(Spoilers) Future technological developments.

JohnRoth wrote:

Theemile wrote:And this so happens in the business world.

Got a chip in my windshield last month and stopped by a glass place to fix it. As I waited for the repair, the scheduler and manager were standing in the nearby office, discussing an inner city dollar store which had their glass door broken overnight - again, making the robbery event a weekly occurrence for several months.

It seems the glass place attempted to sell them roll down shutters to protect the glass doors and windows of the storefront- the place refused. It seems broken glass and theft is covered by insurance, but taking preventative measures like the roll downs would be a capital expense the thrift store would have to pay for out of it's own, meager budget.

So, every tuesday morning.....

Yeah, that'll work until some drone in the insurance company notices the pattern and the company refuses to renew the insurance policy or raises the rates to where it's not economic to keep the store open.

Penny Wise, Pound foolish, as the saying goes.

But dsrseraphin is correct, I can't count the times I've made recommendations and drafted project proposals with rational business cases, complete with long term cost analysis and business rationalization, only to get torpedoed for the smallest of reasons - even when there was a short term cost savings.

But as our good friend Bob McNamara proved - we can save money by bundling 4 or 5 airplanes into one, and that no one needs a chromed firing chamber on the M-16.

Here are some of my suggestions for future technological developments.

Depth Charges

Gwandar & Beta Band GravCom

Gwandar Buoys & Gravlance Semaphores

-David S.

dsrseraphin wrote:In the case of an omni-directional dispersion, such as a bomb blast, the wave front can be simplified as a spherical surface with a radius that increases over time; the energy of the impulse is fixed but the total area of the wave front increases as the wave front travels outward, therefore the mean energy density (unit of energy per unit of area) of the wave front decreases the farther the wave front is from its inception.

In the case of a focused dispersion, such as a laser of any wavelength, the wave front can be simplified as a 'bullet' from a gun barrel with the 'bullet' having the same cross-section as the barrel (for the sake of simplicity, I'm assuming the 'bullet' has negligible depth/length). Upon leaving the barrel the wave front begins to spread (like it is frangible/soft); it spreads as if it were a perpendicular slice of a cone whose axis is collinear with the gun barrel. Now, determining the focal point of that cone is a non trivial task; for e-m wave it involves: the e-m wave lengths being used, the gun barrel aperture, the gun barrel length, the aperture medium interface, the traveling media, and a few other esoteric parameters; suffice to say, a well built 'lasing' apparatus will have a focus point orders of magnitude longer than the barrel. Again the energy of the 'impulse' is fixed in the wave front as it exits the barrel at the aperture and the mean energy density decreases as the wave front travels onward (because the wave front is spreading). Reflective and inflictive apparatus are special cases of focusing - the reflective surface is in essence the aperture and you deal with the 'cone' directly without a 'barrel'; any 'lens' also will behave as an aperture with respect to the resultant wave front.

If you really want to do the math, then omni-directional dispersion will involve spherical formulas and focused dispersion will involve conical formulas.

The key thing in all cases is the amount of energy transferred to the target is ideally the occulted surface area of the target times the energy density of the wave front at the time of impact - the materials and the media involved will always attenuate the transfer (there is no perfect transfer).

One last thing, wave fronts are not just e-m waves, they can be: compression in media, particles in motion, and even grav waves; anything that can transfer energy!

Ok, I'm taking off my lecture's robes! Please forgive me, lecturing is a nee-jerk reaction for me and inspiring critical thinking is my compulsive disorder.

-David S.

I do not think it is helpful to say "it spreads as if it were a perpendicular slice of a cone whose axis is collinear with the gun barrel" and then try to talk about "the focal point of that cone". Isn't the focal point of a cone its origin? What you are saying with the spreading part of your remark is that the beam width increases linearly with distance (the same as TFLYTSNBN) and that is not what the formulas that I have found suggest. The Wikipedia page on a Gaussian beam states that the beam width does at some point become linear with range; but only after the Rayleigh range is exceeded. Now if you would put your lecturing robes back: is that true or not for a laser beam in space? Do you agree with JohnRoth that the formula on the website is for a point source, even though it includes an initial beam width and the article is titled "Attenuation of a laser in space"?

