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Lawnie
Sep 6, 2006

That is my helmet
Give it back
you are a lion
It doesn't even fit
Grimey Drawer

FreshFeesh posted:

A stupid, frustrating question I'd genuinely appreciate any help with:

Occasionally (multiple times per week) my Samsung smart TV will simply stop playing sound from YouTube. The TV itself still has sound (e.g. the menus, interface), all other connected apps and devices have sound, it's just YouTube. Sometimes ads will play just fine and then the main video won't have any sound. Sometimes both the ads and the video have no sound. Sometimes a video will play fine, hit an ad-break which plays fine, and then on return there's silence. It's as if the entire audio track is just missing when loading (or switching to) a video. Once the problem arises, no video (or autoplay ads) will have sound for up to hours at a time. Even trying to cast YouTube from a different device to the TV will still present the issue.

I've verified the issue happens whether or not I have my network-level adblocker enabled (pi-hole), it persists through YouTube app and TV restarts, and there's no discernible pattern as to when the audio shuts off (time of day, other network activity, order of app usage, et cetera). There aren't any firmware or software updates for my TV, and I went so far as to do a complete reset of the device. That last step seemed to resolve the problem for about two weeks until it came back tonight.

How can I make sure that I have sound when playing YouTube videos through/on my TV?

This started happening randomly on my LG TV’s YouTube app in the last few weeks. However, turning the tv off and back on again resolves it. Maybe something on YouTube’s backend SmartTV infrastructure is glitching and in need of some attention. Sorry this isn’t super-helpful, but you’re not alone.

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Busy Bee
Jul 13, 2004
Does anyone have experience sending a small high value package from EU to the US? I want to get a few of my trading cards graded and the estimated value can range from a few thousand dollars to over ten depending on the grade.

1) I will be sending it directly to the grading company - how can I make sure there will be no issues in customs and how do I figure out if there is a tax to be paid?
2) If, for some reason, the package gets lost, how would I prove to the shipping company the value of the cards?

smackfu
Jun 7, 2004

Can you ask the grading company? Seems like a common question.

tuyop
Sep 15, 2006

Every second that we're not growing BASIL is a second wasted

Fun Shoe
I'm trying to figure out if this can be turned into a math question:

You have a roll of toilet paper. Each sheet is 10cm long and .5mm thick. There are 180 sheets on the roll. The cardboard roll around which the paper is wound is 6cm in diameter.

First, is it possible to determine from that information the diameter of the roll?

Second, what expression communicates how many sheets there are on a given "layer" of the roll? The first outermost layer would, of course, have the most with the final innermost layer only being like two.

TooMuchAbstraction
Oct 14, 2012

I spent four years making
Waves of Steel
Hell yes I'm going to turn my avatar into an ad for it.
Fun Shoe

tuyop posted:

I'm trying to figure out if this can be turned into a math question:

You have a roll of toilet paper. Each sheet is 10cm long and .5mm thick. There are 180 sheets on the roll. The cardboard roll around which the paper is wound is 6cm in diameter.

First, is it possible to determine from that information the diameter of the roll?

Second, what expression communicates how many sheets there are on a given "layer" of the roll? The first outermost layer would, of course, have the most with the final innermost layer only being like two.

The number of sheets on a given layer should be a function of the layer's diameter and the length of a sheet: n=pi*D/(10cm). Choose D as the diameter of the middle of the layer.

As for your first question, I haven't found a spiral function that works off of the thickness of each layer, just one that uses the inner/outer diameters, the number of layers, and the overall length of the spiral. Since number of layers and outer diameter are related, that leaves us with two unknowns. However, I'm pretty sure you can solve for outer diameter with the values you provided. It just may require some more complicated algebra or basic calculus. The problem is that the relationship between number of sheets and overall diameter is nonlinear, i.e. as you add more sheets, you get less and less additional diameter. This feels like a situation where you'd apply an integral, but my calculus is super rusty.

tuyop
Sep 15, 2006

Every second that we're not growing BASIL is a second wasted

Fun Shoe
Double post oops

tuyop fucked around with this message at 21:21 on Jan 2, 2023

tuyop
Sep 15, 2006

Every second that we're not growing BASIL is a second wasted

Fun Shoe

TooMuchAbstraction posted:

The number of sheets on a given layer should be a function of the layer's diameter and the length of a sheet: n=pi*D/(10cm). Choose D as the diameter of the middle of the layer.

