Car headlights for growing?

Discussion in 'Growing Marijuana Indoors' started by TheMysticMan, May 17, 2006.

  1. Guess I am off in wonderland sitting here thinking about whether or not you could use car headlamps for a grow. I know that one of the latest technologies in headlamps is HID if I am not mistaken, but other than that I do not not much more about headlamps.

    Is it possible?
  2. Only if you use HPS lights in your car.. which you probably dont :D

    No, it wouldn't work. Sorry. But hell the HID bulb kits usually cost a fuckin' buck or two, might as well get a light made for what you're doing if you're going to try it.
  3. Not going to try it, just a passing thought and was curious;)
  4. Actually, the headlight bulbs in some luxury and almost luxury cars are an HID light. They are different than the HID's that we use in that the HPS has sodium in it while the headlights in cars use a Xenon gas. The Xenon gas actually creates a spectrum similar to that of an MH light (bright blue). I would not be surprised if Xenon headlights could grow MJ because after all they are HID lights. Don't hold your breath on any experiments because I don't think there are any bone heads out there willing to pull their luxury cars into their house to light their plants for 18 hours.
  5. yeah that sounds about right..

    maybe someone will have access to extra parts or something and will try at some point
  6. um but good luck flowering with a HID..
  7. Uhh....
  8. Uhh, he says it himself, the HID produces a blue light spectrum which is what metal halide is (MH) which is used for vegging and not FLOWERING.

  9. HID isn't just Metal Halide. HID light is HPS, and Metal Halide, or... any lamp that uses electricity to excite gases to create light. Therefore, High Pressure Sodium (which is ideal for flowering) AND Metal Halide are HID lights. And I think a lot of people would agree....

    Edit: And by the way, I've seen lots of grows from start to finish with a Metal Halide. It's not impossible... it's just HPS produces better results.
  10. HID does not refer to "any lamp that uses electricity to excite gases to create light". If that was the case, then every CFL would be HID.

    HID refers to teh intensity of the discharge. The reason you wouldn't want xenon HID is cost and efficiency. That's why my answer is still "NO".

    The only advantage I can see to using these, would be because they are 12 volt (or 24). This means you can hook up a small generator and power it with yyour fat ass on a bicycle :D or use a solar panel or 5 or car batteries or something. Hassle.
  11. I didn't know we were getting that technical, sorry bro.

    I was just making a point that HID isn't just Metal Halide. Whatev.
  12. In actuallity low voltage is a disadvantage!
    Low voltage means high currents because you'll need to same Watts for your lumers/feet.

    A 400W HPS @ 220V pull a bit less then 2 AMPS
    A 400W HPS @ 110V pull a bit less then 4 AMPS
    A 400W HPS @ 12V pulls a big fat 33 AMPS.

    You will produce pretty strong magnetic fields and loads of extra heat powering up anything that requires 33 AMPS. Both heat and the magnetic field are the direct result of the current, not the voltage.
    SO with low voltage lighting the overall efficiency will drop like a brick.

    This is the reason all power companies on earth use high voltage installations for transportation of electricity. In the Netherworld...uhhm...lands they use 180.000V for transport between powerplant and city. In cities they use 10.000V all the way to the transformer that powers up the block. (a few dozen houses)
  13. ^^ They also use extremely high voltage because you lose a lot less due to resistance of the wire.

    If you sent just a regular 120V charge over the line, you'd lose about 80% of it from just the wire's resistance. (500 amps)

    120,000V on the other hand, you only lose 0.01% of the voltage. (5 amps)

    Which I guess is exactly what you already said... my post just adds some numbers for no other reason than because I'm bored.
  14. What you say is allmost true my friend!;)

    The loss of energy is NOT the result of the resistance of the wire. The resistance is what minimizes the current. Resistance is the enemy of current.

    If there is sufficient power available and the resistance equals zero the current will be infinite and so will the energy 'loss'.

    Ofcourse this energy is not 'lost' it simply goes where you do not want it.
  15. I = P/V
    P(loss) = I^2 * R
    P(loss) = (P/V)^2 * R

    If resistance was zero, the power lost would be zero. As resistance increases, so does the power lost.

