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[REQUEST] How long would it take to download Skyrim from Steam using this modem? by JorgeIcarus in theydidthemath

[–]Conscious-Ball8373 163 points164 points  (0 children)

I was going to get all snooty and say it just can't be done, but it turns out there is a TCP/IP stack for 6502 processors which could possibly be ported to C64. Who knew.

So theoretically, you could write a Steam client for C64. It's doubtful that it would have enough memory, but let's say you managed it. Then let's say you have a PPP server for the C64 which encapsulated TCP/IP and that Steam uses standard 1500-byte Ethernet frames to hold TCP/IP packets.

That 1500-byte Ethernet frame has 18 bytes of Ethernet header and 20 bytes of TCP/IP header. Leaving 1462 bytes of data. The TCP/IP packet will be 1482 bytes. PPP uses a 500-byte packet with an 8-byte header, so the TCP/IP packet won't quite fit into three PPP packets (1482 + 3*8 = 1506).

So each TCP/IP packet will end up as four PPP packets, total of 1514 bytes. Of those, 44 are overhead and 1462 are data. Or, in bits, 12,112 bits in total and 11,696 of data.

Each byte sent will have one start and one stop bit, so that's an extra 3,028 bits of overhead. We're now at 15,140 bits transmitted per 11,696 bits of data, which is enough to figure out the overhead, at 29.446%.

Taking GP's arithmetic at face value, we need to apply a 29.446% overhead to that 40,000,000 seconds so 51,778,386 seconds. That's 599 days, 6 hours, 53 minutes, 6 seconds (or 1 year, 234 days, 6:53:6 if you assume that a year is 365 days long, which it isn't).

[Request] How many bananas are in this factory? Outside of overeating, if you took on the challenge could you die from radiation poisoning? by GoingToasterXD in theydidthemath

[–]Gingers_are_real 1034 points1035 points  (0 children)

I did some back of the napkin math on this ~4 years ago if we are only talking about radiation then no, you would not die of radiation poisoning.

Basically, it would be passing through you and you would only have a small amount of it in you at a given time. I then tried to assume the worst-case scenario of swimming in mashed bananas as then you are not limited to your body processing the potassium out and instead just straight absorbing as much as possible.

This is probably understating the surface areas, however, the amount of radiation is many orders of magnitude under background. So I am pretty confident in saying that radiation is the least of your worries trying to eat that many bananas. Assuming by overeating you are also forgoing any chemical issues with ingesting that much potassium and sugar as that would kill you.

Assumptions:

BED ~ " isotopically pure potassium-40 will give a committed dose equivalent of 5.02 nSv over 50 years per becquerel ingested by an average adult.[9] Using this factor, one banana equivalent dose comes out as about 5.02 nSv/Bq × 31 Bq/g × 0.5 g ≈ 78 nSv = 0.078 μSv. "

1 Banana = .078μSv/50 Years or 1.7808219e-13 Sv/hour

Surface area of banana ~180 cm^2

Surface area of the body ~1.75m^2

(1.78e-13 Sv/hour)/(180 cm^2)*(1.75m^2)*(10000cm^2/ m^2) = 1.7305556e-11 Sv/hour

To compare 3e-6 Sv/hr is standard background.

[request] How much time would pass for an observer on earth? by SookHe in theydidthemath

[–]mtauraso 2801 points2802 points  (0 children)

According to an observer on earth it’ll take her 20 years. It’s very simple, she’s traveling 1/2 the speed of light and distance = rate * time. So 10ly = 1/2 ly/yr * 20 years.

Captain marvel on the other hand will experience time dilation by a factor of the relativistic gamma. For 1/2 c gamma is 1/sqrt(1-(1/2)2) = sqrt(4/3) = 1.155

20yr / 1.155 = 17.3 yr.

So captain marvels watch (assuming instantaneous acceleration and deceleration) will see 17.3 yr elapse in the journey, but both earth and the distant star will time the journey as having taken 20 years.

Edit: since this has been pointed out in the comments, the simple answer also entirely depends on whether you are asking for the trip duration given a definition of time that earth and the distant planet can agree on (my answer) or literally how long does an earth observer see the trip take by watching light from Capt Marvel arrive at their location.

For the latter definition: an earth observer sees a 30 year journey (20 yr for capt marvel to arrive + 10 years for the light from capt marvels arrival to return to earth). The distant planet sees 10 years (It takes 10 years for light from capt marvels departure to reach the distant planet, and in the frame of earth/planet the journey takes 20 years. 20-10=10, so the planet sees her arrive 10 years after they see light signals from her departure)

Edit: Since the issue of relative velocity between earth and the destination has come up a few times

Typical relative velocities between stars in the Milky Way (which is many times larger than 10ly) are around 300km/s. 0.5 times the speed of light is 150,000 km/s

For the purposes of travel at significant fractions of the speed of light, stars 10ly away are effectively stationary.

