I'm only asking this on here out of interest, this is not homework. I want to know how to calculate exact distances between to poles connected at the top by a cable. The length or height of the objects are not important to me, but if you must use an example please do so. So I want to know what formulas to use, what variables in those formulas mean, and how the slope of the cable will affect the distance. For example, if two poles have a distance between them of 0, and the cable extends 100, it would not mean that the drop of the cable (called a catenary?) would be 50, (it's misleading because you would think that since the distance between the two poles is 0, the cable would drop 50 and rise 50. But this is not taking into account the slope it has to take at the bottom of the catenary. It must lose some drop (perhaps to 48 or 49) in order to bend and rise back up...I want this out of my head, it's bugging me that I can't figure it out.
No sign can commute faster than the fee of sunshine (neglecting tachyons, that have in no way been pronounced interior the genuine international, yet in concept can in no way slow all the way down to or commute under the fee of sunshine). The ropes and poles could flex because of the fact the sign traveled alongside them. because of the fact that they'd not flow right this moment, the sign could commute at a speed nicely under c. right that's a version on the pole concept. think you have 2 very long poles crossed at an attitude. The poles are rotating from their crossed orientation to alter into extra parallel. the factor P the place the poles bypass will commute down the poles as they attitude parallel, shifting faster and faster. Can the action of P exceed the fee of sunshine? particular it could! P would not have any actual manifestation and for this reason no mass; for this reason its action isn't constrained with the aid of relativity. in spite of the undeniable fact that, if P did have mass -- as an occasion, if there became a hoop around the crossing factor -- then the poles could be compelled to flex because of the fact the relativistically increasing mass drew further and further ability from them. * To Michael under: Einstein's particular concept of Relativity states that it could consume an limitless quantity of ability to strengthen up any particle with mass to the fee of sunshine. So that is impossible to strengthen up even an electron to the fee of sunshine, regardless of the certainty that scientists can get very very close now. Photons have not have been given any mass so as that they commute on the fee of sunshine and easily on the fee of sunshine (and is the reason why every person measures an identical fee for c). Tachyons have an 'imaginary' mass (a dissimilar of the sq. root of -a million) and so as that they can't commute as slow or slower than mild... merely faster!
So you want the length along the cable? I honestly don't know how to compute the arc length of a catenary, but I'm sure that the power company knows how to do it because they need to know how much cable it will take to string up a whole line of poles. Your example doesn't work only because you impose a physically impossible constraint (zero distance between the poles - the poles have no thickness) and then try to impose physical reality (the cable can't bend instantaneously) on the problem. In the physically unrealizable problem, yes the cable does drop 50 ft because it is infinitely bendable. You're dealing with hyperbolic sines and cosines. Good luck
