“Appear,” not “Appeared,” rather.
[Edtngi Sklils week tongith]
After a bit of thought, one further post (then I’ll shut the hell up!):
Modern propellants are designed to have a specific burn rate within atmosphere (performance X, between Y bars and Z bars of atmospheric pressure). With no atmosphere, your burn is very different, and unless you’re postulating the chamber of a firearm is a totally sealed environment, you will experience a large gas loss. Most propellants are approximately 20% oxidizer, but taken as with any equation, the presence of atmosphere is a distinct part of it. It may even be opaque to most, e.g. not mentioned, but it’s there nonetheless. In short, a huge decrease in gas-pressure by which to propel your round.
[Different skew if this is not an automatic, of course]
There is no magical appearance of oxidizers needed. The simple formula for the combustion of black powder (comprised of saltpeter, sulphur, and charcoal) for instance:2KNO[sub]3[/sub] + S + 3C => K[sub]2[/sub]S + 3CO[sub]2[/sub] + N[sub]2[/sub]
requires no atmospheric oxidizers. (The actual products are more complex due to incomplete combustion and side products, but this has nothing to do with atmosphere or lack thereof.) The grain size of black powder and smokeless powders is insufficiently small–and deliberately so, lest handling the powder in bulk result in a hazardous atmosphere–to allow rapid combustion at surface using atmospheric oxygen. Taking an equivilent amount of charcoal ground to the same grain size would result in an aggregate that might burn aggressively at STP but wouldn’t serve as an adequate propellant for a firearm. In order to get effective combustion using atmospheric oxygen requires a much finer grain size and a concentration that is roughly stoichiometric to atmospheric oxygen, as occasionally seen in silo or coal mine explosions.
The oxidizer for a solid propellant–the oxygen-bearing nitrate–functions as the oxidizer for this reaction. See Cooper’s Explosives Engineering, Chapter 2, for more details on oxidation reactions in combustion and detonation and oxidizer balance.
Stranger
Smart people arguing arcane topics and even getting a little testy - just one more reason I love the Dope!
Stranger on a Train handled the chemical part of contradicting you. On the physics side – the difference between ground-level atmospheric pressure and a vacuum is only 15 psi. Surely that’s a small fraction of the gas pressure created by the combustion of a round’s propellant. It’s hard to believe that a significantly greater amount of gas will leak out ahead of the bullet (I presume that’s what you’re talking about) in orbit than on the ground. If anything, since most of the gas will stay behind the bullet until it’s expelled from the barrel, the lack of an outside atmosphere should allow for more rapid acceleration (in the barrel) and a higher velocity for the ballistic projectile (after it’s left the barrel) than when you’re shooting in your backyard.
Correct. As the wise, handsome, and eloquent author of this article suggests, the internal gas pressure involved exceeds 3,000 PSI. It’s not particularly relevant for this series of events if the external pressure is 15 PSI or 0 PSI, since the difference would change from 2,985 PSI to 3,000 PSI - a negligible shift.
As Stranger explains, explosives are different than combustibles. An explosive has to contain its own oxidizer, otherwise it won’t explode, it will burn.
This is why gasoline can’t be used as propellant in a gun. It will burn, but it won’t explode unless it is carefully mixed with air first.
-
I believe the question that was asked was “would a gun work in space.” Whether the bullet fires is a subset question. If it is false, that ends the original thread. If it is true that bullets work in space, that still doesn’t answer the asked question. So, in short, I do not agree with you that this topic subject should be changed just because one or more people want to focus on a subset issue.
