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On September 05 2016 10:30 FiWiFaKi wrote:Show nested quote +On September 04 2016 16:36 Cascade wrote: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.
I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0.
The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width.
So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons.
If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point.
I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/
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On September 05 2016 11:06 Cascade wrote:Show nested quote +On September 05 2016 10:30 FiWiFaKi wrote:On September 04 2016 16:36 Cascade wrote: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. I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0. The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width. So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons. If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point. I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/
Your equation assumes the length is infinite. I'm not trying to look at this from a math perspective, but rather a practical one. I want to have a certain thickness at the bottom, Which means I will have a certain radius at the top, and a certain radius at the bottom. So yes, yours would be the case for a never ending string, but it's not what I'm looking for, which like you said, your equation wasn't designed to solve S:... Like you mentioned, it will be on the safe side.
My objective was to calculate a maximum length based on some starting dimensions. Without any treatment, your equation just says, yep, the length of the string will be infinite. With your equation the radius will start at infinity, and approach zero. I want to have the thickest part be say 1m, and thinnest part be 1cm, and calculate how long of a rod I could make. That was the point of obtaining a length vs radius graph. I made a super quick awful looking sketch lol.
![[image loading]](http://i.imgur.com/abqnQch.png)
But yeah, I wanted both, the curve the radius vs length makes, as well as calculating the maximum length of the rod based on some given radii at both ends (and ideally as a function of some weight hanging off of it as well). Also I fucked up labelling the graph, obv e^-x = 1 at 0, opps ^__^
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On September 05 2016 11:55 FiWiFaKi wrote:Show nested quote +On September 05 2016 11:06 Cascade wrote:On September 05 2016 10:30 FiWiFaKi wrote:On September 04 2016 16:36 Cascade wrote: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. I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0. The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width. So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons. If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point. I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/ Your equation assumes the length is infinite. I'm not trying to look at this from a math perspective, but rather a practical one. I want to have a certain thickness at the bottom, Which means I will have a certain radius at the top, and a certain radius at the bottom. So yes, yours would be the case for a never ending string, but it's not what I'm looking for, which like you said, your equation wasn't designed to solve S:... Like you mentioned, it will be on the safe side. My objective was to calculate a maximum length based on some starting dimensions. Without any treatment, your equation just says, yep, the length of the string will be infinite. With your equation the radius will start at infinity, and approach zero. I want to have the thickest part be say 1m, and thinnest part be 1cm, and calculate how long of a rod I could make. That was the point of obtaining a length vs radius graph. I made a super quick awful looking sketch lol. But yeah, I wanted both, the curve the radius vs length makes, as well as calculating the maximum length of the rod based on some given radii at both ends (and ideally as a function of some weight hanging off of it as well). Also I fucked up labelling the graph, obv e^-x = 1 at 0, opps ^__^
It doesn't assume infinite length. It does accommodate it, but it also accommodates finite length solutions where a(L) is 0 after a certain L1. Not what comes out of the equations as I set them up though. It also doesn't assume infinite width at the ceiling. Indeed, the ceiling is at L=0, where the width is 1, or whatever constant you chose to set in front of the exponential when you solve the differential equation.
To find the solution to the new problem you pose, with the constraints of 1m at ceiling and at least 1cm thickness, it is just a small modification to my first solution. You first solve it as normal, setting the constant in front of the exponential to get the right thickness at the ceiling.
so the solution is
a(L) = C*exp(-k*L)
where C is the free parameter from the differential equation. k is a constant depending on the material and probably on g. As a(L) is the cross section, C would have to be like 2*pi*r_0 or something for the desired starting radius r_0. Let's make it easy and set C to 1 for now. That solves the ceiling boundary condition.
To let it not go below a certain minimum radius (let's say a(L) has to be at least a_m), we cut the exponential at the length L_m where it reaches that width:
a(L_m) = a_m
C*exp(-k*L_m) = a_m
L_m = -ln(a_m/C)/k = ln(C/a_m)/k
We can now add extra volume V_e underneath that with a cross section of a_m. The amount it can support is the amount that we just cut off, which is
V_e = integral{x from L_m to infinity} a(L)*dx = A(infinity) - A(L_m) = -C*(exp(-k*infinity) - exp(-k*L_m))/k
= C/k*exp(-k*L_m)
where the primitive function is
A(L) = -C*exp(-k*L)/k.
