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On September 04 2016 22:33 Artesimo wrote:Ok, this is going to be an odd one. After having spend way to much time on r/unexpectedjihad/ I really, really need the song that is used in this video: + Show Spoiler +I find it really relaxing but haven't managed to find it without "alahu akbar" shouts or in acceptable quality. If there is a clean high quality recording of this please send me in its direction. EDIT: Name of the song would also help but propably not that much...
Found it
+ Show Spoiler +
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On September 04 2016 21:13 SPcrusader wrote:
Something I have always wondered: " If you lay down straight on your bathroom floor, could you use the bathroom scale to weigh your own head?" I don`t have a scale at this time, but I have heard that most of the TL staff have bathrooms. Could anyone post the findings ?
It'd be hard, as you would struggle to completely relax your neck and shoulders. And even if you manage to really relax it'd still depend on how close to your neck you is to the pivot point of your head, as illustrated in figure 1, panel 1), where the red and the blue contact points would produce different read outs on the scale. To understand this, consider the thought experiment in panel 2), where a 1 kg weight is attached to a firm (but weightless) rod. The rod is attached to the floor, but the other end is free. If you support the rod at the point of the weight (1m away from the attachment), you will clearly need a force corresponding to 1kg. However, to support it only half a meter from the weight, you'd need twice the force, which can be most easily understood from conservation of energy, together with
E = F*d
where E is the energy, or work done, and F is the force and d the distance. The rod moves twice the angle if moved a given distance at half the distance from the attachment. Similarly, if the rod is supported 2m from the attachment, a scale would read out only 0.5 kg. This effect is the basis of many everyday tools, such as nutcrackers.
Figure 1:
![[image loading]](http://i.imgur.com/Ww7apPJ.png)
So to achieve reliable measurements of your heads mass, you would need to address those two issues:
a) the relaxation of neck and shoulders
b) the contact point (or rather, contact area) between the scale and the head.
This is not a trivial task, so instead, if you want to measure the mass of your head, I am happy give an advice that I have been wanting to give to someone for a long time:
Go and stick your head in a bucket of water.
Slowly and calmly though, don't splash around. You then measure how many liters you need to refill the bucket, and then you've measured the volume of your head. For this purpose, seeing that we cannot aim for an accuracy better than 10% or so, you can the approximate the density of your head with about 1, and then the number of liters needed is the weight of your head in kilos.
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On September 04 2016 23:20 Dan HH wrote:Found it + Show Spoiler +https://www.youtube.com/watch?v=ZQoJvI8XUa0
Thanks a lot, this song is just too mesmerizing 
Now I just have to make sure I never disconnect my headset while listening to this on the train lol.
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See, now I'm curious about the density of a human head. Like, I can imagine it being a fair bit more than 1. They feel dense when you, for instance, hold a baby's head.
But I think you get put on watchlists if you spend too much time googling things like "density of a human head."
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Zurich15365 Posts
You guys are stupid, it's much easier to just step on the scale like any normal person, write down the weight, then step again on the scale without your head on and calculate the difference = the weight of your head. duh.
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On September 05 2016 05:18 zatic wrote:
You guys are stupid, it's much easier to just step on the scale like any normal person, write down the weight, then step again on the scale without your head on and calculate the difference = the weight of your head. duh.
Or you can more easily just have your head severed, then weighed, then sawed back on
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The problem with any head removal is closing the hole quickly enough that no body fluids leak out from the head or collecting only the ones from the head and not the body. Perhaps a very high performance laser such as used for welding could cauterize as it cuts it off and if you take into account where you cut so the burnt away portion isn't part of the area you want weighted.
I also agree with the water theory for an actual easy way to do it with decent margin of error. We aren't that different from it.
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On September 05 2016 05:35 oGoZenob wrote:Show nested quote +On September 05 2016 05:18 zatic wrote:
You guys are stupid, it's much easier to just step on the scale like any normal person, write down the weight, then step again on the scale without your head on and calculate the difference = the weight of your head. duh. Or you can more easily just have your head severed, then weighed, then sawed back on
Isn't this like a cliche? I'm sure I've heard this exchange before somewhere...
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For best weighing results, the scale top should be slightly lower than the surface supporting your shoulders. If it's even or higher you'll end up with your neck supporting your head. If it's more than a little lower you'll end up weighing some neck or failing to make contact with the scale.
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My head seems to weigh about 10 pounds.
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On September 05 2016 07:29 Buckyman wrote:
For best weighing results, the scale top should be slightly lower than the surface supporting your shoulders. If it's even or higher you'll end up with your neck supporting your head. If it's more than a little lower you'll end up weighing some neck or failing to make contact with the scale.
The actual best way to weight your head is to make up a number, and just give a fucking huge margin of error.
"7 lbs, give or take 200-300 lbs"
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United States43991 Posts
Surely there is some way to do this by resting the head on a floating platform and looking at the weight of the water displaced. Fill a pool with water to the brim, have the person lie on the side of it with their head over the edge and support their head with a float.
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On September 05 2016 01:36 Yoav wrote:
See, now I'm curious about the density of a human head. Like, I can imagine it being a fair bit more than 1. They feel dense when you, for instance, hold a baby's head.
But I think you get put on watchlists if you spend too much time googling things like "density of a human head."
I'd guess that it's just below 1. The human body is close to 1 on average. Muscles a bit more, fat a bit less. I can float or sink depending on a few liters of air in my lungs, so my volume is same as my weight within 5% or so. The head also has a few cavities, so I'd guess the density is slightly below 1? Unless brain is super dense of course..
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On September 05 2016 08:56 KwarK wrote:
Surely there is some way to do this by resting the head on a floating platform and looking at the weight of the water displaced. Fill a pool with water to the brim, have the person lie on the side of it with their head over the edge and support their head with a float.
