Converting from Lux to Irradiance? (halp)
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Re: Converting from Lux to Irradiance? (halp)
theckhd wrote:If you're limited to excel, the easiest way is this:
1) Take your data of intensity (I) and distance (d) (or use your old data)
2) make a column that's x=1/d^2
3) Plot I vs. x
The formula for the dropoff of Intensity should be something like I = a+b/(d^2) = a+b*x. So fit this with a linear trendline. That will let you predict the intensity dropoff with distance. This formula will work for all of the LEDs, but the parameters will be a little different. If you want a quick and dirty fit that works for all of them, just assume a=0 and let b=I1*d1^2, where I1 and d1 are the intensity and distance values from one of your measurements with that LED.
I'm confused about the linear trendline part (how do I fit it and what do I do with it).
What exactly are "a" and "b"? I assume I am supposed to solve for b but I'm not sure what that is.

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
mew wrote:theckhd wrote:If you're limited to excel, the easiest way is this:
1) Take your data of intensity (I) and distance (d) (or use your old data)
2) make a column that's x=1/d^2
3) Plot I vs. x
The formula for the dropoff of Intensity should be something like I = a+b/(d^2) = a+b*x. So fit this with a linear trendline. That will let you predict the intensity dropoff with distance. This formula will work for all of the LEDs, but the parameters will be a little different. If you want a quick and dirty fit that works for all of them, just assume a=0 and let b=I1*d1^2, where I1 and d1 are the intensity and distance values from one of your measurements with that LED.
I'm confused about the linear trendline part (how do I fit it and what do I do with it).
What exactly are "a" and "b"? I assume I am supposed to solve for b but I'm not sure what that is.
You fit it in excel, just like you did with your other data. Add Trendline, but instead of choosing "exponential" you choose "linear" (or "polynomial", but the 2nd order correction will probably be too small to matter).
If you click the "Show equation" and "show R^2" boxes, it will give you the equation of the trendline and the R^2 value (a measure of how good a fit it is, R^2 near 1 is very good, R^2 < 0.8 or so is not so good).
The equation it puts on the graph should look something like this:
y = 1.234 + 56.789*x
1.234 is a
56.789 is b
"Theck, Bringer of Numbers and Pounding Headaches," courtesy of GrehnSkipjack.
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty

theckhd  Moderator
 Posts: 7970
 Joined: Thu Jul 31, 2008 3:06 pm
 Location: Harrisburg, PA
Re: Converting from Lux to Irradiance? (halp)
theckhd wrote:mew wrote:theckhd wrote:If you're limited to excel, the easiest way is this:
1) Take your data of intensity (I) and distance (d) (or use your old data)
2) make a column that's x=1/d^2
3) Plot I vs. x
The formula for the dropoff of Intensity should be something like I = a+b/(d^2) = a+b*x. So fit this with a linear trendline. That will let you predict the intensity dropoff with distance. This formula will work for all of the LEDs, but the parameters will be a little different. If you want a quick and dirty fit that works for all of them, just assume a=0 and let b=I1*d1^2, where I1 and d1 are the intensity and distance values from one of your measurements with that LED.
I'm confused about the linear trendline part (how do I fit it and what do I do with it).
What exactly are "a" and "b"? I assume I am supposed to solve for b but I'm not sure what that is.
You fit it in excel, just like you did with your other data. Add Trendline, but instead of choosing "exponential" you choose "linear" (or "polynomial", but the 2nd order correction will probably be too small to matter).
If you click the "Show equation" and "show R^2" boxes, it will give you the equation of the trendline and the R^2 value (a measure of how good a fit it is, R^2 near 1 is very good, R^2 < 0.8 or so is not so good).
The equation it puts on the graph should look something like this:
y = 1.234 + 56.789*x
1.234 is a
56.789 is b
Ah, that's what I thought you might have meant (but I didn't make the connection that those were the a and b variables :V). The problem is that I am getting a super negative value for that line at 6.7inches. Or does that not matter? I'll give it a try and see what happens.

