Back from LA with Honorverse move news

I'll defer to your opinion regarding what the threshold of vision is. Your assumption of a 1 millisecond duration of the flash is not totally unreasonable. However; my estimate of a 1 second duration is based on the knowledge that any emissions in the visible spectrum are not the direct result of fusion reactions but secondary emissions from the plasma cloud formed by the explosion.

The emission power for a black body is 5.67eex-8W/M^2 x T^4.

The calculation for the emission power from a plasma cloud is far more complex, but the above equation defines an upper limit.

I'll leave it to you to calculate what the temperature of the weapons residue after detonation would be. While your at it, calculate what the wavelength of the emission peak would be and what percentage of energy would be emitted in the visible spectrum.

As to your second paragraph, Not Really.

A planet reflects incident light over a much wider semi-spherical wave front than a flat mirror which when illuminated by a near point source will reflect a well collimated beam that you will not even see unless it is aimed at you.

The bottom line because of the nuclear physics and the plasma physics involved, unless you detonate a nuke inside a target or atmosphere that absorbs a large percentage of the weapon energy in a large enough mass so that it can be reradiated at an emission temperature of a few thousand Kelvin, you will not see diddly shit.

SWM wrote:

namelessfly wrote:Astronomical observations are not my strong point so I am not conversant with equating apparent magnitude to a particular power density nor am I into biology enough to know what thevsensitivity of the eye is. However; I'll muddle through.

At what range are you referring to?

A common, 1 MT nuke yields 4.2eex15 J by definition. Assume this energy is releases as light for a duration of one second so the power is then 4eex15 Watts. Now compare to other, natural astronomical bodies that reflect about 4eex15 Watts of sunlight. At one AU out system from
Grayson, the solar constant would be about 700 Watts/M^2. 4eex15W / 700 W/M^2 = 6eex12M^2 or a body on the order of 3,000 Km in diameter.

We can quibble over the probable yield of the laser heads (these were not Mk-16s), the flash duration (briefer = brighter), the energy partition (more energy as Neutrons, Nuclei, Gamma rays, X-rays, UV, IR and microwaves = dimmer) and a host of other details, but this puts the issue in perspective.

I see. I also assumed a 1 Megaton explosion (4e15 Joules), but I used 1 millisecond rather than 1 second. And I also assumed a distance to the explosion of 1 AU. If I use your time of 1 second, then the apparent magnitude of the explosion would be between 2 and -2, still very easily visible. The brightest star, Sirius, has an apparent magnitude of 0.

Your calculation does not actually represent the brightness of a planet 3000 km in diameter. What your calculations are actually saying is that the explosion would look like a flat mirror 3000 km in diameter reflecting sunlight at a distance of 1 AU. That is many, many times brighter than a spherical planet of similar size.

StriderDowding wrote:Hi All

Man, there are a bunch of lurkers like me coming out over this.

Been hoping for this for years, almost as much as LOTR. If this is half as successful I will be very happy.

<<Snip>>
Just pulled my 1994 copy of OBS off the shelf, to be read starting tonight.

Mr. Weber Thanks for many good hours and thoughts. I hope this movie makes you proud!

Welcome to the forums to all of you lurkers who are coming in from the cold. It is always nice to have new opinions and new takes on old opinions. Anything that keeps the discussion lively, without offending anyone.

namelessfly wrote:As to your second paragraph, Not Really.

A planet reflects incident light over a much wider semi-spherical wave front than a flat mirror which when illuminated by a near point source will reflect a well collimated beam that you will not even see unless it is aimed at you.

I'll do a more detailed analysis of a nuclear explosion in my next post. But I would like to address this piece separately.

The formula you used for your calculation is the one for a flat mirror, not a spherical body. Consider light falling on a sphere and being reflected to an observer. Only a small part of that sphere is at an angle to directly reflect the light to the observer. The other parts of the sphere are at more grazing angles, such that much of the light is reflected away from the observer. Yes, a planet is a rough surface, so there will be some light reflected toward the observer. But much of it will not. This results in an effect called "limb darkening". Look it up. A planet will be much darker than your calculation suggested.

You have been using a 1Mton explosion for your calculations. Where did you get that figure? I thought the Honorverse missile warheads were bigger than that. How does increasing the output of the warhead increase the visibility of the fireball?

KNick wrote:You have been using a 1Mton explosion for your calculations. Where did you get that figure? I thought the Honorverse missile warheads were bigger than that. How does increasing the output of the warhead increase the visibility of the fireball?

Let's take the Mk-73, which was the CA/BC laser head first deployed in 1883. This is a 78 ton missile, which matches the spec of the missiles used by Fearless. That's got a 15 Mt warhead.

kzt wrote:

KNick wrote:You have been using a 1Mton explosion for your calculations. Where did you get that figure? I thought the Honorverse missile warheads were bigger than that. How does increasing the output of the warhead increase the visibility of the fireball?