PS. After rereading UH, I find that I have been incorrect in my criticism of Beowulf in that they did assign blocking walls to the orbitals (which did not turn on their wedges once the missiles were seen to be attacking the assembly lines) and those blocking ships did prevent debris damage to the planet after the internal explosions.

tlb wrote:
I do not think it is helpful to say "it spreads as if it were a perpendicular slice of a cone whose axis is collinear with the gun barrel" and then try to talk about "the focal point of that cone". Isn't the focal point of a cone its origin? What you are saying with the spreading part of your remark is that the beam width increases linearly with distance (the same as TFLYTSNBN) and that is not what the formulas that I have found suggest. The Wikipedia page on a Gaussian beam states that the beam width does at some point become linear with range; but only after the Rayleigh range is exceeded. Now if you would put your lecturing robes back: is that true or not for a laser beam in space? Do you agree with JohnRoth that the formula on the website is for a point source, even though it includes an initial beam width and the article is titled "Attenuation of a laser in space"?

PS. After rereading UH, I find that I have been incorrect in my criticism of Beowulf in that they did assign blocking walls to the orbitals (which did not turn on their wedges once the missiles were seen to be attacking the assembly lines) and those blocking ships did prevent debris damage to the planet after the internal explosions.

Since you asked for it, the lecture robes are going back on <shrugging>...

First off, I'm not going to comment on the Wikipedia article since I have not read it (recently), nor did I reference it.

Now on to my imagery - yes I can see how the gun barrel & cone visualization can be problematic; that said, let's go back to it and pay particular attention to the aperture of the barrel.

An ideal Gaussian beam is parallel within the lasing barrel. The 'spreading' occurs when the beam leaves the barrel. The exit aperture is in essence a slit/hole in a grate/cover. The nature of a wave is such that whenever any wave passes through a slit it spreads out. The underlying Uncertainty Principle is at work here. [just a quick refresher: Uncertainty Principle - if you restrict position, momentum gets fuzzed or if you restrict momentum, position gets fuzzed; the Observer Effect is derived from the Uncertainty Principle, if you observe a particle's position with absolute certainty, you now know absolutely nothing about its momentum (e.g. its speed and direction); with regards to wave functions: momentum is related to propagation velocity and its frequency (and in some cases amplitude), while position is related to wave length and displacement] When the wave (front) is passing through the aperture on its exit, it is at an interface between two separate media and it is being localized/constricted across the aperture. Since the energy load (e.g. the momentum) of the wave is fixed upon exiting then the only thing that can be 'uncertain' is the direction & velocity of the wave. Therefore, upon entering a medium that has more freedom than what existed before, if the waves in the wave front were parallel before, their direction can now be something other than parallel. The wave front spreads, and the spreading is in the same plane as the aperture.

Why is the spreading conical? That's assuming there are no other forces interacting with the wave front. The spreading is within a maximum angle from the initial traveling direction of the wave front. Assuming the ideal parallel wave front - then, just paying attention to the edges of the aperture, you will have rays (vector lines) delimiting the maximum possible spread as the wave front propagates onward. To simplify calculations, we assume the spreading is even up to the theoretical edge. From the aperture onward what is described is a (truncated) prismatoid, now the formulas for those can be a bit much - so we adapt and overcome by backtracking the rays until they converge into a single point, and lo & behold we are now dealing with a cone (and conic sections) and we have a pretty good tool box to deal with those things.

So, for lasing, even though the internal pumping may ideally achieve parallel throughput - the practical effect will be a spreading discharge that can be related to a conic section.

This may look like a lot of 'handwavium' but it is real science. Wave theory, while being applicable in our everyday experiences, is NOT exactly transparent in the details; it is one step before and a necessary prerequisite to the really spooky stuff - quantum theories.

My advice, don't get hung up on the math - it can be understood without it. Now if you have a project where you need to build something reliant on detailed knowledge of the behavior of waves, then its off to the Calculus mines for you sir!

-David S.

tlb wrote:What you are saying with the spreading part of your remark is that the beam width increases linearly with distance (the same as TFLYTSNBN) and that is not what the formulas that I have found suggest. The Wikipedia page on a Gaussian beam states that the beam width does at some point become linear with range; but only after the Rayleigh range is exceeded.

Let's look at the equations for Gaussian beam propagation and see what we can see. For Gaussian beams (where the energy is highest in the center and drops rapidly as you measure to the edge) the beam width follows this equation:
w(z) = w(0) * squareroot(1 + (z/R)**2),
where z is the distance from the source and R is the Rayleigh distance. For distances small compared to R the beam width is about the width at the source (which is some fraction of the size the end of the laser) and for distances much greater than R the beam width increases linearly with distance. The equation for the Rayleigh distance is
R = pi * (w(0)**2 / L),
where L is the wavelength of the emitted radiation. This is a simplification based on the wavelength being much smaller than w(0).
The wavelength of an x-ray is in the vicinity of 0.3 nanometers (exponent of -9) and for gamma rays is in the region of 30 femtometers (exponent of -15). If we try those figures we get distances that are nearly a light minute or more. So we should not worry about getting energy to the target, provided we can actually hit it.