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
mew wrote:Ah, that's what I thought you might have meant (but I didn't make the connection that those were the a and b variables :V). The problem is that I am getting a super negative value for that line at 6.7inches. Or does that not matter? I'll give it a try and see what happens.
You shouldn't be. Are you sure you're plotting I vs. 1/d^2 and not I vs. d?
"Theck, Bringer of Numbers and Pounding Headaches," courtesy of GrehnSkipjack.
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty

theckhd  Moderator
 Posts: 7970
 Joined: Thu Jul 31, 2008 3:06 pm
 Location: Harrisburg, PA
Re: Converting from Lux to Irradiance? (halp)
theckhd wrote:mew wrote:Ah, that's what I thought you might have meant (but I didn't make the connection that those were the a and b variables :V). The problem is that I am getting a super negative value for that line at 6.7inches. Or does that not matter? I'll give it a try and see what happens.
You shouldn't be. Are you sure you're plotting I vs. 1/d^2 and not I vs. d?
Ooooh, I see. I must have been doing it wrong last time, I should be able to get this now I'm going back in to the laser guy's lab on Monday to use his power meter again for the new LED.
He didn't want cookies for his help either I'm wondering if I should bring a few anyways or if that would be considered rude.

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
So I did it right this time. My a values are super high, over 1000, so I'm probably going to have to use that a=0 thing you mentioned (although I don't fully understand it).
And you say this is the drop off per distance? So as in if I plug in 6.7in for my d and use the equation I=a+[b/(d^2)] then the I would be the intensity reduction? As in, I subtract that from the I value at d=0 to get the current intensity?
Edit: Wait. I always heard the linear line equation was y=ax+b You are saying to use what looks like y=a+bx. Is this correct or do you have your a and b mixed up?
And you say this is the drop off per distance? So as in if I plug in 6.7in for my d and use the equation I=a+[b/(d^2)] then the I would be the intensity reduction? As in, I subtract that from the I value at d=0 to get the current intensity?
Edit: Wait. I always heard the linear line equation was y=ax+b You are saying to use what looks like y=a+bx. Is this correct or do you have your a and b mixed up?

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
mew wrote:Edit: Wait. I always heard the linear line equation was y=ax+b You are saying to use what looks like y=a+bx. Is this correct or do you have your a and b mixed up?
It's arbitrary, you can define it however you want. There's no standard definition for that. It could be y=qx+z if you wanted it to be.
"Theck, Bringer of Numbers and Pounding Headaches," courtesy of GrehnSkipjack.
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty

theckhd  Moderator
 Posts: 7970
 Joined: Thu Jul 31, 2008 3:06 pm
 Location: Harrisburg, PA
Re: Converting from Lux to Irradiance? (halp)
theckhd wrote:mew wrote:Edit: Wait. I always heard the linear line equation was y=ax+b You are saying to use what looks like y=a+bx. Is this correct or do you have your a and b mixed up?
It's arbitrary, you can define it however you want. There's no standard definition for that. It could be y=qx+z if you wanted it to be.
Alright, I just wanted to make sure that you didn't accidentally them up.

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
mew wrote:So I did it right this time. My a values are super high, over 1000, so I'm probably going to have to use that a=0 thing you mentioned (although I don't fully understand it).
And you say this is the drop off per distance? So as in if I plug in 6.7in for my d and use the equation I=a+[b/(d^2)] then the I would be the intensity reduction? As in, I subtract that from the I value at d=0 to get the current intensity?
It may help if you post an example set of data in excel format, so I can add the trendline and show you what I mean.
If you plot Intensity vs. 1/d^2, and fit it with a linear trendline, you'll have an equation that looks like I = a + b*x, where a and b are constants that the trendline equation gives you. That will let you estimate Intensity for an arbitrary distance d, because you can plug d into that equation and calculate I.
Like I said, it'll be slightly different for each LED, but it may be "close enough" that we can get away with just using one data point because the emission area of the LED is very small.
"Theck, Bringer of Numbers and Pounding Headaches," courtesy of GrehnSkipjack.
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty

