Lumens per SF according to the inverse square law HELP

[t@intshredder]

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.

Excellent summation.
I'll give the kid credit, though, he is very quick on the draw when it comes to Googling answers he need.
I'm just amazed that he expects people to listen to him when he admittedly doesn't grow.
He should take his nonsense to high school physics forum.

 

[mindphuk]

Excellent summation.
I'll give the kid credit, though, he is very quick on the draw when it comes to Googling answers he need.
I'm just amazed that he expects people to listen to him when he admittedly doesn't grow.
He should take his nonsense to high school physics forum.

He would get eaten alive on http://www.advancedphysics.org/ or http://www.physicsforums.com/

The members there have no tolerance for people that think they know what they are talking about and answer questions incorrectly. He could learn a lot by posting there, but I think he would end up arguing with the physicists and get himself banned.

 

[TeaTreeOil]

According to your calculations, 20 feet away from 1600 lm source would be .318 lm/sq ft. According to mine it's 4 lm/sq ft. A difference of 12.57 times, or the square feet of the surface area of a sphere with a radius of 1 foot. A foot candle is not a complete measure of the total output. It is merely 1 steradian @ 1 foot. There are 4pi or 12.57 steradians. Imagine that.

So, .318 ft candles vs 4 foot candles.

Here's an image I took with my lighter @ one foot from a wall versus a 1600 lm 26w CFL 20 feet away from a wall. According to you, the lighter should be brighter than the bulb's light(on the right) in this setup. It's clear to see that's just not the case.

I'm not saying all my measurements are perfect. I did not take into account the actual 3D distance, just the 2D distance from 'wall to wall'(light to wall, using the floor). And I'm not saying my lighter is producing exactly 1 foot candle, but it's probably pretty close.

So, according to my calculations I'm 1 lm/sq ft over from a real world measure. I also mentioned my calculations were maximums(in a vacuum). 3/4 = .75.

Your calculation is 2.682 lm/sq ft under the real world measure. Or 9.4 times off, pretty close to 12.57. 9.4/12.57 = .75!

Imagine that. The error percentages are identical.

Case closed.

 

[mindphuk]

wow, a picture with vague to no description of what you are doing or how or exactly what you are trying to measure. How about some pics of your light meter you are using to make your claim.

Since you don't like the formula for the spherical propagation of light (even though it is the basis for the inverse square law), let's do one for a 120° reflector

Foot-Candles= Lumen / (2 x Pi x d² x (1-cos(A/2)))
d=distance in feet or radius from the point source

Let's do 24" away and see how it compares to the charts.

F-C= 90,000/ (2 x Pi x 2² x (1 - cos(120/2)))
F-C= 90,000/ (25.13 x (1 - cos(60))
F-C= 90,000/ 12.57

F-C=7161 lum/ft²
Chart=7321
(interesting since TTO calculates 43.2" or 3.6' to before it falls off to 7000)

Let's try 10"
Calculation= 41253
Chart= 42170

Any questions?