-
<<an object near earth and in sunlight would reach a reasonable temperature (close to the average temperature on earth)>>
This is incorrect. The Moon is near Earth, and its temperature during the day on its equator is 390 degrees Kelvin, or 242 degrees F. There are several reasons why the Earth is cooler in the full light of the Sun than an object like The Moon or on a geostationary orbit as satellite TVs. (sources: www.Wikipedia.com and Fahrenheit (°F), Celsius (°C), Kelvin - Scales Temperature Converter).
a) The Earth radiates much of the Sun’s energy by reflection off clouds. I don’t know how much of the surface is covered in clouds but if you look at any photo of our planet you’ll notice it is substantially covered in clouds.
b) At the poles, the arctic and antarctic are covered in ice and snow, which further reflect light into space, albeit not efficiently.
c) One of the biggest reasons is that we have a dense atmosphere. Surface winds, convection, and especially the jet stream distribute the heat from the sunlit side to the night side. Without this the night side would be too cold to survive.
d) Another big reason is that 70% of the world is covered by water. The solar energy water receives is dissipated by many ways. Some of the energy warms the water below the surface through convection and conduction. Thermohaline circulation (THC) (of which the Gulf Stream is an example) distribute water from the hot equator towards higher latitudes. That’s why northern Europe, European Russia and especially Ireland and Great Britain aren’t much colder. As the circulation becomes colder and saltier, it sinks and then continues in a southerly direction, until it picks up heat near the equator. This is the world-wide current driven by the Sun’s energy.
There are probably other reasons I’m just not thinking of right now, but hopefully you get the idea. Anything in airless space in view of the Sun gets very hot, and anything in shadow cools down to well below freezing. The reason the Space Shuttle used to fly with its cargo bay doors open and facing the Earth was to help dissipate heat. The reason space suits and other technology in space are uniformly white is to reflect heat rather than absorb it.
Source: Thermohaline circulation - Wikipedia
Ed Isenberg
There was an episode of Firefly where Jayne fires his favorite gun, Vera, at another spaceship just outside of his spaceship. They wrapped the gun in a spacesuit, so it would have the oxygen it needed in order to fire.
I guess the creators of the show didn’t realize that guns don’t need free oxygen to fire, or they simply decided that most people wouldn’t know that and this would keep down the complaints.
And nighttime temperatures on the Moon are much colder than Earthly temperatures. The average, though, is still about the same. And heat transport around the surface of the Earth is negligible compared to the heat that a spot has from being in sunlight just a few hours earlier-- Even the jet stream doesn’t blow at 1000 miles per hour. The Moon’s temperature would be a lot more stable if it rotated rapidly like the Earth.
EDIT:
Or the creators of the show did know, but figured that Jayne wouldn’t, and so had him taking the precaution without realizing that it was unnecessary.
First, you are still incorrect. The average temperature on The Moon at its equator is 220K or minus 64 degrees F.
Second, heat transport around the surface of the Earth (and to lower levels of the oceans) is not insignificant. To the contrary, the atmosphere, its components (e.g., carbon dioxide and methane gas) and oceans are the major mitigating forces behind our climate.
Third, the Moon’s temperature would not be more stable if it rotated rapidly like the Earth. The reason is that conduction of the lunar surface w/o convection to help doesn’t work well at all.
I don’t know where you learned either of these three statements.
Nowhere on Earth does the temperature make swings from 390 K (242 F) during the day to 100 K (-280 F) at night. The reason why is that between our huge oceans and thick atmosphere the Sun has to heat far more than a surface in view of the Sun in outer space. the deep oceans and our atmosphere both act as heat sinks.
www.wikipedia.com is my major source. If you have information to the contrary what is its source, how credible is it compared to wikipedia, and how does it account for the relatively stable temperatures from day to night on Earth’s surface as opposed to The Moon or any object (e.g., Shuttle, EVA suits, satellites, International Space Station) roughly one Astronomical Unit from the Sun?
We’ve gotten a bit far from the topic of the practicality of using a gun in outer space, but I didn’t want to read all the messages after yours until we had settled the issues you raised.
BTW, I am an amateur astronomer with about $5,000 in telescopes, cameras etc., I was for two years President of an Astronomical Society. I’ve given more speeches on the subject of astronomy that I care to remember. All of them were first researched on reputable web-sites. Oh, and on Tuesday I give a major presentation to Southwest Writers (500 members in New Mexico) on Research for Writers.