theckhd  Moderator
 Posts: 7970
 Joined: Thu Jul 31, 2008 3:06 pm
 Location: Harrisburg, PA
Re: Converting from Lux to Irradiance? (halp)
theckhd wrote:mew wrote:So I did it right this time. My a values are super high, over 1000, so I'm probably going to have to use that a=0 thing you mentioned (although I don't fully understand it).
And you say this is the drop off per distance? So as in if I plug in 6.7in for my d and use the equation I=a+[b/(d^2)] then the I would be the intensity reduction? As in, I subtract that from the I value at d=0 to get the current intensity?
It may help if you post an example set of data in excel format, so I can add the trendline and show you what I mean.
If you plot Intensity vs. 1/d^2, and fit it with a linear trendline, you'll have an equation that looks like I = a + b*x, where a and b are constants that the trendline equation gives you. That will let you estimate Intensity for an arbitrary distance d, because you can plug d into that equation and calculate I.
Like I said, it'll be slightly different for each LED, but it may be "close enough" that we can get away with just using one data point because the emission area of the LED is very small.
Yeah, I got it. I just wanted to make sure the stuff was right. I can post my excel sheet later today at work.
I'm having trouble uploading it to google documents Keeps giving me an error.

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
Today I got to use the power meter again (and also the same setup in the same place, I don't know if I posted this already but their lab is a lot colder than mine), so I took a bunch of power meter and lux meter readings.
One thing that I was wondering about, I told him about how I was estimating the distance by a function of distance squared, how I had previously fit my curve with an exponential function (it fit pretty well too) but then I did the distance squared method instead, etc. And he said that I should be fitting it to a quadratic equation? That just seemed kind of odd to me that he would suggest something like that when my curves looked very exponential (he hasn't seen them, I'm still trying to figure out why google documents won't let me upload the file).
Do you think he was going somewhere with this or was he probably just confused?
I think he mentioned something that it would be exponential if it was isotropic but it isn't.
One thing that I was wondering about, I told him about how I was estimating the distance by a function of distance squared, how I had previously fit my curve with an exponential function (it fit pretty well too) but then I did the distance squared method instead, etc. And he said that I should be fitting it to a quadratic equation? That just seemed kind of odd to me that he would suggest something like that when my curves looked very exponential (he hasn't seen them, I'm still trying to figure out why google documents won't let me upload the file).
Do you think he was going somewhere with this or was he probably just confused?
I think he mentioned something that it would be exponential if it was isotropic but it isn't.

