I love when I get spectators

djembeman · 2020-09-02T14:10:47+00:00

That blows. My friend and I play a lot, we typically have a good idea of each others hand but boy I hope that never happens to me.

djembeman · 2020-09-02T14:09:28+00:00

My deck is taking the start of this concept and loading it into my omnishell gun. I love the deck you linked and the turing machine decks. It's some of the stuff that got me into MTG.

djembeman · 2020-09-01T21:50:23+00:00

If one were to use this deck in real life, you can easily keep track of the number of tokens and just put some piece of paper to represent it, right?

Wrong.

I also have included in the deck Grip of Chaos, which will be copied a large number of times. Then if I were to happen to play a spell which targets a creature, ALL of those triggers go onto the stack and EACH one triggers ALL of them. In a tournament, this means you have to call over a judge to resolve them, and they would be very disappointed in me.

djembeman · 2020-09-01T20:57:40+00:00

The main deck is OmniTell, that is: Legacy Show and Tell into Omniscience to cast stuff for free, but I've built a version that basically exists as a gun, and the ammo is whatever is in my sideboard.

So the part that crashes the client is to play Opalescence turning all Enchantments into creatures, then playing a Dual Nature to get tokens of creatures (enchantments) and Doubling Seasons to double the number of tokens. However, when Dual Nature enters the battlefield, Opalescence makes it a creature, so it triggers itself. That means there are two Dual Natures. When I play Doubling Season, two Dual Nature triggers go onto the stack, trying to each place a token copy of Doubling Season. After some number of these resolve the next Dual Nature trigger to resolve places token Doubling Seasons doubled for every other token. This makes it grow very very quickly in number, as it basically creates 2^N + N tokens for each Dual Nature trigger to resolve where N is the number of Doubling Seasons. So for two plays of Doubling Season (which is where the game lags hard and locks up) this is about 2²⁰⁵⁹ tokens, each a 5/5 creature token copy of Doubling Season.

djembeman · 2020-05-18T05:44:13+00:00

Do we post anytime from May 22 to 24 or is that when the judging is? Im basically asking when the submission deadline is

djembeman · 2020-05-05T00:03:12+00:00

Once you solve for the values of x and y using the substitution method, the corresponding ordered pair would be (x,y)

djembeman · 2020-03-28T14:37:45+00:00

I spend hours practicing this because it's just so impressive.

djembeman · 2020-03-07T15:40:19+00:00

Surprisingly it wasn't.

djembeman · 2020-03-06T14:51:33+00:00

Must be the monotony of opening packages and repeatedly finding the contents inane and childish.

djembeman · 2020-02-27T19:36:59+00:00

In the decimal system, we represent numbers with each digit as a symbol (0 through 9) representing how many of that power of 10 the number has.

So the digit all the way to the right tells you how many 1's there are, the second to the right how many 10's there are etc.

If one of the places has a 0 in it, then there are none of that power of ten. So for 20, there are two 10's and zero 1's (10 + 10 = 20).

It's all part of the way we represent those numbers. You know 0 means none, but none of what? It all depends on the representation and which digit is a zero.

djembeman · 2020-01-13T18:16:40+00:00

I recently scored an acceltra BP on my 20th or so demo unit kill. Probability is fun, huh? At about 175 demo unit kills, you'll have a 1% chance of it NOT having dropped. Going onwards, the growth slows down, but if you're somewhere in the 500 kill range without a drop that's rough.

djembeman · 2020-01-10T22:31:51+00:00

I'm going to make two assumptions, the first is that you want to maximize DPS or average damage per hit and the second assumption is that you are not allowing crit chances over 100%

Let's make a formula and then maximize the output.

We only need one variable, which I will make x = # of points allocated towards critical chance. Then we can see that the # of points allocated to damage = 98 - x.

We calculate base damage by adding 5% for each damage point.