This volume corresponds to an extra length L_e of constant cross section a_m where
L_e*a_m = V_e = C/k*exp(-k*L_m)
L_e = C/(k*a_m)*exp(-k*L_m)
replace L_m from above:
L_e = C/(k*a_m)*exp(-k*(-ln(a_m/C)/k)) = C/(k*a_m)*exp(ln(a_m/C)) = C/(k*a_m)*a_m/C = 1/k
which should reproduce your formula for flat wires.
So the longest shape you can have with those constraints is starting from the ceiling as a negative exponential, reaching the minimum allowed cross section a_m at L_m = ln(C/a_m)/k, then followed by a flat wire of minimum cross section of length L_e = 1/k, for a total length L_t of
L_t = ln(C/a_m)/k + 1/k
Replacing C with a_c for the cross section at the ceiling, we see that the length is the same as for a flat wire (1/k), but allowing it to grow near the ceiling will add an extra ln(a_c/a_m)/k. So the extra length you can add is proportional to the logarithm of the ratio of the maximum and minimum allowed cross section. In the case of allowing e times as large cross section in the ceiling, you will get a twice as long shape. With your example of 1m vs 1cm, making the ratio of the cross section go like 100^2 = 10^4, the wire can be 9.21+1 = 10.21 times as long, which is essentially your same result.
Makes more sense?
(edited an uncanceled C at the end.)
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Why foreigners think that every Korean believes city legends like Fan death? Since no one believes that since 1998.
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On September 05 2016 13:15 Thouhastmail wrote:
Why foreigners think that every Korean believes city legends like Fan death? Since no one believes that since 1998.
For the same reason they think that all Australians wrestle crocodiles in the desert on a daily basis: in media you only get the click-baity news. You rarely hear about normal people. There are no articles like "EXTRA EXTRA Completely Normal Korean Doesn't Believe In Fan Death EXTRA EXTRA"
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On September 05 2016 13:20 Cascade wrote:Show nested quote +On September 05 2016 13:15 Thouhastmail wrote:
Why foreigners think that every Korean believes city legends like Fan death? Since no one believes that since 1998. For the same reason they think that all Australians wrestle crocodiles in the desert on a daily basis: in media you only get the click-baity news. You rarely hear about normal people. There are no articles like "EXTRA EXTRA Completely Normal Korean Doesn't Believe In Fan Death EXTRA EXTRA"
Don't all Australians hunt crocodile by riding on kangaroos? 
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On September 05 2016 14:00 Thieving Magpie wrote:Show nested quote +On September 05 2016 13:20 Cascade wrote:On September 05 2016 13:15 Thouhastmail wrote:
Why foreigners think that every Korean believes city legends like Fan death? Since no one believes that since 1998. For the same reason they think that all Australians wrestle crocodiles in the desert on a daily basis: in media you only get the click-baity news. You rarely hear about normal people. There are no articles like "EXTRA EXTRA Completely Normal Korean Doesn't Believe In Fan Death EXTRA EXTRA" Don't all Australians hunt crocodile by riding on kangaroos?
They rather eat kangaroos.
Believe me, I studied in Brisbane
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On September 05 2016 13:15 Thouhastmail wrote:
Why foreigners think that every Korean believes city legends like Fan death? Since no one believes that since 1998.
My impression was that while younger people don't believe in fan death, belief in it is still somewhat common among the older generations.
