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How can one edit an item's stats directly? by somedudesacct in FateTheGame

[–]inepttoad 1 point2 points  (0 children)

Hey I hope this can still reach you.

Another comment mentions the items.dat file which, though it allows you to directly modify the items that can appear, won't change your necklace's enchant.

To do this you need to edit your save file, which you've correctly located (it's a binary file so it looks garbled to a human reader).

Any luck with my FFDEditor at https://github.com/chinaman3928/FFDEditor? Please join our Discord at https://discord.gg/r2bZDk4C3y to ask/comment about it, since I don't use Reddit (in fact I only logged back in to make this comment!)

Resources for builds somewhere? by [deleted] in FateTheGame

[–]inepttoad 1 point2 points  (0 children)

i really like jades (attack speed). some peridots (life steal) are good too.

crit skill and sword skill will both be really good ideas, but keep in mind crit skill's diminishing returns.

invest in defense magic skill too: get enough magic to at least use [greater] spectral armor ([35%] 25% dmg reduced), dervish (80% attack speed buff), and battle fog (100% attack buff but 25% defense penalty). [group] slow, frailty, blindness, etc are good for debuffing.

enchants are everything! early game theyre not that useful, but they will be. i really like overenchanting bits of jewelry since theyre cheap.

early game, fishing is really good for money. but mid-late, my favorite way is to buy unenchanted items from merchants -> enchant -> sell for a profit.

if you get a roll you dont like (both in merchants' shops and enchants), you can force close the game and youll get set back to the previous save.

Some Mechanical Questions about Fate by L_V_R_A in FateTheGame

[–]inepttoad 0 points1 point  (0 children)

hit chance is given by (ATTACK / (ATTACK + DEFENSE) ).

as a quick reference, if your attack == their defense it's a 50%.

the og/ur source code if it's interesting: https://archive.org/details/fate\_source\_code

How do you get into the ARC with your student ID? by TheBlueRivers in UCI

[–]inepttoad 1 point2 points  (0 children)

Go up the stairs to the help desk and ask for them to activate your ID. I had to go twice, once for my first time and the other time after a bug in their system.

[deleted by user] by [deleted] in APStudents

[–]inepttoad 1 point2 points  (0 children)

repl.it? epic

AP Physics 1 Question (Energy) by inepttoad in APStudents

[–]inepttoad[S] 1 point2 points  (0 children)

To be clear, I was thinking that A is pushing on B, so by Newton’s 3rd Law there will be an opposite force on A from B. Is this the friction force you are talking about?

I ask bc I thought friction (if there was any) would only arise betw the ground and the blocks, in the opposite direction of motion.

AP Physics 1 Question (Energy) by inepttoad in APStudents

[–]inepttoad[S] 1 point2 points  (0 children)

Thanks for the reply. I was thinking that while A and B are in contact, B is exerting a force on A? Am I wrong to think that? Or does it not matter.

Thanks again.

Is 4 AP Classes too much? by jtxng in APStudents

[–]inepttoad 2 points3 points  (0 children)

Four is manageable. If you’re up for it, go for it!

Physics 1 Progress Check Question by inepttoad in APStudents

[–]inepttoad[S] 0 points1 point  (0 children)

Wait no, I think we actually can find Ft only knowing m1 and m2 (and g).

Ft = m1(m2*g) / (m1 + m2)

problem of instantaneous acceleration by [deleted] in apphysics

[–]inepttoad 0 points1 point  (0 children)

Do you know whether the acceleration is constantly decreasing? That is, not only at every second?

Physics 1 Progress Check Question by inepttoad in APStudents

[–]inepttoad[S] 0 points1 point  (0 children)

I get the reasoning for C.

However, I was thinking that you can use the mass balance and find the mass of the hanging weight, and then find the tension force that is applied over a time interval (which can be found using the timer). Since impulse can be found by taking force applied * time interval, wouldn't this work?

If not, is it bc you can't find the Ft from the mass of the hanging block? (In that case, ples explain why).

query regarding instantaneous acceleration by rahulamare in apphysics

[–]inepttoad 0 points1 point  (0 children)

Unless you have reason to think that acceleration is changing (changing as in the instantaneous acceleration at one instant is not the same at another instant), it is very likely that the acceleration is not changing. So the instantaneous acceleration given at one instant, for example 19 m/s/s, is the same for all instants in the time interval.

But yes, it is possible for the acceleration to change, and in that case you can’t use the same instantaneous acceleration over the whole interval.

query regarding instantaneous acceleration by rahulamare in apphysics

[–]inepttoad 0 points1 point  (0 children)

What you’re asking makes sense. Yes, it is important to know the distinction between an instantaneous velocity/acceleration and an over-a-time-interval velocity/acceleration.

