Meteor vs. Projectile

[Korvan]

Well there is an explosion and a bright white channel as the shots fire so that could be a potential cover for that aspect of things if we choose that as our explanation of choice.

Found a freeze frame of the shot. It's pretty much what I would expect from a laser channeling through the atmosphere.
too big to show inline

[Jake]

I'm no expert on this, but I did some reading on lighting, and it seems that the channel is not a vacuum, but a channel of ionized air that would be used to produce the conductivity needed to create lightning. The air is still there, it just has fewer electrons. If this was true, it shouldn't change anything for the projectile. I may very well be wrong about this, and if I am, anyone is welcome to set me straight.

[Feil]

Laser blooming does not create a vacuum. It creates an ionized path, which itself absorbs more and more energy from the laser. Dunno what that would do to the mass density of the air, which is what matters for the drag equation.

[Jake]

I didn't think so. The mass density of air doesn't seem to be that major of a term in the drag equation (we have a 131000m/sec squared for the velocity portion), so it would have to decrease significantly from the value of 1.2kg/m^3. Also, ionization in this case removes electrons from the air so they can flow more freely, like a metal, which improves conductivity needed for a lightning strike. I don't think this would affect density, as the amount of particles in the given area should not change.

[Feil]

I didn't think so. The mass density of air doesn't seem to be that major of a term in the drag equation (we have a 131000m/sec squared for the velocity portion), so it would have to decrease significantly from the value of 1.2kg/m^3. Also, ionization in this case removes electrons from the air so they can flow more freely, like a metal, which improves conductivity needed for a lightning strike. I don't think this would affect density, as the amount of particles in the given area should not change.

In itself, ionization has no impact. However, it would generally be associated with intense heating, so the density would fall.

[Norade]

I didn't think so. The mass density of air doesn't seem to be that major of a term in the drag equation (we have a 131000m/sec squared for the velocity portion), so it would have to decrease significantly from the value of 1.2kg/m^3. Also, ionization in this case removes electrons from the air so they can flow more freely, like a metal, which improves conductivity needed for a lightning strike. I don't think this would affect density, as the amount of particles in the given area should not change.

In itself, ionization has no impact. However, it would generally be associated with intense heating, so the density would fall.

Intense enough heating could lower density by a lot.

If wiki is to be believed increasing the temperature of air by 15 degrees Celsius decreases density by 5%. Using the ideal gas law and inputting a temperature of 1000 degrees Celsius we get a new density of 0.000353kg/m3. Now it's idealistic to think that no other factors would result in denser air as the pressure of air rushing to fill the void would be rather impressive yet even if we take my equation to be off by 100 times we still get air with a density of only 3% of what it would be at 20 degrees.

Using the same numbers as Feil would have originally we get a drag force of 1.6E7 Newtons. Originally we would have had a drag force of 5.5E8 Newtons so we have an over order of magnitude drop which is significant. Using my first number for air density we would get a 1.6E5 Newtons which is even lower. Do note that I also used a pretty low temperature for the air just because I wanted to see what sort of effect we would get with less dense air.

I'm a bit lost trying to do the black body radiation stuff so Feil if you would be so kind as to help out once again.

[Korvan]

Going with the 0.75 radius channel with 65 km length, it has a total volume of just under 115,000 m3. Using the ideal gas law assuming 1 barr pressure (1 barr = 100,000 pascals and is close enough to sea level air pressure for our purposes) and temperature of 298 K (25 C), PV = nRT => n = PV/RT, we get n = 4,636,171 mols. This is the number of air particles in the column.

Assuming all the particles are heated to 30,000 K at the same time, at the new temperature the number of particles will now occupy a new volume, V = nRT/P

V = 11,563,538 m3 The ratio of the new volume to the old one is about 100 to 1 (which in fact is the same ratio as the new temperature is to the old one).

So essentially, if we were dealing with ideal gases, the density in the channel would be about 1% compared to the surrounding air. However, we are not dealing with ideal gases. This will come into play when dealing with the air in the channel. Since the air particles are now charged, they will repel each other, further increasing the volume occupied and thus further decreasing the density along the channel.

Van der Waals equations can deal with this, (P + n2a/V2)(V - nb) = nRT
b is the amount of space occupied by the particles and a is a measure of attraction between the particles. In the case of an ionized gas, a would be negative and would be a measure of the repulsion between particles. I am not able to find any examples of Van der Waal's constants for ionized gases, so I cannot calculate how much of an effect this would have.