That's just an elaborate scale.
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United States43991 Posts
On September 05 2016 09:14 Cascade wrote:Show nested quote +On September 05 2016 08:56 KwarK wrote:
Surely there is some way to do this by resting the head on a floating platform and looking at the weight of the water displaced. Fill a pool with water to the brim, have the person lie on the side of it with their head over the edge and support their head with a float. That's just an elaborate scale.
Yeah well so is your face ya reptile fuck.
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I accidentally burned some skin off my middle finger via boiling water, whats the best way to treat/reduce pain
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On September 04 2016 16:36 Cascade wrote:Show nested quote +On September 04 2016 03:46 FiWiFaKi wrote:Important question I need help with because I'm being stupid. Alright, you have rod with a circular cross section made out of some material, hanging from a ceiling. We can take its tensile strength (Pa = kg*m*s^-2*m^-2) and divide it by density (kg*m^-3). We are left with specific strength, kind of like the weight to strength ratio in units of m^2/s^2. Now we can divide this by the acceleration due to gravity, to express the specific strength in meters. This value tells you how long of a single cross section beam you could hang from a ceiling before breaking. For example, steel is around 6km, aerospace aluminium is 20km, the strongest metal is titanium at 29km, so on and so forth. However, you could make the cross section thinner near the end where it doesn't carry much stress, and add it on to the bottom, to make the rod longer. My question is, what would be the function of the radius of the length from the top of the ceiling be (in the axial component of the rod), if given the radius at the bottom and top of the rod, in order to maximize the length of the structure (cross sectional area as a function of length is fine too of course). Please help, my curiousity is getting the best of me, and I'm struggling to obtain an answer  I haven't done this kind of mechanics, but seems like a pretty basic differential equation? I'm on my phone though, so bear with me... The strength scales with cross section. The mass to carry is the cross section integrated over the length underneath. So if you want to be on the carrying capacity at all points, let L be the length from the top, and a(L) the cross section at that point: a(L) proportional to integral{of x from L to infinity} a(x) dx = A(infinity) - A(L), where A(L) is the primitive function of a(L). Set the proportionality constant to 1 for easier notation (I'm on a phone) and take derivate wrt L: a'(L) = -a(L) Which solved by a(L) = exp(-L). The radius, being the square root of the area, is still an exponential. It kindof makes sense that it's an exponential, so that the thickness at any point of the"thread" does not depend on what is happening above it, only below. Doesn't know where the ceiling is. Edit: you can trace all the constants pretty easily if you want, but not doing it on the phone...
Thanks for the help, yeah for some reason I was having trouble starting it, although your answer is slightly away from what I was looking for, as you intergrated from L -> infinity, and you assumed the infinity term equals zero, which you can't do, as between x2 and infinity there is no mass.
As for the result with your help:
sigma = F/a, F = mg, rho = m/V
Combine: a(L) = (rho*g/sigma)*V, V = integral L->infinity a(x)*dx -> A(infinity) - A(L)
Since A(infinity) is a horizontal asymptote to zero, both its value and derivative is zero.
so a(L) = -(rho*g/sigma)*A(L), take derivative: da(L)/dL = (rho*g/sigma)*a(L)
Rearrange: da(L)/a(L) = -(rho*g/sigma)*dL
Integrate: ln(a(L)) = -(rho*g/sigma)*L -> a(L) = -exp((rho*g/sigma)*L)
Although I find L as a function of r more useful, so:
L= -(rho*g/sigma)*ln(pi*r^2), but I need two r's, and the difference between that, so set r1 > r2, and do r2 - r1.
L = -(rho*g/sigma)*ln(pi*r2^2) - -(rho*g/sigma)*ln(pi*r1^2), use log properties:
L = (rho*g/sigma)*ln([r1/r2]^2), lastly normalize it with standard definition of breaking length (divide by (rho*g/sigma))
And you're left with the extremely simple equation of Length coefficient = ln([r1/r2]^2)
Hence the result is that if for example the diameter of the rod of the ceiling is 100x as wide as at the hanging end, the cable can be created to be 9.21x longer than if it was made same diameter throughout. This of course assumes the stresses will remain purely axial and whatnot, but probably not an awful approximation.
But the result is against my intuition, and something just doesn't look right. In the step: V = integral L->infinity a(x)*dx -> A(infinity) - A(L)... It's not supposed to be A(infinity), but A(x2)... There's nothing in there that accounts for x2, and A(x2) - A(L), and you're only taking the derivative w.r.t. L and not x2... And I don't know, it's been a little while, so I'm just unsure how to treat this situation. To me it seems like after you add the x2 term, it's going to act like a bit of a constant there, making the differential equation non-linear... I know next to nothing about non-linear differential equations, since vibrations and controls is all linear, and my 2nd year differential class I don't remember much from, but I feel like we didn't have too many fancy things, and it was a lot of the y''''+p(q)*y'''+r(q)*y'' + ... + t*y = function here, and then doing the homogeneous and particular solution separately. Not only that, even if I define the end of the rod to be say 1cm in diameter, there's nothing constraining the rest of the beam to be r >= 1cm. Looks like it should be some piecewise function to me, where at the very bottom it will be a constant radius beam, and then eventually it'd become a sqrt(exponential) radius with some constant added to it, due to the the cutoff making it not act like a perfect exponential.
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On September 05 2016 10:23 Zambrah wrote:
I accidentally burned some skin off my middle finger via boiling water, whats the best way to treat/reduce pain
50mg of morphine every 2 hours until pain isn't felt.
Or just run it under cool to lukewarm water for 10 minutes, wrap a rubber band over your finger to lower blood circulation. Apply Aloe Vera.
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EPT
