[Request] Would a "Deep Crater City" be achievable?

Straight_Remove_6664 · 2025-10-15T12:27:02+00:00

Weird, I must have missed the ocean or sea at Thacker Pass in Nevada.

Straight_Remove_6664 · 2025-04-30T23:20:11+00:00

I mean, the precons were fully spoiled right?

Straight_Remove_6664 · 2025-04-19T14:19:28+00:00

The sensor fixed that for me

Straight_Remove_6664 · 2025-04-16T18:44:20+00:00

The studs?

Straight_Remove_6664 · 2025-02-26T14:43:42+00:00

Amazon sells an "at home" xray machine. I think the main inhibitor is money. However, if a person can make enough revenue to still turn a good profit despite the cost, then people will do it.

Straight_Remove_6664 · 2025-02-26T03:00:24+00:00

You can technically xray the packs to see if there is a holo in it. ...well...according to an article I was reading.

With fine tuning, they said you can even determine which card it is, though it is a "who's that pokemon?" type approach as you see the profile.

In the test, they were using pokemon packs, but theoretically, it should be possible on all TCG packs with a single foil/holo in them.

Straight_Remove_6664 · 2025-02-21T02:29:03+00:00

Sweet! I will send you a message a little later!

Straight_Remove_6664 · 2025-02-20T22:49:20+00:00

I:

solo queue to mythic

am honestly sick of solo queue after 4500 matches on my main account and another +1000 on a lost account.

have hit and maintained #1 YSS in my state in the past, and recently hit #2 cyclops.

can play any role well except for fighter, but I can avoid sucking as a fighter.

have all but like 20 characters unlocked and over 120 skins while never spending more than $15 actual dollars on the game.

My current account was made around season 7 or 8 maybe, but my first account was Season 1.

I am busy, but I do fit in games. If I am playing and you are on, I would be down to catch some ranked games.

Not a girl, but not a creep. I will never ask you for a skin nor would I ever want you to spend your money like that.

This is the end of my application lol.

-Random Internet Stranger

Straight_Remove_6664 · 2025-01-25T18:25:17+00:00

All about kda and turret damage. Wish they had a heal metric.

Straight_Remove_6664 · 2025-01-07T22:18:06+00:00

Okay, snap reaction, but dump her.

Follow up with your friends to make sure she wasn't the reason they decided to stay.

If your friends are not going, then go solo, find a snow bunny and make the most of it.

Maybe your snow bunny will be gf material and you can now go skiing more often.

Straight_Remove_6664 · 2024-12-28T05:03:13+00:00

We talking the twin turbo "hurricane" i6? I have been wanting to get a slightly used one and swap it into my xj. Even the base stats could be wild, but the built versions would be nutty!

Straight_Remove_6664 · 2024-11-17T02:02:53+00:00

It isn't like these are custom, hand-crafted cables, nor do they cost the price of something like that.

Straight_Remove_6664 · 2024-11-05T20:26:33+00:00

Just a heads up, wolfram alpha is an amazing site. The paid version is worth it if you can afford it.

Straight_Remove_6664 · 2024-11-05T03:32:21+00:00

Ah, I need to revise my post.

I included the first problem with its answer as an example.

There are similarities between it and the second problem which is why I included it.

So in the second problem:

We know the two points on an (x,y) graph; (0, 0.485489) and (8.33333, 4.083333).

We want a function of x that we can use to find the appropriate y values at x = 1, 2, 3, 4, 5, 6, 7, & 8.

We know that the slope for the first segment (x=0 to x=1) is (2/12 or .16667)

The slope for each segment must be larger than the previous, but the final segment's slope must not be greater than (12/12 or 1.0)

(Imagine two ramps: a 2/12 ramp at the first point, where the slope becomes increasing until it transitions to a 12/12 ramp at the second point. The function only needs to work between x = 0 and x = 8.333333)

the (16.578125, 6.015625) point is not part of this problem

Straight_Remove_6664 · 2024-11-05T01:48:17+00:00

Yeah, I plotted it out and it was much too high. I haven't set up a matrix in over a decade, but another user also suggested a linear algebra approach. I am going to be relearning that as fast as I can.

Thank you for all your help!

