## FANDOM

25,507 Pages

• What the title says. Can you apply ISL to a laser beam way bigger than the thing it destroys? Wouldn't the logic be similar to that of an explosion? Like the explosion, only a small part of the laser beam would be destroying the target compared to the entire thing.

• implementingÂ power (total energy expended over time eg joules per second)Â orÂ pressure (joules per square meter) is something everyone on this wikia is too lazy to do tho

everyone just works with joules, which is sorta faulty because surface area and wattage is a big dealio in rl

• Mephistus wrote:
implementingÂ power (total energy expended over time eg joules per second)Â orÂ pressure (joules per square meter) is something everyone on this wikia is too lazy to do tho

everyone just works with joules, which is sorta faulty because surface area and wattage is a big dealio in rl

Would the forumla be different for large differences? I think there are differentÂ formulas for explosions specifically since they get weaker as they spread out.

• You can't apply the Inverse Square Law itself, but you would reduce the result somewhat.

Inverse Square Law applies when the energy is being spread out in all directions; laser beams are concentrated so that doesn't work.

• Agnaa wrote:
You can't apply the Inverse Square Law itself, but you would reduce the result somewhat.

Inverse Square Law applies when the energy is being spread out in all directions; laser beams are concentrated so that doesn't work.

Yes, but this is for laser beams vastly bigger than the target they destroy. Also, do you know what formula I would use to reduce the result?

• you'd just compare the flat surface area of the frontal part of the beam that hit the object to the csa of the object. youd multiply the energy yield of what his the object by that ratio to get the total power

still, energy intensity for surface area is something that is sorta downplayed lolÂ Â

• I would assume it's something like, you'd figure out the surface area of the laser beam that actually hits the target, find what percentage that area is of the total area of the laser, and then apply that ratio to the laser's energy.

It'd be too strange to write the formula out, since you'd need to pixel scale to get the surface areas anyway.

• Mephistus wrote:
you'd just compare the flat surface area of the frontal part of the beam that hit the object to the csa of the object.

still, energy intensity for surface area is something that is sorta downplayed lolÂ Â

So frontal area of beam / frontal area of target? Thats an oof. My estimates took into account the beams entire surface area.

• yeah, basically unless the beam was hitting a character / object larger than it, not all the energy would hit the target.

even with explosion calcs, the real life value is the intensity or joules/m^3 that the target takes.

• Mephistus wrote:
yeah, basically unless the beam was hitting a character / object larger than it, not all the energy would hit the target.

Yeah, but wouldn't calculating the entire beam take into account all of it's energy rather than the energy near the front? Or is most of a beams energy just naturally highest at the front?

• This seems to be getting into some really feat-specific stuff right now.

A beam's energy isn't higher anywhere in the beam, but if it can destroy a mountain in a second, then we know that tanking it for a second tanks mountain-tier energy. If they're only hit by a quarter of the beam's area, we know that they tanked 1/4th of mountain-tier energy.

It's gonna be hard to give much more information without knowing the feat (why do you know the beam's entire energy but not its output/s?)

• Agnaa wrote:
This seems to be getting into some really feat-specific stuff right now.

A beam's energy isn't higher anywhere in the beam, but if it can destroy a mountain in a second, then we know that tanking it for a second tanks mountain-tier energy. If they're only hit by a quarter of the beam's area, we know that they tanked 1/4th of mountain-tier energy.

It's gonna be hard to give much more information without knowing the feat (why do you know the beam's entire energy but not its output/s?)

Chakravartins beam is pretty fairlyÂ larger than these planets it hits https://youtu.be/6aOpa15szaQ?t=110. This isn't about the AWverse specifically. it's just an example of what I mean.

Well yeah, but I was confused as to why I'd only use the beams frontal surface area.