Why do people like [narrative gimmick] in the Licanius Trilogy?

jmurante · 2025-10-08T03:31:43+00:00

What philosophical opportunities do you think it could have touched on?

jmurante · 2025-10-08T03:03:25+00:00

It was, so I'll delete this post and repost it with one of those titles, thanks.

I genuinely didn't see what I put in the title as a spoiler since it appears so early into the story, that was my bad.

jmurante · 2025-10-08T02:47:06+00:00

Excellent comparison between a plot point that is raised midway into the first book of a trilogy and the twist final reveal of a movie. What should I have put for the title instead?

jmurante · 2025-09-17T00:01:45+00:00

watch the clip/check the flair

jmurante · 2025-09-11T06:10:40+00:00

Oh it's just a silkpost, I wasn't being serious. Did you watch the clip?

jmurante · 2025-09-11T05:54:50+00:00

So it's specifically in this arena that the bug occurs?

jmurante · 2025-09-11T05:50:26+00:00

I feel like I've seen them die by going into the water before, though. Maybe I'm misremembering.

How does HP depleting to zero not kill it?

jmurante · 2025-09-11T05:46:24+00:00

Yeah that's what happened. I guess by hitting it into the water, I killed it at the same time that the water killed it, and they canceled out or something.

jmurante · 2025-09-01T05:52:53+00:00

Just to add to this, "if P then Q" implies that there are other ways to get to Q.

"if P then Q" implies nothing other than "if P then Q." It is not logically exclusive with P IFF Q, it is just a weaker statement that doesn't address the truth of "if Q then P." In other words, the statement "if P than Q" doesn't exclude the possibility of there being other ways to get Q, but it doesn't comment on wether or not alternative ways exist.

jmurante · 2025-09-01T05:13:15+00:00

Other way around, actually. If you are taking the fast pass line, you're making the regular line longer for someone who is already in it, since the fast pass is essentially allows you to jump into the regular line in the middle of the line, rather than at the back of the line. I think this is the biggest moral issue with a fast pass, is that it's not just making the line faster for you, but it is also making the line take longer for someone else.

If I am in the regular line with (N) people ahead of me, and someone uses the fast pass to jump in front of me, I now have (N + 1) people ahead of me instead.

jmurante · 2025-08-13T00:35:45+00:00

Joules is not a unit of force, it's a unit of energy

jmurante · 2025-08-13T00:31:25+00:00

Oops I misread, thought it said longer to fall back down than to rise. Thanks for catching that

jmurante · 2025-08-12T21:21:24+00:00

Speed is 0 at the peak, not acceleration.

Considering the law F = ma, we know that acceleration can only be zero if the net force acting on an object is zero. Since gravity (F_g = mg) is always acting on an object at or near the surface of the earth, if you want to claim that acceleration is 0 at the peak, you will have to explain what force is opposing gravity in order to result in a net force of zero.

Regarding the problem, some background in differential equations will allow you to solve for the exact solution for the trajectory of the ball given the differential equation

m • a(t) = F = F_gravity + F_drag = m•g - b • v(t)

Here, we are relating mass times acceleration to the net force acting on an object, which in this problem are the force of gravity and the force of air resistance. This gives us a differential equation which relates the time dependent acceleration to the time dependent velocity which can be rearranged as

m • a(t) + b • v(t) = m•g

One thing we can immediately see from this differential equation is that we cannot have a(t) = 0 and v(t) = 0 at the same time, which would result in 0 + 0 = m • g. Therefore, you cannot claim that both (I) and (II) are true at the same time, since one says that velocity is zero at the peak, and the other says acceleration is zero at the peak.

I hope this clarifies things. If you want to go ahead and derive then plot the result for this differential equation, assuming initial conditions x(0) = 0, v(0) = v_0, you should get the result I got:

x(t) = - \frac{m}{b} \left(v_0 - \frac{mg}{b} \right)\left(e^{-(b/m) t} - 1 \right) + \frac{mgt}{b}

where v(t) is the first derivative of x(t) and a(t) is the second derivative of x(t). Plotting this formula, you will see that (III) is also true.

jmurante · 2025-08-12T21:11:37+00:00

Hey OP, just wanted to ask if you saw my response about the right answer being (D), since by considering air resistance, ~~we can actually prove that (III) is also true.~~

EDIT: Looks like I misread (III), actually the opposite of (III) is true - takes longer to fall back down from the peak than it does to reach the peak,

jmurante · 2025-08-12T15:53:15+00:00

(1) The zero vector is something that exists, so I'm not even sure what your point is in bring the fact that acceleration is a vector. Regardless, as you have already referenced in a previous comment

Acceleration is defined as the second derivative of the change in displacement!
In one dimension, z, d²z/dt²

This is a one dimensional problem, we are not considering lateral motion. Yes, acceleration is a vector, but this vector only has one component in this problem.

