Math behind

[NoFucks2Give]

Have you tried your formula on well known SPDs?

I know my formulas are correct, I tested them against measured values in the link I had in my previous post. Some were off like the low 400's where the signal was weak. That example was a Luxeon CoB.

For the Vero Decor 1750K 97 CRI, 63 lm/W
I measured 3000 lm and 65 µMoles at 700 mA at a distance of 19".
So my conversion factor would be 0.021 (62/3000). 3000 x 0.021 = 62.

But PFFD is standardized to 1 meter, not 19" so 62 µMole will become 15 µMole:
1 meter = 39.37"
Inverse Square Law => (19"/39.37")² x 62µMole = 15 µMole.

The apply Inverse Sq to 3000 lm
And (19"/39.37")² x 3000 lm = 1380 lm
Normalized conversion factor is 15 /1380 = 0.01086

Vf = 28.8
My measurements were made at 700mA
28.8v x 700mA = 20.16 watts

20.16W x 63 lm/W = 1270 lm.

1270 lm x 0.01086 = 13.7µMole
13.7µMole / 15µMole = 8%, Within tolerance of datasheets and reasonable error.

To calculate PPFD for a 28.8v Vero 1750K:

28.8v x current x 63 lm/W x 0.01086

Proof
28.8v x 700mA x 63 lm/W x 0.01086 = 13.8

There is no number anything like 4.65

I am now able to see the formula in the spreadsheet where I could not last night.

I will try and track down the error.

He is using:
E•((λ•10-9)/h•c) =
E [W/m²]•λ•10-9[m]•/ (1.988•10-25) [J/s•m/s]
E•λ•5.03•1015 [1/(m²•s)] (with Irradiance E [W/m2])

E = NP/ 6.022x10-34
E•λ•0.836•10-2 [µmol/(m²•s)]

Which I reduced to PPFD = ((Lux / 683) * (wavelength * 0.00836)) / cie[wavelength]
I'm going to look further for an error, I will explain in another post following this one.

 

[Rahz]

I don't think a spot reading is going to provide the data necessary to derive umols per radiant watt, but I'm curious what you come up with using the SPD.

 

[Stephenj37826]

I know my formulas are correct, I tested them against measured values in the link I had in my previous post. Some were off like the low 400's where the signal was weak. That example was a Luxeon CoB.

For the Vero Decor 1750K 97 CRI, 63 lm/W
I measured 3000 lm and 65 µMoles at 700 mA at a distance of 19".
So my conversion factor would be 0.021 (62/3000). 3000 x 0.021 = 62.

But PFFD is standardized to 1 meter, not 19" so 62 µMole will become 15 µMole:
1 meter = 39.37"
Inverse Square Law => (19"/39.37")² x 62µMole = 15 µMole.

The apply Inverse Sq to 3000 lm
And (19"/39.37")² x 3000 lm = 1380 lm
Normalized conversion factor is 15 /1380 = 0.01086

Vf = 28.8
My measurements were made at 700mA
28.8v x 700mA = 20.16 watts

20.16W x 63 lm/W = 1270 lm.

1270 lm x 0.01086 = 13.7µMole
13.7µMole / 15µMole = 8%, Within tolerance of datasheets and reasonable error.

To calculate PPFD for a 28.8v Vero 1750K:

28.8v x current x 63 lm/W x 0.01086

Proof
28.8v x 700mA x 63 lm/W x 0.01086 = 13.8

There is no number anything like 4.65

I am now able to see the formula in the spreadsheet where I could not last night.

I will try and track down the error.

He is using:
E•((λ•10-9)/h•c) =
E [W/m²]•λ•10-9[m]•/ (1.988•10-25) [J/s•m/s]
E•λ•5.03•1015 [1/(m²•s)] (with Irradiance E [W/m2])

E = NP/ 6.022x10-34
E•λ•0.836•10-2 [µmol/(m²•s)]

Which I reduced to PPFD = ((Lux / 683) * (wavelength * 0.00836)) / cie[wavelength]

I found the error, I will explain in another post following this one.

