...All Things Vero...

Would you consider buying a VERO after reading through some of the posts?


  • Total voters
    357

bicit

Well-Known Member
My mates just given me one of them heatsinks. Will be ordering the veros next week. Might stick one on eBay as I won't need it
Just had to go and buy a new airpump as well -_-
Everything always needs ordering together lol
Any preference of thermal paste? On PCs I always used Arctic silver 5. Anything new and superior worth getting?
The quality of the thermal paste has been found to be of lesser importance than the application of it.

I like arctic silver 5 because in a warm room it's very easy to get spread super thin and it's cheaper. Prolimatch PK-3 is one of the best AFAIK, but it's very thick and hard to work with IMO. I've got lights with both and I couldn't tell you which was which anymore.
 

UKpeanuts

Well-Known Member
sorry dude, KB are a company based in china, a few people from here have got cree through them, bit more choice available, slightly better price than RS. delviery might be steep if you only need 2. I'll pm you his emauil
 

UKpeanuts

Well-Known Member
Well first off I need to apologize for the typos I can no longer edit....
Well I'm not as far along in my mathematics as some on this board. So forgive any mistakes, but my line of thinking is that we can use trig to calculate the area that a cob will cover at a given height. Then we can calculate the PPFD of the individual cob over that area. Then by plotting the cobs position in the grow room we can calculate how much the light footprints of each cob will overlap and hopefully predict the increased intensity at a given position within the grow space.

I'm no mathematician so it may take me a minute to properly explain my line of thought.

How uniform is the light output from a single cob with a reflector? Is it relatively even?
Could this be achieved easier with a model rather than a calc? If you know the emitter angle of the cob/refector combo and you know you grow room dimensions, could you know not draw a model, estimate area of coverage for 1 cob, covert to m2. divide by umol output.


Ahhh shit I'm talking bollocks
 

bicit

Well-Known Member
Could this be achieved easier with a model rather than a calc? If you know the emitter angle of the cob/refector combo and you know you grow room dimensions, could you know not draw a model, estimate area of coverage for 1 cob, covert to m2. divide by umol output.


Ahhh shit I'm talking bollocks
More or less that's basically what I'm proposing. The tricky part is accounting for the overlapping footprint between two cobs and how that intensity scales.
 

AquariusPanta

Well-Known Member
My mates just given me one of them heatsinks. Will be ordering the veros next week. Might stick one on eBay as I won't need it
Just had to go and buy a new airpump as well -_-
Everything always needs ordering together lol
Any preference of thermal paste? On PCs I always used Arctic silver 5. Anything new and superior worth getting?
MX-2 from arctic is affordable and effective (imo).

Well first off I need to apologize for the typos I can no longer edit....






Well I'm not as far along in my mathematics as some on this board. So forgive any mistakes, but my line of thinking is that we can use trig to calculate the area that a cob will cover at a given height. Then we can calculate the PPFD of the individual cob over that area. Then by plotting the cobs position in the grow room we can calculate how much the light footprints of each cob will overlap and hopefully predict the increased intensity at a given position within the grow space.

I'm no mathematician so it may take me a minute to properly explain my line of thought.

How uniform is the light output from a single cob with a reflector? Is it relatively even?
I believe we covered this area awhile back, if my memory serves me correctly, by using law of sine to find the distance at which light was to be displaced.

It's fairly uniform/circular. Your approach would be worth testing.

Curious as to why you didn't spread the outer COBs further out to more evenly distribute the light? Might get you closer to that theoretical 800 PPFD.
I'll test pushing the outer COBS further out and see if that doesn't make a difference but the point that I was making was that it wasn't only until after reflectors were attached and the fixture was lowered to 12" that I came close to the calculation. This worries me because without reflectors (or lenses), some of us may not be receiving the photons that we are drafting up on paper. There are a few other factors that I will alter tomorrow when I return to the 'lab', to see if there's more to the setup.
 

rosswaa

Well-Known Member
Nice one guys, I'll just order 3 off digikey and have either one as backup or sell it on fleabay or something. Should be quicker than China and a decent price tbh

I'll give the Intel heatsinks a go and if they need upgrading then I'll swap them out seeing as they're cheap as chips :)

Hopefully they're not to intense at 1500ma on the heatsinks or the plant
 

Rahz

Well-Known Member
As far as PPFD not adding up, it should be considered that the arrays aren't going to be running at 25C so the actual PPFD will be slightly lower than the math will indicate.

As for reflectors, I think they do offer some improvement and are easier to wipe clean than a tent, but as a guess I'll bring up the possibility that there is a drop off on the measuring device in regard to the angle of approach? Reflectors would correct the light into more of a direct path which might make more difference to a reading than to a plant.
 

Greengenes707

Well-Known Member
I'm really ripped now so take this as a disclaimer. I'm only sharing here to help the discussion and possibly some other thoughts. Much of this are my own GG theories that I would not go trying to publish, so don't take it as dictionary...but I have developed a good understanding of light, intensity, and plants over the years...so you might have an apiphany or two if you can hang on for the ride.