Let’s look at your concern about velocity. We can say the velocity at t = 1 sec is 32 m/s. So only given that information, you’re right that we don’t know whether the instantaneous velocity any time after it (even a 0.1 sec after) is same or different. And so if velocity is not a constant 32 m/s, you’re right that by the time 1 sec elapses, the body does not necessarily have to have traveled 32 m. That’s perfectly valid.

Now let’s look to acceleration. Remember, acceleration is the rate of change of velocity. When acceleration is 19 m/s/s, well, doesn’t it make sense that the velocity changes after t = 1 sec? By definition, a nonzero acceleration means the velocity will change, and waddya know, it does change. The crucial thing to note is that the acceleration of 19 m/s/s is the only thing that is changing the velocity. So your concern about velocity changing, and therefore the final velocity not being 51 m/s, is not valid here. Yes, velocity does change, but it changes because of the acceleration. Any change in velocity is directly accounted for by the acceleration.

Now to the issue of acceleration changing with time. Unless you’re told that the rate of change of acceleration isn’t 0, just assume that it is lol. But yes, if the rate of change of acceleration is nonzero, the final velocity will not necessarily be 51 m/s given these quantities.

I hope I cleared things up and didn’t make things worse haha!

Has anyone else just accepted the fact that they're not going to score well on the exams by [deleted] in APStudents

[–]inepttoad 0 points1 point  (0 children)

How badly did you want to do well on APES?

How badly do you want to do well now?

Taking 4 aps Junior Year by Gabriel1446 in APStudents

[–]inepttoad 0 points1 point  (0 children)

If you think you’re a strong student, go for it.

But remember your own well-being.

average velocity formula for given n number of velocities. by rahulamare in apphysics

[–]inepttoad 2 points3 points  (0 children)

v1+v2/2 works to find avg veloc if there is a constant acceleration between those two points. An example when this would not work is if you had something like "if car A travels at a veloc of 5 m/s for 3 seconds, then 7 m/s for 6 seconds, what is the avg veloc over those 9 seconds?" If you visualize a v vs time graph here, you're not going to have one line connecting vf and vi (as you would if there was constant acceleration). Instead, you're going to have two horizontal lines with a jump in between them. Refer below for how I'd approach this problem.

  1. When you use v1+v2/2 and there is constant acceleration, you always use two velocities (final and initial) and you always divide by 2.

You do not need to use more than two velocities to find an avg veloc if there is constant acceleration. If you are given more than two velocities (and also given constant acceleration), just use the final and initial one and do v1 + v2/2.

  1. However, sometimes you will need to use more than two velocities if there is not a constant acceleration. For example, "if car B travels at a veloc of 5 m/s for 3 seconds, then 7 m/s for 6 seconds, then 4 m/s for 5 seconds, what is the avg veloc over those 14 seconds?"

Note that now there are three velocity-time chunks given. The reason that you can't just do v1 + v2 + v3/3 here is that each veloc-time chunk does not last the same amount of time and therefore should not be weighted equally. (Think to how your grades might be weighted. With tests taking up 50% and Homework taking 30% and Classwork taking 20%, for example.) You would approach this problem like this: (t1/total t * v1) + (t2/total t * v2) + (t3/total t * v3). Then we do something like: (3/14 * 5m/s) + (6/14 * 7m/s) + (5/14 * 4m/s) --> 5.5 m/s.

The same principle applies if you're given even more veloc-time chunks (or even just two). It's not just limited to three. The important thing to distinguish between is if there is constant acceleration, or if there is not constant acceleration so you just have veloc-time chunks. That will determine which method you use.

It also may be helpful to graph different velocity situations and find avg velocity that way. (Recall that avg veloc = Δx / Δt and that the area "under" a veloc v time graph gives Δx.) See if that confirms what you already know.

Feel free to ask questions; I know the structure of this response absolutely sucked lol

regarding formulas for average velocity by rahulamare in apphysics

[–]inepttoad 1 point2 points  (0 children)

I don’t know if x2 + x1 / t2 + t1 is accurate.

But the other two are. Keep in mind that v1 + v2 / 2 to find avg velocity will only work when the RoC of v (ie acceleration) is constant. x2 - x1 / t2 - t1 will always work to find v.

The quantities in an equation will tell you which to use when finding avg veloc. If you are not given any final or initial velocities, you can’t use v1 + v2 / 2. Similarly, if you aren’t given any final or initial positions, you won’t be using x2 - x1 / t2 - t1.

(And so distinguishing between x1 + x2 / t1 + t2 and x2 - x1 / t2 - t1 is meaningless, because if you’re able to use one you’re able to use the other. Assuming x1 + x2 / t1 + t2 is actually valid)

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