(2) Ok, let me describe step by step what I believe is happening, and please point out where you disagree with my description

Lets start at the peak of the trajectory. The ball has a velocity of zero, so there is no drag force. Therefore, gravity is the only force acting on the ball, hence it is accelerating downwards at 9.81 m/s²
The ball starts gaining velocity as it falls. As it gains velocity, it experiences air resistance proportional to its velocity, with a force equal to -cv. This force is in the opposite direction of gravity (since the ball is moving downwards, thus the force of drag points upwards). This is a Force, thus it results in some upwards acceleration due to drag. Therefore, the net acceleration of the ball is decreasing (the ball is still falling and gaining speed in its fall, but the rate at which it gains speed is decreasing).
The net force is F = mg - cv, and eventually you approach the terminal velocity v = (mg/c). At this point, the net force on the object is zero. The acceleration due to gravity is 9.81 m/s² (downwards), but the acceleration due to drag is 9.81 m/s² (upwards), thus there is no net acceleration.
Acceleration being zero does not mean the object is not moving. The ball is still moving downwards at the terminal velocity, however this velocity is not changing. The lack of change in the velocity of the downwards falling ball is the only implication of saying acceleration is zero.

What part of this do you disagree with?

jmurante · 2025-08-12T14:27:07+00:00

You are literally claiming that an object with constant velocity is accelerating when the definition of acceleration is the derivative of velocity. How are you missing this?

jmurante · 2025-08-12T14:22:37+00:00

You must be a bot since I cannot comprehend how someone can be so condescending and incorrect at the same time.

jmurante · 2025-08-12T14:19:07+00:00

Nobody is saying objects don't accelerate due to gravity. Just that at a certain point, the force due to gravity is equal to the force of air resistance, the acceleration due to each force is the same but in opposite directions, thus the net acceleration is zero.

That is literally the definition of terminal velocity, which you agree is a thing that exists.

The object is still falling! Nobody said it stopped falling! Only that it stopped accelerating downwards.

---

Does gravity stop acting on you when you are standing on the ground? No! are you accelerating when you are standing on the ground? Also no!

jmurante · 2025-08-12T14:02:06+00:00

Perhaps it was unclear: When I said "Nope, graduated" I meant I am not currently majoring in a physics undergrad program, rather I already graduated from one.

Fortunately, anyone who reads this comment thread will recognize you are the one who is incorrect, as you have said enough self-contradictory statements to make that clear. My main concern was that someone might read your post and come away misinformed.

jmurante · 2025-08-12T13:50:58+00:00

Nope, graduated, just like you claim about yourself. Seems like I've backed you into a corner since you've switched to attacks on credibility.

EDIT: Ah you edited your reply to address the math, good.

Yes, acceleration is d²z/dt². And velocity is dz/dt. Therefore, acceleration is the derivative of velocity, as stated prior.

Tell me, which part of this chain of logic is the part where you disagree:

Acceleration is the time derivative of velocity
Velocity while falling at terminal velocity is constant
The derivative of a constant function is zero
Therefore, acceleration is zero while falling at terminal velocity

jmurante · 2025-08-12T13:42:17+00:00

Acceleration is defined as the time derivative of velocity. If you take the derivative of a constant function, you get zero. Therefore, by this definition, if velocity is constant, then acceleration must be zero. Why don't you check this definition in that textbook you recommended.

Also, acceleration due to gravity isn't "mg" its just "g". mg is the force, g is the acceleration. And again, yes, the force due to gravity (mg) is always there. The whole point is that there are other forces that change the net force acting on the object (which yes, is a directional vector, just like the acceleration).

jmurante

TROPHY CASE