Lights for auto plants, very low budget.

you can patchwork something with what you have, or you can save up and do it right. In the long-run it's probably best to invest in some lights.. keep it inexpensive for now to get you through. The efficacy of the midpower 3030 diodes, like the samsung LM301B or H, will save you a bunch of money while providing same amount of light. but in your context I'm guessing you'll provide more light and spend equal or less than you do now. lumens don't translate equally for all lights. they 'can' but need to account for spectral flux density or some fancy term like that. lumens are direclty related to intensity, so an inespensive light meter can be useful in that sense relative to any 1 light fixture. Bulbs are nowhere near as efficent as the smaller midpower 3030 led diodes. It seems like a large expenditure up front, but your monthly bill will save you money over several years of use, apples to apples comparison. Even less espensive products will provide an efficacy around 2.5umol/J. Beware false advertising is rampant when it comes to umol/s, PPFD, PAR, efficacy. so, do what yo must as a stopgap while not wasting too many resources. Save up for a DIY light or even some array of entry level quantumboards and such. All yo need to do for LM301-based fixtures-- bulk will be "white" samsung diodes. IR and UV will lower efficacy of what i say below. At the least you want white and 660nm reds. the other stuff needs more research before spending big bucks on it or reducing efficacy, which has the greatest potential to add up over 7-10 years of use. First, divide number of watts by didoes. If it is approaching 0.4watts per diode or lower, you know it is a high-end light. You don't wan tthis to be significantly greater than 0.5watts per diode. The key here is to not drive the diode too hard. It wears it down sooner (dims) and isn't as efficient. Though this curve does have diminishing returns. 0.4watts per diod and the sammy diodes are ~2.7-2.8umol/J efficacly. other diodes will reduce this value, but they are typicaly 4-5% of total diode count. 0.2watts/diode and you are pushing 3.0umol/J -- beyond that it doen'st improve much while doubling # of diods.. diminishing returns. 0.5watts and you are around 2.5umol/J -- this will directly and proportionately relate to your monthly electric costs. Recurring costs are far, far greater than initial investment. if providing same umol/s of PAR to your plants, 2.5umol/j will cost ~12% more than the fixture running .4w/diode... and ~20% more than the 3.0umol/J Run them in that .5-.2 rnage and it'll last 5-7-10 years before 80-90% loss of intensity. how long do cfls and led bulbs last? 10khours? 1/5th. multiple bulb cost x5 and add the electricty too... it costs more to stick to the CFL/LED bulbs. bin of the diodes will make specs sway a few percent either way. if you can guesstimate a fairly accurate umol/J from # of diodes and # of watts, you can multiple watts x umol/j value and that equals umol/S. You can cross reference any LM301B-based light with reasonable accuracy. 2.5 * 150w = 375umol/s -- you'll see most light manufacturers are either really bad at math or simply know they can lie to most people with zero repercussions. now, how much do you need? ambient co2, 850-900ppfd is likely at or a bit above ceiling - ppfd is par light relative to/filling 1m^2. A little extra never hurts, as you can dim initially and counteract loss of intenisty of 5-10 years and never lose an inch of coverage nor needed intensity of light. umol/s of PAR light is a rate. We need to udnerstand that rate per area too. PPFD simply standardizes that relative to 1m^2, You have 2.625m^2 2.625 * 850 = about 2,250umol/s 2250 / 2.5 = about 900watts of a lower priced quantum board setup with LM301 chips. If you did that, you'd be providing a roughly what a plant can handle at ambient co2 levels. Shooting 10-20% over that isn't a bad idea, if you can easily dim the units a bit. As i said, it allows you to countract natural diming of the led diodes over 50-60,000 hours of use. 2250 / 2.8 = about 800watts. you'd save 11.1% off your electricity bill every month. how long does that take to equal the cost difference with the option above? is that near the end of 50-60,000 hours of use? 12 hours/day * .1 kw-hour * 365 / 12 = 36.5kw-hr per month average. if :text.25 per kw-hour, that's per month saved. 8 per year of use. Unless the cost difference is >0-00 range, it's actually cheaper to spend more today than be bleed each month for 5-10 years. True cost of ownership needs to consider time of use and cost of electricty. bwahah feel free to DM me.