etc, etc. This will be important, momentarily.

Now, the density of air at sea level is about 0.0012 grams per cubic centimeter. So, if you inflate your hypothetical average-sized party balloon to a diameter of 1 foot (0.3048 meters, which means 30.48 centimeters), you get a sphere (for the sake of not making me crazy) containing 14,826.7 cubic centimeters of air. The inflated balloon weighs a total of (14,826.7 x 0.0012) + 1.7 = roughly 19.5 grams, and displaces 17.8 grams of air. So, it sinks. If you inflate the balloon to 2 feet in diameter (60.96 centimeters), you get a balloon containing 116,613 cubic centimeters of air. It weighs 141.6 grams, and displaces 139.9 grams of air.

Clearly, both balloons sink. And, in a vacuum, both would sink at the same rate because they have the same lift (-1.7).

But they don’t fall at the same speed. Not the ones I’ve played with, anyway.

Nope. Because we live in an atmosphere. And atmospheres create air resistance. I won’t go into the math there, because it made my head hurt a little, but it works like this: an object produces drag (a resistance to acceleration) based on the cross-section of the object perpendicular to the direction of movement. As the cross-section gets larger, the power needed to overcome the drag increases. How much? Well, it’s based on the cube of the cross-section. If you double it, you need 8 times as much power. If you triple it, you need 27 times as much power. And so on.

For the balloon, acceleration is down towards the ground and the cross-section is the diameter of the balloon. Doubling the diameter of the balloon means you would need 8 times the power to make it fall at the same speed as the smaller balloon. Since gravity (roughly) stays the same, that means you would expect to see it fall 8 times as slowly.

So, getting back to the wish to “fly into the sky”…

Sure. See, to make a balloon fly, you need something less dense than room-temperature air. That’s why hydrogen and helium are so popular. They’re gaseous at “room temperature”, and they weigh far, far less. Hydrogen weighs 0.000089 grams per cubic centimeter, and helium weighs 0.00018 grams per cubic centimeter. So, looking at the two balloons from the earlier example, we get the following information:

  • The 1 foot balloon weighs 3 grams if you fill it with hydrogen, and 4.4 grams if you fill it with helium. It displaces 17.8 grams of air.
  • The 2 foot balloon weighs 12 grams if you fill it with hydrogen, and 23 grams if you fill it with helium. It displaces 139.9 grams of air.

Regardless of which gas you fill the balloon with, it weighs less than the gas it displaces. So it has positive lift and it goes up. In fact, it could even lift additional weight – the 2 foot balloon filled with hydrogen would have neutral buoyancy with a 127.9 gram weight attached to it, so you could attach two Hershey’s chocolate bars (1.55 oz, or 44 grams each) to the balloon and still watch it go skyward.

Heating the air will also work, as gasses become less dense with heat. Sadly, I don’t have a good equation (that I understand) to show how much you’d have to heat the air to make it lift.

How many balloons would I need to fly to the sky, then?

Well, that’s more or less easy. How much do you weigh, and what gas are you using? I’ll illustrate with my son. He weighs around 60 pounds right now. That’s 27.2155 kilograms, or 27,215.5 grams. Looking at the two foot balloons, the lift for the hydrogen balloon is 127.9 grams per balloon and the lift for the helium balloon is 116.9 grams. So, it would take 27,215.5/127.9 = 213 2 foot hydrogen balloons to give him neutral buoyancy. 233 2 foot helium balloons would be required to achieve the same effect. You’d need another 18 hydrogen (20 helium) balloons to offset the weight of his clothes (maybe more if he’s planning on flying high). And I have no idea how many balloons would be required to offset the weight of the lines he’s holding on to or the harnesses to keep the balloons attached to him. And, of course, he’d need more to actually go up.

By contrast, I weight 316 pounds. So I’d need 1,121 hydrogen balloons or 1,227 helium balloons to achieve the same effect. That’s 37,553.5 cubic feet of hydrogen balloons, or a sphere roughly 42 feet in diameter. Oh, and it could explode.

Don’t do this at home.

No kidding.

You’d think so, but at least one person did.  Larry Waters used 45 8-foot weather balloons filled with helium to lift himself, his lawn chair, his parachute, his pellet gun (so he could pop balloons and descend), his CB radio, his camera, and sandwiches and beer to a height of 16,000 feet.  He flew for 45 minutes, and got fined $1,500 by the FAA after an appeal.  But, because he was a trained pilot and lucky, he didn’t die.

