5 ms·

The losses when transmitting through vacuum are pretty negligible; it's only the last fifteen miles that really cost you.

• dllthomas 13y ago
  That certainly might be the case, but I don't immediately see why.
  • csense 13y ago
    I'm not a physicist, but my understanding is that there are two ways a wireless photon-based energy transmission system loses energy in transit. First, each molecule has a specific set of frequencies where photons are "easy" to absorb because they correspond to transitions between quantum states. Any system whose goal is to deliver energy wirelessly through the atmosphere will obviously not use those frequencies. The other way for a photon to lose energy to the medium it's traveling through is, if the medium contains molecules like water that have different parts with positive and negative charges, the electromagnetic wave will move the differently charged parts in opposite directions. The energy to do that is lost [1]. [1] "Lost" is an application-specific notion. This mechanism is also how a microwave heats food, and in this case the energy imparted to the water molecules in the food can't be considered "lost" because it's supposed to go into the food!
    • dllthomas 13y ago
      Ah, you intend to keep them photons. I had been imagining converting to electrons and running down cabling attached to the elevator cable. The other approach might very well be better, but doesn't really require a space elevator does it?
      • adwf 13y ago
        I'd imagine they'd go with super-conducting cables personally. Space is extremely cold, so achieving a super-conducting state is simply a matter of shielding the cable from the sun. Shouldn't be too difficult... At least, not in comparison to building a space elevator in the first place!
        • dllthomas 13y ago
          Space is cold, but getting rid of heat is not easy in a vacuum. Not sure what that means with respect to this particular project, though.
          • lutusp 13y ago
            > Space is cold, but getting rid of heat is not easy in a vacuum. On the contrary, getting rid of heat in a vacuum is easier than getting rid of heat in an atmosphere. Here's a diagram of human heat loss at the surface: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.htm... According to this source, at the surface of earth, under the atmosphere, a person loses: Perspiration: 17 watts Conduction: 11 watts Radiation: 133 watts Without an atmosphere, the direct radiation of heat energy into space becomes more efficient (no greenhouse effect), and it's always the most efficient way to radiate heat. It is the efficiency of direct radiation of heat energy that explains why objects at the surface can fall below air temperature overnight under a clear sky, as they surely do.
            • dllthomas 13y ago
              Hm, that's contrary to what I'd heard, but I'll certainly defer to you on the question.
              • lutusp 13y ago
                > Hm, that's contrary to what I'd heard I think some of the confusion arises because of vacuum thermos bottles, which are really efficient at holding onto their heat. But how they work is a bit complicated. They deal with conduction and convection losses by having the vacuum barrier between the contents and the outside. As to radiation, they rely on a reflector that's part of the vacuum bottle, which has the effect of greatly slowing the rate of heat loss by radiation. So the vacuum thermos avoids radiation heat loss, not because of the vacuum, but with a reflector. That works in space too -- many orbiting telescopes use reflectors to keep the sun's heat energy from heating up the sensors and spoiling their performance. But a vacuum is a pretty good medium for heat loss by radiation. The moon's surface, heated to several hundred degrees Celsius during the lunar daytime, drops to 26 Kelvins after a few (earth) days of darkness (26 degrees above absolute zero). That's a new figure, lower than had been realized, and seven degrees colder than the surface of Pluto.
    • gliese1337 13y ago
      A third way is by diffraction- any beam, including a laser, will spread out as it travels. You get less spread with short wavelengths and large transmitters, but both of those imply expense. If you use long wavelengths and small transmitters, you need huge receivers to get all the beam energy back.