Ed Isenberg
From NASA’s website, they have an estimate for the temperature of a flat black plate in space:
(You can ignore the particle flux energy most of that link is concerned with. At the end, he determines that it’s negligible.)
If I’m reading that correctly, that assumes the radiating area is the same as the area receiving sunlight. So it would be correct for a flat plate facing the Sun, with an insulated back side.
Since radiation goes as T^4, if both sides of the plate could radiate, the temperature would be smaller by (1/2)^(1/4) = 0.84, or 331 °K, or 136 °F. For a sphere (small enough to have a roughly uniform temperature), the radiating area would be four times the area receiving solar radiation, and the temperature would be smaller by (1/4)^(1/4) = 0.707, or 279 °K or 43 °F. For those last two temperatures, the radiation from the Earth should also be included, so they’d be somewhat higher.
A gun won’t be a true black body, and also is neither a flat plate nor a sphere, so there’s some large error bounds, but until someone models an actual gun, using it’s actual absorption and emissivity, and for a range of orientations, assuming the gun will be somewhere in the range of 40 to 140 °F seems the best we can do.
I’d expect a gun to operate over that range of temperatures.
Not just Jayne’s favorite gun but his most prized single possession. Maybe he just didn’t want Vera suffering the aforementioned mechanical abuse and degradation.
The whole thread is just a physics exercise anyway, true space cowboys pack Gyrojets.
**Your conversion of Kelvin (°K) to °F in incorrect. ** You can find the correct conversion by going to Fahrenheit (°F), Celsius (°C), Kelvin - Scales Temperature Converter which shows the true temperature is not 136 °F but 249 °F. This is in complete agreement with the maximum temperature on The Moon’s equator of 390 °K or 242.3 °F. Actually, I was surprised at the similar numbers, since The Moon is not a “black body” but has many different colors, including shades of white (in mountainous terrain). I don’t know what contribution energy reflecting off the mountains back down to nearby flat surfaces, but given the light color of those nearby surfaces I wouldn’t expect it to be significant.
**Whether a gun would fire even once, never mind multiple times, is more complicated, and the effectiveness of the gun is still more complicated. **
Would a gun work in outer space? Sometimes yes, sometimes no, and sometimes it will explode on you, ruining your whole day. Don’t ask what would happen to an astronaut with a dozen holes in his/her space suit and, under that, a dozen holes in his/her body. It is very grisly.
The problem is because a gun may be white or black, may be in sunlight or shadow, and the bullet (not the casing) is inside the gun, and may be made of a lead/tin alloy or jacketed coated lead. The shell (casing) can be made of lead, copper, brass, steel, or other materials including combinations of the above. This variety means it is hard to predict which is more affected by heat, the bullet, the casing, or the gun barrel. Guns and ammunition are made so that upon firing the bullet will eject from the casing and then the muzzle. If the bullet expands by heat but the gun barrel doesn’t, you could easily have an explosion in the chamber where the bullet sits (see above). Or it may work perfectly. It is impossible to generalize.
Whether guns can be fired accurately or repeatedly is another issue, depending on physics. Most fully-automatic “assault rifles” or machine guns have three settings: single shot, 3-shot, and continuous fire. The reason is that after 3 shots are fired on Earth, your gun will rotate and lift in your hands, making additional shots on automatic completely uncontrollable. “Continuous” is only used for “covering file” while another person tries to advance toward the target. It keeps the target’s head down. Now, that’s on Earth with Earth’s gravity. If you are on The Moon, with only one-sixth Earth gravity, firing two shots in quick succession will result in the second shot going wild.
In zero or microgravity (not the same) or even on a low-gravity environment like The Moon, the recoil from a gun aimed by sight will make the shooter tumble backwards head-over heels. It will also apply some thrust to the shooter, so that he/she will be thrust backward in space. There is no easy recovery when this happens, and when multiple shots are fired from the eye of the shooter, the tumbling can cause all the blood to run to his/her feet, making the shooter pass out and even die of oxygen-starvation to the brain. The impact of the gun hitting the shooter’s helmet might even leave a serious crack, so that the shooter dies both from decompression and oxygen-starvation.