As for your first question, I haven't found a spiral function that works off of the thickness of each layer, just one that uses the inner/outer diameters, the number of layers, and the overall length of the spiral. Since number of layers and outer diameter are related, that leaves us with two unknowns. However, I'm pretty sure you can solve for outer diameter with the values you provided. It just may require some more complicated algebra or basic calculus. The problem is that the relationship between number of sheets and overall diameter is nonlinear, i.e. as you add more sheets, you get less and less additional diameter. This feels like a situation where you'd apply an integral, but my calculus is super rusty.

I think it goes the other way, more sheets is more diameter, but yeah that’s what I was thinking. Definitely beyond my students and me then, though I am wondering what will be involved.

Mr. Nice!
Oct 13, 2005

bone shaking.
soul baking.

tuyop posted:

I'm trying to figure out if this can be turned into a math question:

You have a roll of toilet paper. Each sheet is 10cm long and .5mm thick. There are 180 sheets on the roll. The cardboard roll around which the paper is wound is 6cm in diameter.

First, is it possible to determine from that information the diameter of the roll?

Second, what expression communicates how many sheets there are on a given "layer" of the roll? The first outermost layer would, of course, have the most with the final innermost layer only being like two.

I don't know exactly how to make it into a succinct question, but I solved it in excel without too much trouble.

I converted everything to mm for simplicity. The roll is 60mm in diameter which means it has a circumference of 188.5mm. This will use 1.88 sheets (at 100mm per). After the first pass, there are 148.12 sheets left and a new diameter of 61mm. This has a circ. of 191.64mm which takes 1.92 sheets and leaves 146.2. I just filled down the cells until I got to negative sheets remaining. After 54 passes, the diameter is 114mm and there are 3.25 sheets of TP remaining that will overlap.

There is definitely a way to write this up as a simple function, but I don't know offhand the easiest way to do that. Solving in excel is trivial, though. Here's my numbers if anyone wants to check. I used =pi() for pi.

code:
Pass	dia.	Circumference	Sheets used	sheets rem
Pass	diameter	Circumfrence	Sheets used	sheets rem
1	60	188.4955592	1.884955592	148.1150444
2	61	191.6371519	1.916371519	146.1986729
3	62	194.7787445	1.947787445	144.2508854
4	63	197.9203372	1.979203372	142.2716821
5	64	201.0619298	2.010619298	140.2610628
6	65	204.2035225	2.042035225	138.2190275
7	66	207.3451151	2.073451151	136.1455764
8	67	210.4867078	2.104867078	134.0407093
9	68	213.6283004	2.136283004	131.9044263
10	69	216.7698931	2.167698931	129.7367274
11	70	219.9114858	2.199114858	127.5376125
12	71	223.0530784	2.230530784	125.3070817
13	72	226.1946711	2.261946711	123.045135
14	73	229.3362637	2.293362637	120.7517724
15	74	232.4778564	2.324778564	118.4269938
16	75	235.619449	2.35619449	116.0707993
17	76	238.7610417	2.387610417	113.6831889
18	77	241.9026343	2.419026343	111.2641626
19	78	245.044227	2.45044227	108.8137203
20	79	248.1858196	2.481858196	106.3318621
21	80	251.3274123	2.513274123	103.818588
22	81	254.4690049	2.544690049	101.2738979
23	82	257.6105976	2.576105976	98.69779197
24	83	260.7521902	2.607521902	96.09027006
25	84	263.8937829	2.638937829	93.45133224
26	85	267.0353756	2.670353756	90.78097848
27	86	270.1769682	2.701769682	88.0792088
28	87	273.3185609	2.733185609	85.34602319
29	88	276.4601535	2.764601535	82.58142165
30	89	279.6017462	2.796017462	79.78540419
31	90	282.7433388	2.827433388	76.9579708
32	91	285.8849315	2.858849315	74.09912149
33	92	289.0265241	2.890265241	71.20885625
34	93	292.1681168	2.921681168	68.28717508
35	94	295.3097094	2.953097094	65.33407799
36	95	298.4513021	2.984513021	62.34956496
37	96	301.5928947	3.015928947	59.33363602
38	97	304.7344874	3.047344874	56.28629114
39	98	307.8760801	3.078760801	53.20753034
40	99	311.0176727	3.110176727	50.09735362
41	100	314.1592654	3.141592654	46.95576096
42	101	317.300858	3.17300858	43.78275238
43	102	320.4424507	3.204424507	40.57832788
44	103	323.5840433	3.235840433	37.34248744
45	104	326.725636	3.26725636	34.07523108
46	105	329.8672286	3.298672286	30.7765588
47	106	333.0088213	3.330088213	27.44647058
48	107	336.1504139	3.361504139	24.08496644
49	108	339.2920066	3.392920066	20.69204638
50	109	342.4335992	3.424335992	17.26771039
51	110	345.5751919	3.455751919	13.81195847
52	111	348.7167845	3.487167845	10.32479062
53	112	351.8583772	3.518583772	6.806206849
54	113	354.9999699	3.549999699	3.256207151
55	114	358.1415625	3.581415625	-0.325208474
e: someone corrected my math and I've updated the numbers.

