7 ms·

Omg I need to make an led version

bleepblorp 6y ago
Probably not possible. Searchlights work by focusing light from a source to infinity. Physics dictates that small light sources focus better. Getting searchlight-level brightness from an LED source will require an LED array, which will be too big to focus effectively. LEDs make excellent floodlights, but if you want reach, xenon discharge lamps are still the best option.
- rocqua 6y ago
  What about focusing each LED source separately?
  - bleepblorp 6y ago
    LED car headlights do use this principle; it does have some applications. However, the diameter of each LED optic would need to at least equal the diameter of the single optic used for a point source light. The result would be an enormous searchlight with a very broad beam. I can't find any solid information on the quantitative light output of the AN/TVS-3, but xenon arc lamps typically produce 30 lumens per watt. Therefore, the 20KW AN/TVS-3 should produce around 600,000 lumens. Readily available high-power LEDs are capable of 1700 lumens, so equaling the AN/TVS-3 would take--as a very rough ballpark that does not take into account the different light distribution between LEDs and discharge lamps--350 LEDs and optics. The AN/TVS-3 uses 30" optics, so those optics would also need to be at least 30" in diameter. 350 30" optics is a lot more like a wall of light than a searchlight. Aiming that many separate optics would also be a nightmare.
    - MisterTea 6y ago
      > I can't find any solid information on the quantitative light output of the AN/TVS-3, but xenon arc lamps typically produce 30 lumens per watt https://en.wikipedia.org/wiki/Luminous_efficacy https://en.wikipedia.org/wiki/Luminous_efficacy What I'm not sure of is if the lumins/watt rating is given for the light output or the power input. The input power rating is 450 Amps * 200 Volts * 1.732 = 155880 electrical watts. That's nearly 7.75 times as much electrical power as light output. That makes sense since the peak efficiency is about 7.3 for a xenon arc lamp.
    - rocqua 6y ago
      Why does the optic need the same diameter? For gathering all of the light, with a mirror at least, I think you could place a smaller optic closer to the light and still capture the same fraction of the light? I probably am missing something about optics though.
      - bleepblorp 6y ago
        The ability of an optical system (a lens, mirror, or combination) to collimate the output of a light source into a narrowly diverging beam is directly proportional to the diameter of the optical system's final element and inversely proportional to the size of the light source. Bigger optics = less beam divergence = more useful range. This is a function of the diffraction limit[0] and is a property of physics. If you want to match the beam divergence of a 30" optical system, you have to use 30" optics; there are no shortcuts. [0] https://www.edmundoptics.com/knowledge-center/application-notes/imaging/diffraction-limit/ https://www.edmundoptics.com/knowledge-center/application-no...
        rocqua 6y ago
        I know that we can create very large virtual apertures for receiving signals by using arrays of many receivers. The basic diffraction physics presume a single source, so it isn't totally obvious that they generalize to a multi-source system. I can, however, imagine that a multi source system would require in-phase light sources to kind of mimic a phased-array receiver on the send side.
- avian 6y ago
  If anyone wants to dive into physics behind the reasons why you can't efficiently collimate light from a large emitter like a LED array, this Stack Exchange answer [1] and the Wikipedia article about etendue [2] is a good start. [1] https://physics.stackexchange.com/a/91536 https://physics.stackexchange.com/a/91536 [2] https://en.wikipedia.org/wiki/Etendue https://en.wikipedia.org/wiki/Etendue
  - ramses0 6y ago
    There was a very interesting article about how you can't focus 100's of magnifying glasses using light from the moon to start a fire, probably similar in principles? XKCD explainer here: https://what-if.xkcd.com/145/ https://what-if.xkcd.com/145/

Probably not possible. Searchlights work by focusing light from a source to infinity. Physics dictates that small light sources focus better. Getting searchlight-level brightness from an LED source will require an LED array, which will be too big to focus effectively. LEDs make excellent floodlights, but if you want reach, xenon discharge lamps are still the best option.

rocqua 6y ago
What about focusing each LED source separately?
- bleepblorp 6y ago
  LED car headlights do use this principle; it does have some applications. However, the diameter of each LED optic would need to at least equal the diameter of the single optic used for a point source light. The result would be an enormous searchlight with a very broad beam. I can't find any solid information on the quantitative light output of the AN/TVS-3, but xenon arc lamps typically produce 30 lumens per watt. Therefore, the 20KW AN/TVS-3 should produce around 600,000 lumens. Readily available high-power LEDs are capable of 1700 lumens, so equaling the AN/TVS-3 would take--as a very rough ballpark that does not take into account the different light distribution between LEDs and discharge lamps--350 LEDs and optics. The AN/TVS-3 uses 30" optics, so those optics would also need to be at least 30" in diameter. 350 30" optics is a lot more like a wall of light than a searchlight. Aiming that many separate optics would also be a nightmare.
  - MisterTea 6y ago
    > I can't find any solid information on the quantitative light output of the AN/TVS-3, but xenon arc lamps typically produce 30 lumens per watt https://en.wikipedia.org/wiki/Luminous_efficacy https://en.wikipedia.org/wiki/Luminous_efficacy What I'm not sure of is if the lumins/watt rating is given for the light output or the power input. The input power rating is 450 Amps * 200 Volts * 1.732 = 155880 electrical watts. That's nearly 7.75 times as much electrical power as light output. That makes sense since the peak efficiency is about 7.3 for a xenon arc lamp.
  - rocqua 6y ago
    Why does the optic need the same diameter? For gathering all of the light, with a mirror at least, I think you could place a smaller optic closer to the light and still capture the same fraction of the light? I probably am missing something about optics though.
    - bleepblorp 6y ago
      The ability of an optical system (a lens, mirror, or combination) to collimate the output of a light source into a narrowly diverging beam is directly proportional to the diameter of the optical system's final element and inversely proportional to the size of the light source. Bigger optics = less beam divergence = more useful range. This is a function of the diffraction limit[0] and is a property of physics. If you want to match the beam divergence of a 30" optical system, you have to use 30" optics; there are no shortcuts. [0] https://www.edmundoptics.com/knowledge-center/application-notes/imaging/diffraction-limit/ https://www.edmundoptics.com/knowledge-center/application-no...
      - rocqua 6y ago
        I know that we can create very large virtual apertures for receiving signals by using arrays of many receivers. The basic diffraction physics presume a single source, so it isn't totally obvious that they generalize to a multi-source system. I can, however, imagine that a multi source system would require in-phase light sources to kind of mimic a phased-array receiver on the send side.
avian 6y ago
If anyone wants to dive into physics behind the reasons why you can't efficiently collimate light from a large emitter like a LED array, this Stack Exchange answer [1] and the Wikipedia article about etendue [2] is a good start. [1] https://physics.stackexchange.com/a/91536 https://physics.stackexchange.com/a/91536 [2] https://en.wikipedia.org/wiki/Etendue https://en.wikipedia.org/wiki/Etendue
- ramses0 6y ago
  There was a very interesting article about how you can't focus 100's of magnifying glasses using light from the moon to start a fire, probably similar in principles? XKCD explainer here: https://what-if.xkcd.com/145/ https://what-if.xkcd.com/145/