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How Long Would It Take To Get To The Moon?

“Dad?” my son asked while we were playing with his Legos. “How long would it take to get to the moon?”

“I think that depends on how fast you’re going,” I replied.

“No,” he says, sounding exasperated as only a 6-year-old can, “I mean, if you were going as fast as the Death Star!” Because that was entirely clear from the context, right?

“I don’t know,” I tell him. “I don’t know how fast the Death Star is.”

“It’s really fast,” he assures me.

Where to start?

There are a couple of things we need to know here, in order to answer the question. How far away is the moon? How fast do we have to go at minimum to make it? Oh, and how fast is the Death Star? So, let’s dig in.

How far is it to the moon?

The distance from the Earth to the Moon varies based on the time of the month, because the Moon orbits us in an ellipse – so it gets closer and then moves further away. At apogee (the farthest it gets from us), it’s 405,400 km away, while it gets as close as 362,600 km at perigee. So, clearly, how long it takes will really depend on how fast we’re going – just like any other trip we can take.

How fast do we need to go?

How fast you need to go to get to the moon will depend on the method you’re using to get there, and the amount of time you want to take. So, let’s start with the concept of escape velocity. This is the minimum speed required to “out-pull” gravity and leave an object behind. If you launch at that speed or greater, you fly away. If you don’t, you fall back to the surface. Eventually. Escape velocity varies with the gravity of the object and is approximately 11.2 km/s, or 40,320 kph on Earth. Assuming there is no friction, which is a popular physics assumption to keep equations simple. If you launch at that speed, you fly away from the earth – you slow down over time, as Earth’s gravity pulls on you, but you never actually stop moving. Ever.

There’s a down side to trying to get to the moon by launching at escape velocity (say, by using a variant of Project HARP’s big gun): Earth’s force of gravity is 9.807 m/s2, so you’re pulling around 1,142 gravities at the instant of launch. You would be a thin, wide smear on your pilot’s chair well before you reached the moon.

Clearly, we didn’t send a gelatinized melange of Neil Armstrong, Michael Collins and Edwin Aldrin to the moon on Apollo 11 – those three men made it to the moon and back with bones and organs intact, after all. So, how did they do it? Well, the important thing to remember is that escape velocity is only needed if you have an initial push and then add no additional thrust after that. This isn’t how the Saturn V – or any other rocket for that matter – works. They lift themselves at a slower pace, but apply a constant (or near-constant) thrust by carrying fuel. There’s a point of diminishing returns on this, because you have to lift your fuel as well as the ship (something described in the Tsiolkosky rocket equation, which I discussed when I tried to describe how to make a house fly).

The Saturn V was a multi-stage rocket, with the first stage burning for 2 minutes 41 seconds and pushing the rocket about 68 km into the air (hitting a velocity of 2,756 meters per second). Then it ditched the first stage and started the second stage burn. This pushed it another 107 km (for a total of 175 km) into the air over the course of 6 minutes, reaching a velocity of 6,995 meters per second). Stage 3 burned for about 2 minutes 30 seconds, reaching a velocity of 7,793 meters per second and putting it in orbit at an altitude of 191.1 km. Stage 4 burned for six minutes, pushing the ship to a velocity of 10,800 meters per second once it was time to head for the moon.

So, how long would it take?

How fast are you going?

Let’s say you just boosted off Earth with a canon, firing you straight up at escape velocity. Let’s also say you timed things so that you’d intersect with the moon at perigee. That’s 362,600 km, or 362,600,000 meters. At 11.2 meters per second, that’s 32,375,000 seconds to reach the moon. This translates into 8,993 days, or 24 years, 7 and one half months. Approximately. Your gelatanized corpse has a long trip ahead.

Apollo 11 was moving at 10.8 kilometers per second, which (mathematically) means you’d expect the trip to the moon to take 33,574.07 seconds. In theory, this means 9.326 hours. It actually took three days. Why? Well, there’s two reasons and they’re both gravity. See, the Apollo 11 wasn’t maintaining constant thrust. It had fuel that it used for course corrections and orbital insertions and the like, but it coasted most of the way. Earth’s gravity pulled on the ship the whole time, slowing it down. In addition, the ship didn’t fly in a straight line. It was in a long, figure-eight-shaped orbit with the Earth and the Moon – like so:

But what about the Death Star?