Mr. Nice! fucked around with this message at 14:20 on Jan 3, 2023

tuyop
Sep 15, 2006

Every second that we're not growing BASIL is a second wasted

Fun Shoe

Mr. Nice! posted:

I don't know exactly how to make it into a succinct question, but I solved it in excel without too much trouble.

I converted everything to mm for simplicity. The roll is 60mm in diameter which means it has a circumference of 188.5mm. This will use 1.88 sheets (at 100mm per). After the first pass, there are 148.12 sheets left and a new diameter of 60.5mm. This has a circ. of 190mm which takes 1.9 sheets and leaves 146.2. I just filled down the cells until I got to negative sheets remaining. After 63 passes, the diameter is 91mm and there is 0.57 sheets of TP remaining that will overlap. So for 57mm it will be 91.5mm thick but the rest of the roll will be 91mm.

There is definitely a way to write this up as a simple function, but I don't know offhand the easiest way to do that. Solving in excel is trivial, though. Here's my numbers if anyone wants to check. I used =pi() for pi.

code:
Pass	dia.	Circumference	Sheets used	sheets rem
1	60	188.4955592	1.884955592	148.1150444
2	60.5	190.0663555	1.900663555	146.2143809
3	61	191.6371519	1.916371519	144.2980093
4	61.5	193.2079482	1.932079482	142.3659299
5	62	194.7787445	1.947787445	140.4181424
6	62.5	196.3495408	1.963495408	138.454647
7	63	197.9203372	1.979203372	136.4754436
8	63.5	199.4911335	1.994911335	134.4805323
9	64	201.0619298	2.010619298	132.469913
10	64.5	202.6327262	2.026327262	130.4435857
11	65	204.2035225	2.042035225	128.4015505
12	65.5	205.7743188	2.057743188	126.3438073
13	66	207.3451151	2.073451151	124.2703562
14	66.5	208.9159115	2.089159115	122.1811971
15	67	210.4867078	2.104867078	120.07633
16	67.5	212.0575041	2.120575041	117.9557549
17	68	213.6283004	2.136283004	115.8194719
18	68.5	215.1990968	2.151990968	113.667481
19	69	216.7698931	2.167698931	111.499782
20	69.5	218.3406894	2.183406894	109.3163751
21	70	219.9114858	2.199114858	107.1172603
22	70.5	221.4822821	2.214822821	104.9024375
23	71	223.0530784	2.230530784	102.6719067
24	71.5	224.6238747	2.246238747	100.4256679
25	72	226.1946711	2.261946711	98.16372122
26	72.5	227.7654674	2.277654674	95.88606654
27	73	229.3362637	2.293362637	93.5927039
28	73.5	230.90706	2.3090706	91.2836333
29	74	232.4778564	2.324778564	88.95885474
30	74.5	234.0486527	2.340486527	86.61836821
31	75	235.619449	2.35619449	84.26217372
32	75.5	237.1902453	2.371902453	81.89027127
33	76	238.7610417	2.387610417	79.50266085
34	76.5	240.331838	2.40331838	77.09934247
35	77	241.9026343	2.419026343	74.68031613
36	77.5	243.4734307	2.434734307	72.24558182
37	78	245.044227	2.45044227	69.79513955
38	78.5	246.6150233	2.466150233	67.32898932
39	79	248.1858196	2.481858196	64.84713112
40	79.5	249.756616	2.49756616	62.34956496
41	80	251.3274123	2.513274123	59.83629084
42	80.5	252.8982086	2.528982086	57.30730876
43	81	254.4690049	2.544690049	54.76261871
44	81.5	256.0398013	2.560398013	52.20222069
45	82	257.6105976	2.576105976	49.62611472
46	82.5	259.1813939	2.591813939	47.03430078
47	83	260.7521902	2.607521902	44.42677888
48	83.5	262.3229866	2.623229866	41.80354901
49	84	263.8937829	2.638937829	39.16461118
50	84.5	265.4645792	2.654645792	36.50996539
51	85	267.0353756	2.670353756	33.83961163
52	85.5	268.6061719	2.686061719	31.15354991
53	86	270.1769682	2.701769682	28.45178023
54	86.5	271.7477645	2.717477645	25.73430259
55	87	273.3185609	2.733185609	23.00111698
56	87.5	274.8893572	2.748893572	20.25222341
57	88	276.4601535	2.764601535	17.48762187
58	88.5	278.0309498	2.780309498	14.70731237
59	89	279.6017462	2.796017462	11.91129491
60	89.5	281.1725425	2.811725425	9.099569486
61	90	282.7433388	2.827433388	6.272136098
62	90.5	284.3141351	2.843141351	3.428994747
63	91	285.8849315	2.858849315	0.570145432
64	91.5	287.4557278	2.874557278	-2.304411846

Oh that's a clever solution, I think that works and I teach Excel as well so great idea!

