Lumens per SF according to the inverse square law HELP

[t@intshredder]

Did you forget this thread?
https://www.rollitup.org/indoor-growing/173926-yet-another-light-quiz.html


O TeaTreeOil, O Tea Tree!
How are thy leaves so sticky!
O TeaTreeOil, O Tea Tree,
How are thy leaves so sticky!

Not only in the summertime,
even under CFL's is thy are so shiney.

O TeaTreeOil, O Tea Tree,
How are thy leaves so sticky!
O TeaTreeOil, O Tea Tree,
Much pleasure your threads bring me!

O TeaTreeOil, O Tea Tree,
Much pleasure doth thou rendering bring me!
For every year the TeaTreeOil,
Brings to us all both the bullshit and glee.

O TeaTreeOil, O TeaTree,
Much your bullshit is boaring me!

O TeaTreeOil, O TeaTree,
Thy CFL's shine out so brightly!

O TeaTreeOil, O TeaTree,
Thy candle power burn so powerfully!
Each tiny lumen doth hold its PAR light,
That makes my toy HPS to sparkle like a fairy light.

O TeaTreeOil, O TeaTreeOil,
Thy shit shine out so brightly!

(thank you 9inch Bigbud)

Ha! Is TTT getting schooled in another thread today? Damn, son ...you're a straight-up GLUTTON for punishment.

 

[jigfresh]

WOAH! I didn't know I was getting in the middle of a cute little spat.

I'll put myself in the corner where stupid people belong, you guys can fight over who belongs here with me.

Hey bluex. I'm glad you got your question answered. Sorry for being an asshole.

 

[TeaTreeOil]

No one is claiming anything different. Why must you obfuscate the issue with irrelevant information?

Yes, and you can use this to approximate the lux or FC, but it will be for a lamp with no reflector so your measured value will be different.

It's not irrelevant. Trying to make up calculations for <all reflector here> is full of fallacy. I don't see any value in doing such error-prone calculations.

I blatantly stated the figures were for a bulb without a reflector.

For example, take that 90,000 lumen lamp and say that the light will radiate away from that point and at one foot, the energy will have dissipated in a sphere. That calculation is 4/3 Pi x r^2= 4.188 recurring. 90,000 lumens divided by 4.188= 21,489 F-C, lower than the chart but that didn't take into account a reflector, so the higher value in the chart is probably more accurate if using a reflector.

Absolutely wrong. Lumen takes into account the sphere's area through the use of the steradian.

My equations and results are correct in a perfect vacuum.

At 2 feet, the F-C would be 5372 F-C, much lower than you 7,000 F-C @ 3.24 feet as you claim.

You absolutely did apply it incorrectly. Did you or did you not claim that the sun would be 10,000 F-C over 9sf or not?

It'd be higher with a good reflector.

Yes, the sun's light on earth's surface is around 10,000 foot candles. Or 10,000 lm/sq ft. Which would be 90,000 lm for 9 sq ft. FC does not change within the same area when the distance is the same. The area is defined by the distance. It already takes into account I = P / r ^ 2. Where r = 1, P is lumens. This is why 1 lumen per square foot(P / r ^ 2) is 1 foot candle(I).

 

[TeaTreeOil]

WOAH! I didn't know I was getting in the middle of a cute little spat.

I'll put myself in the corner where stupid people belong, you guys can fight over who belongs here with me.

Hey bluex. I'm glad you got your question answered. Sorry for being an asshole.

Hahaha. You haven't done shit compared to most assholes here.

 

[mindphuk]

Absolutely wrong. Lumen takes into account the sphere's area through the use of the steradian.

No, lumens have nothing to do with area. That is the domain of lux and foot-candle

Yes, the sun's light on earth's surface is around 10,000 foot candles. Or 10,000 lm/sq ft. Which would be 90,000 lm for 9 sq ft. FC does not change within the same area when the distance is the same. The area is defined by the distance. It already takes into account I = P / r ^ 2. Where r = 1, P is lumens. This is why 1 lumen per square foot(P / r ^ 2) is 1 foot candle(I).

Sorry, I took the wrong quote.
Here's the correct one:

Remember that the sun gives out 10,000 lumens per square foot on a sunny summer day. If you had 100,000 lumens that would be highly sufficient for 10 ft by 10 ft

That's 10' *10' = 100 sq ft. Which is 1000 lm/sq ft.