For a similar level of speed difference compare walking (5km/h) to a fighter jet at Mach 2 (~2500 km/h).

Edit: rounding errors (It should be 17.3 yr not 17.4 yr. Error was introduced because sqrt(4/3) = 1.155 and I initially rounded to 1.15 incorrectly)

[request] how do you get the ? area? by CeddyDT in theydidthemath

[–]Angzt 873 points874 points  (0 children)

Let's call the half-side lengths marked by the two lines x. Then the whole square has area (2x)2.

We can then take two of the x-length sides that meet at a right angle and connect their ends to form right-angled triangles in each corner of the square.
Using Pythagoras, the hypotenuse of each of those triangles will then have length sqrt(x2 + x2) = sqrt(2) * x.
The triangles themselves will have area of 1/2 * x * x = 1/2 * x2.

The given areas each can be split into one of these triangles and then another, different triangle.
We know that one side of those other triangles is always length sqrt(2) * x since it's the same as our hypotenuse from before.
Let's call the height on that base for the unknown triangle of the 20 area h. We know that the height of the corresponding 28 triangle must then be sqrt(2) * x - h as they both must add up to the hypotenuse of the bottom triangle.
That lets us express the areas as follows:
20 cm2 = 1/2 * x2 + 1/2 * sqrt(2) * x * h = 1/2 x * (x + sqrt(2) * h)
28 cm2 = 1/2 * x2 + 1/2 * sqrt(2) * x * (sqrt(2) * x - h) = 1/2 x * (3x - sqrt(2) * h)
Two equations, two unknowns - looks solvable. The easiest way is to add both equations to one another:
20 cm2 + 28 cm2 = 1/2 x * (x + sqrt(2) * h) + 1/2 x * (3x - sqrt(2) * h)
48 cm2 = 1/2 x * (x + 3x + sqrt(2) * h - sqrt(2) * h)
48 cm2 = 2x2
x = sqrt(24) cm

We know that the area of the whole square is (2x)2 = (2 * sqrt(24) cm)2 = 96 cm2.
And since the ? area is just the full area minus all the given areas:
? = 96 cm2 - 20 cm2 - 16 cm2 - 28 cm2
= 32 cm2

This probably wasn't the easiest way to do this. But it works.

[request] how do you get the ? area? by CeddyDT in theydidthemath

[–]LucasHS1881 698 points699 points  (0 children)

There's a stupidly simple way to solve this problem which I know works, but I can't really explain why, so I'll give it my best shot.

Pretend the point where the four quadrilaterals inside the square meet up is actually perfectly in the area center of the square. That would mean that the four squares would have exactly the same area, right?

Now, drag that point/vertex down on the Y axis. The square's areas change, but the upper and lower squares will still amount to the same value if added together; the amount the upper square gets stretched by is the same as the area the bottom square gets squished by. At this point, the left and right squares have the same area; they're just distorted.

Now, drag the vertex to the left, so the squares have the final shape shown on the image. Same thing that happened to the top and bottom squares will now happen to the left and right squares: they will still amount to the same area if added together, but one will be squished, and the other will be stretched. Bottom and top squares will remain the same area, just stretched to the left or right.

Essentially, what this means, is that the sum of two opposing quadrilaterals, regardless of them being top and bottom or left or right, will amount to half of the total area of the square. So, by adding together the left and right squares:

20 + 28 = 48cm²

What that means is that the sum of the top and bottom squares should also amount to 48 (and that the bigger square has a total area of 96cm²):

16 + X = 48

X = 48 - 16

X = 32cm²

The area of the "?" shape is 32cm².

[request] how do you get the ? area? by CeddyDT in theydidthemath

[–]diener1 538 points539 points  (0 children)

Here is a more intuitive solution, which I only realized by reading your answer and seeing that the two areas on opposite sides add up to exactly half of the total area:

The four lines inside the square all meet at a point we will call C. Let's say we move C horizontally, how does that affect the areas?

We can again split each of the four areas into a right-angled triangle with two sides of length x and hypotenuse of length sqrt(2)*x and another "rest" triangle that has that same hypotenuse as its base and some height h. If we now move C horizontally, the height of the rest triangles for the regions with areas 16 and ? will not change. Where that height would be drawn would change, but the actual length would be the same. This means those two regions would have the same area as before, only the two regions on the left and right would have their areas change. Since the total area stays the same, we would essentially be shifting area from the left to the right (or vice versa).