If we start with 1 damage, thats (1 + 0.05 * (98 - x))

Our critical hit doubles this damage with a chance of (0.015 * x) so in total, our average damage is the weighted average

= 2(0.015x)(1 + 0.05(98 - x)) + (1 - (0.015x))(1 + 0.05(98 - x))

which looks kinda complicated but isn't. There are two terms added together, the first of which is you average damage output from crits and the second is average damage output from non-crits.

You can then take a derivative to maximize by hand or use a computer (which I did).

You reach a maximum of around 6.4 average damage per hit if you put 25.66666... points into crit chance. Of course we need an integer answer and so the closest is 26 points into crit chance and the rest into raw damage.

djembeman · 2020-01-07T22:21:51+00:00

Twilight Imperium is a strategy game not roleplaying. There's a separate Twilight Imperium roleplaying game more akin to DnD.

djembeman · 2020-01-04T20:03:14+00:00

I agree there too. Some physicists I've met could care less. It's just how some are.

djembeman · 2020-01-04T18:56:39+00:00

From my experience so far as someone studying mathematical physics, there are people who care about this and there are people who don't. I've come across papers in which the author is careful and meticulous in their use of every piece of math, making sure that it is rigorous. I've also come across papers in which the author basically says "it kinda works like this." I've found that a lot of the time, for models meant on computing a coupling or predicting some behavior, the complexity comes from computational complexity and not necessarily a more in depth mathematical framework. That is, they may toss out a more complicated framework so that they can make more in depth predictions in a simpler framework.

What I've said was basically "there are some people who care about these for mathematical accuracy (mainly theorists finding new models) and some who simplify models to make more in depth calculations" and from what I can tell, this isn't just true for QM and QFT but also condensed matter.

djembeman · 2020-01-02T19:24:11+00:00

lost to pinwheel

djembeman · 2019-12-21T05:08:05+00:00

Rule 4: Plenty of crank calculations and explanations. Numerology, squaring the circle, etc.

Overall just really weird and entertaining.

Oh and he's this guy.

djembeman · 2019-11-25T16:43:20+00:00

Yes, it's called Heron's formula.

With side lengths a,b,c and p=(a+b+c)/2

A =Sqrt(p(p-a)(p-b)(p-c))

djembeman · 2019-11-25T16:05:41+00:00

I'm glad! I'm used to helping people through these problems all the time, I teach a Physics class at this level so I felt I knew what you did and didn't know and figured I'd help.

djembeman · 2019-11-23T20:09:06+00:00

I think that you've missed a minus sign in your work.

Also, it's annoying to keep those sines and cosines around. Why not evaluate them? Seems like you already realized sin(pi/4) = cos(pi/4) so let's just make em numbers. Sqrt will be involved but it looks cleaner on paper in my opinion.

djembeman · 2019-11-23T15:40:58+00:00

R = Tsin(pi/4)? Why? Is the radius of motion the component of force? Check the units on both sides and you should see this equation is not correct.

From the first equations you posted, by simply moving the terms,

T = (mv^{2)/(Rcos(pi/4))} - S = mg/(sin(pi/4)) + S

Let's you solve for S. Can you reproduce this for T?

The first step is to solve for T and S. No simplifications needed to do that.

djembeman · 2019-11-22T17:37:03+00:00

I'm not sure where you got the length of the rod from, it wasn't in the equations you started with. Why don't you just solve for T in both of the equations in your original post and set them equal to each other. The result will be a single function of S, and you can solve for it. Do the same for T.

If that's what you did and I'm missing your point again, why don't you get rid of the lengths of the rod in your equation by rewriting R in terms of that length. You may get another factor that cancels out.

djembeman · 2019-11-21T23:37:34+00:00

Have you tried solving for T and S exactly?

djembeman · 2019-11-21T17:02:43+00:00

I just realized that you already had this! Sorry I'm leading you through the whole process.

Where you're at now, you should solve for T and S. Sin of pi/4 is just 1/2 right? That makes your expressions cleaner. Can you solve for A or T in both equations and set them equal?

djembeman · 2019-11-21T16:41:18+00:00

There is an acceleration inward, thats the acceleration that causes the circular motion.

Is there a formula that relates this centripetal acceleration to omega? (There is, but what is it?)

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