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[Lenglon]

@ new tunnel - still what I would have done. I'm operating entirely off of guesswork regarding what she's got and is doing. sometimes I'm going to get it wrong.
@ orbital fire - laser spread is a ratio, not a fixed value. a smaller laser fired at the moon has a MUCH smaller spread than a large one. take a picture of the moon, then aim a laser pointer at it. the fraction of the moon that the dot covers is the amount it would spread to when fired at the moon from the picture location. now take that picture and aim a larger laser at it - the same is true for it - much more spread. in the case of an orbital target, I agree that I need a needle-thin laser... like, say, the kind of needle-thin laser I would use on a regular basis for long-distance power transfer from node to node. with even less precision as that required to efficiently strike another node for the power transfer. the precision and focus needed is negligable compared to standard procedure.

oh, and the moon is farther away than a normal orbit by several orders of magnitude, making it a really bad example except to demonstrate an extreme case and explain the concepts involved. As an example it is utterly useless to give a sense of scale for how extreme an issue this is (hint: it isn't one at all).

atmospheric scatter also is a negligable factor as long as I avoid firing at extreme angles. the atmosphere is only about 11km in depth before it becomes thin enough that it doesn't matter anymore. Obviously, that distance is going to increase as I get farther and farther from firing directly downwards, but assuming the nodes are mostly firing down, then the scatter isn't going to be signifigant enough to matter. those military lasers you're mentioning are able to be effective at over 45 degree angles, and I don't need anything remotely that extreme. again, we're talking optics. If I can see it without issue, then I can laser it without issue. the issues you're mentioning also affect normal optics to an equal degree. and I can see my targets.

[_DivideByZero_]

@ orbital fire - laser spread is a ratio, not a fixed value. a smaller laser fired at the moon has a MUCH smaller spread than a large one. take a picture of the moon, then aim a laser pointer at it. the fraction of the moon that the dot covers is the amount it would spread to when fired at the moon from the picture location. now take that picture and aim a larger laser at it - the same is true for it - much more spread. in the case of an orbital target, I agree that I need a needle-thin laser... like, say, the kind of needle-thin laser I would use on a regular basis for long-distance power transfer from node to node. with even less precision as that required to efficiently strike another node for the power transfer. the precision and focus needed is negligable compared to standard procedure.

Umm, diffraction angles decrease as you make the aperture size wider. A needle-thin laser will diffract far more than a wide one, actually. The formula is theta = wavelength/(slit width) for small angles. The angle is smaller if the slit width is larger or the wavelength is smaller.


You also can't see your targets, any more than you can see trees from space. Yeah sure, the green light reaches you, but precision is a thing.

[Lenglon]

stop mixing diffraction from distance and diffraction from air particles, you responded to my distance argument with a particle one. that is something completely different and has jack-all to do with what you quoted.

[_DivideByZero_]

EDIT: I think you need to read up on Diffraction. I'm not understanding any of your terms. Light doesn't diffract due to distance, it diffracts at the aperture. Once your light is scattered it doesn't matter how it was scattered, the distance just increases the size of the image.
Atmospheric scattering is based on distance, but travelling through space does not further change the angle. Although I see what you mean--if it gets scattered in the atmosphere it won't spread far. That I can agree with. But you still have a puny laser.

...diffraction and atmospheric scattering are completely different things. I didn't mention scattering in the last post (but the atmosphere caused a 1 arcsecond deviation during the lunar ranging experiment, if we're getting to that).

A needle-thin laser doesn't work like you said it would. Your lasers have to be wide by necessity, because MEGAWATTS through a needle-thin point on a node would vaporize the node material.

[H4zardZ1]

((Who wants music? This))

[Yourmaster]

bring balloons back.

[_DivideByZero_]

bring balloons back.

You can start on the next one if you'd like.

[Lenglon]

so in other words, you've only been talking about atmospheric scatter this entire time, correct?
then I refer right back to what I said on the subject, the part you didn't quote or respond to, the atmosphere is only about 11km deep and the local atmosphere is unusually thin. it's a non-issue.

[H4zardZ1]

((Put my charsheet into the Current Character Sheets, Please.))

[_DivideByZero_]

so in other words, you've only been talking about atmospheric scatter this entire time, correct?

I'm talking about diffraction right now. I'm saying you can't have a needle-thin laser to transmit power over long distances because it would diffract a lot over astronomical... heck... orbital distances. I do concede the point that scattering isn't as significant as with the moon experiment.

The problem is that you can't just pick a tiny beam width because it would diffract too much to be helpful. To get the minimum diameter of the beam on the ground we can solve the equation x + 1.22*2*0.000000300/x*200000 (1.22 multiplier is for circular aperture, 200000 is the height we're firing from, 2x is because the beam diffracts on both sides). Using Google we get a minimum ground size of about .76m with an aperture of 0.4m. The area of this is 0.45mē.

Now, 100 MW over half a square meter is ~20 KW per square CM. That's deadly. Except that to boil a cubic centimeter of water at that wattage it takes 2 KJ. So far so good, it takes 0.2 seconds to wipe out a square meter. Now, Seattle is 369.2 kmē, or 369200000 mē. Divide by 5 and 24*3600*365 to get (very) roughly 2 years for one city-sized area of the planet. This lifeform was said to cover entire coasts.

then I refer right back to what I said on the subject, the part you didn't quote or respond to, the atmosphere is only about 11km deep and the local atmosphere is unusually thin. it's a non-issue.

I never said the atmosphere was thin. Marie said it wasn't thick enough to provide appreciable UV protection without UV-absorbing gases.

99% of our atmosphere's UV resistance comes from the ozone layer.

Anyway~

Marie said it barely penetrated the atmosphere (though I assume your character would have used a UV laser in the frequencies 200-300 nm). That's not a statement of the GM. That means she either lied to you, or she knows something that you don't.

[Lenglon]

...
valid
okay.
gimmie a sec, action incoming. busy IRL as well.

[H4zardZ1]

((not changing my dedicion, but what amount of turns are spent going to beta pic?))

[_DivideByZero_]

...
valid
okay.
gimmie a sec, action incoming. busy IRL as well.

No problem, still waiting on Pancaek and a continued response from Yourmaster.

((not changing my dedicion, but what amount of turns are spent going to beta pic?))

((
0.

The way time works is that when you are out of combat and nothing can interrupt you, you just timeskip.

If you go to the same star as someone else you will retroactively have taken off at the right time to meet them. That way I don't have to deal with causality issues such as predicting what the players are going to do. Although a time travel RtD... hmm...))

[H4zardZ1]

((Could i check radio signal at other stars without using a turn?))

[Lenglon]

I need more nodes to work with... a lot more.
3 chains - intercept the three orbs and drain them. assuming *nasty* things happen when they are intercepted I will be dropping everything to manage what happens.
orbital node: stop relaying power to planetary node, switch back to collection.
additionally: make as many nodes as I comfortably can without using Greater potions. and spread them out in a cloud around me.