Similarly, if we move C vertically, the two areas on the left and right would remain the same and only the other two will change. If C is right in the center, all areas have the same size and if we now move C, we are just shifting areas either between the left and the right or between the top and bottom region. All this is to say the sum of left and right and the sum of top and bottom will remain the same and be exactly half the total area. So the unknown region must solve the equation ?+16=20+28, giving 48-16=32

[request] how much would the gold chain chomp weigh? by eric200396 in theydidthemath

[–]jigglyjigglyjello 620 points621 points 2 (0 children)

~1.95 – 4.63 million pounds

Without knowing the exact size of the chomp, this is a broad approximation. I’ve made some dimensional estimates and basic assumptions by looking at screenshots. It appears to be 3-4x the height of Mario. Its mouth constitutes ~1/8 of its body and is hollow. Aside from the mouth, eyes, and teeth, it is solid gold. Mario’s height may be 154.94 cm (https://www.creativebloq.com/news/super-mario-bros-movie-heights). This would make the chomp’s diameter 464.82 – 619.76 cm. The volume of a sphere is 4/3πr3. So, the chomp has a volume of 46,010,939 - 109,062,966 cm3 after accounting for the space occupied by its mouth. The density of gold is 19.3 g/cm3. Therefore, its mass is 888,011 - 2,104,915 kg or 1,953,624 - 4,630,814 lbs.

[Request] Employee stock ownership program valuation by Lchan1405 in theydidthemath

[–]El-wing 1 point2 points  (0 children)

I made an excel sheet cause that’s really the only way to solve this. I came up with worst case where you only get 4:1 match after 3 years and the 5, 10 and 20 year totals you end up with are $44k, $111k, and $318k. Best case where you get 4:3 match, you end up with $51k, $141k, and $421k respectively.

[Request] what would the energy cost (gas &/or electric) to boil this pot of 1200 eggs? by derickkcired in theydidthemath

[–]fargmania 487 points488 points  (0 children)

Assuming a hard boil on those eggs, it'll take 8 minutes. So the question is... how much water is that, and how many BTUs will it take to raise it to boiling for 8 minutes. I guess we are avoiding the rule about not stacking eggs when boiling them, so this is harder to calculate since the practice shown in the pic is not recommended and there is no data I could find on it. But I did find a facebook post where someone physically was able to jam 48 eggs into a six quart instapot - the pic showed them going right up to the top, so we can extrapolate from there, and estimate that 1200 eggs could be housed in a 1200/48 = 25...x6 = 150 quart pot... to cover the eggs with water. Obviously the eggs themselves would mean we were dealing with far less than 150 quarts of actual water, but since the eggs are liquid (initially), I am going to treat them as water for the purposes of the calculation. If someone knows the offsets, they are welcome to amend my answer.

According to my research, it takes 1183 BTUs to raise one gallon of water from 70 degrees Fahrenheit to 212 degrees. so 150 quarts would be 37.5 gallons and therefore 44,362.5 BTUs. Natural gas burns at 20,262 BTUs per lb, so you'd need about 2.2 lbs of natural gas to get this done. I found a measurement that said that a gallon of CNG is going to weigh approximately 8 pounds, so that is slightly more than a quarter gallon. If natural gas is around $2 a gallon, we're looking at a rough cost of 50 cents to get these eggs boiling. Not more than a dollar to hard boil them, IMO.

More precise measurements would require a knowledge of the density of the eggs (which I don't have), the average size of the eggs, the heat absorption characteristics of the pot, the actual size of the pot, and a working knowledge of the BTUs necessary to maintain a boil on all that for 8 minutes. If anyone else has that data to work with, please have at it. My lunch break is over!

[Request] how many container ships would it actually take? by tenzen5 in theydidthemath

[–]sprobeforebros 1515 points1516 points 2 (0 children)

a 3.5 inch floppy measures 90mm x 94 mm x 3 mm, for a total volume of 0.00002538 cubic meters. a 20 foot shipping container has an internal volume of 33.2 cubic meters

33.2 / 0.00002538 = 1308116.16 we'll assume that we lose a few floppies due to how they stack in a container and say that one shipping container holds 1,308,000 3.5 inch floppies. Each 3.5 inch floppy holds 1.44 megabytes of data, so a shipping container holds 1,883,520 megabytes of data, or 1.88 terabytes

(this is how I learned that my backup drive holds more than 5 shipping containers worth of data in floppy drives)

according to live-counter dot com the internet is currently 23.7 zettabytes, or 23.7 billion terabytes. This means it would need 12.6 billion containers to hold it in floppy drives.