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On September 05 2016 13:05 Cascade wrote:Show nested quote +On September 05 2016 11:55 FiWiFaKi wrote:On September 05 2016 11:06 Cascade wrote:On September 05 2016 10:30 FiWiFaKi wrote:On September 04 2016 16:36 Cascade wrote: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. I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0. The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width. So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons. If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point. I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/ Your equation assumes the length is infinite. I'm not trying to look at this from a math perspective, but rather a practical one. I want to have a certain thickness at the bottom, Which means I will have a certain radius at the top, and a certain radius at the bottom. So yes, yours would be the case for a never ending string, but it's not what I'm looking for, which like you said, your equation wasn't designed to solve S:... Like you mentioned, it will be on the safe side. My objective was to calculate a maximum length based on some starting dimensions. Without any treatment, your equation just says, yep, the length of the string will be infinite. With your equation the radius will start at infinity, and approach zero. I want to have the thickest part be say 1m, and thinnest part be 1cm, and calculate how long of a rod I could make. That was the point of obtaining a length vs radius graph. I made a super quick awful looking sketch lol. But yeah, I wanted both, the curve the radius vs length makes, as well as calculating the maximum length of the rod based on some given radii at both ends (and ideally as a function of some weight hanging off of it as well). Also I fucked up labelling the graph, obv e^-x = 1 at 0, opps ^__^ It doesn't assume infinite length. It does accommodate it, but it also accommodates finite length solutions where a(L) is 0 after a certain L1. Not what comes out of the equations as I set them up though. It also doesn't assume infinite width at the ceiling. Indeed, the ceiling is at L=0, where the width is 1, or whatever constant you chose to set in front of the exponential when you solve the differential equation. To find the solution to the new problem you pose, with the constraints of 1m at ceiling and at least 1cm thickness, it is just a small modification to my first solution. You first solve it as normal, setting the constant in front of the exponential to get the right thickness at the ceiling. so the solution is a(L) = C*exp(-k*L) where C is the free parameter from the differential equation. k is a constant depending on the material and probably on g. As a(L) is the cross section, C would have to be like 2*pi*r_0 or something for the desired starting radius r_0. Let's make it easy and set C to 1 for now. That solves the ceiling boundary condition. To let it not go below a certain minimum radius (let's say a(L) has to be at least a_m), we cut the exponential at the length L_m where it reaches that width: a(L_m) = a_m C*exp(-k*L_m) = a_m L_m = -ln(a_m/C)/k = ln(C/a_m)/k We can now add extra volume V_e underneath that with a cross section of a_m. The amount it can support is the amount that we just cut off, which is V_e = integral{x from L_m to infinity} a(L)*dx = A(infinity) - A(L_m) = -C*(exp(-k*infinity) - exp(-k*L_m))/k = C/k*exp(-k*L_m) where the primitive function is A(L) = -C*exp(-k*L)/k. This volume corresponds to an extra length L_e of constant cross section a_m where L_e*a_m = V_e = C/k*exp(-k*L_m) L_e = C/(k*a_m)*exp(-k*L_m) replace L_m from above: L_e = C/(k*a_m)*exp(-k*(-ln(a_m/C)/k)) = C/(k*a_m)*exp(ln(a_m/C)) = C/(k*a_m)*a_m/C = 1/k which should reproduce your formula for flat wires. So the longest shape you can have with those constraints is starting from the ceiling as a negative exponential, reaching the minimum allowed cross section a_m at L_m = ln(C/a_m)/k, then followed by a flat wire of minimum cross section of length L_e = 1/k, for a total length L_t of L_t = ln(C/a_m)/k + 1/k Replacing C with a_c for the cross section at the ceiling, we see that the length is the same as for a flat wire (1/k), but allowing it to grow near the ceiling will add an extra ln(a_c/a_m)/k. So the extra length you can add is proportional to the logarithm of the ratio of the maximum and minimum allowed cross section. In the case of allowing e times as large cross section in the ceiling, you will get a twice as long shape. With your example of 1m vs 1cm, making the ratio of the cross section go like 100^2 = 10^4, the wire can be 9.21+1 = 10.21 times as long, which is essentially your same result. Makes more sense? (edited an uncanceled C at the end.)
You are wonderful. Thank you very much for the explanation, exactly what I was looking for. Not overly complicated when looking at it, but piecing it together for me was a real struggle for me. Merci!