Ah, yes. That. Well, it still depends on the speed the ship can manage.

How fast is the Death star?

This is… questionable. According to the DS-1 Orbital Battle Station entry on Wookieepedia, the Death star had a speed of 10 megalight (MGLT).

So, what’s a megalight? Well, also according to Wookeepedia, a megalight “was a standard unit of distance in space”. Which is entirely unhelpful, although it does indicate that when it was used in the Star Wars: X-Wing Alliance instruction manual, it appeared to be a unit of distance and that when used as speed it should imply “megalights per hour”.

In all likelihood, “megalight” is a word that got made up because it sounded cool and had no actual meaning attached to it. But if we try to break it down, “mega” as a metric prefix means million. So, one megalight could be a million light seconds. However, this would mean that the Death star flies at 10 million light seconds per hour, or 2,777.7 times the speed of light – meaning that it could reach Alpha Centauri from earth in less than 14 hours of cruising on its “sublight” drives.  So I’m going to assume that this is not what was intended.

The Star Wars Technical Commentaries on TheForce.net speculate in “Standard Units” on what MGLT means in terms of real world [i]anything[/i]. The author of the article comes to the conclusion that 1 MGLT is “at least 400 m/s2” acceleration, which is roughly 40 gravities of acceleration.

One thing we also know about ships in Star Wars is that constant acceleration isn’t an issue – they have something close to the “massless, infinite fuel” I mentioned above. The Death Star isn’t fast, compared to the other ships in Star Wars, but it can accellerate at a constant 4 kilometers per second. Now Dummies.dom provides us with a simple formula for determining the distance (s) covered for a given time (t) at a particular acceleration (a), and that formula is s = 0.5at2. Which means we can reverse engineer, because all we need is the time. The equation looks like this:

362,600 = 0.5(4)t2
362,600 = 2t2
181,300 = t2
t = square root of 181,300 = 425.7933771208754 seconds

So, assuming that the Death Star didn’t engage it’s hyperdrive, it would take a little over 7 minutes to reach the Moon at a velocity of approximately 1,703.17 kilometers per second. And it would keep going, because it can only slow down at 4 kilometers per second. So, if the Death Star wanted to stop at the Moon, it would need to slow down about halfway there (yes, I know that orbital mechanics are a little more complex than this, but we’re talking about a 160 kilometer diameter ship that can accelerate at 4 kilometers per second. So cut me some slack, would you?). That it would have to accelerate to halfway to the moon, and then decelerate the rest of the way. So, that would look something like this:

2(181,300 = 0.5(4)t2)
2(181,300 = 2t2)
2(90,650 = t2)
2(t = square root of 90,650 = 301.0813843464919)
t = 602.1627686929838 seconds, or slightly over 10 minutes.

“All your tides are belong to us, now.”

Why Is It Called A Helicopter?

Not too long ago I happened to have my son (who is six) and his cousin (who is also six) in the back seat of my car as I was running some errands. Which, as you may have noticed, is how so many of these articles start. The car, it turns out, is a wonderful place to start getting questions. Why? Maybe watching the world roll by makes him ask questions. Or maybe it’s the science and history podcasts I listen to. Or maybe he’s just six, and questions get asked.

Anyway, back to the story. I’m driving, and they’re in the back seat chatting and playing and fighting as six year olds tend to do. Then my son says “Look! A helicopter!” They both ooh and aah and stare, watching it fly overhead. “Why is it called a helicopter?” my nephew asks.

Helicopter?

I have no idea, of course. I think that, way back when I realized that English words were often made up of other words, I constructed an etymology in my head that derived it from the Greek word for ‘sun” – which I was pretty sure was “Helios”, because a little knowledge is a dangerous thing. But I’ve never actually bothered to look. Also, in the interests of fair and full disclosure, it turns out that I’ve been spelling it wrong my whole life. I always thought it was “helecopter”, when it’s actually spelled “helicopter”.

