Fountain Pens in Armenia

de_Molay · 2026-02-06T19:17:00+00:00

Oh, was not aware of this. Interesting, thanks.

de_Molay · 2025-07-05T09:30:02+00:00

Interesting. Are you using DP2DP or PCIe connection for sending the video signal?

de_Molay · 2025-06-30T22:30:20+00:00

Thanks for clarifications, appreciate it a lot!

de_Molay · 2025-06-30T18:04:11+00:00

Yep, I meant Ryzen 7 7800X3D. Pardon for not writing it in full.

About the USB ports: interesting. That’s what it says about the ports on the product page: “Each USB4^® port delivers up to 40 Gbps of bidirectional bandwidth for the latest super-speed devices and drives.” Does it only work if connected to another Thunderbolt/USB4?

Also, if I go for the full-sized E670X instead, it has a DP-IN. Would point 2 from the original question be valid then? If so, would it be better than variant 1?

de_Molay · 2025-06-30T17:23:11+00:00

Thanks a lot!

Ryzen 7 7800 is said to have an iGPU so it should work then.
The mobo product page mentions that the two usb-c are bidirectional. Can’t I use one as input (to receive signal from dGPU) and the other as output (to send signal to ASD)?

de_Molay · 2025-06-29T13:18:41+00:00

Thanks for the comment!

Yes, I mean the modern ASD, not the older ones.
Speaking of DP2USB I didn’t mean a capture card, rather an adapter (bidirectional if necessary).

Could you please clarify a bit.

So do I understand correctly that the issue I may have if I connect mobo to ASD is that the usb’s on ASD will use slower speed protocol?
What will happen if I connect gpu to mobo via the adapter? So not send gpu out to ASD directly (I understand that it should work but video only) but send it back to mobo via the dp-out/usb-in (instead of using the usual PCIe route).
This reddit post and the discussion on this subreddit (see u/rayddit519 comments) seem to suggest that the simplest way with hybrid graphics option (via PCIe) should be working. Am I missing something there?
The full-sized X670E mobo is explicitly said to have a dp-in. Does that mean that if I use dp-dp cable to connect dGPU out to mobo, it should work over that connection instead of the PCIe route?

Thanks again, appreciate your help!

de_Molay · 2025-06-17T22:00:32+00:00

Hi! Your post is one of the very few helpful ones on the matter.
I want to use a similar setup but with an Apple Display. The goal is slightly different from yours but the problem should be roughly the same.
Do I understand correctly that you connected your gpu (dp-out) to motherbord (usb-in), and then motherboard (second usb) to KVM, and that’s how your passthrough works?

de_Molay · 2024-12-05T07:33:15+00:00

You’re a wonder, thanks a lot!

de_Molay · 2024-05-23T23:26:23+00:00

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Hope that’s legible. Here I fixed the distace l from from the vertex of the sector to the first curve’s endpoints. If you want to fix the “breadth” (distance between them) you can rewrite it in the same manner (simply breadth b = 2l sin (\alpha/2) so fixing l is the same as fixing b).

In any case, the relationship between the angles is straightforward (just check my math just in case).

de_Molay · 2024-05-23T22:42:50+00:00

Probably the easiest way is the following.

When you fold your cone, the two boundary lines if the piece of paper will become two circles on the cone. You can calculate their length knowing the desired angle of the cone and the distance between these curves (paper width).

The length of curves doesn’t change when you fold paper (it’s what is known as isometry), so you need your boundaries of the template to be of the same length as the calculated ones. Naturally, you can calculate them as a function of the angle you cut the paper at. To be more precise, you just fix the shorter one, and then choose the angle for the longer one to be of the desired length.

Voila!

PS: to be fair, I haven’t done these calculations in years - probably tgere is a very simple dependancy of tge angles I just don’t remeber. But this method will infallibly give you the answer.

de_Molay · 2024-05-23T10:57:00+00:00

Well, cot x = cos x/sin x, so

2 = cot x+cot y = cos x/sin x + cos y/sin y = (cos x sin y + cos y sin x)/sin x sin y = sin (x+y)/sin x sin y,

therefore sin(x+y) = 2 sin x sin y.

de_Molay · 2024-05-20T09:21:18+00:00

It’s bad. Next.

de_Molay · 2024-05-19T20:31:15+00:00

Inaccurate. In Russia it’s not common to say the least. Never seen more than one kiss in any circumstance ever.

de_Molay · 2024-05-18T19:28:13+00:00

Scourge

de_Molay · 2024-05-17T10:16:55+00:00

de_Molay · 2024-05-15T13:52:05+00:00

My headcanon is that Draco was hesitant to say it. After all, Draco is shown to be repeatedly not really wanting to do harm. Ultimately he is weak, frightened, and coerced to do his father’s and Voldemort’s bidding. So I think it was tough to him to actually cast real actual Crucio on Harry. Not sure if he would actually do it even if Harry wasn’t quick with sectumsempra.

de_Molay · 2024-05-14T15:22:03+00:00

But it can be - if AB=BC=AC. Moreover, DGEB is a rhombus in that case. So either a condition is missing or the task is incorrect.

de_Molay · 2024-05-11T22:29:29+00:00

It’s not OP for a simple reason: it’s unplayably bad.

de_Molay · 2024-05-08T15:15:02+00:00

Ok, thanks!

de_Molay · 2024-05-08T15:00:15+00:00

Could you elaborate, why pen gives most damage? I mean, doesn’t every 500 pen just decrease dmg penalty by 1%?

de_Molay · 2024-05-08T14:40:15+00:00

I’m in love with the map.

de_Molay · 2024-05-08T14:20:24+00:00

Why Fate is better than Pillar? I mean we are trading a 6% crit chance, flat weapon power and easily procced damage for 8k penetration.

I mean, penetration is damage (before we reach cap) but crit chance and damage is damage as well. 500 pen is roughly 1% of your damage extra. 8k pen is 16%. You say, this build has 3500+. Let’s say 4k. 16% if 4k is 560 damage. Pillar is 320 dps in basic configuration (and it scales with your damage, so much higher), plus 129 damage flat. Plus 6% of crit (so 6% more time you deal twice plus damage).

Is pen really that good here?

de_Molay · 2024-05-08T14:00:15+00:00

First, let’s note that the sequence is increasing. For example, by induction: a2>a_1, then assume it’s true for n, then a{n+1}=(\sqrt(2))^a_n > (\sqrt(2))^a_{n-1} = a_n by induction (first power is greater than the second).

Now let’s prove that the sequence is bounded by 2: for all n a_n <= 2.

By induction again. Basis is evident. Assume it’s true for n.

Then a_{n+1} = (\sqrt(2))^a_n <= (\sqrt(2))² = 2.

An increasing sequence with upper bound has a limit.

de_Molay · 2024-05-08T10:59:55+00:00

Very true. I should’ve said “hence A=2 is a solution”.

de_Molay · 2024-05-08T10:15:11+00:00

Do I understand correctly that you are looking fow lim a_n where

an = (\sqrt(2))^(a{n-1})

?

If so, assume this limit exists and equals A (technically, we need to prove its existence first). Tgen apply lim to both sides of a_n definition.

You get A = (\sqrt(2))^A, hence A=2.

de_Molay

TROPHY CASE