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On September 05 2016 17:29 FiWiFaKi wrote:Show nested quote +On September 05 2016 13:05 Cascade wrote:On September 05 2016 11:55 FiWiFaKi wrote:On September 05 2016 11:06 Cascade wrote:On September 05 2016 10:30 FiWiFaKi wrote:On September 04 2016 16:36 Cascade wrote: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. I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0. The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width. So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons. If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point. I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/ Your equation assumes the length is infinite. I'm not trying to look at this from a math perspective, but rather a practical one. I want to have a certain thickness at the bottom, Which means I will have a certain radius at the top, and a certain radius at the bottom. So yes, yours would be the case for a never ending string, but it's not what I'm looking for, which like you said, your equation wasn't designed to solve S:... Like you mentioned, it will be on the safe side. My objective was to calculate a maximum length based on some starting dimensions. Without any treatment, your equation just says, yep, the length of the string will be infinite. With your equation the radius will start at infinity, and approach zero. I want to have the thickest part be say 1m, and thinnest part be 1cm, and calculate how long of a rod I could make. That was the point of obtaining a length vs radius graph. I made a super quick awful looking sketch lol. But yeah, I wanted both, the curve the radius vs length makes, as well as calculating the maximum length of the rod based on some given radii at both ends (and ideally as a function of some weight hanging off of it as well). Also I fucked up labelling the graph, obv e^-x = 1 at 0, opps ^__^ It doesn't assume infinite length. It does accommodate it, but it also accommodates finite length solutions where a(L) is 0 after a certain L1. Not what comes out of the equations as I set them up though. It also doesn't assume infinite width at the ceiling. Indeed, the ceiling is at L=0, where the width is 1, or whatever constant you chose to set in front of the exponential when you solve the differential equation. To find the solution to the new problem you pose, with the constraints of 1m at ceiling and at least 1cm thickness, it is just a small modification to my first solution. You first solve it as normal, setting the constant in front of the exponential to get the right thickness at the ceiling. so the solution is a(L) = C*exp(-k*L) where C is the free parameter from the differential equation. k is a constant depending on the material and probably on g. As a(L) is the cross section, C would have to be like 2*pi*r_0 or something for the desired starting radius r_0. Let's make it easy and set C to 1 for now. That solves the ceiling boundary condition. To let it not go below a certain minimum radius (let's say a(L) has to be at least a_m), we cut the exponential at the length L_m where it reaches that width: a(L_m) = a_m C*exp(-k*L_m) = a_m L_m = -ln(a_m/C)/k = ln(C/a_m)/k We can now add extra volume V_e underneath that with a cross section of a_m. The amount it can support is the amount that we just cut off, which is V_e = integral{x from L_m to infinity} a(L)*dx = A(infinity) - A(L_m) = -C*(exp(-k*infinity) - exp(-k*L_m))/k = C/k*exp(-k*L_m) where the primitive function is A(L) = -C*exp(-k*L)/k. This volume corresponds to an extra length L_e of constant cross section a_m where L_e*a_m = V_e = C/k*exp(-k*L_m) L_e = C/(k*a_m)*exp(-k*L_m) replace L_m from above: L_e = C/(k*a_m)*exp(-k*(-ln(a_m/C)/k)) = C/(k*a_m)*exp(ln(a_m/C)) = C/(k*a_m)*a_m/C = 1/k which should reproduce your formula for flat wires. So the longest shape you can have with those constraints is starting from the ceiling as a negative exponential, reaching the minimum allowed cross section a_m at L_m = ln(C/a_m)/k, then followed by a flat wire of minimum cross section of length L_e = 1/k, for a total length L_t of L_t = ln(C/a_m)/k + 1/k Replacing C with a_c for the cross section at the ceiling, we see that the length is the same as for a flat wire (1/k), but allowing it to grow near the ceiling will add an extra ln(a_c/a_m)/k. So the extra length you can add is proportional to the logarithm of the ratio of the maximum and minimum allowed cross section. In the case of allowing e times as large cross section in the ceiling, you will get a twice as long shape. With your example of 1m vs 1cm, making the ratio of the cross section go like 100^2 = 10^4, the wire can be 9.21+1 = 10.21 times as long, which is essentially your same result. Makes more sense? (edited an uncanceled C at the end.) You are wonderful. Thank you very much for the explanation, exactly what I was looking for. Not overly complicated when looking at it, but piecing it together for me was a real struggle for me. Merci!
Happy to help. Yeah, each step is easy, but it was more fiddly than I expected towards the end. May I ask where this problem came from? You just made it up?