Too many variables here. for starters par meters aren't 100% accurate. To the sphere with yah lol. Seriously you can request a par multiplier and the numbers we have gotten back so far basically coincide with the numbers using digitized SPD.

 

[alesh]

These numbers do not look correct.
The digitized numbers, are you converting them from Luminance to quantum PPFD?
Common sense says that 4 or 5 µMoles/J is way too high. 2 would be very good. 5 is like printing your own photons.

They are correct, though.
Most white spectrum LEDs do have 4.5 - 5 µmol/J. It's actually 3.76 µmol/J for monochromatic 450 nm light and 5.85 µmol/J for 700 nm. Basically any LED is between these 2 values. However, it's micromoles per joule of light (radiation) not per joule dissipated.

Not sure where did you get the 0.00836 constant you're using. Seems fishy.

 

[alesh]

For the Vero Decor 1750K 97 CRI, 63 lm/W
I measured 3000 lm and 65 µMoles at 700 mA at a distance of 19".
So my conversion factor would be 0.021 (62/3000). 3000 x 0.021 = 62.

Guess you meant lux and µmol/s/m^2. Doesn't matter for the conversion, though.
Anyway, I calcuted LER = 209.2 lm/W and QER = 5.23 µmol/J (or 4.59 in the 400-700 nm range). Guess what. (3000 lm / 209.2 lm/W) * 4.59 µmol/J = wait for it = 65.82 µmol/s

But PFFD is standardized to 1 meter, not 19" so 62 µMole will become 15 µMole:
1 meter = 39.37"
Inverse Square Law => (19"/39.37")² x 62µMole = 15 µMole.

The apply Inverse Sq to 3000 lm
And (19"/39.37")² x 3000 lm = 1380 lm
Normalized conversion factor is 15 /1380 = 0.01086

Sorry this section doesn't make any sense.

You're also confusing luminous flux with illuminance, PPF with PPFD.

 

[wietefras]

But PFFD is standardized to 1 meter, not 19" so 62 µMole will become 15 µMole

Good to see that you are finally trying to get the hang of this, but there is no need to "standardize PPFD on 1 meter". So no need to apply inverse square corrections.

Problem also is that you are doing this incorrectly. If the math was applied correctly you would be dividing the 3000lux and 62umol/s/m2 by the same number (19/39.37)² = 0,2329 and then you would get exactly the same conversion factor (3000/62=48.4) before and after the inverse square obfuscation.

 

[NoFucks2Give]

The problem with do the conversion from lumens to PPFD is there is no simple way to know if your result is correct.

When I did this years ago I searched the internet for days looking for a formula. Nothing. The question had been asked on a number of physics forums and help sites, but no answers. Today it's not much better but there are certainly more pieces of the puzzle out there now.

I had the radiospetrometer to check my answers against measured values.

I do not know if there is an error in the calculations but someone in another thread came up with some unrealistic numbers saying he got them here. In my previous post I showed my method I cannot see what is going on here. @alesh posted some formulas from Wikipedia and said it's easier than it looks. A cursory look at the spreadsheet only brought new questions and no answers.

So I'm going to go through the spreadsheet in detail and see what I can find.




Now let's break down the spreadsheet formula. This is the formula in cell E2

=B2/(($H$2*$G$2*$I$2)/(A2*10^-3))
​

$H$2 = cell H2 = planck's constant h Planck came up with this number many years ago and everyone goes along with it.
$G$2 = cell G2 = Avogadro number which is the number of photons in a Mole. Where a mole is a fixed number of particles and photons are considered particles in quantum physics. And makes as much sense as Planck's constant.

$I$2 = cell I2 which is the speed of light. How fast does a photon of light travel?

There is only one thing to do with those three numbers that I know of, calculate the energy of a photon (Ep).