So, let’s check it out. The Online Etymology Dictionary gives the following derevation for helicopter: “1861, from French hélicoptère “device for enabling airplanes to rise perpendicularly,” thus “flying machine propelled by screws.” From Greek helix (genitive helikos) “spiral” (see helix) + pteron “wing” (see ptero-).”

Clearly, nothing to do with the sun at all.

Why ‘spiral wing’?

Because of Leonardo da Vinci.

leonardo-da-vinci-helicopter

Leonardo wasn’t the first to come up with a device that could take off vertically. That credit goes to the creator of the bamboo dragonfly or Chinese top, a toy created around 400 BCE that consists of a propeller on a stick. You spin it the correct direction, and it leaps into the air. However, Leonardo is credited as being the first to try and design a device that could carry a human being (although he never constructed it, and since it would have been human-powered it would never have worked), and his design used a spiral-shaped wing – hence the name “aerial screw” and the origin of the word we now know as “helicopter”.

How Do You Get To Space?

We’ve got another contextless question here. Just the question “how do you get to space?” listed in my notes. I strongly suspect, though, that it has to do with my son’s current obsession with Star Wars. There’s a lot of space ships in those movies, after all, and he’s got a toy Millennium Falcon that he abuses in true five-year-old fashion. So he thinks about space – or at least movie “space” – a lot.

What Is Space?

Generally speaking, “space” is “outside Earth’s atmosphere”. Specifically, though, this gets problematic. There’s six different layers of the atmosphere, after all:

  • The troposphere, which averages about 7 miles (11 km) thick, and ranges from 8 km thick at the poles to 16 km at the equator. This is what most of us think of as “the atmosphere”, with breathable air and weather and most of clouds. The temperature drops with altitude in the troposphere.
  • The stratosphere, which sits on top of the troposphere and extends upwards to 30 miles (45 km) above the Earth’s surface. You find the ozone layer here, as well as temperature increasing slightly with altitude – up to a high of 32 degrees Fahrenheit (0 degrees Celsius).
  • The mesosphere, which sits on top of the stratosphere and extends upwards to about 53 miles (85 kilometers) above the surface. Temperature begins dropping with altitude again in this layer, reaching a low of -130 degrees Fahrenheit (-90 degrees Celsius), and it’s the atmospheric layer in which meteors begin to burn up as they fall towards Earth.
  • The thermosphere, which sits on top of the mesosphere and extends upwards to about 372 miles (600 kilometers) above the surface of the Earth. Temperatures can reach thousands of degrees (Celsius or Fahrenheit), but it’s measured in the energy of the molecules of gas in this layer and there’s not a lot of molecules that high up (the average molecule would have to travel 0.62 miles/1 kilometer to collide with another molecule), so it doesn’t feel hot.
  • The exosphere, which sits on top of the thermosphere and extends upwards to about 6,200 miles (10,000 km).
  • The ionosphere, which starts 30 miles (48 km) above the surface and stretches to 600 miles (965 km) above the surface. That’s an average figure, though, as it grows and shrinks based on solar conditions. It’s also divided into sub-regions, based on what wavelength of solar radiation it absorbs. Note that the ionosphere, although considered a seperate layer of the atmosphere, overlaps the mesosphere, the thermosphere, and the exosphere.

So, where’s space? The definition “outside Earth’s atmosphere” could mean that you have to exit the exosphere to be “in space”, but this would put us in the ridiculous position of having to consider the International Space Station (which orbits at an average height of 249 miles) within the atmosphere. Clearly, that’s ridiculous.

The commonly accepted altitude definition for space is the Kármán line, which is 62 miles (100 km) above sea level. This line, sitting in the lower reaches of the thermosphere, is approximately the altitude above which wings no longer provide lift and below which atmospheric drag makes orbital paths fail without forward thrust.

Ways to get to space

Clearly, then, to get into space “all” we have to do is travel at least 62 miles (100 km) straight up. Simple, right? Well, maybe not. At present, we have only a handfull of ways to leave the surface of the Earth: balloons, really big guns, rotor wing aircraft, fixed-wing aircraft, and rockets. Each has some advantages and disadvantages.

Balloons

For balloons, the current record-holder for unmanned flight is the Japanese BU60-1, which reached an altitude of 33 miles (53 km) and carried a 10 kg payload. The altitude record for a manned balloon was the StratEx, which reached an altitude of 25.7 miles (41.4 km). Both were impressive achievements, but neither made the Kármán line.