Let's take the Mk-73, which was the CA/BC laser head first deployed in 1883. This is a 78 ton missile, which matches the spec of the missiles used by Fearless. That's got a 15 Mt warhead.

That is more along the lines of what I remember. I tried looking for some kind of information in HotQ, but couldn't find anything. Now, how does that change the visibility of the fireball Does that make it more or less likely that it would be visible to Abigail on Grayson?

kzt wrote:

KNick wrote:You have been using a 1Mton explosion for your calculations. Where did you get that figure? I thought the Honorverse missile warheads were bigger than that. How does increasing the output of the warhead increase the visibility of the fireball?

Let's take the Mk-73, which was the CA/BC laser head first deployed in 1883. This is a 78 ton missile, which matches the spec of the missiles used by Fearless. That's got a 15 Mt warhead.

Sorry to nit, especially since it's irrevalant to the conversation, but you mean the mk13 cruiser weight missile, not the mk73. I would hate to see misquotes perpetuated.

Full details of the missile were in IFF, KNick.

Possibly throwing in a curve ball, but its mentioned that the warheads are NOT pure nuclear warhead. Its mixed with another type of explosion material. Will that affect how much light is made?

Theemile wrote:

kzt wrote:Let's take the Mk-73, which was the CA/BC laser head first deployed in 1883. This is a 78 ton missile, which matches the spec of the missiles used by Fearless. That's got a 15 Mt warhead.

Sorry to nit, especially since it's irrevalant to the conversation, but you mean the mk13 cruiser weight missile, not the mk73. I would hate to see misquotes perpetuated.

Full details of the missile were in IFF, KNick.

It is easy to see how the confusion got in. The Mark-73 Laser assembly on the Mark-13 missile body makes it easy to interchange numbers.

Now for a closer look at a nuclear explosion. I've done some more reading. The tricky part is extracting the atmospheric effects from the descriptions.

When the bomb first explodes, the first thing that comes out is a burst of gamma rays, followed shortly by x-rays. This, of course, is what laserheads take advantage of to produce their x-ray lasers. The x-rays uniformly heat up the residue of the bomb.

In the first microsecond, the temperature of the residue is around 60 million degrees C or more. The residue radiates a nice blackbody spectrum, of course. This residue expands very rapidly. In an atmosphere, this would turn into blast and thermal effects, but not in space. 80% of the energy yield initially comes out in photons, and without an atmosphere, it is not converted to other effects.

Since there is no atmosphere, there is no fireball. But the residue radiates. As it expands, it cools rapidly. At a few microseconds into the explosion, the explosion is a few meters across and a temperature of tens of millions of degrees. By the time the explosion is 13 meters across, the temperature is down to 300,000 degrees. In an atmosphere, this would happen 100 microseconds into the explosion, but it will happen much faster without an atmosphere. Since the thermal radiation goes as the fourth power of the temperature, the total power being emitted has dropped by a factor of more than a thousand.

It is easy to see that the great majority of the energy is emitted in the first 100 microseconds. By 1 millisecond, the power emitted is a tiny fraction of what was emitted initially. It is fair to say that the flash occurs in the first millisecond or less. In an atmosphere, the radiated power is spread out over a much longer period, because it is absorbed and reradiated by the atmosphere, and much of the radiation comes from the hot shock wave. So I would agree that in an atmosphere, a flash of 1 second is reasonable. But without an atmosphere, even 1 millisecond is probably too long.

If we use a time of 1 millisecond and say that 80% of the yield is in a black-body spectrum, and assuming a 1 Megaton bomb, we get an effective power of 3.4e18 W.

Nameless is correct that much of this will be in non-visible wavelengths. It is a bit hard to guesstimate how much of the power would be in the visible range, especially with the temperature dropping so precipitously in the first millisecond. So let us assume that only 0.1% of the energy is emitted in the visible spectrum. Thus, the power in the visible spectrum is 3.4e15 W.

The luminosity of the sun is 3.8e26 W, and an absolute magnitude of 5. Thus, the sun is about 1.1e11 times brighter than the bomb. So we go through the calculations I demonstrated before. A factor of 10^11 is a difference of 28 magnitudes. So the bomb has an absolute magnitude of 33. Absolute magnitude is the apparent magnitude at 10 parsecs. The explosions in HoTQ took place at 150 million km. 10 parsecs is 3.1e14 km, or 2.0e6 times further, so the explosions in HoTQ will be 4.0e12 times brighter than magnitude 33. 4.0e12 is more than 31 magnitudes. That makes the explosions magnitude 2.

The explosions are quite easily visible on Grayson, even if we assume only a 1 Megaton bomb, and only 1/1000 of the radiated energy is in the visible spectrum. A 15 megaton bomb would be about 3 magnitudes brighter, or brighter than Sirius.