Ep= h•f = h•(c/λ)
​

WHERE:
h=Plancks' constant
f=the frequency of a photon. frequency is the inverse of wavelength 1/λ
c=speed of light
λ = wavelength, but not in nanometers but whole meters. 555nm = 0.000000555 meters

The energy of a deep red 660 nm photon is 3.012 • 10-19 watts (feel free to check my math)
The energy of a blue 470 nm photon is 4.229 • 10-19 watts or 0.0000000000000000004229 watts
My point being a blue photon has more energy than a red photon. That is not too relevant because plants like all photons and don't care about the energy. That's why we use quantum measurements for grow lighting rather than watts.

Why do we care about the energy of a photon?
If we have x number of photons then we will know how many watts of flux (aka energy) they all have.

Just thinking out loud here.

This means the formula in cell E2 ($H$2*$G$2*$I$2) hold the value to ????? I do not yet know. but it's got some of the numbers required to calculate photon energy.

($H$2*$G$2*$I$2) / A2 * .001

A2 x .001 = .380nm

If it were A2 x 0.000000001 then D2 would be the energy of a 380nm photon.
For now, it's the energy of 100,000 photons. Why 1000,000 photons? No idea. Not my formula. Probably a decimal place thing.

On to Cell F2 =$J$2*SUM(D$1$1048576)/SUM(B$1:B$1048576)

Oh. This is interesting.
$j$2 = cell J2 = 683

683 is from the definition of a candela. Something to do with the maximum efficiency of something. I'll have to get the definition. Wikipedia was useless they kind of ignored it then at the very end the called it an arbitrary number. That is not true. I'm getting old and can't remember jack shit any more. Except that 683 is from candla, I remembered that. Wait. The most sensitive wavelength for the human eye is 555nm. 683 lumens/watt is based upon the sensitivity of the eye of 1 watt of luminance at 555 nm, Sometimes I'm a little slow.

But anyhow back in the day when I come up with my simplified formulas 683 is there. In my previous post.

1 converted 100 lm to Watts and PPFD for wavelengths 400-699nm
$ppfd = ((100/ 683) * ($wavelength * 0.00836)) / $cie[$wavelength];
$watts = $ppfd / 0.00836 / $wavelength;

2 converted 10 µMole to Lumens and Watts
$lux2 = 10 * $cie[$wavelength] / ($wavelength * 0.00836) * 683 ;
$watts2 = 10 / 0.00836 / $wavelength;

​

I'm not a spreadsheet guy so I do not know exactly what this means:

SUM(D$1$1048576)/SUM(B$1:B$1048576)
​

Okay now this is looking better. Maybe I'm wrong.
Totaling column B the digitized SPD numbers and column D = CIE x SPD

the 1048578 could/should have been 402 the last row rather than the bottom of the spreadsheet.

Now I am beginning to see familiar numbers. Earlier I took the first two columns and was trying various formulas looking for some number that match the numbers here. One thing I did was treat column B as watts and used my watts to PPFD and watts to Lux formulas. I then summed the new columns. One column summed to 985.665522. I just summed column D and it sums to 985.5885560317.

This is good news. I have a much better idea what is going on here now.

By summing column B it appears the SPD numbers are being treated as watts. Whereas up until now it appeared they were going to be treated as lumens.

Column D is the energy of the photons for each wavelength. When they get summed you have the total energy for the CoB with a mysterious decimal point. I did not see that coming, looking better.

So now we have the total radiometric wattage SUM(B$1:B$1048576)
and the total quantum number of photons: SUM(D$1$1048576).
So SUM(D$1$1048576)/SUM(B$1:B$1048576) gives us the ratio between luminous and radiometric.

That is multiplied by $J$2 which is... oh yeah, 683.

But column E is labeled Energy per a mole of photons. I think it was only 100,000 photons.

So let me calculate the energy per mole then go back and see what is happening.

Previously I calculated the Energy Ep for 660nm and 470nm. All we have to do is multiple Ep by Avogadro's number

Red Ep 3.012 • 10-19 watts x 6,022•10^23 = 0.181 watts/mole
Blue Ep 4.229 • 10-19 watts x 6,022•10^23 = 0.255 watts/mole

Well that didn't help any. So again D2 =B2/(($H$2*$G$2*$I$2)/(A2*10^-3))

Oh! Column E is labeled wrong. It's not energy per mole, it's energy per wavelength.

