B How to understand the steradians equation for measuring a sphere of light?

AI Thread Summary
The discussion focuses on understanding how to measure light using lumens and candelas, specifically in the context of an 8'x8'x8' room aiming for 60 foot-candles of illumination. The total output of 4800 lumens is noted, with an explanation provided that candela is less relevant for this scenario, as the focus is on illumination per unit area. The calculation shows that at the edge of the room, the illumination would be approximately 5.9 foot-candles, derived from dividing the total lumens by the surface area of the sphere created by the light source. The conversation emphasizes the importance of using lux instead of foot-candles for future measurements. Overall, the thread serves as an educational exchange on light measurement principles.
squeekymouse
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I am trying to understand the steradians equation for measuring a sphere of light.
Hello and please know I am very greatful for your help!
I am wanting to learn how to measure light. I have chosen a specific light for this to help me better understand.
Lm- 7800
CD-620.7
So, I got that far, lol. I don't really know how to input the numbers for the Steradians equation, I have an idea, but I would like to see how it is solved so I can understand it better.
1(cd) • 4 pi (sr) = 4 pi (cd•sr) = 12.57 (lm)
Thank you again for helping me learn 😁
 
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Can you explain what Lm and CD are? It would help if you had units attached to the numbers. Also, what is the steradians equation that you are referring to? A steradian is a unit of solid angle so it is not clear what you are asking about.
 
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kuruman said:
Can you explain what Lm and CD are? It would help if you had units attached to the numbers. Also, what is the steradians equation that you are referring to? A steradian is a unit of solid angle so it is not clear what you are asking about.
I apologize, I am trying to understand how Lumens and Candelas are measured within a given space.
 
Let me try this another way. If it is an 8'x 8'x 8' room and I wanted 60 foot candles of light and I know that it will equal 4800 lumen. What would that equation look like? I can use an internet calculator for the information, but I am interested in seeing the actual equation, so I can change the size of the room or foot candles as needed.
 
squeekymouse said:
Let me try this another way. If it is an 8'x 8'x 8' room and I wanted 60 foot candles of light and I know that it will equal 4800 lumen. What would that equation look like? I can use an internet calculator for the information, but I am interested in seeing the actual equation, so I can change the size of the room or foot candles as needed.
For your problem, it seems that candela (lumens per unit of solid angle) are largely irrelevant to your problem. If a source is isotropic, meaning equally bright in all directions, then the number of candela will just be equal to the total number of lumens divided by 4pi steradians, which is the total solid angle of the entire sphere (all directions into which the source is emitting the light).

You want to know something different, which is the brightness or illumination (in lumens per unit area) at a certain viewing distance from the source. You are using foot-candles (lumens per ft2), but in the future I would strongly encourage you to use lux (lumens per m2) instead, because US/Imperial units are a pain.

Anyway, you've stated that you know for sure that the source emits a total output of 4800 lumens into the room. The question is what area that output will be spread out over at a given distance. Again, we'll assume that the source is isotropic (emits the same in all directions) so that we can just divide. (Otherwise you are dealing with differentials i.e. calculus). At the very edge of the room, the total output is spread out over a sphere of radius 8 ft. This sphere has total surface area 4*pi*(8 ft^2), which works out to about 804 ft^2. The number of foot-candles at the edge of the room will just then be

(total output in lumens)/(total surface area) = (4800 lm)/(804 ft^2) = 5.9 lm/ft^2

Or in other words, just under 6 foot-candles of illumination at the very edge of the room. If you need more, you need to either get closer to the light source or increase its total output (in lumens). I hope that helps.
 
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LastScattered1090 said:
For your problem, it seems that candela (lumens per unit of solid angle) are largely irrelevant to your problem. If a source is isotropic, meaning equally bright in all directions, then the number of candela will just be equal to the total number of lumens divided by 4pi steradians, which is the total solid angle of the entire sphere (all directions into which the source is emitting the light).

You want to know something different, which is the brightness or illumination (in lumens per unit area) at a certain viewing distance from the source. You are using foot-candles (lumens per ft2), but in the future I would strongly encourage you to use lux (lumens per m2) instead, because US/Imperial units are a pain.

Anyway, you've stated that you know for sure that the source emits a total output of 4800 lumens into the room. The question is what area that output will be spread out over at a given distance. Again, we'll assume that the source is isotropic (emits the same in all directions) so that we can just divide. (Otherwise you are dealing with differentials i.e. calculus). At the very edge of the room, the total output is spread out over a sphere of radius 8 ft. This sphere has total surface area 4*pi*(8 ft^2), which works out to about 804 ft^2. The number of foot-candles at the edge of the room will just then be

(total output in lumens)/(total surface area) = (4800 lm)/(804 ft^2) = 5.9 lm/ft^2

Or in other words, just under 6 foot-candles of illumination at the very edge of the room. If you need more, you need to either get closer to the light source or increase its total output (in lumens). I hope that helps.
Thank you so much for taking the time to explain this to me Especially with me not being very clear about what I am asking. You're awesome!
 
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squeekymouse said:
Thank you so much for taking the time to explain this to me Especially with me not being very clear about what I am asking. You're awesome!
No problem, glad to be of help! :)
 
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