Lux & Lumens -- how do they work?

[Xenon02]

TL;DR Summary

How to calculate lux here ? And what are Lumens, when I have 800 Lumen Light bulb does it mean that in everydirection I have 800 Lumens ?

Hello !

I've got a problem understanding what are Lumens and how do I look at lux as well.

For example here :

It makes 800 Lumens in 1 m^2. Does it mean that in every direction in 1m^2 there are 800 Lumens ?
Or when I have a light bult with 800 lumens :

Those 800 Lumens in light bult means that all those arrows in this picteres are 800 Lumens ?

Also how do I calculate lux in those pictures :

All I know is that all of them are 600 Lumens or 640 Lumens but some of them have 100 Lux or 300 Lux, in the picture Lux = Luks.

How does it work ? I know the definition but I don't really understand. Can someone tell me more in simple words ?

 

[jim mcnamara]

Lumens:
"The lumen (symbolized lm) is the International Unit of luminous flux. It is defined in terms of candela steradians (cd multiplied by sr). One lumen is the amount of light emitted in a solid angle of 1 sr, from a source that radiates to an equal extent in all directions, and whose intensity is 1 cd."
-- From techtarget.com

I think this answers part of your question, if it does not please ask us.

 

[Xenon02]

Lumens:
"The lumen (symbolized lm) is the International Unit of luminous flux. It is defined in terms of candela steradians (cd multiplied by sr). One lumen is the amount of light emitted in a solid angle of 1 sr, from a source that radiates to an equal extent in all directions, and whose intensity is 1 cd."
-- From techtarget.com

I think this answers part of your question, if it does not please ask us.

I'm not that bright at physics so is it something like that ?

800 lumens in every direction ? Or is it divided ? if so
then why here it's 800 lumens

Thanks for the answer :)

It's not an easy subject for me because it creates a lot of misundestanding in my head. Like intensity, where the light is 800 lumen or something like that.

The only thing I know is that lux = Lumen/m^2.

 

[Janus]

Lumens are a measure of total light output the source per unit of time.
Lux is lumens per square meter.
So, in the picture of the light bulb, the total output of the bulb is measured in lumens, while the lux is the total lumens/ m^2
So, let's say the total surface of the bulb is 0.01 square meter, then the lux at the surface is 80000 lux.
If we enclose the bulb in a 1 meter radius shell, the total light striking its interior side is still 800 lumens, but the lux (since the lumens are spread out over a larger area) drops to 63.66 lux

 

[Xenon02]

So, let's say the total surface of the bulb is 0.01 square meter, then the lux at the surface is 80000 lux.
If we enclose the bulb in a 1 meter radius shell, the total light striking its interior side is still 800 lumens, but the lux (since the lumens are spread out over a larger area) drops to 63.66 lux

So I can have more lux than lumens like for example 800 lumens and 80000 lux. The surface still has 800 lumens but it's much tighter ? The same goes with larger area, the area still has 800 lumens but it is spread all over.

I thought that total brightness of the light is lumen not lux. So 80000 lux is really bright. I thought that 800 Lumens is a total brightness of a light. Or maybe I am wrong ?

So may I ask that. My picture with a bulb and arrows around it means that I have 800 lumens but every single one of this arrow is like 1 lumen or lower ? Or how does it work with single arrow in the picture?

 

[Steve4Physics]

Edit - corrections.

Hi @Xenon02. This may help, though it’s not very scientific/rigorous/complete.

Think of light as water coming out of a spray-nozzle.

The ‘luminous flux’ is the total amount of water coming out of the spray-nozzle per second. It is measured in units of lumens.

(The water coming out of the spray-nozzle could be a thin jet or a wide spray. But let's not worry about that as it complicates matters.)

The water then hits a wall some distance from the nozzle, covering an area 'A' of the wall. The amount of water per second hitting each square metre of the area is the luminous flux illuminance, measured in units of lux.

Example: a 100 lumen torch produces a beam of light which hits a wall square-on and makes a 4m² bright patch on the wall. The illuminance of the patch is 100lumen/4m² = 25lux.

 

[Xenon02]

Example: a 100 lumen torch produces a beam of light which hits a wall square-on and makes a 4m² bright patch on the wall. The illuminance of the patch is 100lumen/4m² = 25lux.

So when I get closer even if the wall is still 4m^2 it will make a smaller bright patch witch increases lux ?

But how should I look at those arrows here ?

They have lumens or not ?

Also when I have a light bulb without a lampshade like here :

the lux is equal 100 lux.

But when I add lamp shade :

It increases to 300 lux. But some light is "hitting" lampshade. So why is it 300 lux ? Lampshade is also a
surface. Part of light at the surface and part at the table. But here 300 lux is just how 800 lumens are distributed. But some of this 800 lumens are distributed on the lampshade so I don't know ...

 

[Steve4Physics]

So when I get closer even if the wall is still 4m^2 it will make a smaller bright patch witch increases lux ?