That was a trip. Unnecessary trip.
Column E is much simpler than what we have here.

I would remove this ugly formula in column E =B2/(($H$2*$G$2*$I$2)/(A2*10^-3))
And replace it with =A2*B2*0.00836
It is just as reduced and simplified equation that gives the same result.

So on to cell F2. This is not good. This takes the ratio of the sums of SPD and Energy and multiples it times 683, our candela number.

That may be okay. In my formula for calculating lux I use 0.00836 and 683.

$lux = PPFD * $cie[$wavelength] / ($wavelength * 0.00836) * 683
​

So I'll have to reconstruct how I came up with 0.00863. I do not remember.

It was a reduction of the conversion between irradiance and quantum flux.

Energy Quantum Flux = Number Photons / Avogadro number

= (E•λ•5.03•1015[1/(m²•s)]) / (6.02•10^17[1/µmol])
= E•λ•0.836•10^-2 [µmol/(m²•s)]
​

Okay! There it is. It goes back just a little further to the number of photons

The number of photons Np can be calculated by

Np= E/Ep
= E•((λ•10^-9)/h•c)
= E [W/m²]•λ•10^-9[m] / (1.988•10^-25) [J/s•m/s]​

= E•λ•5.03•10^15 [1/(m²•s)] (with Irradiance E [W/m²])

reciprocal of speed of light 1/1.988 = 5.03 times Avogadro number
5.03 / 6.02 = 0.835548
And that where the 5.03 comes from that gave me the 863 => 0.863 => 0.00863
So 863 is used in the conversion between the number of photons and the energy of photons
So 863 is only used in the conversion of quantum to radiometric or luminous or vice versa


So 5.03 / 6.02
Now in the spreadsheet I need to find Avogadro number in the numerator and speed of light in the denominator

How about E2?

​

=B2/(($H$2*$G$2*$I$2)/(A2*10^-3))
=Wavelength / Planck x Avogradro x speed of light/ SPD

Nope, not there Avogadro and speed of light are side by side.


How about F2?

Continued in next post, this post exceeded the max characters.

 

[NoFucks2Give]

Continued from previous post, the post exceeded the max characters.

How about F2?

=$J$2*SUM(D$1$1048576)/SUM(B$1:B$1048576)

683 x sum column D / sum of column B

Oh boy, not there either.

Okay column D is the wavelength x CIE photopic efficacy, that would be a step from radiometric toward luminous.
If I were converting watts to luminous flux...
I don't ever do that. I go from quantum and luminous and vice versa.


lux = PPFD x cie[wavelength] / (wavelength x 0.00836) x 683

So if I do not need quantum conversion I can remove (wavelength x 0.00836)?????

lux = watt x cie[wavelength] / 683

But is this correct? Let's say we have 22.4 lux at a wavelength of 680

CIE 680 => 0.017

22.4 = watt x 0.017 / 683
watts = 22.4 x 683 / 0.017
900
no, not even close to 1.93, no short cuts. Fuck!

680 and 22.4 were not random
I got them from the table where I verified my equations at http://growlightresearch.com/ppfd/convert.html
At 680 the calculated and measured values were close.

M Lux C Watts M Watts C PPFD M PPFD
22.4 1.927 1.93 10.96 10.94

I convert Lux to PPFD and then PPFD to watts

$ppfd = ((Lux/ 683) * ($wavelength * 0.00836)) / $cie[$wavelength];
$watts = $ppfd / 0.00836 / $wavelength;

combining the two above
$watts = (((Lux/ 683) * ($wavelength * 0.00836) / $cie[$wavelength])) / 0.00836 / $wavelength
$watts x $wavelength = (((Lux/ 683) * ($wavelength * 0.00836) / $cie[$wavelength])) / 0.00836
$watts x $wavelength x 0.00836 = (((Lux/ 683) * ($wavelength * 0.00836) / $cie[$wavelength]))
$watts x $wavelength x 0.00836 x $cie[$wavelength] = (((Lux / 683) * ($wavelength * 0.00836) ))

Now we can get rid of the $wavelength * 0.00836 on both sides

$watts x $cie[$wavelength] = Lux / 683
$watts x $cie[$wavelength] x 683 = Lux
​

so it is x 683, not / 683 as in my guess.