Mr. Nice!
Oct 13, 2005

bone shaking.
soul baking.

tuyop posted:

Oh that's a clever solution, I think that works and I teach Excel as well so great idea!

Awesome! Definitely double check my math and such. I'm glad I could help.

E: someone did double check me and I didn’t increase diameter enough! Thanks, goons.

Mr. Nice! fucked around with this message at 13:53 on Jan 3, 2023

Hyperlynx
Sep 13, 2015

Hyperlynx posted:

Any clever ideas for using great big garbage bags with tiny bins?

My bin is only around 15L, which is the right size for most plastic shopping bags, but I don't always have those, and they're easy to overfill. I can easily find drawstring ones, which is ideal, but they're always the great big 56L ones for the big wheely bins.

I'm pretty sure there's no way around this, and I've just got to waste bag capacity, really really search until I find little bags, or cram a big kitchen bin into my tiny kitchen, but just in case I'm missing something I thought I'd ask. Unless I can somehow seal the bag closed, cut the excess off, seal the bottom of that to form a new, smaller bag, like those heat-sealing knives on factory assembly lines that seal a continuous stream of products using a continuous tube of plastic...

E: maybe I'm asking the wrong question, and what I actually need is a way to have a big bin in a very small space. That would be even better!

Ha, the real real question was "where do I find smaller bin liners" and I just realised the answer is "an office supplies shop"!

https://www.officeworks.com.au/shop/officeworks/search?q=bin%20liners&view=grid&page=1&sortBy=bestmatch

Inceltown
Aug 6, 2019

Hyperlynx posted:

Ha, the real real question was "where do I find smaller bin liners" and I just realised the answer is "an office supplies shop"!

https://www.officeworks.com.au/shop/officeworks/search?q=bin%20liners&view=grid&page=1&sortBy=bestmatch

Get a bin that fits the colesworth fruit bags and use them as your bin liner. No point buying more plastic.

Mario
Oct 29, 2006
It's-a-me!

Mr. Nice! posted:

I don't know exactly how to make it into a succinct question, but I solved it in excel without too much trouble.

I converted everything to mm for simplicity. The roll is 60mm in diameter which means it has a circumference of 188.5mm. This will use 1.88 sheets (at 100mm per). After the first pass, there are 148.12 sheets left and a new diameter of 60.5mm. This has a circ. of 190mm which takes 1.9 sheets and leaves 146.2. I just filled down the cells until I got to negative sheets remaining. After 63 passes, the diameter is 91mm and there is 0.57 sheets of TP remaining that will overlap. So for 57mm it will be 91.5mm thick but the rest of the roll will be 91mm.

There is definitely a way to write this up as a simple function, but I don't know offhand the easiest way to do that. Solving in excel is trivial, though. Here's my numbers if anyone wants to check. I used =pi() for pi.
Diameter increases by 1.0mm per pass (paper is added to two "sides" of the circle).

As an algebraic approach, consider a top-down view of the filled roll. It is a circle with some portion in the center occupied by the empty roll, and the rest occupied by paper.

let f = radius of filled roll
let r = radius of empty roll = 30
let n = number of sheets = 180
let l = length of one sheet = 100
let w = width of one sheet = 0.5

Area of filled roll = Area of empty roll + Area of paper
pi(f^2) = pi(r^2) + nlw

Divide out pi
f^2 = r^2 + (nlw)/pi

Isolate f
f = (r^2 + nlw/pi)^0.5

f = (30^2 + (180*100*0.5/pi))^0.5

f = ~61.36mm radius (122.72mm diameter)

Hyperlynx
Sep 13, 2015

Inceltown posted:

Get a bin that fits the colesworth fruit bags and use them as your bin liner. No point buying more plastic.