Lab measurements and calculations vs field measurements and reality...I've barked up this tree a few times.

I don't even know where to begin...get ready for a long one, I'm super irie...

We need to actually understand what they are expressing in order to use them in an effective manner. And also where/how/why they were developed.
They were developed based on field or outdoor situations where the sun, for all intents and purposes, is like a blanket of even light that isn't affected by source distance. You get 2000µmols on the ground or 100ft in the air...move it left 5 miles or right 5 miles...still 2000µmols. And the algorithms used in PAR measurements are based off this. And so is "avg PPFD". This is also similar to what a big facility with many lights creates...even spread is created, and variations are minimized over a majority of the crop. For single lights and/or smaller grows, you need to think deeper about what is going on.

What is avg PPFD, the calculation?
It's the PPF(total possible output) of a lamp in a square meter. And is very easy when working with a m^2. But, what if the area is not a m^2?? Then you would confine(divide) the light into the area you are working with(3x3=.836m)...then projects/predicts the rest of the square meter with imaginary light...because it's based off being outside where no fall of would happen.

Ex.
So 720 ppf...confined to a 3x3...is 720ppf÷.836meter^2=~861ppfd. So that means that if that lights' intensity within the 3x3 were to continue out to a full m^2, as it would outdoors, it would be putting out 861µmols over that m^2. But it doesn't...no matter what, there is only 720µmols in the actual grow area for the plants to use...so remember and think about that. It is just LIKE growing in a m^2 area with 861ppf over it.

At first seems off...but when you have that, and can see the next step/style, you really can know the possibilities of a lamp.

What is a PAR meter showing?
A PAR meter is an instantaneous field measurement of µmols of photons per second in a m^2(square meter). It is a photodiode that creates a current. An algorithm is applied to come up with a figure for a m^2. But's it's based on the intensity at the sensor ...in that single point in space where the sensor is held...then extrapolated based on outside light behavior using the algorithm, to a figure expressing how many µmols of photons per second are falling in a m^2 based on that sensor as if the whole m^2 was covered in that intensity.

Ex.
So an 1000µmols dead center reading is saying...that if the sensor(1sqin in size) was the center of a m2, and the sensor is getting .645µmols...it assumes every 1550sqin of the m2 is roughly the same, and would give you 1000µmols/sec/m2. (.645 x 1550=~1000)
IMG_6543.jpg

Lets keep the whole sensor is the center of a m^2 thing in our heads...
Now we move to the outer edges ofthe canopy. We get a low reading with our PAR meters...say 400µmols. Well that means you imagine the m2 around the sensor...half the m2 is now off of the canopy by almost 20" and outside the reach of many of the photons that it was sensing while in the center.
IMG_6544.jpg
And then the corners are even worse...
IMG_6545.jpg

Now tangent to the concept of 700-1000µmols per m2 is ideal. This is super easy to relate and cross check with "avg ppfd"...it is either in there or it isn't. But I believe this gets misunderstood when using instantaneous meters(PAR meters) because of what I just showed above. And what I am about to say almost the reverse of confining ppf to make "avg ppfd". Take pic#2...400µmols on the edge...but half of the expressed meter is outside the light...well what if we confine it to the lighted/canopy area...what would it show.? 800µmols is my theory and what I have seen worth of growth in my gardens over the years.

Time to recharge
 

testiclees

Well-Known Member
ymmv, should have put that little disclaimer in there. I do belive it could take up too 30 days.

So long as my package get's here before the end of the germination period for the solo cup comp I'll be happy :P:D
Thanks. I wasnt counting on a quick delivery and i dont need the added light right now. Its more that ive got everything else in place and im stoked to see if my first diy works propely.

Thanks for the caveat.
 

AquariusPanta

Well-Known Member
Ex.
So 720 ppf...confined to a 3x3...is 720ppf÷.836meter^2=~861ppfd. So that means that if that lights' intensity within the 3x3 were to continue out to a full m^2, as it would outdoors, it would be putting out 861µmols over that m^2. But it doesn't...no matter what, there is only 720µmols in the actual grow area for the plants to use...so remember and think about that. It is just LIKE growing in a m^2 area with 861ppf over it.

At first seems off...but when you have that, and can see the next step/style, you really can know the possibilities of a lamp.
In my eyes, this here is legendary input on your part. Everything is organized and laid out in a easy to follow fashion - I'm impressed! :clap:

I did get caught up on your math though, as it is late here and I've been working over time.

If a lamp were producing 720 PPF and were positioned in a square meter space/tent, then it would produce 720 PPFD. If you placed the same lamp in a two square meter space/tent, the PPFD would be 360 PPFD (720PPF / 2 m^2).

Your saying that if I put the same lamp in a .836 square meter space/tent, the PPFD is going to be 720PPFD? In other words, PPFD can never exceed PPF in value?

Edit:

As far as the PAR sensor goes, it's reading are in umol per square meter, every second. Taking in your concepts and beliefs, I question whether this sensor is any good in areas less than a square meter in size and even then, the corners and sides wouldn't be correctly measured unless the tent size was bigger but then you run into walls, which would need to be corrected in order to account for the square area being less than a square meter.