​

So let's try 1.93 watts @ 680

1.93 x 0.017 x 683 = 22.4
​

formula proven, that's a relief.

​

and back to cell F2 =$J$2*SUM(D$1$1048576)/SUM(B$1:B$1048576)

​

Column D could/should have been SPD x CIE x 683 and then column D would have been Illuminance, but this works.

This means F2 is correct.

On to cell F4 = col E / col B or photon energy / watts

wait... E already has B included, does that matter?

E is not the energy per mole it's the energy of a photon at that wavelength if I was previously correct on that.

But none the less if we substitute the formula for D in the formula in F4 we have

=B2/(($H$2*$G$2*$I$2)/(A2*10^-3)) / B
​

It appears B cancels out B2

Is that wrong? I think it's wrong. Which give me an idea on how to find out if I'm wrong.
@wietefras loves to point out where I am wrong

@wietefras, am I wrong? or is F4 correct? Is Column E Energy per mole or the energy of 100,000 photons for that wavelength?

 

[NoFucks2Give]

They are correct, though.

I hope they are. You can see where I got caught up. Help me out.

Not sure where did you get the 0.00836 constant you're using. Seems fishy.

And I do not like stuff that smells like fish.
But I think you may have seen where it is a simplification reduction with speed of light and Avogradro's number.

LEDs do have 4.5 - 5 µmol/J. It's actually 3.76 µmol/J for monochromatic 450 nm light and 5.85 µmol/J for 700 nm.

No I don't think I agree. I like to drink beer. I am usually fairly buzzed when I am on this site. So I can be wrong. But I'm most often wrong when I think I'm wrong.

You cannot just pick an arbitrary number without looking at the luminous factors.

Further more µMole/J has to do with the architecture of the LED die. It's not about wavelength. Wavelength comes into play with lumens and µMoles.

I do not comprehend your correlation between µMole and wavelength. I can understand the correlation with lumen and µMole.

450nm I'm going with .14495 µMoles per Lumen.
700nm I'm going with 1.9 µMoles per Lumen.

If you want radiometric:
450nm I'm going with 0.03853 watts per Lumen
700nm I'm going with 0.32444 watts per Lumen

But µMole/Joule per wavelength, that is just Wrong.
µMole/Joule is efficacy and not wavelength related. At least not at this level. If you want to get into the photoelectron quantum efficiency spectrum of silicon, that's a whole different ball game. Bring it on.

 

[NoFucks2Give]

Sorry this section doesn't make any sense.

there is no need to "standardize PPFD on 1 meter". So no need to apply inverse square corrections.

Can we agree that as the LED gets closer to the canopy the µMoles will increase?

So when someone states their fixture output 1000 µMoles do they just pick whatever distance they want. Or is there some sort of standard saying what the distance should be e.g. 1 meter?

So yes there is a standard and the stand is 1 meter.
If you state the PPFD µMoles at a photon travel distance other than 1 meter you must state the distance.
Actually you should always state the distance.

If there was no standard distance I would use 1 micron as my distance.

 

[NoFucks2Give]

Guess you meant lux and µmol/s/m^2.

That is just fuckin' hilarious. You were okay with Lux but had issue with µMole vs. µMole/m²/s.

If I use µMole what the fuck else would I mean besides µMole/m²/s. I do not use PPF because it is completely useless.

But Lux, I use that just because I like short words and I like words with X in it. And I use it because no one will ever know I use in incorrectly.