I did also find https://biobagworld.com.au/ just now, when I thought "duh, why not actually just search for 'biodegradable kitchen garbage bags'"

RPATDO_LAMD
Mar 22, 2013

🐘🪠🍆

tuyop posted:

I'm trying to figure out if this can be turned into a math question:

You have a roll of toilet paper. Each sheet is 10cm long and .5mm thick. There are 180 sheets on the roll. The cardboard roll around which the paper is wound is 6cm in diameter.

First, is it possible to determine from that information the diameter of the roll?

Second, what expression communicates how many sheets there are on a given "layer" of the roll? The first outermost layer would, of course, have the most with the final innermost layer only being like two.

Rather than doing it iteratively, layer-by-layer, you can also just solve it numerically by looking at the volume. (or in this case the area of a 2d cross-section, since we don't know how wide the sheets are)

With w as the unknown width:
The volume of tp is 0.10m * 0.0005m * w * 180 = 0.009 m^2 * w
the volume of the cardboard tube is pi * (0.06m )^2 * w = pi*0.0036 m^2 * w

the resulting cylinder's volume has to be the sum of those two, so we set pi * r^2 * w = 0.009 m^2 * w + .0036 pi m^2 * w
simplify and cancel out the w and we get r = sqrt ( 0.009 m^2 + pi*0.0036m^2 ) / pi or approximately r ~= 0.0454 m
so the roll is 4.54 cm thick, or 9.08cm in diameter.

this is pretty close to Mr Nice's result!

(Of course, it is not actually accurate unless you make your toilet paper out of a magic material that can never stretch or compress.... but neither is any other solution. Such is life!)

RPATDO_LAMD fucked around with this message at 01:10 on Jan 3, 2023

Delphisage
Jul 31, 2022

by the sex ghost
What's the joke with :eng101:? I see it used in a disparaging sense a lot.

alnilam
Nov 10, 2009

Delphisage posted:

What's the joke with :eng101:? I see it used in a disparaging sense a lot.

To me it's just like, someone being educational about something, sometimes sarcastically though

PiratePrentice
Oct 29, 2022

by Hand Knit
It's used sarcastically and genuinely, it just depends on the context of the post. It usually comes up when someone has an interesting but not super relevant fact that they'd like to add, bit they don't expect it to change the course of the discussion.

kedo
Nov 27, 2007

If I put a frozen piece of meat in a ziplock bag into a bowl of hot water that isn’t deep enough to fully cover it, and then run hot water over it out of a kitchen faucet, will it heat up faster than if I ran the hot water into the bowl directly without hitting the meat?

I’m having an admittedly silly debate with someone about thawing meat. My theory is that running water over the meat will be faster because it increases the surface area that is being heated (while the water-into-bowl method only directly heats the submerged meat), and to a lesser extent creates some minor convection as the water runs off the meat and into the bowl. The other guy claims the water going into the bowl will be faster because the water isn’t cooled first by hitting the frozen meat (so the water in the bowl will be warmer), and it creates much stronger convective currents.

Will the same hold true for a material that conducts heat better than meat, say an ice cube (frozen cooking stock)?

Flash Gordon Ramsay
Sep 28, 2004

Grimey Drawer
Don’t thaw meat with warm water. Use cold water. It still works fast. Not sure which of your proposed methods is the fastest, but in a bowl with just a trickle of water to encourage convective currents and keep the water temp up works drat well. And it isn’t nearly as wasteful as just running a bunch of water on the meat endlessly.

CzarChasm
Mar 14, 2009

I don't like it when you're watching me eat.

Busy Bee posted:

Does anyone have experience sending a small high value package from EU to the US? I want to get a few of my trading cards graded and the estimated value can range from a few thousand dollars to over ten depending on the grade.

1) I will be sending it directly to the grading company - how can I make sure there will be no issues in customs and how do I figure out if there is a tax to be paid?
2) If, for some reason, the package gets lost, how would I prove to the shipping company the value of the cards?

Are there literally no grading companies in the EU? If these items can be said to have any true monetary value, certainly there's not only one country on this planet that determines that value.

In my experience, if you are truly worried about a package being lost, you insure it ahead of time. You say that you are shipping a package worth $10,000 (as an example). If you have to declare it as anything, you can call it memorabilia. Yes, it will cost you more to ship, but it's not going to cost you $10k to ship it. As a quick test, I just added $10K insurance to ship something from WI to TX. It bumped the total from about $15 to $150 on a 2lb package. That's a lot of money, but if I was shipping something irreplaceable, that I wanted to be sure was going to it's destination, I'd buy the insurance. Doubly so if it was something fragile.

