1000w v 600w

[rangersroy]

i know 1000w are better than 600w but if you had 100m2 and 60000w would it be better to use for yeild 100 600w lights or 60 1000w in 60m2 for a better yield or would it end up the same so equipment wise your still better with 1000w

 

[nibec]

most 600 watt hps bulbs acutally perform more efficiently than a 1000 watt hps.. more lumens per watt

you can get 600's closer, and overall, overlapping 600 watt bulbs will result in better yeilds.

running 10 - 600 watt hps bulbs, is better than 6 - 1000 watt bulbs, although setup would cost more.

however, you could go really expensive, and get some dual arc 1000 watters.

that would be (per bulb) basically, 600 watts of hps, and 400 watt MH in ONE bulb on ONE 1000 w hps ballast! ... now thats cool... veg with the same lights as you flower with, and get the best of both worlds plus more!

 

[Mr White]

I concur, if cost is not an issue go with the multi 600s.

 

[Tricks]

most 600 watt hps bulbs acutally perform more efficiently than a 1000 watt hps.. more lumens per watt

you can get 600's closer, and overall, overlapping 600 watt bulbs will result in better yeilds.

running 10 - 600 watt hps bulbs, is better than 6 - 1000 watt bulbs, although setup would cost more.

however, you could go really expensive, and get some dual arc 1000 watters.

that would be (per bulb) basically, 600 watts of hps, and 400 watt MH in ONE bulb on ONE 1000 w hps ballast! ... now thats cool... veg with the same lights as you flower with, and get the best of both worlds plus more!

WOW! I wish the grow shop guy would have said something to me about that! My friend has a 400 W MH and was going to go with a 600 HPS for flower, that would have been perfect! we are using the old style ballast because there are only a couple bulbs made for digital ballast, and growers are reporting burnout rates of bulbs in as little as 6 months. So anyway, now he has to buy another ballast and bulb. 2 questions, how much do this cost and what about using a single 600 HPS in a 5X5 area? we thought about switching to a 1000w hps, but I read somewhere that a 600 would work better...what do ya think? Peace

 

[chuckinc]

600 watts can easily cover the 5'x5' area if you use the right reflector!

 

[too old]

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[symbiote420]

1k for sure! I used to grow under 600s I just got my 1k in Nov and I gotta say my overall bud size and yields have doubled! Same strains, results don't lie, plus the fact that lumens don't add up 140,000 lumens penetrates deeper into the canopy than 90,000. I have two homies that have switched from using (2) 600s to 1k(s) because I now yield more than both of them, I am using a light rail but I grow organically and they use Advanced Nutes Connie line. 600 = nice yields/ 1k = bigger yields!!

 

[drdanks]

1k for sure! I used to grow under 600s I just got my 1k in Nov and I gotta say my overall bud size and yields have doubled! Same strains, results don't lie, plus the fact that lumens don't add up 140,000 lumens penetrates deeper into the canopy than 90,000. I have two homies that have switched from using (2) 600s to 1k(s) because I now yield more than both of them, I am using a light rail but I grow organically and they use Advanced Nutes Connie line. 600 = nice yields/ 1k = bigger yields!!

Lets not forget the brand of lamp your using. There are many different variables that could play into this. I actually use 1k's for my 2bdrm apt grow and 600's for my 1bdrm apt grow. Now the 1'ks actually perform the same as the little ones, it really comes down to HOW your using them; distance, positioning for maximum overlapping, air cooled or not, etc. The brand of the lamp is what your going to best determine the power your going to receive. I believe they can be equal, just depends on the grower and the ability he/she has to maximize!

 

[GanjaDin]

I've got to weigh in on this one, I love the 600. I use a GE Lucalox with an Adjust a wing reflector. I have been using it in a perpetual grow for about a year and a half. The yeilds are not spectacular but they are ok. I figure with the amount of money I am saving on electrical cost due to exhausting and refreshing / cooling I'm better off in the long run. I use 2 8" duct fans to exhaust and 2 6" fans for intake, 2 recirculating fans for mixing and 1 tiny fan to cool the ballast. My room is 5' wide by 8' deep and 8' tall. I started off with 4 plants 24" tall and was yeilding 1 to 1 1/2 oz per plant. I have 8 plants under it now, each about 30 to 36" tall in 3 gal pots, 5 different strains. I do a version of lollypopping and some bending on the sativa doms and am yeilding about an ounce per plan and harvest about every 10 dayt. I'm not in it to break the bank so the outcome is acceptable.
I would like to try another 600 watter overlapping but space and money say no right now. In my humble opinion the 600 is a great tool.

 

[Batcave]

600 watts can easily cover the 5'x5' area if you use the right reflector!

I go by:

2x2 = 250w hps
3x3 = 400w
4x4 = 600w
5x5 = 1000w

Now on the other hand if you have your setup dialed in with area and ventilation you can up the power. I ran a 600w in my 4x4 for a year and did good, but swapped to a 1000w in the same tent and now have great yields. The 600w is good for the 4x4 but I just didnt feel I got full amount of light to the edges of the 4x4. My buddy runs a 1000w dimmable dimmed to 750w in his 4x4 and does as well as I do from my 1000w at 100%. 750w is just about ideal for a 4x4 tent.

 

[eye 1 der]

4 x 4 tents....4 x 250 w every time, double the light of a single 1000w and half the heat...

 

[Batcave]

4 x 4 tents....4 x 250 w every time, double the light of a single 1000w and half the heat...

Double the light and half the heat??? How so?

 

[eye 1 der]

inverse square law...I = 1/d2
light intensity (I) =1 divided by the distance from the light source(lamp) squared (d2)

 

[growcraze]

eye 1 der got real technical on us with that one... lol What's the recommendation for a 400W area 2x6?