Going back to your original break down of the origin of the PPFD calculation, the Sun faces no walls. Even if I'm totally delirious, this is one of the best mind-scratchers I've had in awhile.
 
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Rahz

Well-Known Member
The few vids I've seen show the readings being taken in a tent, though I'm not sure if there is a more proper way to do it that's considered standard. If readings are to be taken outside a tent, I think a large room would be fine and produce reasonably accurate measurements. Standard or not, such a method would be useful when determining ideal emitter spacing based on a desired output density since a tent wall will artificially inflate the numbers. The COBs would end up being placed too far apart. OTOH, if a particular light of whatever size is going to be used in a particular tent, then the tent readings would be the most useful whether it's a meter or a 2x2. It just depends on what one is trying to accomplish with a product - being suggested for a warehouse where most lamps won't see the walls, or suggested for a tent. Whether there's a difference between honest and useful is a matter for discussion.

Side note: When taking readings outside center, doesn't it make sense to angle the meter towards the emitter? That's what leaves do.

As far as PPFD, it's estimated density in a given area as far as I understand it. If the area shrinks the density goes up. PPFD is valid when given along with the recommended growing area. IE- a 600 watt lamp that provides 800 PPFD in a 4x4 would provide 3200 PPFD in a 2x2 since it's based on the density of light in a given area.
 

AquariusPanta

Well-Known Member
The thing about the meter/sensor is that it's typically unobstructed, where as canopies are commonly uneven in height, width, and thickness, all of which leads to shadow. Leaves seem to reach for the light but if their behind a cola or something, then they aren't getting many photons. This can be remedied by introducing multiple lights from multiple points but is somewhat cost intensive in some scenarios.

@Rahz

My understanding is quite similar, if not identical, in respect to how PPFD works although it seems GG has a different take on this and he's been measuring light for longer than I've been a COB-DIY guy; I'm interested in what more he has to say on this subject.
 

SupraSPL

Well-Known Member
In my eyes, this here is legendary input on your part. Everything is organized and laid out in a easy to follow fashion - I'm impressed! :clap:

I did get caught up on your math though, as it is late here and I've been working over time.

If a lamp were producing 720 PPF and were positioned in a square meter space/tent, then it would produce 720 PPFD. If you placed the same lamp in a two square meter space/tent, the PPFD would be 360 PPFD (720PPF / 2 m^2).

Your saying that if I put the same lamp in a .836 square meter space/tent, the PPFD is going to be 720PPFD? In other words, PPFD can never exceed PPF in value?
If you focused 720PPF into a 3X3 tent (.836m²), it would measure 861 PPFD average, assuming all the light emitted reached the canopy. There would be some losses from reflector/lens/walls.

To exaggerate that example, if you put a 720PPF lamp in a 2X2 (.37m²) tent, the canopy would measure 1946 PPFD average (minus reflection losses)
 

AquariusPanta

Well-Known Member
If you focused 720PPF into a 3X3 tent (.836m²), it would measure 861 PPFD average, assuming all the light emitted reached the canopy. There would be some losses from reflector/lens/walls.

To exaggerate that example, if you put a 720PPF lamp in a 2X2 (.37m²) tent, the canopy would measure 1946 PPFD average (minus reflection losses)
Ok, thank you for reinforcing my previous understanding of this. Does this mean I misinterpreted GG's example?
 

SupraSPL

Well-Known Member
This part is correct:

If a lamp were producing 720 PPF and were positioned in a square meter space/tent, then it would produce 720 PPFD. If you placed the same lamp in a two square meter space/tent, the PPFD would be 360 PPFD (720PPF / 2 m^2).
This part is incorrect, your PAR meter would measure 861 PPFD:

Your saying that if I put the same lamp in a .836 square meter space/tent, the PPFD is going to be 720PPFD? In other words, PPFD can never exceed PPF in value?
 

Greengenes707

Well-Known Member
Ok, thank you for reinforcing my previous understanding of this. Does this mean I misinterpreted GG's example?
Yes misinterpreting. It's not a different take. It is just a look at what is actually going on to end up with those numbers and what they mean. In all those examples(mine and supra)...there is one constant...720PPF...and in the end, no matter how much space you either spread or confine it into...there is 720µmols of photons per second for the plants to work with.

The increased density is saying that if it was a full m2 at the density you created by confining(2x2, .37m2), it would have 1946µmols over that imaginary m2.
Lets go the otherway and look at a 4x4(1.49m2) with the same 720ppf. That's 720ppf÷1.49=483µmols ppfd. This is saying that is you spread the 720 over 1.49m...then imagine cookie cuttering a m2 out of it, there would be 483µmols in the m2. While the remaining 237mols is on the other .49m2 that wasn't cookie cuttered.
 

OneHitDone

Well-Known Member
So, ppfd calculations will get us the total photons in a space. Can't a foot candle or lux meter be used to just check relative intensity across that space?
Sounds like all of the par graph's we have seen on all of the websites even using spectroradiometers pretty much just got tossed in the trash? - unless those are calulated by the square inch rather than meter
 
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