To put it another way, your cards have no value, either before or after their valuation. The value in this transaction is the box itself. If you don't get the insurance UPS might throw you $100, and that's if you can prove that it was in their possession last, before it disappeared. But if you pay for insurance, it gives you a leg to stand on.

kedo
Nov 27, 2007

Flash Gordon Ramsay posted:

Don’t thaw meat with warm water. Use cold water. It still works fast. Not sure which of your proposed methods is the fastest, but in a bowl with just a trickle of water to encourage convective currents and keep the water temp up works drat well. And it isn’t nearly as wasteful as just running a bunch of water on the meat endlessly.

Oh I don’t. I put meat in a bowl of cool water a few hours before I need to use it and let it come up to room temp slowly.

This is only one of the reasons why this argument is entirely ridiculous. We’re arguing physics at this point, and neither of us know enough to prove the other wrong. Substitute cool water for hot water if it makes ya feel better ;)

smackfu
Jun 7, 2004

Best thing I’ve found for thawing is putting it in a cast iron pan. Lots of thermal mass and no chance of leaks.

RPATDO_LAMD
Mar 22, 2013

🐘🪠🍆

kedo posted:

If I put a frozen piece of meat in a ziplock bag into a bowl of hot water that isn’t deep enough to fully cover it, and then run hot water over it out of a kitchen faucet, will it heat up faster than if I ran the hot water into the bowl directly without hitting the meat?

I’m having an admittedly silly debate with someone about thawing meat. My theory is that running water over the meat will be faster because it increases the surface area that is being heated (while the water-into-bowl method only directly heats the submerged meat), and to a lesser extent creates some minor convection as the water runs off the meat and into the bowl. The other guy claims the water going into the bowl will be faster because the water isn’t cooled first by hitting the frozen meat (so the water in the bowl will be warmer), and it creates much stronger convective currents.

Will the same hold true for a material that conducts heat better than meat, say an ice cube (frozen cooking stock)?

Running water is better at transferring heat than still water.
The other guy's argument pretty much makes no sense. He says the water will be worse at warming up meat because touching the meat first means it's colder, but the only way it can be colder is if has already transferred more heat to the meat. Nothing just gets hotter or colder on its own, there's always heat transferring from and to something.

Trapick
Apr 17, 2006

kedo posted:

This is only one of the reasons why this argument is entirely ridiculous. We’re arguing physics at this point, and neither of us know enough to prove the other wrong. Substitute cool water for hot water if it makes ya feel better ;)
Do an experiment with a certain number of ice cubes in a ziplock, whichever method melts them first will be obvious. This is like the Mpemba effect, explanations may be hard to find but results speak for themselves.

Baron Porkface
Jan 22, 2007


https://www.youtube.com/watch?v=ZxJVaQTWQ24

How does signaling a hole card from a different table help the cheaters?

RPATDO_LAMD
Mar 22, 2013

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Baron Porkface posted:

https://www.youtube.com/watch?v=ZxJVaQTWQ24

How does signaling a hole card from a different table help the cheaters?

You need 2+ cheaters working together.

Alice reads the dealer's cards at Bob's table and signals that info to Bob somehow.
Bob reads the dealer's cards at Alice's table and signals that info to Alice somehow.

Xenoborg
Mar 10, 2007

Glasses guy is at the table with the bad dealer. Beard guy sees the dealers card from another table and sends a signal to glasses what his dealer has.

fartknocker
Oct 28, 2012


Damn it, this always happens. I think I'm gonna score, and then I never score. It's not fair.



Wedge Regret

Baron Porkface posted:

https://www.youtube.com/watch?v=ZxJVaQTWQ24

How does signaling a hole card from a different table help the cheaters?

They're playing blackjack. Guy with the beard at the other table sees the card the dealer hasn't flipped and signals the guy with the glasses, who can see the dealers first card. Guy with the glasses now knows the dealers exact hand so far. Guy with the glasses is himself playing multiple hands at one time, betting on all of them aggressively, and always winning, which is what draws attention to him and eventually gets them both caught.

Busy Bee
Jul 13, 2004

CzarChasm posted:

Are there literally no grading companies in the EU? If these items can be said to have any true monetary value, certainly there's not only one country on this planet that determines that value.

In my experience, if you are truly worried about a package being lost, you insure it ahead of time. You say that you are shipping a package worth $10,000 (as an example). If you have to declare it as anything, you can call it memorabilia. Yes, it will cost you more to ship, but it's not going to cost you $10k to ship it. As a quick test, I just added $10K insurance to ship something from WI to TX. It bumped the total from about $15 to $150 on a 2lb package. That's a lot of money, but if I was shipping something irreplaceable, that I wanted to be sure was going to it's destination, I'd buy the insurance. Doubly so if it was something fragile.

