Why do people use watts?

[Lockedin]

Like all of us, I'm always researching new equipment to make my grows better.
But bad information makes decision making difficult at best.

Why do we use watts INCORRECTLY to compare lights?
Watts are a measure of throughput to power a device (for us a light) to a certain output.
WATTS = how fast the meter spins and how big your elec. bill wil be.
PPFD or PAR seem like they would be much better indicators.

Here's an analog argument:

2 cars - each has an OUTPUT of 300hp (horsepower for non-car ppl)
Car 1 is a 1969 Chevelle with a 350 cubic inch gas V-8 motor
Car 2 is a 2007 Lotus Elise with a 1.8 litre supercharged I-4 engine.

Again ------------------------------------ Both cars output --- 300 hp each. BUT:
Car 1 has a 1 gallon THROUGHPUT of fuel every 12 miles --- 12 mpg
Car 2 has a 1 gallon THROUGHPUT of fuel every 31 miles --- 31 mpg

It makes no sense to compare the power output of a car by the mileage does it?
So why do people compare light power by how much it costs to run?

IMHO - watts at the wall - or the fixture, are a really poor indicator of power levels reaching the canopy.

 

[spek9]

Way back in the day, long, long, long before LED and other alternate lighting choices, you really only had one... High Intensity Discharge (HID). Lamp comparison was done by the power draw (ie. watts drawn) by the lamp.

It has stuck ever since. There's nothing incorrect about it, it just doesn't tell you a whole lot other than power consumption.

 

[Lockedin]

Exactly.

But, while power consumption is defo a consideration (esp. in SoCal), Watts are not as big of a consideration as PPFD / PAR.
IMHO that's a lot of the confusion with LEDs - they are very efficient, so comparing the output of HID & LED by consumption is almost useless.

 

[printer]

Lights do not generally have PAR ratings on them. Also it is not bad to know how many 'real' wats you are pulling from the wall as not to trip breakers.

 

[Lockedin]

btw - motivation for this is that I'm planning to convert a spare room in the house and use my 4x4 for veg.

LED will keep elec. and heat down, but figuring coverage / consumption is frustrating. HID is a very well known appliance.
*** dishonest advertising doesn't help - my rectangular LED shows a square coverage map on its amazon page.....

 

[printer]

Maybe others would know what you have in terms of performance if you showed the item. I used regular screw in bulbs, no idea on the plant friendly output. I just got a cheap meter that has pH, light intensity and moisture setting on it. I stuck it outside in the sun and noted where the needle ended up, adjusted the height of my lights to correspond to it. The plants seem to be doing well with the light level.

 

[Lockedin]

Lights do not generally have PAR ratings on them. Also it is not bad to know how many 'real' wats you are pulling from the wall as not to trip breakers.

Consumption is a big part of the equation when designing for sure - that and heat are why I have 3 LED in my tent instead of 1 HID!

I worked the studios most of my career, lighting guys always spoke in lux / lumens (what cameras see), and then discussed the watts needed.

It seems like a lot of discussions get lost in watts - how much power is required by a certain fixture - when trying to figure out coverage / ppfd

 

[printer]

Isn't PAR not some form of watts per meter? I don't remember. I looked into getting a cheap PAR meter but the cost was out of my price range.

 

[Lockedin]

Maybe others would know what you have in terms of performance if you showed the item.

Trying to plan for a 10' X 15' X 8' room.
interior will most likely include my current 4x4 tent for vegging.

IMHO the LED industry needs to come up with a real way to do apples to apples comparisons for output and coverage.

My 4x4 currently has (2) "300w" burples and a Viparspectra VA-1200 (265w).

I'm content with how they work, but as I upgrade, I'd like to do it more intelligently.

 

[Lockedin]

I think DIY will be the only way to do this accurately --- square shaped coverage maps for rectangular shaped fixtures and advertisements saying "replaces a million watt growlight / tanningbed" just doesn't cut it.

 

[nc208]

Isn't PAR not some form of watts per meter? I don't remember. I looked into getting a cheap PAR meter but the cost was out of my price range.

No.
Photosynthetically active radiation (PAR) is light of wavelengths 400-700 nm and is the portion of the light spectrum utilised by plants for photosynthesis.

 

[Lockedin]

No.
Photosynthetically active radiation (PAR) is light of wavelengths 400-700 nm and is the portion of the light spectrum utilised by plants for photosynthesis.

Makes sense - measuring the actual radiation from a light rather than expressing it in visible spectrum (lumens, lux, etc)

PAR - Defo a useful measurement; especially for the burple, UV & IR fixtures that can appear dimmer.
Maybe PPFD then? Just overall photons hitting the canopy?

 

[Airwalker16]

WATTS = how fast the meter spins and how big your elec. bill wil be.

This is why.

 

[Lockedin]

This is why.

Agreed.

Hi @Airwalker16 - I was hoping you'd weigh in. You've been at this longer than I have!

But what if how much I pay at the pump isn't as much concern as how much output I have?

Still staying with internal combustion:

A studio grade diesel generator is usually a diesel engine with an output of 480amp / 3 phase power (and matched ac waveforms - CLEAN power)

Ask anybody in the lighting dept. how many HP the genny makes. Answer ---------- "No idea."
Horsepower is an nearly meaningless number here.

The people using the generator are concerned about 2 numbers:
GPH - gallons per hour - or consumption - watts to us and a big deciding factor.
AMPs - how much power is available for use.