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On September 05 2016 18:06 Cascade wrote:Show nested quote +On September 05 2016 17:29 FiWiFaKi wrote:On September 05 2016 13:05 Cascade wrote:On September 05 2016 11:55 FiWiFaKi wrote:On September 05 2016 11:06 Cascade wrote:On September 05 2016 10:30 FiWiFaKi wrote:On September 04 2016 16:36 Cascade wrote: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. I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0. The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width. So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons. If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point. I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/ Your equation assumes the length is infinite. I'm not trying to look at this from a math perspective, but rather a practical one. I want to have a certain thickness at the bottom, Which means I will have a certain radius at the top, and a certain radius at the bottom. So yes, yours would be the case for a never ending string, but it's not what I'm looking for, which like you said, your equation wasn't designed to solve S:... Like you mentioned, it will be on the safe side. My objective was to calculate a maximum length based on some starting dimensions. Without any treatment, your equation just says, yep, the length of the string will be infinite. With your equation the radius will start at infinity, and approach zero. I want to have the thickest part be say 1m, and thinnest part be 1cm, and calculate how long of a rod I could make. That was the point of obtaining a length vs radius graph. I made a super quick awful looking sketch lol. But yeah, I wanted both, the curve the radius vs length makes, as well as calculating the maximum length of the rod based on some given radii at both ends (and ideally as a function of some weight hanging off of it as well). Also I fucked up labelling the graph, obv e^-x = 1 at 0, opps ^__^ It doesn't assume infinite length. It does accommodate it, but it also accommodates finite length solutions where a(L) is 0 after a certain L1. Not what comes out of the equations as I set them up though. It also doesn't assume infinite width at the ceiling. Indeed, the ceiling is at L=0, where the width is 1, or whatever constant you chose to set in front of the exponential when you solve the differential equation. To find the solution to the new problem you pose, with the constraints of 1m at ceiling and at least 1cm thickness, it is just a small modification to my first solution. You first solve it as normal, setting the constant in front of the exponential to get the right thickness at the ceiling. so the solution is a(L) = C*exp(-k*L) where C is the free parameter from the differential equation. k is a constant depending on the material and probably on g. As a(L) is the cross section, C would have to be like 2*pi*r_0 or something for the desired starting radius r_0. Let's make it easy and set C to 1 for now. That solves the ceiling boundary condition. To let it not go below a certain minimum radius (let's say a(L) has to be at least a_m), we cut the exponential at the length L_m where it reaches that width: a(L_m) = a_m C*exp(-k*L_m) = a_m L_m = -ln(a_m/C)/k = ln(C/a_m)/k We can now add extra volume V_e underneath that with a cross section of a_m. The amount it can support is the amount that we just cut off, which is V_e = integral{x from L_m to infinity} a(L)*dx = A(infinity) - A(L_m) = -C*(exp(-k*infinity) - exp(-k*L_m))/k = C/k*exp(-k*L_m) where the primitive function is A(L) = -C*exp(-k*L)/k. This volume corresponds to an extra length L_e of constant cross section a_m where L_e*a_m = V_e = C/k*exp(-k*L_m) L_e = C/(k*a_m)*exp(-k*L_m) replace L_m from above: L_e = C/(k*a_m)*exp(-k*(-ln(a_m/C)/k)) = C/(k*a_m)*exp(ln(a_m/C)) = C/(k*a_m)*a_m/C = 1/k which should reproduce your formula for flat wires. So the longest shape you can have with those constraints is starting from the ceiling as a negative exponential, reaching the minimum allowed cross section a_m at L_m = ln(C/a_m)/k, then followed by a flat wire of minimum cross section of length L_e = 1/k, for a total length L_t of L_t = ln(C/a_m)/k + 1/k Replacing C with a_c for the cross section at the ceiling, we see that the length is the same as for a flat wire (1/k), but allowing it to grow near the ceiling will add an extra ln(a_c/a_m)/k. So the extra length you can add is proportional to the logarithm of the ratio of the maximum and minimum allowed cross section. In the case of allowing e times as large cross section in the ceiling, you will get a twice as long shape. With your example of 1m vs 1cm, making the ratio of the cross section go like 100^2 = 10^4, the wire can be 9.21+1 = 10.21 times as long, which is essentially your same result. Makes more sense? (edited an uncanceled C at the end.) You are wonderful. Thank you very much for the explanation, exactly what I was looking for. Not overly complicated when looking at it, but piecing it together for me was a real struggle for me. Merci! Happy to help. Yeah, each step is easy, but it was more fiddly than I expected towards the end. May I ask where this problem came from? You just made it up?
Yeah, I thought up of it on my own. I just ask myself a silly question, for example if I drip down 1L of water on a flat surface made of x material, how much will it spread, what's the fastest speed a shaft can rotate at before breaking apart (magnetic bearing, vacuum), what's the tallest free standing structure made out of x material (this will definitely help me with that, but its quite a bit different due to the inherent instability and buckling effects), etc.
Just little things that spark my curiousity, and it makes me think a bit, so it's a nice time. I enjoy going through the math and deriving these things, unlike in my past courses where it's mostly given to you. From a practical sense it's often not super useful, but I like having some physical meaning behind the math I'm trying to do
Often times it provides a lot of insight to the limitations of certain designs. For example, if you have a belt driving two rollers, you're told you can't go above 30m/s or so in some machine component design course, for the belt speed because of the centrifugal effects... But then doing the math and figuring out the effect and how much is feasible, how it scales, how it varies with the drum diameter all these things are interesting to me (simple example).
