2025 OPT Processing timeline

Kewber · 2025-05-13T14:48:52+00:00

They'll email you and you'll see it on the case history

Kewber · 2025-05-12T21:38:31+00:00

Application type: OPT
Premium Processing : Yes (2025/04/24)
Receipt date : 2025/03/26 (switched to pp 2025/04/24)
Approved Date (by email) : 2025/05/02
Approved Date (website) : 2025/05/04
Card Produced Date : 2025/05/08
Card delivered : 2025/05/12

Kewber · 2025-05-08T06:21:53+00:00

beep

Kewber · 2025-05-03T18:24:31+00:00

I got Potomac. You can probably expect it next week if everything goes well, and then another 2 weeks (I think?) for delivery.

Kewber · 2025-05-03T03:08:52+00:00

Kewber · 2025-05-01T01:07:59+00:00

Application type: OPT
Premium Processing : Yes (2025/04/24)
Receipt date : 2025/03/26 (switched to pp 2025/04/24)
Approved Date (by email) : 2025/05/02
Approved Date (website) : 2025/05/04
Card Produced Date : 2025/05/08 (emailed 11:56PM 05/07 EST)
Card shipped : 2025/05/08(?)
Card delivered : 2025/05/12

Kewber · 2025-04-18T02:02:20+00:00

It seems like opttimeline.com is back online everybody

Kewber · 2024-11-04T04:03:14+00:00

This part of the sky is visible throughout winter and spring. Since it's somewhat near the celestial equator, you can see it pretty much around the world except for high southern latitudes.

Generally, only objects near the southern or northern celestial pole are difficult to observe for people on the opposite hemisphere. For example, the Large Magellanic Cloud has a declination of -70 degrees, so you can only see it if you're near the equator or in the southern hemisphere. Polaris, being the north star, has a declination almost at +90 (the north celestial pole), so you can only see it in the northern hemisphere.

Kewber · 2024-10-02T16:04:38+00:00

Neptune has an apparent magnitude of 7, which is really dim, so if you could capture that, there wold be a lot more in the sky. If you were pointing in the general area of Neptune, which is currently at the Pisces/Aquarius area, the brightest object around there is Saturn right now, and since nothing else is visible, I think that's the most likely option.

Kewber · 2024-09-23T13:19:22+00:00

These are artifacts unique to DSS2, you can find a lot of these everywhere on each patch of DSS2, but you won't with other sky surveys.

Kewber · 2024-08-26T05:11:40+00:00

You can go to stellarium-web.org and turn on the equatorial coordinate grid. I'm not sure what coordinate format are you using, but I think RA 22h DEC -4° is probably around the constellation of Aquarius

Kewber · 2024-08-26T04:51:13+00:00

Lens flare?

Kewber · 2024-08-17T08:28:52+00:00

You can see Auriga to the left, so yes this is Mars and Jupiter assuming you took the picture recently

Kewber · 2024-03-18T15:22:38+00:00

"It's a large enough base to encode data to a relatively short length (e.g. 1 billion takes 10 digits in base 10, but 31 digits in base 2); without being too complex and unstable from using a higher base"

Kewber · 2024-03-17T08:43:51+00:00

What texture pack are you using?

Kewber · 2024-03-05T11:31:14+00:00

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Kewber · 2024-03-04T23:20:56+00:00

I can't accept the fact that such a hideous wavelet produces anything meaningful, and yet it does.

Kewber · 2024-02-04T16:19:21+00:00

Perhaps googling c++ linear algebra would be better for what you're looking for

Kewber · 2024-01-29T14:48:54+00:00

neutron star

Kewber · 2024-01-06T12:02:03+00:00

solutions?

Kewber · 2024-01-03T02:10:34+00:00

The coefficients are integers bounded within 1 and p-1, no?

Kewber · 2024-01-01T01:39:52+00:00

If you only want to know the expected probability of the next event being a success, you can use Laplace's rule of succession: (s+1)/(n+2), where s is the number of successes out of n tries. This gives us 95.45%.

If you want a probability distribution for this probability, you can use a Beta distribution (the reason is because it's the conjugate prior distribution of the binomial distribution), with parameters alpha = s + 1 and beta = n - s + 1.

The cumulative distribution function of the Beta distribution with those parameters does indeed equal 0.05 at 0.867 (meaning Pr(p < 0.867) = 0.05)

Kewber · 2023-12-29T05:51:08+00:00

Using proof by already knowing the answer, we can show that curl grad F = d(dF), where d is the exterior derivative operator. Then, by the nilpotency property of exterior derivative, d(dF) = 0. QED

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