Balloons function because of Archimedes’ Principle:

When a body is fully or partially submerged in a fluid, a buoyant force Fb from the surrounding fluid acts on the body. The force is directed upwards has a magnitude equal to the weight, mfg, of the fluid that has been displaced by the body.

In other words, if an object is submerged in a fluid but is lighter than that fluid, the fluid pushes it up until the object it is the same weight as the surrounding fluid. And while we don’t tend to think of it in this fashion, our atmosphere behaves like a fluid. We don’t float in the air because we’re denser then the air. A balloon inflated with air doesn’t float, because it’s actually slightly more dense than the atmosphere (because you forced more air in to inflate the balloon). But balloons filled with hot air, or with a gas that is less dense than our atmosphere (hydrogen or helium, say) will float.

Because of this, you could theoretically float a balloon into space – all you need is for the balloon to be less dense than the surrounding environment, after all. There’s a catch, though. The material of the balloon needs to be strong enough to not burst if the interior is pressurized, strong enough not to be crushed by the denser exterior, and light – because the mass of the balloon is added to the mass of the interior to determine if Archimedes’ Principle will lift it. A vacuum would be the ideal interior, but so far we don’t have anything simultaneously strong enough to keep the atmosphere from crushing a vacuum balloon and light enough for the air to push it up.

Big Gun

Jules Verne proposed this in From the Earth to the Moon way back in 1865. All you need is enough power, and you can launch something (or someone) into space from the ground. How much power? Well, using a ballistic trajectory calculator and assuming that the projectile is fired straight up from sea level, you would need a muzzle velocity of 4,595.52 feet per second (1,400.715 meters per second) to launch something 100 kilometers into the air – ignoring the effects of atmospheric drag. If you’re curious, that’s 142.78 times the force of gravity.

This is clearly not a good way to put people into space, but it would work well for launching non-fragile items. And in 1966, Project HARP demonstrated that it would work. A 16-inch (and 119 feet long) gun constructed by the US Department of Defence in Yuma, Arizona fired a 165 pound shell 590,000 feet (179,832 meters) into the air on November 19, 1966 – that’s 111.74 miles (179.8 kilometers), putting it well past the Kármán line

Plane

Planes seem an obvious choice, right? We all know what they are, they take off and land, so why not fly into space? Well, there’s a good reason for that – the Kármán line (meaning that by the time you reach space your wings aren’t working) and the propulsion (both jets and propellers require air to function). So, any plane that could reach space would need to be a rocket as well as an airplane. Right now, the world altitude record for a airplane was set on August 31, 1977 by Alexander Fedotov, who reached 23.4 miles (37.66 kilometers) in a MiG-25.

Rocket

Right now, this is the way we get to space. A rocket engine carries stored fuel and utilizes Newton’s third law by expelling that fuel in one fashion or another from one end of the craft to push the opposite direction. Most rockets in operation are combustion rockets, meaning that the fuel is ignited and burned in some fashion. These can be quite expensive, as the rocket has to lift all of its fuel at the time of launch. However, with sufficient fuel and engineering and money, you can build a rocket capable of reaching any altitude.

Other Strategies

Any number of launch methods that do not rely on rockets have been proposed – the space gun technically is counted in this category, but it differs from the others in the fact that it has actually been constructed. Most of the others have a US Department of Defense technology readiness level of 2, meaning that they are dependent on the invention of the materials and technologies needed to support them, and may not actually be feasable. They include:

  • Space tower: a tower that reaches above the Kármán line, possibly as far as geosynchronous orbit (22,369 miles or 36,000 km).
  • Skyhook: A satellite that lowers a lift cable (or the equivalent) and then reels in the payload.
  • Space elevator: A tether attached to the Earth at one end and a geosynchronous satellite at the other, which can then allow vehicles to climb into space.

How do you stay in space?

To quote Randall Monroe, “getting to space is easy. The problem is staying there.” Why? Well, gravity is still pulling you down. So, you have to go ‘sideways’ fast enough that you keep missing the Earth as you fall (which is pretty much what an orbit is). Using the Earth Orbit Velocity calculator on Hyperphysics, an orbit at the Kármán line would require an orbital speed of 25,745.41 feet per second (7847.2 meters per second), which translates to 17,549.1 mph (28,249.64 kph).