Fluid dynamics problems are particularly interesting to me, but they are tough to get nice solutions for, as not only is some of the stuff empirical (and stuff like assuming constant friction coefficient throughout the whole flow is just wrong from a what happens in nature standpoint), but most equations can only be solved numerically.
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Which font do you prefer, Times New Roman or Helvetica? maybe Comic Sans?
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Norway28798 Posts
I have never understood why people hate comic sans so much. it's FUN!!
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Ok, so what if you have somebody dig out a hole in the floor to put the scale in, such that the surface of the scale is flush with the surface of the ground, and then you rest your head on it?
![[image loading]](http://i.imgur.com/pLClqCo.png)
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On September 06 2016 05:49 Liquid`Drone wrote:
I have never understood why people hate comic sans so much. it's FUN!!
Anyway. Helvetica. And in ss I like Android's font a lot.
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Helvetica, Verdana, Tahoma.
Rest is shit (not sure if the last one is actually called like this).
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On September 05 2016 18:21 FiWiFaKi wrote:Show nested quote +On September 05 2016 18:06 Cascade wrote:On September 05 2016 17:29 FiWiFaKi wrote:On September 05 2016 13:05 Cascade wrote:On September 05 2016 11:55 FiWiFaKi wrote:On September 05 2016 11:06 Cascade wrote:On September 05 2016 10:30 FiWiFaKi wrote:On September 04 2016 16:36 Cascade wrote: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. I don't set A(infinity) to 0, I just say that it doesn't depend on L, so when I take the derivative of L, it goes away. A(infinity) will set the total volume of the structure, if we set the integration constant so that A(0) = 0. The equation I set up was to be on the limit everywhere, which turns out to allow for infinitely long string in the mathematically perfect world where you can have infinitely thin strings. Of course you will run into a single atoms width pretty quickly in practice with exponentially decreasing width. So to answer your original question: the distribution should be exponential, which allows infinitely long strings, or at least up to where you can't make it thinner due to practical reasons. If you want to set the width to 0 after a certain L, that is of course fine. It will no longer solve the equation I set up that was to be on the limit at all points, but it will indeed be on the safe side of the limit, so will not break the string. It still fulfills the first equation if you replace the equality with an inequality, saying that the mass below any point needs to be equal OR SMALLER that a constant times the cross section. You can still run the integral to infinity, but you set a(L) to 0 for L larger than the cutoff, so it'll not contribute to the integral after that point. I don't follow the last part of your calculation. Why do you need two r? What's going on there? Why the difference? The width of the structure is already described at all point in the function a(L). You describe the length L of the structure at a distance r from the center, and then for some reason you introduce two different structures r1 and r2 and take the difference in length L between the two? What is that achieving? I don't follow at all. :/ Your equation assumes the length is infinite. I'm not trying to look at this from a math perspective, but rather a practical one. I want to have a certain thickness at the bottom, Which means I will have a certain radius at the top, and a certain radius at the bottom. So yes, yours would be the case for a never ending string, but it's not what I'm looking for, which like you said, your equation wasn't designed to solve S:... Like you mentioned, it will be on the safe side. My objective was to calculate a maximum length based on some starting dimensions. Without any treatment, your equation just says, yep, the length of the string will be infinite. With your equation the radius will start at infinity, and approach zero. I want to have the thickest part be say 1m, and thinnest part be 1cm, and calculate how long of a rod I could make. That was the point of obtaining a length vs radius graph. I made a super quick awful looking sketch lol. But yeah, I wanted both, the curve the radius vs length makes, as well as calculating the maximum length of the rod based on some given radii at both ends (and ideally as a function of some weight hanging off of it as well). Also I fucked up labelling the graph, obv e^-x = 1 at 0, opps ^__^ It doesn't assume infinite length. It does accommodate it, but it also accommodates finite length solutions where a(L) is 0 after a certain L1. Not what comes out of the equations as I set them up though. It also doesn't assume infinite width at the ceiling. Indeed, the ceiling is at L=0, where the width is 1, or whatever constant you chose to set in front of the exponential when you solve the differential equation. To find the solution to the new problem you pose, with the