Trying to Determine if Proper Illumination is Provided

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The discussion focuses on calculating illumination using the formula c / (4*π*distance^2), where c is 5.4 watts and distances are specified in feet for three problems. Participants clarify that the distances should be treated as feet and that the resulting illumination units are ft-candles (lux). There is confusion about using a scientific calculator, particularly with the squared operator. The required illumination levels for emergency lighting are also mentioned, emphasizing the need for at least 1 ft-candle on average. The user seeks guidance on converting from watts per square foot to ft-candles.
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Homework Statement



Can someone help me solve the following? I trying to determine how much illumination is applied to a given space based on the following equation. I have no math solving abilities whatsoever.

Homework Equations



Where c = 5.4 watts

Where ft = Problem 1: 22; Problem 2: 27; Problem 3: 30

(c) / (4*π*distance (ft) ^2)


The Attempt at a Solution



Tried to use a science calculator - Multiplied 4 x pi x the distance. Couldn't figure out how to multiply in the caret 2.
 
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Shag-X said:

Homework Statement



Can someone help me solve the following? I trying to determine how much illumination is applied to a given space based on the following equation. I have no math solving abilities whatsoever.

Homework Equations



Where c = 5.4 watts

Where ft = Problem 1: 22; Problem 2: 27; Problem 3: 30

(c) / (4*π*distance (ft) ^2)


The Attempt at a Solution



Tried to use a science calculator - Multiplied 4 x pi x the distance. Couldn't figure out how to multiply in the caret 2.

Welcome to the PF.

First, you need to be careful to work in consistent systems of units. When you typed distance (ft), is that meant to say that the units of the distance are in feet? Farther up in the problem you assigned values to "(ft)", which doesn't make sense if that's just the units of "distance".

And what are the units of the 3 distances? Feet? Like:

1) distance = 22 feet
2) distance = 27 feet
3) distance = 30 feet

And do you know what the units are for the illumination results that you are trying to calculate? The way the equation is set up now (assuming I've interpreted your use of "(ft)" right), you would get a result in Watts/Square Feet.

And finally, the "^2" is not something you multiply by. It represents the "squared" operator. Like, 10^2 = 10*10 = 100.
 
1. When you typed distance (ft), is that meant to say that the units of the distance are in feet?

Answer 1: Yes. (ft) is the distance, in feet, that the area to be illuminated is from the light
source.

2. What are the units of the 3 distances?

Answer 2: Yes. The units of the three distances are feet. The area is a gradually riser theater
floor, with aisles that are required to be illuminated along their entire length.

3. What are the units for the illumination results?

Answer 3: The units for the illumination results are ft-candle (lux).

Here is the requirement:

Emergency lighting facilities shall be arranged to provide initial illumination that is not less than
an average of 1 ft-candle (10.8 lux) and, at any point, not less than 0.1 ft-candle (1.1 lux),
measured along the path of egress at floor level.

Made an attempt at solving and came up with - .00088785. But, I don't think that correct.

Thank you,
 
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Shag-X said:
1. When you typed distance (ft), is that meant to say that the units of the distance are in feet?

Answer 1: Yes. (ft) is the distance, in feet, that the area to be illuminated is from the light
source.

2. What are the units of the 3 distances?

Answer 2: Yes. The units of the three distances are feet. The area is a gradually riser theater
floor, with aisles that are required to be illuminated along their entire length.

3. What are the units for the illumination results?

Answer 3: The units for the illumination results are ft-candle (lux).

Here is the requirement:

Emergency lighting facilities shall be arranged to provide initial illumination that is not less than
an average of 1 ft-candle (10.8 lux) and, at any point, not less than 0.1 ft-candle (1.1 lux),
measured along the path of egress at floor level.

Made an attempt at solving and came up with - .00088785. But, I don't think that correct.

Thank you,

How does one convert from Watts per square foot to ft-candles?
 
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