[Jake]

Plasmas are an example of a completely non ideal gas. The chemistry of the air definitely becomes involved in any calculations, so you can not use the ideal gas law to get an answer with any form of accuracy. In fact, in my thermodynamics class, we used charts to determine properties of non ideal gases, as the ideal gas law will not give an appropriate approximation. Unfortunately, I sold the book with said charts back a long time ago, and I don't know if they make charts for plasmas, but if someone else has some available, maybe we can get more information. Also, at the speed of our projectile, the pressure will not be ambient. I'm pretty sure it will be significantly higher the ambient, which will also have an effect. For more info on re-entry level speeds, here's the wikipedia article: http://en.wikipedia.org/wiki/Atmospheric_reentry. Some of it, including the gas physics, is pretty complicated.

[Feil]

Irradiance (radiant flux per unit surface area) varies proportionate to force.
Temperature varies according to the x^0.25 of irradiance hence also according to the x^0.25 of force.
Since I started you off with some values already, you should be able to scale them to whatever you need with those proportionality factors.

If you want the peak frequency (which will correspond to the light's color), Planck's law solves handily to v_peak=2.82kT/h where k is the boltzmann constant and h is the planck constant, v_peak is peak frequency, and T is temperature.

[Korvan]

Looks like I was wrong about the charged particles of air repelling each other, further decreasing the density. I neglected to consider the effect of the electrons stripped from the air particles. As it turns out, it looks like these electrons shield the positively charged air particles from each other and once fully ionized, a plasma actually becomes almost perfectly ideal.

[Jake]

So plasmas are ideal then? Then all we really need to know is the pressure, as we can extrapolate the temperature (it should be similar to the lightning, the volume should be a cylinder of 131 km length and radius equal to that of the round, then use
P=pRT, where p=m/v=density to get the density of the gas.

[Jake]

Edit:
65.5 km length

[Jake]

another edit:
Actually, now that I think about it, the volume would just be that of the cylinder since that's where the pressure will be localized.

[Korvan]

I think I worked that out in my post a ways back. Simply put, the ratio of densities will be the inverse of the ratio of temperatures, so if the initial temperature is 298 K (about 25 C) and the final temp is 30000 K (about that produced by lightning) then the final air density will be about 1/100th of the initial density. I think this is still much to high to avoid the projectile going boom in the atmosphere.

Just realized that if the gun barrel is open to the atmosphere, a fair bit of energy is going to be shed inside the gun itself. I don't think the warranty will cover that!

[Norade]

I think I worked that out in my post a ways back. Simply put, the ratio of densities will be the inverse of the ratio of temperatures, so if the initial temperature is 298 K (about 25 C) and the final temp is 30000 K (about that produced by lightning) then the final air density will be about 1/100th of the initial density. I think this is still much to high to avoid the projectile going boom in the atmosphere.

Just realized that if the gun barrel is open to the atmosphere, a fair bit of energy is going to be shed inside the gun itself. I don't think the warranty will cover that!

That does drop the force of friction down by two orders of magnitude though, which is significant, but possibly not enough to make allow the round to survive. It does make it much easier to rationalize the round surviving than if we assumed to other mechanisms involved.

[Jake]

I just realized, for this to work the air has to be heated to 30,000K. The highest melting point in the alloy, that of tungsten, is 3,695K. This projectile will most likely vaporize due to the heat of the channel that is produced for it.

[Norade]

I just realized, for this to work the air has to be heated to 30,000K. The highest melting point in the alloy, that of tungsten, is 3,695K. This projectile will most likely vaporize due to the heat of the channel that is produced for it.

That is retarded, there are less particles in there now at low density and in that state they will suck at transferring energy. Just goes to show why we don't listen to you when it comes to heat transfer...

[Jake]

That is retarded, there are less particles in there now at low density and in that state they will suck at transferring energy. Just goes to show why we don't listen to you when it comes to heat transfer...

Uh, plasmas have a similar electron configuration to metals, which are excellent conductors of heat. Conductivity of gases also increases with temperature, and 30,000K is pretty fucking hot. Care to explain in detail, acknowledging both of these points, to this retard why you expect there to be no conduction between two materials that are around 30,000K apart?

[Norade]

That is retarded, there are less particles in there now at low density and in that state they will suck at transferring energy. Just goes to show why we don't listen to you when it comes to heat transfer...

Uh, plasmas have a similar electron configuration to metals, which are excellent conductors of heat. Conductivity of gases also increases with temperature, and 30,000oK is pretty fucking hot. Care to explain in detail, acknowledging both of these points, to this retard why you expect there to be no conduction between two materials that are around 30,000oK apart?

Because while each particle is 30,000oK there are only 1% the number of them and they will be moving so fast that they will only be in contact with the slug for very limited amounts of time. An example that may apply here is the Therosphere which is comprised of plamsa recorded at 2,500oC yet you would still freeze to death because you wouldn't ever touch enough particles. Yet that is also way less dense than this channel would be.

However, I failed to account for the speed of the projectile, so they may impart more heat that I think they would.