You clearly took the value of the sun's output and incorrectly tried to spread it out over a specified area.

 

[TeaTreeOil]

No, lumens have nothing to do with area. That is the domain of lux and foot-candle
Sorry, I took the wrong quote.
Here's the correct one:You clearly took the value of the sun's output and incorrectly tried to spread it out over a specified area.

What?

100,000 lumens / 10'*10' = 1000 lm/sq ft.

A 10'x10' Earth surface area lit by the sun would receive the equivalent of 1,000,000 lm.

Like I said, you misunderstood. What that guy posted didn't make sense.

Lumen is defined as 1 candela steradian.

 

[mindphuk]

What?

100,000 lumens / 10'*10' = 1000 lm/sq ft.

A 10'x10' Earth surface area lit by the sun would receive the equivalent of 1,000,000 lm.

Like I said, you misunderstood. What that guy posted didn't make sense.

Lumen is defined as 1 candela steradian.

Fail! That's merely a conversion when you know the candela and distance you need for the calculation.
Lumen is defined by amount of radiation (luminous flux) per unit of time. Candela is a derivative of lumen (= to lumen/steradian), not the other way around. Lumen is a measured value, not calculated. Of course you can convert back and forth, as long as you have all of the correct information, including distance and area.

 

[mindphuk]

So are you going to stick with your answer that a 90,000 lumen lamp will provide 7000 fc at 3.24 feet?

90,000 lm / 3.24 sq ft = 27,778 lm/ sq ft.

Yes, 90,000 lumen over 3.24 sf will equal 27K foot-candle. That information is useless to the OP without a distance.

If a 90,000 lumen lamp is placed 3.24 feet away from a specified area (which is exactly what you claim to have calculated, it will not be close to 27,000 f-c which is what you would need for a 9sf area to average 7000 f-c.

90,000 lumen/ 4/3 Pi x 3.24^3 =631 f-c
Even with a reflector, there's no way to get close to 7000 f-c

 

[TeaTreeOil]

Fail! That's merely a conversion when you know the candela and distance you need for the calculation.
Lumen is defined by amount of radiation (luminous flux) per unit of time. Candela is a derivative of lumen (= to lumen/steradian), not the other way around. Lumen is a measured value, not calculated. Of course you can convert back and forth, as long as you have all of the correct information, including distance and area.

Yes, candela is also defined by lumen/steradian. One does not exclude the other.

Too bad you seem to completely fail to understand any of these terms. And keep regurgitating the same nonsense.

90,000 lumen/ 4/3 Pi x 3.24^3 =631 f-c

That isn't even close to correct.

If you wanted to calculate the average over a square plane, which seems is what you want to do.

You could take the average intensities between a corner(square) * 4 + center and divide the result by 5.

So 2 feet away, over 4 sq feet. Is sqrt(2^2+2^2+2^2) = 3.46'. 90,000/3.46^2 or 90,000 lm/12 sq ft = 7500 lm/sq ft.

2 feet below is: 90,000/2^2 or 90,000 lm / 4 sq ft = 22,500 lm/sq ft.

7500*4+22,500/5 = ~10,500 average lm/sq ft.

To get a better(higher) average we could also use the midpoints.
sqrt(2^2+2^2) = 2.83', 2.83^2=8, so:

90,000/8 = 11,250

There are also four of these. 11250*4+7500*4+22500/9 = ~10,833 lm/sq ft average.

But the spot directly 2 feet below is still going to be 22,500 lm/sq ft.

I don't see the value in an average that varies from 7,500 to 22,500 lm/sq ft. Maybe you do.

 

[t@intshredder]

TTT - since you worked so hard on my new avatar, I figured I would return the favor.
This one might be a little more appropriate given your current argument track record.

 

[mindphuk]

That isn't even close to correct.

If you wanted to calculate the average over a square plane, which seems is what you want to do.

Did I say that? No, I calculated the foot-candle from a 90,000 lumen point source at 3.24 feet away. I said nothing of the area except the foot-candle being 1 square foot. But you are partially correct, I should have used sphere surface area rather than sphere volume to get my answer.

You could take the average intensities between a corner(square) * 4 + center and divide the result by 5.