mew  Posts: 1903
 Joined: Sat Sep 27, 2008 5:42 pm
 Location: US
Re: Converting from Lux to Irradiance? (halp)
I don't think he knows what he's talking about (what area of physics does this guy work in?). If it were an isotropic source, it would drop off as 1/r^2. That's not exponential or quadratic, it's an inverse square law decay.
In your situation, it's closer to a gaussian beam in the farfield, which is basically a cone. As you go farther away the radius of the spot increases linearly with distance (i.e. radius is proportional to d^2). The average intensity is just the power emitted into this cone divided by the area, or I/A. That will be proportional to I/d^2, so you'd expect your data to decrease as 1/d^2, just like one should expect.
When you fit it with an exponential in Excel (you were fitting I vs. d), it forced it into the form d^n. The fitting algorithm just found the value of n that best matched the data. You'll notice that the values it settled on were very close to 2; that's because the dropoff is a 1/d^2. The difference is just due to a limitation of that fit form  it has no way to account for a constant offset, which is present due to the annoying details of the source (it's not a perfect point, and it's likely not exactly a cone because of imperfections in a variety of things in the LED itself).
By plotting I vs. x, where x=1/d^2, you've now transformed the equation to I = a*x + b. Now you can fit it with a linear equation in Excel, which gives you a little more freedom (and thus gives better fits). The downside is that it's a little less obvious what's going on, but it will give you much better results this way.
This is common practice in Physics by the way. We had an undergraduate lab where students would measure the period of a pendulum for a variety of arm lengths. The period (T) should be proportional to sqrt(L), but the students generally don't know that (or pretend not to for the purposes of the lab). To figure out what the relationship is, they plot T vs L, T^2 vs L, and T vs L^2 to see which one looks linear. They can then use the slope of the linear fit to figure out the earth's gravitational acceleration g.
In your situation, it's closer to a gaussian beam in the farfield, which is basically a cone. As you go farther away the radius of the spot increases linearly with distance (i.e. radius is proportional to d^2). The average intensity is just the power emitted into this cone divided by the area, or I/A. That will be proportional to I/d^2, so you'd expect your data to decrease as 1/d^2, just like one should expect.
When you fit it with an exponential in Excel (you were fitting I vs. d), it forced it into the form d^n. The fitting algorithm just found the value of n that best matched the data. You'll notice that the values it settled on were very close to 2; that's because the dropoff is a 1/d^2. The difference is just due to a limitation of that fit form  it has no way to account for a constant offset, which is present due to the annoying details of the source (it's not a perfect point, and it's likely not exactly a cone because of imperfections in a variety of things in the LED itself).
By plotting I vs. x, where x=1/d^2, you've now transformed the equation to I = a*x + b. Now you can fit it with a linear equation in Excel, which gives you a little more freedom (and thus gives better fits). The downside is that it's a little less obvious what's going on, but it will give you much better results this way.
This is common practice in Physics by the way. We had an undergraduate lab where students would measure the period of a pendulum for a variety of arm lengths. The period (T) should be proportional to sqrt(L), but the students generally don't know that (or pretend not to for the purposes of the lab). To figure out what the relationship is, they plot T vs L, T^2 vs L, and T vs L^2 to see which one looks linear. They can then use the slope of the linear fit to figure out the earth's gravitational acceleration g.
"Theck, Bringer of Numbers and Pounding Headaches," courtesy of GrehnSkipjack.
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty

theckhd  Moderator
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Re: Converting from Lux to Irradiance? (halp)
Alright, thanks! I just wanted to make sure that it was nothing I should be concerned about.
http://www.nhn.ou.edu/~abe/research/pubs.html The guy seems to work with temperature (cold stuff), nitric oxide, and lasers. I asked him about if the temperature difference between his lab and mine would have an effect and he said that it's not big enough of a change to worry about.
http://www.nhn.ou.edu/~abe/research/pubs.html The guy seems to work with temperature (cold stuff), nitric oxide, and lasers. I asked him about if the temperature difference between his lab and mine would have an effect and he said that it's not big enough of a change to worry about.

mew  Posts: 1903
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 Location: US
Re: Converting from Lux to Irradiance? (halp)
mew wrote:I asked him about if the temperature difference between his lab and mine would have an effect and he said that it's not big enough of a change to worry about.
It probably isn't, but there's no way to be sure without knowing how the lux meter is built internally. Better safe than sorry IMO (though it's probably at most a 1015% variation).
"Theck, Bringer of Numbers and Pounding Headaches," courtesy of GrehnSkipjack.
MATLAB 5.x, Simcraft 6.x, Call to Arms 6.0, Talent Spec & Glyph Guide 5.x, Blog: Sacred Duty
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theckhd  Moderator
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Re: Converting from Lux to Irradiance? (halp)
interesting discussion
but now, can anybody help me to convert the other way around
w/m2 to luxes
Global Irradiance to Illuminance
please, please, please
Thanks in advance!
but now, can anybody help me to convert the other way around
w/m2 to luxes
Global Irradiance to Illuminance
please, please, please
Thanks in advance!
 artcygne
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