    I just checked those formulas in an old physics text book. Also "power lost" is a universally recognized term for power that gets wasted in whatever system you're working with. Of course it doesn't dissappear, that's impossible, but the resistance of the wire uses up some of the power, and if you don't give the line enough voltage, the current becomes very high, which causes the power loss to increase (see second formula).

    Also, just so we're clear, when I say "lose about 80% of it" I wasn't talking about loosing 80% of the voltage--I meant you lose 80% of the power. With only 120V, you can start with 120kW on the origin and only have 20kW on the other end of the line (assuming resistance = 0.4). Conversly, if you start with 120,000V, you can have 120kW on one end and 119.990kW on the other end, a loss of only 10W. That's probably where we're getting mixed up.


    I'm too fucking high to talk about physics :) I just re-read my previous post and saw where I said "voltage" when I ment to say "power". I understand what you're saying now, and yes, my post that you quoted is incorrect. If I were to change "it" to "power" (for clearification) and "of the voltage" to "of the power" it would be correct. My bad :D Also the amps in brackets aren't ment to imply anything other those would be the approx amps in a standard power line given the voltages I put (and the obsurdity of using low voltages: you get extremely high amps!)
  16. Now thats what I'm talking about Willis, a thread with some posts that are meaty and make you wanna reach for that next hit:)
  17. Dude!
    I'm stunned.
    There is something really wrong here,

    Especially the P(loss) formula must be wrong. It bothers me. Where did you get that science from? US government perhaps?

    For a spark in a vacuum experience zero resistance, there is nothing stopping them, once it gets going.
    V=IR so I=V/R so 1 volt / 0 resistance = infinite current.
    Infinite current means infinite magnetic/electric fields which means infinite 'loss'. (if there is infinite power)

    Also, the wires do not 'consume' the power as you put it. They transform it into another form of energy : magnetic/electric fields.

    Now I realize whats wrong with the P(loss) = I^2 * R formula!!!
    It does not incorporate TIME.
    Any energy loss must be measured over a period of time, because 0 time would mean 0 change.

    w000000000t : genius mode engaged!!!!!:hippie:
  18. I just copied it from an example problem on calculating power loss in an power line, verbatim, from a college engineering physics book. I'll give you the setup, maybe that will clear something up.

    Power: 120kW
    Resistance: 0.4 Ohms

    Calculate power loss if the power is transmitted at (a) 240V, and (b) 240,000V (note: I know I changed 240V to 120V in my prev post because I'm lazy, baked, and didn't read the problem close enough the first time. changing 240V to 120V would double the current)

    I = P/V = 500A
    P(loss) = I^2 * R = (500A)^2 * .4 = 100kW
    Since P(in) is 120kW and P(loss) is 100kW, you've lost 80% of the power.

    I = P/V = 5A
    P(loss) = I^2 * R = (5^2)*(.4) =~ 10W

    -- end copy --

    As far as I can remember, time is not a factor in P, I, or R, with the exception of some AC problems. With AC,
    I(ave) = I' sin(wt)
    P(ave) = 0.5 * I'^2 * R
    But those are for averaging when dealing with AC, which is not constant. Regardless, P(ave) does not depend on time, only peak current (I') and resistance.

    The rest of the formulas:
    P = IV
    P = I^2 * R
    P = V^2 / R
    V = I * R (re-arrange as needed)

    The point is, the P(loss) formula is correct and time is not needed for it's calculation.


    I'm not saying that I'm 100%, without a doubt, correct. But I am coping this info from a college textbook, and a recent one at that. Also I'm looking over my page of cliff notes for this class I made that was stuffed in the textbook, and I'm not seeing many formulas that involve power that use time, with the exception of the one above. Ohms law is pretty straightforward: V = IR. Resistivity of a wire is [ (constant based on material) * (length) ] / (cross sectional area). Now I = (dQ/dt), which clearly uses the change in time, but if you're given V and R, or V and P, or R and P, or I and R, you don't need to use time.
  19. OK then, I now know two things:

    1) Don't grow MJ with car headlights.

    2) I made a good decision not to be an engineer 'cuz I don't know WTF you guys are talking about. ; -)

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