REALSTYLES
Well-Known Member
Then I've seen this .0920 instead of .09290304 and figuring out if I should use ft2 or m2
Hydrofarm par meter
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Abiqua
Well-Known Member
No, I have discussed this way and have dropped lots of links for those concepts....but I am on to a new one using the reverse bias of LED's, so using them as an input...instead of an output....
Forest G. Mims seems to have pioneered this method...but still researching...
Although in the end, I will probably need to calibrate with an outside meter[like the apogee], my ideal meter will still be the Li-Cor, which has more of a linear response than the Apogee... ...but I think there is a work around with a spectrometer.....also an issue on the reliability of diodes that truly radiate their printed nm....
Abiqua
Well-Known Member
m^2
the dopest
Well-Known Member
Using that formula, I get 557.41 Lumens, that's even less than the online calculator. Does that sound right? CXA3070 AD bin on a 1500ma driver at 8" distance in a 2.4 square foot cabinet. I would think it would have been a shit load more.
EDITED, if I use sq meters instead of sq ft, it's even less at 51.78 Lumens. What is wrong with my math here? I thought this COB should be putting out thousands of Lumens instead of 50-500?
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Abiqua
Well-Known Member
Link to the calculator? Was it this one? http://www.rapidtables.com/calc/light/lux-to-lumen-calculator.htm
More than likely your meter is calibrated to use m^2 as its defined area, NOT the size of your grow area....just use the meter^2 calcs below, and see if your answer is different....or just use the link above and enter 1m for surface area and your lux amount....don't mess with spherical radius, it will calculate when you enter SA........
Code:
Lux to lumens calculation formula
Calculation with area in square feet
The luminous flux ΦV in lumens (lm) is equal to 0.09290304 times the illuminance Ev in lux (lx) times the surface area A in square feet (ft2):
ΦV(lm) = 0.09290304× Ev(lx)× A(ft2)
The luminous flux ΦV in lumens (lm) is equal to 0.09290304 times the illuminance Ev in lux (lx) times 4 times pi times the squared sphere radius r in feet (ft):
ΦV(lm) = 0.09290304× Ev(lx)×4·π·r(ft)2
Calculation with area in square meters
The luminous flux ΦV in lumens (lm) is equal to the illuminance Ev in lux (lx) times the surface area A in square meters (m2):
ΦV(lm) = Ev(lx)× A(m2)
The luminous flux ΦV in lumens (lm) is equal to the illuminance Ev in lux (lx) times 4 times pi times the squared sphere radius r in meters (m):
ΦV(lm) = Ev(lx)×4·π·r 2(m2)the dopest
Well-Known Member
This is the link to the online calculator I used.
http://www.ledrise.com/shop_content.php?coID=19
I see the problem here, I was multiplying the LUX measurement by 10 but the meter says it should be by 100. My mistake, that puts it a lot closer to where it should be. Sorry for te confusion guys and thanks a million for the help!
http://www.ledrise.com/shop_content.php?coID=19
I see the problem here, I was multiplying the LUX measurement by 10 but the meter says it should be by 100. My mistake, that puts it a lot closer to where it should be. Sorry for te confusion guys and thanks a million for the help!
Abiqua
Well-Known Member
the reading was 55K lumens right? sounds about right if you were 8" directly under @ 1.5 amps....about 1000 umol/s.....
the dopest
Well-Known Member
It measures about 50,000 LUX at 8", the online calculator I linked to says that's about 6485.86 Lumens.
Using the formula above it comes to about 1035 Lumens...okay now I'm confused again.
I have about 2.4 square foot.
Using the formula above it comes to about 1035 Lumens...okay now I'm confused again.
I have about 2.4 square foot.
Abiqua
Well-Known Member
Yeah, I get readings all over the place...and I am trying to rack my brain, but I don't see how the height is introduced?...
I do know this..and it states it in your calculator link too....
The luminous flux ΦV in lumens (lm) is equal to the illuminance Ev in lux (lx) times the surface area A in square meters (m2):
ΦV(lm) = Ev(lx)× A(m2)
The surface area of your meter is calcuated to 1m^2, so in the above equation...1 lux @ 1m^2 is equal to 1 lumen....so 55K lux/ lumens you are measuring is for that one spot only...
Disregard your cab size...what you could do... is take readings at 12"....start dead center under your light source and them move away on the floor every 8-12" and take a measurement but still keep yourself 12" from the light....you can get an idea of how bright the light is over various spots in the cabinet, then you can determine what your 2.5 sq ft space is....
Here are more conversions to umol/s
http://www.apogeeinstruments.com/conversion-ppf-to-lux/
OneHitDone
Well-Known Member
Have any of you diy cob guys thought about angling them like the led's are on the inda-gro pontoon?
I wonder if coverage/penetration could be improved even further with something like that?
I wonder if coverage/penetration could be improved even further with something like that?
the dopest
Well-Known Member
I gave the height just in case anyone with a similar setup has already been through this could compare numbers. Also, how could I be getting 55K lumens from a LED that is supposed to produce about 6-9k? I know this is not an accurate way to measure, just trying to get a ballpark. I mainly use the meter to make sure I'm getting decent canopy coverage at light height from canopy. With my old HPS, if I lowered it too close, the sides of my cabinet were shaded and when I used dual HPS, I could get even light spread making sure there was not too much overlap.
AquariusPanta
Well-Known Member
Thank you for sharing your experience and knowledge over this subject, GreenGenes; this was a really great post.
I would like to utilize their meter but am a little concerned about the deep blue and red drop offs, as I'm led to believe that the sensor only reads 410nm to 650nm. I'm looking for clarification on whether or not this lack of reading can be accounted for.
You wrote that one could give Apogee the spectral power distribution (SPD) curves/chart and that they would calibrate something perfectly. Is that something to do with the actual handheld meter that they offer at ~$300 w/sensor or do they give the user a correction factor to solve for a better measurement through the use of the sensor?
heckler73
Well-Known Member
No, I have discussed this way and have dropped lots of links for those concepts....but I am on to a new one using the reverse bias of LED's, so using them as an input...instead of an output....
Forrest G. Mims seems to have pioneered this method...but still researching...
I guess I'm reinventing the wheel?It's not reverse-biased, BTW. It's still forward biased