**In summary, unmodified earth guns, rifles and shotguns may work, might not work, or might explode. Firing a single shot in low or zero gravity can also be very dangerous to the shooter, never mind making further shots go completely wild. For these reasons, the answer to “Would a gun work in space” is unpredictable but certainly dangerous to the shooter.
I call that answer a qualified “No.”
**
Ed Isenberg
amateur astronomer
soon to be a published science fiction novelist.
Compared to the Earth’s rotation, it is. Heat transfer via air or water currents from the light side of the Earth to the dark side could only occur as fast as the speed of the current. The jet stream tops out at around 400 km/h, and is usually only a quarter of that. The Earth’s rotation at those latitudes, meanwhile, is well over 1000 km/hr. So by the time heat could reach the dark side from the light side, the dark side would itself be the light side.
The atmosphere does have a significant effect on the average temperature, via the greenhouse effect on one hand, and reflective clouds on the other. But it does not have a significant effect on keeping the planet’s temperature uniform from day to night.
And any gun at risk of exploding in space would also be at risk of exploding on Earth.
According to your link, my conversions are correct. 331 Kelvin is 136.1 Fahrenheit, as I had written.
Also, please drop the red text.
I don’t believe the bolded statement is correct. Clouds and humidity tend to hold in the Earth’s heat overnight. Clear, dry nights allow more heat to escape.
[QUOTE=Chronos
And any gun at risk of exploding in space would also be at risk of exploding on Earth.[/QUOTE]
I won’t keep repeating myself for your entire response, because you make statements without any citation, and without responding with why you feel my remarks are incorrect. I do want to point out your statement quoted above.
The reason a gun can explode in outer space or on The Moon is due to different expansion and contraction of the bullet, the cartridge casing, and the barrel of the gun. Since on Earth we don’t have such a different temperature between sunlight and shadow (or day and night), the expansion & contraction doesn’t have much effect, and are well within the tolerances of the gun and ammunition specifications.
Ed Isenberg
You are correct if the temperature is 331 Kelvin. However, referring back to the excellent site on NASA (http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/estimating_the_temperature.htm), the predicted equilibrium temperature of a black body in full sunlight and in outer space is not 331 but 394 K, which is indeed 249.5 F.
The exact wording of the NASA article includes the following.
OK, you didn’t read or didn’t understand my first post. I’ll go over it again:
At the NASA site, the author calculated the temperature of a flat plate facing the Sun. He assumed that the area of that plate radiating as a black body was the same as the area capturing sunlight. Thus, in his calculation, only the top surface is radiating energy. Plates have two sides, and he has no contribution for energy radiating from the back side. This can be appropriate if he, for example, is calculating how hot the skin of a satellite will get on the side facing the Sun, with insulation between the skin and the inside of the satellite. For this case he gets a temperature of 394 °K. Call this T[sub]0[/sub].
For a thin flat plate, uniformly heated and able to radiate energy from both sides, if the plate were still 394 K, it would be radiating twice as much energy as it receives. Obviously this would not be correct. The temperature of the plate will be less, so that it only radiates the amount of energy it receives from the Sun. The energy from the Sun is the same, so each side needs to radiate half the energy. Call the temperature of the plate in this case T[sub]1[/sub]. Since the energy radiated is proportional to T[sup]4[/sup], we have that 2*T[sub]1[/sub][sup]4[/sup] = T[sub]0[/sub][sup]4[/sup], or that T[sub]1[/sub] is only 331 °K.
Likewise for the sphere. Call its temperature T[sub]2[/sub]. The cross-sectional area of a sphere is pir[sup]2[/sup], and the area of the surface of the sphere is 4pir[sup]2[/sup]. Since the surface area in this case is 4 times the cross-sectional area, we have that 4T[sub]2[/sub][sup]4[/sup] = T[sub]0[/sub][sup]4[/sup], or that T[sub]2[/sub] is 279 °K.