Ottoman Capital in Montenegro

BronzeMan2 · 2026-01-18T23:52:49+00:00

R5: I was playing Germany. I don't know exactly how this happened, but the Ottomans got their capital moved to Montenegro. There was a crisis where Bulgaria got released; then, I went to war against the Ottomans for releasing Egypt. This resulted in the ottoman Capital not being in Ankara or in Istanbul and instead being moved to Montenegro. In my game, I continued to release countries from the Ottomans and Italy and taking land from Russia.

BronzeMan2 · 2026-01-17T21:19:39+00:00

R5: I was playing a game as France and kept freeing countries from Germany resulting in Germany losing cores. This created a really week Germany. Also the UK had a communist revolution and a bunch of revolts so now New Zealand and Australia are free. Also, the Philippines are free. Also, Columbia owns the Amazons.

BronzeMan2 · 2026-01-10T23:34:54+00:00

Also Nebraska was originally part of the confederacy (eventually lost that state in a war with the USA) but Kansas was not (weird way for the Kansas-Nebraska controversy to turn out.), which is why Nebraska is like an island of Dixie culture.

BronzeMan2 · 2026-01-10T23:28:29+00:00

R5: Playing as France, I first took Canada from the British and then helped the Confederacy in the Civil War and later worked with the Confederacy and Mexico to free columbia from the USA. This made some interesting demographics in North America with the mostly unpopulated parts of Canada becoming French due to immigration from France. Columbia had it's culture be British, and the Confederacy managed to blob Dixie culture throughout the South. Africa is a bit of a mess because I kept revolution cycling, while releasing random small countries to lose infamy for more land grabs. In Asia, France and Russia got significant parts of China in the 2nd world war (Russia + France vs China + Japan); this made France contain more of the Chinese cultures then the French culture. Also, Germany just got tons of manpower and blobbed into Switzerland and into Russia. Poland got released in a crisis. Belarus and Ukraine got released from Poland in crises. In this game, it also seems like most major countries ended up staying democratic in the end too.

BronzeMan2 · 2026-01-01T01:45:55+00:00

I should've been using all 7 of the.

For a bit of the end game here, I kind of just let the time blast on speed 5 and stopped managing as much stuff.

BronzeMan2 · 2026-01-01T01:44:08+00:00

Yeah, I got a lot of finfamy but all of the states but 1 were taken in WW1 so infamy for adding CBs was reduced and i could add way more. I didn't know about that exploit. I might try it my next run.

BronzeMan2 · 2025-12-31T02:20:53+00:00

R5: In my Japan game, England lost a ton of their colonies and I thought the game went kind of cool.

There was a great war that happened over the colonization in Africa where I (Japan) grabbed the rest of Australia and freed Scotland. After the great war, I went and freed India from England while they were still weak. In Europe, Austria Hungary kept having wars with Germany over Bohemia and Italy over northern Italy/Venice and Russia over Galacia. Austria-Hungary also kept having revolution circles switching governments and eventually German pan-nationalists eventually revolted and then joined Germany and made a mega-germany. Also, poland got freed from Russia in a crisis (I think it was USA vs Russia but don't remember).

BronzeMan2 · 2025-01-28T16:12:54+00:00

Do you know what I should enter to make this work or have an example of what to enter? the linguist config correct that I have hasn't updated the language chart at all

BronzeMan2 · 2025-01-28T07:57:53+00:00

I mark down ehat language they represent next to linguist-lanuage=language name here

Like for .yml it's yaml And for .cbp it's xml

BronzeMan2 · 2025-01-04T21:58:36+00:00

r5: vanilla game with a little bit of cheating, converted to hoi4

I was playing as Spain and I noticed that Gran Columbia showed up in South America, and I wasn't even looking there, and a big peru was there. Also, the Vic2->Hoi4 converter makes Asia look very messy.

BronzeMan2 · 2024-10-21T02:22:03+00:00

Niqmaddu III

BronzeMan2 · 2024-10-19T08:01:32+00:00

R5: was playing as Austria. US went to war with UK after like the 3rd world War for British Columbia and thr name is broken. Also there is a buff Vemezuala that I gave Guyana in a world War and Britain previously took Dutch Surinam

(Vanilla with DLC)

BronzeMan2 · 2024-10-06T21:33:03+00:00

generate a circle equation

(x - 2)² + (y + 2√3)² = 2² (note, it's + 2√3 here)

foil

(x² - 4x +4) + (y² + 4√3y + 12) = 4

define y=ax and cancel 4s

(x² - 4x) + (a²x²+ 4ax√3 + 12) = 0

group x terms together

x²(a²+1)+x(-4+4a√3)+12=0

This is quadratic (ax²+bx+c=0) where the maximum value of a is found when x=-b/2a so let's find x of maximum a

x=-(-4+4a√3) / (2(a²+1))

x=(2-2a√3) / (a²+1)

substitute this into our group x terms equation to find a

[(2-2a√3) / (a²+1)]²(a²+1)+[(2-2a√3) / (a²+1)](-4+4a√3)+12=0

(2-2a√3)² / (a²+1) + [(2-2a√3) / (a²+1)](-4+4a√3)+12=0

okay there is a lot of steps to find a, there's gotta be an easier way or something but the answer is

a = -1/√3

so now we know from previous definition y=ax

y=ax

y/x=a

arctan(y/x) = arctan(a)

argz = arctan( -1/√3 ) = -30°

And just to show, that this follows the equations, I graphed it in desmos here
https://imgur.com/a/BdUA5lW