 

[Tom Bombodil]

I think ganjadin said a 600w makes a great tool... My opinion is that there's the real answer, all this equipment is just that, tools in a toolbox. And like someone else said, it depends on so many variables. Certainly some tools fit some situation better, and its the creative mind that applies those tools and materials that maximizes any particular set of circumstances. After all, that's the fun part right? That's the REAL therapy, the therapy of creation!

But, for the hypothetical question posed at the onset, holding all variables constant (i.e., you can cool the suckers down enough and not counting the logistical nightmare of venting 100 lights as opposed to 60), my guess would be the 600's, cause you can get them closer, cause like homie said, the upside -down square and all that science and stuff.

 

[Tom Bombodil]

Although i don't necessarily agree that light would be doubled whilst t heat, halved. The light would be closer, but would it be THAT much closer to increase by a factor of 2? I'd have to do the math i guess. And that closer proximity would result in proportionally MORE heat reaching the plant, correct? So despite the smaller fixture producing less heat, the fixture is nevertheless contributing more heat energy to the plant, mitigating or potentially cancelling the proposed heat benefit, would it not? So maybe the heat it reduced by some fraction, and light intensity is increased by some fraction, still netting a positive benefit in both directions?

 

[eye 1 der]

considering all costs.initial setup costs, running costs, cost of adequate ventilation system required.. then yes in a large garden 600w fittings would be most economical. spaced at 1 per meter square but...
consider this example in a 3 x .5 m garden

3 x 600 in standard reflectors
mounting height above the plant canopy to prevent scorching and complete coverage would be 45cm(initial lumen(L) output is 92000 lumens

(all lamp outputs are measured at 300mm from the lamp) so at 45cm the lumens intensity would be 40903 lumens or 44.46% of the initial intensity we can look at it this way:-

area covered by each light =0.5 x 1.0m(40,903 L)
power consumed =1.8kw
total lumen output =122.709 L per 1.5m2
this equates to 68,172 lumens per kw of power or 81.81 L per cm2
three light sources offering moderate illumination consistency

or we may use 4 x 400w mounted 40cm above the canopy to ensure adequate even coverage of light to all areas lumen output would be 31,500 L

area covered by each light =0.5 x 0.75(31,500 L)
power used =1.6 kw
total lumen output =125,000 per 1.5m2
this equates to =78,750 L per kw or 84 L per cm 2
4 light sources offering acceptable illumination consistency

in order to get the most advantageous illumination in our area of 1.5 x 1.0m we should use 6 x 250w lights, while this approach requires the most lights(consider the lower price of 250w units and the much smaller ventilation system required due to lower heat output the initial set up cost will be just a small amount greater than the 600w setup)we can demonstrate this to be the most economical and efficient way of lighting our example area.
the lights are mounted 30cm above the canopy giving the maximum output of light to the plants and the most even light distribution from 6 sources

area covered by each light =0.5 x 0.5(33,000 L)
power used =1.5 kw
total lumen output =198,000 per 1.5m2
this equates to =132,050 L per kw or 132 L per cm 2
six points of light giving excellent consistency of illumination

looking at the above analysis we cab see that the 6 x 250w array works out the most effective, efficient,and cost effective lighting solution for this area. the area covered a the optimum mounting height(30cm) is 50 x 50 cm per lamp

oh..initial lumen output of most 1000w lamps is around 132000 L.

 

[Tom Bombodil]

Ok I said I would need to do the math. So here it is:

If intensity equals initial lumens divided by the squared distance, then the ratio of the intensity of a second light to the first is the PRODUCT of the (RATIO between the initial lumens of the seond light to the first) and (the RATIO between the distance of the first light to the second). So if light1 has intensity1=L1/d1^2, and intensity of light2=L2/d2^2, then intensity2/intensity1 = [L2/L1]x[d1^2/d2^2].

You could use this formula to plug in different number and solve for the unknown of interest. For example, assuming a 1000w with 120,000 lumens at 1" and a 400w with 40,000 lumens at 1", then at 12" the 1000w puts out 833lumens/in^2, and the 400w puts out 277lumens/in^2 at the same distance. If you moved the 400w to half the distance, 6", then it result in 1,111lumens/in^2, or a ratio of 1.33 the output of the 1000w at 12"

If you want to know how close to put the second light so it would match the intensity of the first, then substitute 1 for the ratio "e2/e1" and 12" for d1 and solve for d2. In thatthat case, d2 would equal 6.29, so in order for a 40,000 lumen light to be the same intensity as a 1000w light at 12"you would need to place the 400w approx 6.29" from the plants.

 

[eye 1 der]

tom, at optimum spacing the light intensity at the mid point would not be the product, how can the intensity of the light falling on the area be increased by adding an identical source at the same distance in the opposite direction!

 

[Tom Bombodil]

I think there may be some misunderstanding. To the question posed in the thread, is it more efficient to have more lamps of lesser wattage, one response invoked the inverse square law. I agree that because of this principle, I think more lights of lesser wattage would be preferrable. However, I wasn't sure if one could get them close enough to double or triple, etc the intensity. I got curious as to how much the light intensity would be increased as a function of some lesser wattage at a lesser distance, and how those variables would affect one another. The way of comparing the two intensities falling on a given surface seemed a natural comparison, like being able to answer the question "how close would i need to place a light of this wattage to have the same intensity of a light with that wattage?" Or "how close would i have to place this light with this wattage to double the intensity of a light with that wattage?", etc.

It's a little less computation to go ahead and calculate the ratio of thr second light to the first light (l2/l1 in the equation above), that way you can substitute that number and figure any combination of distance and final intensity you are interested in. For the example above, that would be 40,000/120,000 = 1/3. you can plug 1/3 in for l1/l2 and then you know that the ratio of the intensity of the second light to the first will be be t