Technically lux is the unit of measure for Photometic Illuminance which is the equivalence of lumens/m²

But if you want to get nit picky I will use µMole/m²/s if you stop using lumens without the qualifying bullshit.
That means you must use lumen/m²/sr, lumen/m², or lumen/sr when lumen is not referring specifically to Luminous Flux

If using lumen in the context of Luminous Intensity, Illuminance, or Luminance then use the proper units and I will do likewise for µMole.

 

[NoFucks2Give]

You're also confusing luminous flux with illuminance, PPF with PPFD.

Please point out specifically where I confused luminous flux with illuminance. If I call them both lux I get away with it.

Let's see if I got it correct? In my mind luminous flux is the definition of lumen. Illuminance (aka Lux) is the photometric equivalent of PPF so I would only use illuminance very rarely because, like PPF, it is useless. Maybe you meant Luminance.

But whichever, I am very careful to use the correct terminology. I am trying to set a good example because no one gives a shit and they use the terms interchangeably. They are not interchangeable.

 

[NoFucks2Give]

Too many variables here. for starters par meters aren't 100% accurate.

And that is why I'd never use a PAR meter. I have found my radiospectrometer to be very accurate within 1% of calculated parameters.

Johnson Grow Lights... I have been on your website. There is one minor thing that bugs me about your site.

You have a graph of McCree I do not remember if it was action or absorbance spectra, but which ever it is labeled, it's the other one.

. Seriously you can request a par multiplier and the numbers we have gotten back so far basically coincide with the numbers using digitized SPD.

How do you know the PAR multiplier is correct?
How do you know the digitized SPD is correct?
What if the PAR multiple was generated with the digitized SPD and they are both wrong?
How do you know?

 

[Rahz]

Luminous flux is measured with an integrating sphere. Most people don't have those so we use information from the datasheets or product simulators. Those results can be used to determine the radiometric efficiency which can then be factored with the SPD from a spectrometer to arrive at PPF.

Anyway, here's a calculator that will confirm the umol/j range we've been suggesting.

http://www.calctool.org/CALC/chem/photochemistry/power_photons

 

[NoFucks2Give]

I don't think a spot reading is going to provide the data necessary to derive umols per radiant watt,

Radiant Watt is an ambiguous term. Radiant Flux is measured in Watts. Radiance is the radiometric equivalence of PPFD.

If you have Radiance (W/m²/sr) PPFD µMole (not W/m² Irradiance) is a simple calculation and is independent of luminous efficiency.

Now if you had Luminous Flux rather than Radiant Flux then it is much more complex.


Measuring the Photon Radiance (i.e. PPFD) and measuring the Luminance (lm/m²/sr) to get the multiplier factor works very well if the measuring device is accurate. I use a spectrometer that has a calibration traceable to NIST.

 

[NoFucks2Give]

Anyway, here's a calculator that will confirm the umol/j range we've been suggesting.

That is only a small piece of the solution and independent of Luminosity.
It provides little help with converting luminous quantities to quantum.
It can convert radiometric to quantum, but I prefer to stay away from radiometric because it includes the wavelength energy.
The reason horticulture uses µMoles rather than radiometric watts is the plants do not care about the energy of the photons only the number of photons.

And if interested here is the conversion formula from PPFD to Watts
watts = PPFD / 0.00836 / wavelength

 

[Rahz]

That is only a small piece of the solution and independent of Luminosity.
It provides little help with converting luminous quantities to quantum.
It can convert radiometric to quantum, but I prefer to stay away from radiometric because it includes the wavelength energy.
The reason horticulture uses µMoles rather than radiometric watts is the plants do not care about the energy of the photons only the number of photons.

You were disputing the figures we mentioned. I supplied the link to the calculator to help clear that up.

 

[NoFucks2Give]

You were disputing the figures we mentioned.

I'm sorry. I am not sure what is in dispute. I provided in my post the formulas that that calculator uses.

I got stuck at the formula in cell F4. Got anything to help me with that?

 

[Rahz]

Common sense says that 4 or 5 µMoles/J is way too high. 2 would be very good. 5 is like printing your own photons.

 

[Rahz]

I got stuck at the formula in cell F4. Got anything to help me with that?

The relative SPD values factored with the mole values.