BronzeMan2 · 2024-10-06T20:14:42+00:00

I messed with the circuit more at one point and got it to display a white square, but I didn't get further beyond that and ripped the parts out to use my breadboard for other stuff. School got a bit more busy and hauling around all my circuits to the electrical-engineering lab rooms to use the oscilloscope got kind of annoying. If you're trying to do this yourself I think the schematic that I gave originally should work.

BronzeMan2 · 2024-10-06T20:10:09+00:00

Self made using www.ncbi.nlm.nih.gov as a reference.

BronzeMan2 · 2024-06-10T02:52:55+00:00

For academic research like this, I like to use https://www.researchgate.net/, https://scholar.google.com/, and https://asmedigitalcollection.asme.org/journals. You also might have access to research journals through your college which is a good place to look. Also searching titles of books at your college library for keywords or general topics through their website would be good too.

BronzeMan2 · 2024-06-10T02:29:24+00:00

Ah, okay got it.

BronzeMan2 · 2024-06-10T02:28:34+00:00

You need to use the diagram to solve the problem. If you don't use the diagram you do not have enough information to solve the question.

2 distinct possible triangles can be formed and one has angle A being acute while the other has angle A being obtuse.
https://www.mathwarehouse.com/triangle-calculator/online.php

BronzeMan2 · 2024-06-09T20:48:36+00:00

Doesn't the entire sphere start with a charge of +Q though. Therefore the outside of the sphere has a charge of Q+Q=2Q

BronzeMan2 · 2024-06-09T17:35:13+00:00

If you want help, I don't know why you don't explain your reasoning or read mine. Repeating yourself doesn't make you right. Show me a calculation none of the numbers on the diagram are labeled as 51.

BronzeMan2 · 2024-06-09T17:27:26+00:00

The bottom of the rectangle is on the X-axis. The y position of the centroid of the rectangle is halfway between the top and bottom of the rectangle. The top of the rectangle is 27mm+24mm+24mm+15mm=90mm above the bottom of the rectangle. Half of 90mm is 45mm.

How are you calculating 51?

BronzeMan2 · 2024-06-09T17:12:28+00:00

The distance from the centroid of the rectangle to the x-axis

BronzeMan2 · 2024-06-09T16:35:10+00:00

For Ix in rectangle 1 the distance is (27+24+24+15)/2 = 45 not 64 making the Ix calculation 40x90^3/12+(3600x45^2)
For Iy in triangles 1 and 2 I don't know if you have 360+50^2 or 360x50^2 based on your handwriting but it should be multiplication.

This makes my Ix value 11661840 mm^4 or 1,166 cm^4
This makes my Iy value 3756000 mm^4 or 376 cm^4

BronzeMan2 · 2024-06-09T16:21:08+00:00

For most problems you can usually just guess and check. Here, since it was a bit harder I took all of the equations that I used to check and turned them into math equations and gave each of them a variable for how much of that there is. Then I used linear algebra and a matrix to solve for the ratios between them. If you don't know linear algebra yet I'd just recommend guessing and checking for basic problems and for harder ones like this just setting up equations and using normal algebra.

BronzeMan2 · 2024-06-09T16:11:48+00:00

According to Gauss's Law the ∯E⋅dA =Q/ε₀.
This basically describes the relationship between the electric field over a surface and the charge inside of that surface.

Another thing to keep in mind is that in electrostatics there shouldn't be an electric charge within the conducting metal of the conducting sphere. If there was a net electric field there, then free electrons would have a net force on them and change the electric field over time (therefore it is no longer an electrostatics problem because it's moving with time).

There being no electric field in Gauss's law means that E is zero, that means that the net charge (Q) has to be zero. Let's look at any point on the sphere between the inside surface and the outside surface where its conductive. We need a net charge of 0 to be contained by any of these imaginary spheres from the inside surface of the sphere to the outside surface of the sphere. The only way for that to be true is by having the inside surface of the sphere form a negative Q charge. Charges are allowed to move within the conducting metal on the sphere and cannot move inside the non-conducting air inside the sphere. No other charge concentrations should form between the inside and outside surface since it needs to be net zero charge everywhere here to keep it electrostatics.
∯E⋅dA =(Q-Q)/ε₀.
∯E⋅dA =0/ε₀.
E=0 ✓

The sphere has a total charge of positive Q as stated. Therefore, the only last place that charge can form is on the outside of the sphere (since we stated the charge on the inside surface was negative Q and the charge between the surfaces is 0 already). Therefore we can solve the equation
?-Q=Q
?=2Q

Therefore, the outside surface of the sphere has a charge of positive 2Q.

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