constraints of 1m at ceiling and at least 1cm thickness, it is just a small modification to my first solution. You first solve it as normal, setting the constant in front of the exponential to get the right thickness at the ceiling. so the solution is a(L) = C*exp(-k*L) where C is the free parameter from the differential equation. k is a constant depending on the material and probably on g. As a(L) is the cross section, C would have to be like 2*pi*r_0 or something for the desired starting radius r_0. Let's make it easy and set C to 1 for now. That solves the ceiling boundary condition. To let it not go below a certain minimum radius (let's say a(L) has to be at least a_m), we cut the exponential at the length L_m where it reaches that width: a(L_m) = a_m C*exp(-k*L_m) = a_m L_m = -ln(a_m/C)/k = ln(C/a_m)/k We can now add extra volume V_e underneath that with a cross section of a_m. The amount it can support is the amount that we just cut off, which is V_e = integral{x from L_m to infinity} a(L)*dx = A(infinity) - A(L_m) = -C*(exp(-k*infinity) - exp(-k*L_m))/k = C/k*exp(-k*L_m) where the primitive function is A(L) = -C*exp(-k*L)/k. This volume corresponds to an extra length L_e of constant cross section a_m where L_e*a_m = V_e = C/k*exp(-k*L_m) L_e = C/(k*a_m)*exp(-k*L_m) replace L_m from above: L_e = C/(k*a_m)*exp(-k*(-ln(a_m/C)/k)) = C/(k*a_m)*exp(ln(a_m/C)) = C/(k*a_m)*a_m/C = 1/k which should reproduce your formula for flat wires. So the longest shape you can have with those constraints is starting from the ceiling as a negative exponential, reaching the minimum allowed cross section a_m at L_m = ln(C/a_m)/k, then followed by a flat wire of minimum cross section of length L_e = 1/k, for a total length L_t of L_t = ln(C/a_m)/k + 1/k Replacing C with a_c for the cross section at the ceiling, we see that the length is the same as for a flat wire (1/k), but allowing it to grow near the ceiling will add an extra ln(a_c/a_m)/k. So the extra length you can add is proportional to the logarithm of the ratio of the maximum and minimum allowed cross section. In the case of allowing e times as large cross section in the ceiling, you will get a twice as long shape. With your example of 1m vs 1cm, making the ratio of the cross section go like 100^2 = 10^4, the wire can be 9.21+1 = 10.21 times as long, which is essentially your same result. Makes more sense? (edited an uncanceled C at the end.) You are wonderful. Thank you very much for the explanation, exactly what I was looking for. Not overly complicated when looking at it, but piecing it together for me was a real struggle for me. Merci! Happy to help. Yeah, each step is easy, but it was more fiddly than I expected towards the end. May I ask where this problem came from? You just made it up? Yeah, I thought up of it on my own. I just ask myself a silly question, for example if I drip down 1L of water on a flat surface made of x material, how much will it spread, what's the fastest speed a shaft can rotate at before breaking apart (magnetic bearing, vacuum), what's the tallest free standing structure made out of x material (this will definitely help me with that, but its quite a bit different due to the inherent instability and buckling effects), etc. Just little things that spark my curiousity, and it makes me think a bit, so it's a nice time. I enjoy going through the math and deriving these things, unlike in my past courses where it's mostly given to you. From a practical sense it's often not super useful, but I like having some physical meaning behind the math I'm trying to do Often times it provides a lot of insight to the limitations of certain designs. For example, if you have a belt driving two rollers, you're told you can't go above 30m/s or so in some machine component design course, for the belt speed because of the centrifugal effects... But then doing the math and figuring out the effect and how much is feasible, how it scales, how it varies with the drum diameter all these things are interesting to me (simple example). Fluid dynamics problems are particularly interesting to me, but they are tough to get nice solutions for, as not only is some of the stuff empirical (and stuff like assuming constant friction coefficient throughout the whole flow is just wrong from a what happens in nature standpoint), but most equations can only be solved numerically.
Such an engineer. 
Do you do research? Sounds like you'd like it.
Unless you like money of course.
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What would stop me from using the huge freezers in supermarkets to freeze certain types of food that I can pick up (and buy if they still need to be bought), if I hide them well enough?
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On September 06 2016 11:18 Uldridge wrote:
What would stop me from using the huge freezers in supermarkets to freeze certain types of food that I can pick up (and buy if they still need to be bought), if I hide them well enough?
Takes a couple of hours for most food to freeze solid. I, personally, have better things to do than hang out near the pizza section of my local supermarket, but to each his own.
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EPT