Fortunately, you don’t have to maintain constant acceleration at that speed – one of the defining features of space is that it’s pretty empty, meaning there isn’t a whole lot out there to slow you down once you get going. But still, you have to get going really fast to stay there – orbiting at the Kármán line means you circle the earth every 1.45 hours.

How do you get back down?

Oh, that’s easy. You fall. Whether or not you die is a matter of how you land.

Would a Helicopter Work On Mars?

This is one of those questions that I wish I could remember the context for. But I don’t. It’s just sitting there in the master list of the questions my son has asked, sandwiched between “What’s plankton?” and “What if the oceans froze?”, and I have no idea when or why he asked me this. Just “would a helicopter work on Mars?”

Usually I try to jot down something about the context. But not this time.

So, let’s get to the question. Would a helicopter work on Mars?

The simple answer appears to be “yes”. The Jet Propulsion Laboratory, on January 22, 2015, put out a press release on that very subject. They’ve designed a small proof-of-concept drone with the imaginitive name Mars Helicopter. It’s a 2.2 pound (1 kg) device, cubical in shape, with a 3.6 foot (1.1 meter0 rotor span. As designed, it would be deployed to work in conjunction with a future Mars rover.

The helicopter would fly ahead of the rover almost every day, checking out various possible points of interest and helping engineers back on Earth plan the best driving route.

Scientists could also use the helicopter images to look for features for the rover to study in further detail. Another part of the helicopter’s job would be to check out the best places for the rover to collect key samples and rocks for a cache, which a next-generation rover could pick up later.

If you prefer visuals, JPL also has a video about Mars Helicopter. It’s primarily CGI, since they haven’t constructed a fully-functional model, but it includes a few shots of a prototype trying to fly in a simulated Martian atmosphere.

But, what would it take to make a helicopter work on Mars? Not that I don’t believe JPL when they say it can be done – they are literally rocket scientists, and their entire job revolves around building things for space and other planets – but I’m curious. Fortunatly, NASA has me covered again. Let’s turn now to a paper describing the actual experiments NASA has done on rotary-wing aircraft for Mars, titled Experimental Investigation and demonstration of Rotary-Wing Technologies for Flight in the Atmosphere of Mars.

The paper begins by describing the challenges faced in constructing a rotary-wing aircraft for the Red Planet:

The Martian atmosphere is 95% CO2 with the remaining 5% comprised of N2 and other trace gases. Mars’ gravity is slightly greater than a third of Earth’s. The atmosphere of Mars is extremely cold and thin (approximately 1/100’th of Earth’s sea-level atmospheric density). Further, a seasonal variation of approximately 20% of the planetary atmospheric mass occurs on Mars (a consequence of polar CO2 condensation and sublimation). Given the thin, carbon-dioxide-based Martian atmosphere, developing a rotorcraft design that can fly in that planetary environment will be very challenging.

After that, the article goes into a lot of cool descriptions of various hypothetical rotorcraft designs and a whole lot of math that flies right over my head. It’s all stuff I’d love to understand, but it gets into some advanced-looking physics and aviation engineering concepts that I’ve never studied. Here’s what I could pick out:

* NASA is investigating both coaxial and quad-rotor configurations.
* Experiments have been done with battery power, electricity generated by fuel cells, and engines fueled by hydrazine. At present, some variety of electrical power seems to be the best option.
* It’s really hard to test an actual prototype under conditions that replicate Martian gravity, atmosphere pressure, and composition. At present, we’ve only managed it with the pressure.
* The rotors will have to be really big and spin really fast, and the rotorcraft will have to be really light.

A reasonable layman’s description of what a Martian helicopter can be found in this quote from the second page of the paper:

From an aeromechanics perspective, Mars rotorcraft will be very different from their terrestrial counterparts. Mars rotorcraft will have very large lifting-surfaces and will be required to have ultra-lightweight construction. For example, in order to lift ten kilograms of vehicle mass on Mars, a single main rotor (at a disk loading of 4 N/m2) would have to have a radius of approximately 1.7 meters.

For added fun, check out Interplanetary Cessna on Russel Monroe’s What If? site, which answers the question “What would happen if you tried to fly a normal Earth airplane above different Solar System bodies?” It’s not specifically on the topic, but it’s close.