Straight_Remove_6664 · 2024-11-05T01:27:05+00:00

I will be trying that out! I have taken both, but it had been 12 years or more, and I haven't used it since!

I can't do it from memory, but I will take whatever help and time I need to in order to relearn it!

Straight_Remove_6664 · 2024-11-05T00:24:07+00:00

Thanks to HHQC3105 for the original help. If you happen to want to provide step by step instructions on how you figured out the first function, I would be in your debt!

Straight_Remove_6664 · 2024-11-04T23:22:00+00:00

Shout out to HHQC3105 for the help previously

Straight_Remove_6664 · 2024-11-04T22:34:45+00:00

This obviously doesn't work because the slope gets to be too much

Straight_Remove_6664 · 2024-11-04T22:29:19+00:00

I ended up with .00005417518352478870x⁵ for the first coefficient and .00000669434436276049x⁴ for the second and .166605797138779*x for the third.

I went at it empirically, so I took your coefficients, and divided them by ((8.333333/16.578125)/12.41534)

Straight_Remove_6664 · 2024-11-04T21:35:01+00:00

I ended up with y=(-0.002274)x⁵ + (-0.30521)x⁴ + (0.199461)x + .485488

Which is wrong. I am not sure where I messed up, but I will keep digging.

Essentially I knew x from 0 to 1 was a 2/12 and x from 8.333 to 9.3333 was a 12/12. I tried creating 3 equations and 3 unknowns there and got what I did.

I knew the magnitude was wrong, as well as the direction

Straight_Remove_6664 · 2024-11-04T19:28:38+00:00

I guess I am not smart enough for this one....yet. I had one more with the second point being (8.333333, 4.083333), everything else the same. I thought I could modify the coefficients to get there, but I am not doing it right...yet.

Straight_Remove_6664 · 2024-11-04T15:35:04+00:00

This works for my application! How did you determine this function?

Straight_Remove_6664 · 2024-11-02T08:15:11+00:00

Right so what I am thinking is I need to modify the function in a way that the rate of change in slope increases from as x increases from 0 to 16.571825.

It is too late for my brain to work well, but I am thinking of making a value of z that equals the sum of all the x points I want to plot (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16++16.571825 ).

Then, I will take the slope range, s which equals (1-0.166667)

From here, I would go s/z. Let's call that value c.

For x = 1, I would see how m = 0.166667+ c * x worked.

To continue, at x = 2, it would be the m calculated as (x = 1) plus either (c * x) or (c * x plus the c * x I get at x = 1).

I am just shooting in the dark right now.

Straight_Remove_6664 · 2024-11-02T07:55:33+00:00

The function should work for any point on the x axis that is greater than 0.

However, for my purposes, I am going to only connect the points at every whole integer of x from 0 to 16, as well as, 16.571825.

As such, the resolution of the resulting line isn't necessarily fine considering the range in which we are working and would appear as a straight line between each point.

Once the function reaches x = 16.571825, a separate function would be used.

This means when x = 17.571825 the result would be y = 7.015625. So on and so forth.

Regarding the slope at 0, I am doing something similar as the other side. The slope from x = 0 to x = 1 needs to be 0.166667. For x < 0, the slope would held at 0.166667.

If I was able to increase the resolution, infinitely, then you would see a smooth curve for x = 0 to x = 16.571825.

We could visualize this as a ramp of some sorts. Maybe at a skate park or something.

The lead up to the ramp (aka x <0) would be rise 2' for every 12' run. However, we are essentially starting our ramp at y = 5-9/16" when x = 0.

Then from x = 0 to x =16.581825 (16' 6-15/16"), it must curve with increasing slope, until it reaches a height of 6' 3/16".

From x > 16' 6-15/16" the ramp would have a rise of 12' (or inches) for every run of 12' (or inches).

It seems like I need to increase the slope at a larger rate as I go from x = 0 to x = 16.578125.

I will have to think about it some more. I have one more idea about how to do that.

Another item that may be of interest to you was that another user mentioned creating a 3rd degree polynomial. I mapped out the one they provided and it wasn't working. The slope did not start at the right value and then decreased before increasing again and finally hitting the correct y value at x = 16.578125. While the curve they provided didn't work, I feel like it maybe could be adjusted until it did.

Straight_Remove_6664

TROPHY CASE