So yes, watts (true watts, not potential) are a major number to factor into grow design - they represent the monthly cost of operation per unit.
But there does not seem to be any true representation of photon availability or losses - some kind of real measurement is needed to get a real calculation of how many units and what type will be appropriate for a given room or space.

 

[Fister3id]

You should look into quantum boards. HLG has a really good one but they're kind of pricey.

 

[wietefras]

Well it's sort of in direct relation by factor of the efficacy.

Say you aim for a light production of 800umol/s per m2 and the light you are thinking about has an efficacy of 2.5 umol/s/W then you'd need 320W per m2.

If you know you need 800umol/s per m2 then you still don't know what driver to buy. If you know you need 320W then you know what drivers to look for.

 

[Lockedin]

I like HLG's spec sheets - total output in PPF, Efficiency and a realistic "footprint". They're on my short list if money is good when I go to upgrade.

The total PPF output is a more useful number to me.
I want to know what a light will do on the canopy before I wonder how it will affect the bill (within reason).
eg: how does the output of a HLG-650r compare with a 600w HID?

If watts are the main consideration, it doesn't matter - they are roughly equal.
What is the PPF on the canopy (the output) side?

 

[Lockedin]

Well it's sort of in direct relation by factor of the efficacy.

Say you aim for a light production of 800umol/s per m2 and the light you are thinking about has an efficacy of 2.5 umol/s/W then you'd need 320W per m2.

If you know you need 800umol/s per m2 then you still don't know what driver to buy. If you know you need 320W then you know what drivers to look for.

Thanks - that makes sense and will help a lot in designing a suitable LED rig.
But a 320W LED fixture will have a different output than a 315W HID - they process electricity into light in different fashions, so wattage comparisons such as "replaces a 600w HID" are misleading.

Guess it just irks me when marketing (my lights included) inflate numbers and confuse decision making.

I would rather make a decision based on umol/s per m2 than "replaces a XXXw HID" or forum posts, "Duuuude, LED sucx cumpred to HPS and your stupid fer asking!

The above has certainly given me better information to run with DIY!

 

[PJ Diaz]

For one thing it's much easier for any common grower to measure wattage consumption using inexpensive tools than it is to measure PAR or ppfd, which requires expensive equipment. By this point, enough research has been done to know the general range of wattage needed for HID or LED lighting per sq foot to know approx how many watts you need to adequately fill a space. Another factor is that you really can't trust the ppfd numbers put out by many manufactures. A great example of this was shown just last night on the GML show, where they tested a Chill LED board themselves only to discover that Chill was putting out false claims (they apparently only tested one strip but not the fixture, and didn't account for driver loss). For $15 you can get a tool that will tell you real wattage, whereas you will need to spend thousands on a good test of light output.

 

[printer]

No.
Photosynthetically active radiation (PAR) is light of wavelengths 400-700 nm and is the portion of the light spectrum utilised by plants for photosynthesis.

For some reason I seem to feel light hitting a surface was in watts. I found this,

Can I convert PAR (photo active radiation) value of micro mole/M2/S to Solar radiation in Watt/m2?

I'm working with a quantam meter that gives PAR (photo active radiation) value of micro mole/M2/s. I'm using this for measuring sunlight PAR. Is there anyway I can covert this to W/m2? I've seen some approximate conversion like 1800 micro mole/M2/s equals 427 W/m2. In another lecture its 2200 micro mole/M2/S equals to 500 W/m2. Can anyone provide references?

Global Solar Irradiance is usually measured in W/m2, by instruments called Pyranometers, in the waveband from 1100 nm (IR) to 400 nm (UV).
PAR can be also measured in W/m2, but is usually expressed in μmole.m2/s (moles of photons), which is called photosynthetic photon flux density (PPFD), and is not a SI unit. However, PAR and PPFD consider only the waveband from 400 nm to 700 nm.
The conversion is actually not exact and depends on the instantaneous spectrum of the radiation, which means, on the instantaneous atmospheric conditions.
Some researchers use the approximation 1 W/m2 ≈ 4.6 μmole.m2/s.

University of Georgia
First of all, μmole.m2/s is definitely an SI unit. The approximation 1 W/m2 ≈ 4.6 μmole.m2/s comes from the Plant Growth Chamber Handbook (chapter 1, radiation; https://www.controlledenvironments.org/wp-content/uploads/sites/6/2017/06/Ch01.pdf). Note that the value of 4.57 converts Watts/m2 to μmole.m2/s, assuming that the W/m2 is for radiation from 400 to 700 nm. However, I don't know that anyone ever measures solar radiation in W/m2 in the 400 - 700 nm ranhe. Pyranometers measure TOTAL solar radiation. Since only about 45% of the energy of solar radiation is actually in the 400 - 700 nm range. the conversion of TOTAL solar radiation to PAR is ~2.1, rather than 4.57. As mentioned by others, that is an approximation, but a pretty good one.

Wonder if you can then rate the leds, watts out (light) / watts in (electrical)

Led me to this.

https://www.researchgate.net/profile/Mariana_Reis8/publication/339896036_Conversion_factors_and_general_equations_applied_in_agricultural_and_forest_meteorology/links/5e6b7bd992851c6ba7fdedd3/Conversion-factors-and-general-equations-applied-in-agricultural-and-forest-meteorology.pdf?origin=publication_detail

Not that useful for us but I just needed to know.