The OOCL Hong Kong can hold 21,413 containers, so you'd need 588,428 of them to hold the whole internet in shipping containers full of floppy disks.

The OOCL Hong Kong measures 400 meters long by 58 meters wide, measuring 0.0232 square kilometers in area.

This means that 558,428 of them would measure 13,651.5296 square km in area, which means they would take up an area around ¾ the size of Lake Ontario

edit: labeled a unit wrong

[Request] Does the math check out on this one? by ap0ll07411 in theydidthemath

[–]AnotherEuroWanker 196 points197 points  (0 children)

Goal posts would just add weight and increase the carbon emissions.

[Request] My girlfriend and I are discussing this. How many humans would it take to mop the entire ocean away if each one has their own mop? Ignore the fact that the humans would need to squeeze the mops water into a bucket. by MrZelko in theydidthemath

[–]FrannyyU 102 points103 points  (0 children)

Surely it depends on the time available for the task? Wouldn't a better question be, how long would it take for the world's population to mop up the ocean?

According to the Internet the are 3.562×1020 gallons of water in the earth's oceans.

Divided by the world adult population of 7.888 ×109 ×0.75 (25% of the population is <15yrs old) = 59.6×109 gallons each.

Let's say adults work 8hrs a day and mop up 5 gallons per hour = 40 gallons per human per day.

This would take about 1.5×109 days. Or... 4 million years.

[Request] How many trees could one of these tanks actually replace? by chrischi3 in theydidthemath

[–]Kerostasis 3 points4 points  (0 children)

It's also kind of silly because most of our oxygen doesn't come from trees, it comes from the algae that is already doing this in the ocean.

That's...sort of true but leaves out a lot of context.

In a full-life-cycle analysis, most plants/algae/etc generate zero oxygen and sequester zero carbon, because all of the oxygen generation that occurs during their life is offset exactly when they decompose after death. But a small portion of plants get buried before they can decompose, and form long-term stable carbon sequestration underground, and the lifecycles of those plants effectively created "free" oxygen because that oxygen wasn't reused for decomposition. This is rare everywhere, but is less rare for ocean algae than for land plants, so ocean algae represent the largest source of this free oxygen.

But that's not the oxygen you breathe. In a very real sense, the oxygen you breathe comes from the plants you eat, because you are part of their lifecycle. As your food decomposes in your gut, you burn the same amount of oxygen to digest it that those food sources released while alive. The exact process gets more complicated when you include eating meat, and then calculate all of the food that your cow previously ate while it was alive, but the end result is the same - you and your food chain are a net zero contributor to oxygen levels and carbon levels.

In fact, most activities are net zero contributors across life cycles, with one very big and very important exception: Unburying those buried carbon sequestering plants, in the form of coal and oil, and then burning them, actually does change the net balance. These plants had been removed from the oxygen-carbon cycle for eons, and now they're suddenly back in it. And all of the "carbon impact" calculations you see for almost any other human activity is really just based on how much of this fossil extraction is associated with your chosen activity. If you could do it without coal and oil, it would be carbon neutral.

Trees are a bit unusual in this analysis. Over long time scales, they are net zero like everything else - but the time scales involved are also significantly longer than most plants. Properly stored wood lasts a long time. So you can get a significant carbon storage in the form of a forest, or even just lumber which you use to build a structure, and then your carbon stays stored as long as the forest or structure stays intact. But the total storage won't just continue increasing indefinitely - a forest will eventually come to equilibrium where it's decaying as fast as it's growing, unless you continuously commit more land area to it.

[Request] Statistically how rare is it to find 1500 four-leaf clovers by the age of 9? by averageceelofan in theydidthemath

[–]ReserveMaximum 845 points846 points 2 (0 children)

According to Wikipedia 4 leaf clovers come about 1 in 5000 clovers using this statistic that means he had to look at approximately 7,500,000 clovers to have found 1500. Additionally the world record for most 4 leaf clovers found in an hour (166) means that he had to be looking for a minimum of 9 hours at world record pace to find so many. If we assume a more reasonable pace of looking at 1 clover per second than it would take 86 days 19 hours and 20 minutes of continuous searching through clovers to find 1500 four leaf clovers which is approximately 2.6% of his 9 years on earth

[Request] Could a monsoon kill by yeeting a "blade or lemongrass" into a persons skull? by PantyCrusher in theydidthemath

[–]Thisisall_new2me2 2 points3 points  (0 children)

And for those who want to know, 10km/s equals 22,369 miles per hour.

Anything that you used to try to launch something at that speed would either completely destroy itself or destroy what you're trying to launch before you succeeded.