To put it another way, your cards have no value, either before or after their valuation. The value in this transaction is the box itself. If you don't get the insurance UPS might throw you $100, and that's if you can prove that it was in their possession last, before it disappeared. But if you pay for insurance, it gives you a leg to stand on.

The two grading companies that I would like to have my cards graded at are PSA and BGS - both do not have any services in Europe so the cards have to be sent to the US.

I guess my question was more around if I estimate the value to be $10,000 - wouldn't either UPS / FedEx ask me how I came to this value and if I can show any receipts? Obviously, I wouldn't have any receipts from cards that are over 20+ years old and since they were not graded, the value of the cards can vary by hundreds, if not thousands of dollars.

Busy Bee fucked around with this message at 11:07 on Jan 3, 2023

Platystemon
Feb 13, 2012

BREADS
Do the shipping companies actually care how accurate your estimate is?

You paid them one percent of ten thousand dollars for an event that has significantly less than one percent chance of occurring.

In general overstating an asset’s value to insurer’s is done so that someone can commit insurance fraud, by causing a “freak accident” that wasn’t an accident at all. In the case of shipping, though, the shipper has control the parcel the whole time. You could only defraud them if you somehow caused them to lose your package while it was in their care.

Baron Porkface
Jan 22, 2007


Since Protons are what gives matter mass, is a bar of sliver 67% as massive as a bar of gold?

Since Carbon and diamonds are the same element, why does coal have the property of being black and diamonds transparent?

Baron Porkface fucked around with this message at 10:12 on Jan 3, 2023

PiratePrentice
Oct 29, 2022

by Hand Knit
Protons and neutrons are what give matter mass, not protons alone. Silver is 54% the density of gold, not only because of the number of protons and neutrons in the atoms but also because of the arrangement of the atoms in the way they fit together, which is complicated based on the properties of the atom including the electron orbitals and stuff that I'm not 100% read up on.

Coal is an arrangement of carbon, oxygen, nitrogen, hydrogen, and sulfur, it is very complex and made up of ancient rear end plants that got compressed and heated up in a specific temperature-pressure regime for long enough to make it into coal. Diamonds are made of pure carbon, and have not only been subjected to a specific temperature-pressure regime for an adequate amount of time, but also have been rapidly brought closer to the earth's surface by kimberlite pipes from which they are mined.

Other forms of pure carbon like graphite have different crystal structures because they were subject to different temperature-pressure regimes in different parts of the Earth's crust. The crystal structure of diamonds is made of covalently bonded carbon, which makes it harder than any other naturally occuring mineral (as far as I'm aware). Graphite crystal structure is a bunch of flat little hexagons formed together into sheets stacked on top of each other, which is why they slide off so easily and make for good pencils.

PiratePrentice fucked around with this message at 10:38 on Jan 3, 2023

RPATDO_LAMD
Mar 22, 2013

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Baron Porkface posted:

Since Protons are what gives matter mass, is a bar of sliver 67% as massive as a bar of gold?

Since Carbon and diamonds are the same element, why does coal have the property of being black and diamonds transparent?

Atomic mass is based on both protons and neutrons, which have about equal mass. Silver has 47 protons and 61 neutrons, so it has an atomic mass of about 108 (actually 107.868 but let's ignore that). Gold meanwhile has 79 protons and 118 neutrons for an atomic mass of 197. So an atom of silver has only ~54% the mass of an atom of gold. In this case, their actual densities follow the same ratio.
But one atom is not always the same 'size' as one atom. For example, helium is a gas, and has its one-atom molecules packed much more loosely than the atoms in a bar of gold.

A "mole" is a unit for measuring a number of atoms.

A cubic meter of helium has a mass of 178.5 grams, or 0.1785 kg, meaning it contains about 45 moles or 2.7 * 10^25 atoms.
Meanwhile, a cubic meter of gold would have a mass of 19,300 kg, and would contain 98,000 moles, or 5.9 * 10^28 atoms, about 2000 times as many.

Even though helium has an atomic weight of 4 and gold has an atomic weight of 197, gold is way more than just 50x as dense as helium. It's actually 100,000x as dense. Because it turns out there's just thousands of times more atoms in the same space.

RPATDO_LAMD fucked around with this message at 10:43 on Jan 3, 2023

Platystemon
Feb 13, 2012

BREADS
Mass comes more or less* all from protons and neutrons.

For example, nickel has one more proton than cobalt

The only stable isotope of cobalt has thirty‐two neutrons. This, in combination with its twenty‐seven protons, gives cobalt approximately fifty‐nine atomic mass units.