So 2 feet away, over 4 sq feet. Is sqrt(2^2+2^2+2^2) = 3.46'. 90,000/3.46^2 or 90,000 lm/12 sq ft = 7500 lm/sq ft.

2 feet below is: 90,000/2^2 or 90,000 lm / 4 sq ft = 22,500 lm/sq ft.

You're doing it again. You are mixing up the distance with the area illuminated. Stick with a single square foot since that is the unit of measure for foot-candle. Yes, 90,000 lumens shining on 4 square feet = 22,500 f-c but since there is no specified distance, we have to conclude that is measured at zero distance. Once you begin to talk about moving away from that point source, then you are unable to continue. You keep saying that is at 2 feet, yet the only time you use 2ft in your calculation was in the area illuminated, you completely disregarded the distance.

7500*4+22,500/5 = ~10,500 average lm/sq ft.

To get a better(higher) average we could also use the midpoints.
sqrt(2^2+2^2) = 2.83', 2.83^2=8, so:

90,000/8 = 11,250

There are also four of these. 11250*4+7500*4+22500/9 = ~10,833 lm/sq ft average.

But the spot directly 2 feet below is still going to be 22,500 lm/sq ft.

I don't see the value in an average that varies from 7,500 to 22,500 lm/sq ft. Maybe you do.

The rest of your wild rantings are irrelevant since you failed to use the correct calculations to begin with.

 

[TeaTreeOil]

You're a complete moron. I'm ignoring you now.

Cheers.

When you goto college and take some physics classes and actually pass them.... Ahh, fuck it. You won't.

 

[t@intshredder]

You're a complete moron. I'm ignoring you now.

Cheers.

When you goto college and take some physics classes and actually pass them.... Ahh, fuck it. You won't.

Wow ...act like an adult much?
He never actually ignores anyone. Not sure he knows how to.

 

[TeaTreeOil]

The other moron on my ignore list.

You two should get along nicely.

 

[t@intshredder]

People I [suggest that you] ignore: t@intshredder and mindphuk. Who's next?

And why would anyone ignore either of us? Unlike you the 2 of us actually grow plants and help people with their problems instead of creating new ones an belittling people.
Shouldn't you be simulating some magnesium deficiencies or some heating problems, on your PC?
I love how you don't even grow and you expect people to respect you. FAIL

 

[t@intshredder]

The other moron on my ignore list.

You two should get along nicely.

How are you responding to my messages if you ignored me?
Well you are just 100% failure all around, aren't you?

 

[mindphuk]

You're a complete moron. I'm ignoring you now.

Cheers.

When you goto college and take some physics classes and actually pass them.... Ahh, fuck it. You won't.

Nice. Ad hominem attacks rather than read and respond.

You said in an earlier post:

Light decays according to Iavg = P / A. That's: average intensity = power / area.

The power, lumen, divided by the area, which in a non-shielded point source is the surface of the sphere at any given distance. Do you doubt that? Does a point source not radiate in a spherical pattern?
If the area is 4x pi x r^2, then you have yet to show me why my math is incorrect. You agree with my starting point as your quote above confirms.

The only reason real world data varies from the calculated value is because of good reflectors redirecting light that would normally be radiating away from the target.

 

[TeaTreeOil]

Yea, like I said. Ignoring you.

Good luck with your Never-never-land physics or whatever.

 

[t@intshredder]

Yea, like I said. Ignoring you.

Good luck with your Never-never-land physics or whatever.

Your ignoring skills could use some work, son!

 

[mindphuk]

Yea, like I said. Ignoring you.

Good luck with your Never-never-land physics or whatever.

Funny coming from someone that has yet to take a college level physics course.

"my wife is a science teacher. I have college level textbooks here"

That was TTO's claim in https://www.rollitup.org/grow-room-design-setup/145941-more-reflective-tin-foil-flat-2.html thread

So, to sum up:
mindphuk- graduated from a nationally recognized, accredited university in a science/technical oriented field and received high grades in physics and engineering courses.

TeaTreeOil- has some physics books on his kitchen table. Continually misuses scientific formulas, won't admit when he's wrong and whose advice runs counter to every person on RIU with practical experience. Then when people attempt to help him by correcting his mistakes, resorts to name-calling and personal attacks.