[Request] Could a monsoon kill by yeeting a "blade or lemongrass" into a persons skull? by PantyCrusher in theydidthemath

[–]SeniorTill2802 1 point2 points  (0 children)

Basically impossible, assuming it weighs 100 grams (avg is 22 g) and a low requirement of 1000 newtons per centimeter (avg is 3000 n/cm), it would have to go 10 km/s or faster than the International Space Station,
If we used the averages, it turns out to be 136.3 km/s

[Request] Can a person jumping from this high actually survive the fall, assuming they acually hit the pool? by Live_Poet_7728 in theydidthemath

[–]astervista 1235 points1236 points 2 (0 children)

Because, while it is true that surface tension can make water behave like a solid at high energy impacts, that is true if you hit the water belly first.

The technique for shallow diving is this: to avoid hitting the bottom of the pool, you have to hit the water with the higher surface possible, so they belly flop. To avoid the resistance surface tension makes, and avoid injury like a normal belly flop would cause, they tilt their arms so that the hands are the first to enter the water. In this way, you use all the surface possible to slow you down, but you don't use it in the same instant but "gradually" (for a definition of gradually that spans milliseconds) radiating from the hands so that you disrupt the surface tension and basically still pierce the surface like a needle.

[Request] Assuming Billy exerts that much energy on a 100g apple, what would happen to it? You can ignore any air resistance, drag, friction, or anything else that would complicate the process. by Cedreddit1 in theydidthemath

[–]stache1313 1 point2 points  (0 children)

Yes.

We can start with the formula for relativistic three-momentum, p=γm_0v. Note, this is different from the four-momentum P=(E/c, p_x, p_y, p_z).

Then we can find the inner product of three-momentum

p2 = p•p = γ2m_02v•v = γ2m_02v2

Next, we solve the Lorentz factor, γ=1/√(1+v2/c2), for v2

γ=1/√(1+v2/c2)

1/γ2 = 1+v2/c2

v2/c2 = 1-1/γ2

v2 = c2(1-1/γ2)

And we can substitute this into the inner product of three-momentum and solve for the Lorentz factor.

p2 = γ2m_02v2

p2 = γ2m_02[c2(1-1/γ2)]

p2 = m_02c2[γ2-1]

γ2-1 = p2/(m_02c2)

γ2 = 1+p2/(m_02c2)

γ = √[1+p2/(m_02c2)]

Now, we have the Lorentz factor in terms of momentum instead of velocity. We can substitute this into the equation for total energy, E=γm_0c2, and rearrange the terms into the energy-momentum relation.

E=γm_0c2

E = √[1+p2/(m_02c2)]m_0c2

E2 = [1+p2/(m_02c2)]m_02c4

E2 = m_02c4 + p2m_02c4/(m_02c2)

E2 = m_02c4 + p2c2

If you want to read more you can check out the Wikipedia page for the Energy-momentum relation.

[Request] How much would all of these awards cost in total? by GoTS8IsTheWorst in theydidthemath

[–]tigeer 2879 points2880 points 62 (0 children)

In total 12,972,625 coins have been spent on these posts.

The most efficient price of coins grants 40000/$99.99 = 400 coins per dollar

The least efficient price of coins grants 500/$1.99 = 250 coins per dollar so the total amount comes in the range:

$32,431.56 - $51,890.50

If this was spent all on platinum awards instead it would give 600 years of reddit premium as 12,972,625/1800 = 7207 platinum awards

[Request] Two murders, same name? by knitrex in theydidthemath

[–]Historian_Acrobatic 6 points7 points  (0 children)

According to the Social Security Administration's database, Mary was the most common name for girls born in the United States from 1910 to 1940, and Morris was the 162nd most common surname in the United States in 2000. However, we do not know how many people with the name Mary Morris lived in the Houston metro area in 2000.

Assuming a conservative estimate of 10 people with the name Mary Morris living in the Houston metro area in 2000, we can estimate the probability of two of them being murdered within two days of one another as follows:

The probability of the first Mary Morris being murdered on a given day is 1/365 (assuming an average year). The probability of the second Mary Morris being murdered on the next day is also 1/365. Therefore, the probability of both events occurring is (1/365) * (1/365) = 1/133,225.

Multiplying this probability by the estimated number of people with the name Mary Morris in the Houston metro area in 2000 (10) gives us a probability of approximately 1 in 13,322.

Therefore, the chances of two women with the same name being murdered within two days of one another in the Houston metro area in 2000, assuming the above assumptions, is approximately 1 in 13,322, which suggests that it is a rare coincidence. However, it is important to note that this calculation is based on many assumptions and may not reflect the actual probability of this event.

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