Nickel has five stable isotopes, but more than two thirds of them have a mere thirty neutrons. On average, nickel atoms have less mass than cobalt atoms.

I am confused as to where you got the figure of 67% for the mass ratio between silver and gold. In proton numbers, they are 47/79, or 57.5%. In atomic mass, they are 107.87/196.97, or 54.8%.

To answer the question of how much more massive the bar of gold is, however, we must know how closely the atoms pack. Gold and silver happen to be quite similar, both face‐centred cubic with about the same distance between atoms. This is by the way something suggested in the periodic table, by silver and gold existing in the same column, or period. Copper is also there.



So the density of silver to gold is almost the same as the ratio of their atomic mass, 54.4% at room temperature.


*This “more or less”, the tiny discrepancy between what an atom’s mass “ought” to be if simple addition of protons and neutrons held, versus the actual mass of the atom, yields nuclear energy according to Einstein’s e = mc2.

Platystemon fucked around with this message at 10:55 on Jan 3, 2023

dokmo
Aug 27, 2006

:stat:man
Is there a formula to calculate the time of sunrise/sunset using only location coordinates and date/time, or is more information needed?

vv thanks

dokmo fucked around with this message at 14:20 on Jan 3, 2023

PiratePrentice
Oct 29, 2022

by Hand Knit

dokmo posted:

Is there a formula to calculate the time of sunrise/sunset using only location coordinates and date/time, or is more information needed?

Yep

https://gml.noaa.gov/grad/solcalc/solareqns.PDF

It's easier to just google it though, or keep an almanac on your phone if you got no internet

PiratePrentice fucked around with this message at 14:27 on Jan 3, 2023

Tiggum
Oct 24, 2007

Your life and your quest end here.


Is there a simple android app that will let me manually track how long I sleep for (ie. I enter in the times I went to bed and woke up)? I don't want it to try to figure out if I'm asleep or not or measure how much I snore or any of that stuff, just let me keep track of when I went to bed and roughly how much sleep I get each night. Preferably without any other features (like fitness tracking, etc.) but they're ok so long as I can just ignore them and I don't have to click through several screens to get to the one thing I want.

tuyop
Sep 15, 2006

Every second that we're not growing BASIL is a second wasted

Fun Shoe

Mario posted:

Diameter increases by 1.0mm per pass (paper is added to two "sides" of the circle).

As an algebraic approach, consider a top-down view of the filled roll. It is a circle with some portion in the center occupied by the empty roll, and the rest occupied by paper.

let f = radius of filled roll
let r = radius of empty roll = 30
let n = number of sheets = 180
let l = length of one sheet = 100
let w = width of one sheet = 0.5

Area of filled roll = Area of empty roll + Area of paper
pi(f^2) = pi(r^2) + nlw

Divide out pi
f^2 = r^2 + (nlw)/pi

Isolate f
f = (r^2 + nlw/pi)^0.5

f = (30^2 + (180*100*0.5/pi))^0.5

f = ~61.36mm radius (122.72mm diameter)

RPATDO_LAMD posted:

Rather than doing it iteratively, layer-by-layer, you can also just solve it numerically by looking at the volume. (or in this case the area of a 2d cross-section, since we don't know how wide the sheets are)

With w as the unknown width:
The volume of tp is 0.10m * 0.0005m * w * 180 = 0.009 m^2 * w
the volume of the cardboard tube is pi * (0.06m )^2 * w = pi*0.0036 m^2 * w

the resulting cylinder's volume has to be the sum of those two, so we set pi * r^2 * w = 0.009 m^2 * w + .0036 pi m^2 * w
simplify and cancel out the w and we get r = sqrt ( 0.009 m^2 + pi*0.0036m^2 ) / pi or approximately r ~= 0.0454 m
so the roll is 4.54 cm thick, or 9.08cm in diameter.

this is pretty close to Mr Nice's result!

(Of course, it is not actually accurate unless you make your toilet paper out of a magic material that can never stretch or compress.... but neither is any other solution. Such is life!)

Very cool and simple solutions, thank you! I thought I was probably just thinking about this wrong

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Inceltown
Aug 6, 2019

Tiggum posted:

Is there a simple android app that will let me manually track how long I sleep for (ie. I enter in the times I went to bed and woke up)? I don't want it to try to figure out if I'm asleep or not or measure how much I snore or any of that stuff, just let me keep track of when I went to bed and roughly how much sleep I get each night. Preferably without any other features (like fitness tracking, etc.) but they're ok so long as I can just ignore them and I don't have to click through several screens to get to the one thing I want.

Google Sheets?

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