If you walk towards the wall, shining the torch at it, the area of the patch of light gets smaller.

The area of the patch of light would have reduced from 4m². So it doesn't make sense to say "still 4m^2".

Suppose you are very close and the lit-up area is now only 1m², the illuminance of the lit-up area would now be 100lumen/1m² = 100lux.

But how should I look at those arrows here ?

View attachment 305686
They have lumens or not ?

No! Not! 800 lumens is the total amount of light given off by the bulb.

Suppose you have a garden-sprayer that gives out a total of 2 litres of water per seond. If the spray-head actually produces 1000 very thin streams, does it make sense to label each of the very thin streams '2 litres/second'?

Also when I have a light bulb without a lampshade like here :

View attachment 305687
the lux is equal 100 lux.

But when I add lamp shade :

View attachment 305688

It increases to 300 lux. But some light is "hitting" lampshade. So why is it 300 lux ? Lampshade is also a
surface. Part of light at the surface and part at the table. But here 300 lux is just how 800 lumens are distributed. But some of this 800 lumens are distributed on the lampshade so I don't know ...

The '100lux' and '300lux' refer to how well the table is lit.

Without the lampshade, a lot of the light misses the table. But when the lampshade is added, it reflects this light onto the table. The table receives extra light because of the lampshade. This make the illuminance of the table increase from to 100lux to 300lux. The table is lighter (though other parts of the room will be darker).

 

[Xenon02]

No! Not! 800 lumens is the total amount of light given off by the bulb.

Suppose you have a garden-sprayer that gives out a total of 2 litres of water per seond. If the spray-head actually produces 1000 very thin streams, does it make sense to label each of the very thin streams '2 litres/second'?

Okey so the arrows are like part of the lumens. Just an example 800 lumens is total so like 1 arrow is like 1 lumen. It's just an example in real life it could be smaller maybe even a lot.

The '100lux' and '300lux' refer to how well the table is lit.

Without the lampshade, a lot of the light misses the table. But when the lampshade is added, it reflects this light onto the table. The table receives extra light because of the lampshade. This make the illuminance of the table increase from to 100lux to 300lux. The table is lighter (though other parts of the room will be darker).

The table ?
Isn't it referring to the whole room ?

Same goes here :

It's 200 lux but it's not the part of this light but the whole light.
The table is just the part.

Then how to calculate the whole room the lux ?
Also the lamp shade doesn't reflect everything so some light will be "in" the lampshade.

And one more thing. If I want to have for example 800 lux the light bulb on the first picture :

I have to make a box that sum of all walls is equal 1m^2 ? And how about distance ? Is it also important in that box 1m^2 ?

 

[Steve4Physics]

Okey so the arrows are like part of the lumens. Just an example 800 lumens is total so like 1 arrow is like 1 lumen. It's just an example in real life it could be smaller maybe even a lot.

How many arrows should you draw: 6, 10, 100, 70,000? What about the gaps between the arrows? Labelling arrows with a number of lumens doesn't really mean anything and could be misleading. I would advise against it.

The table ?

As shown in the 3rd diagram in Post #1

Isn't it referring to the whole room ?

No. The illuminance (in lux) of any area depends on how far the area is from the lamp. The values of 100lux and 300lux must refer to some specific region; here it is the table-top.

Same goes here :

View attachment 305693
It's 200 lux but it's not the part of this light but the whole light.

The lamp produces 800 lumen as a beam of light. The beam widens as you get further from the lamp.

In the part of the beam with area 1m², the light is spread across 1m², so the illuminance at this part of the beam is 800lumens/1m² = 800lux.

In the part of the beam with area 4m², the light is spread across 4m²,the illuminance at this part of the beam is 800lumens/4m² = 200lux.

Outside the beam, the illuminance is zero.

Then how to calculate the whole room the lux ?

You can't. Different parts of the room get different illuminances (difference numbers of lux). The calculation would be extremely complicated and depend on reflection and absorption by different surfaces, as well as the size andshape of the room.

A lux value applies to a specific region (surface). The lux value changes depending on the distance to the light source and other factors (e.g. reflections).

Also the lamp shade doesn't reflect everything so some light will be "in" the lampshade.

The key point is that using the lampshade increases the illuminance of the region below it. The figures are for illustration only. E.g. if the lampshade were a perfect reflector, maybe you would get 390lux instead of 300lux - it's not important.

And one more thing. If I want to have for example 800 lux the light bulb on the first picture :

View attachment 305694

I have to make a box that sum of all walls is equal 1m^2 ? And how about distance ? Is it also important in that box 1m^2 ?

The 'first picture' (in Post #1) shows an 800 lumen bulb and this is different to the diagram you refer to. So I can't tell what you mean. The shape of the box would depend very much on the distribution of light from the bulb. Draw a diagram showing your solution and we can comment on it.

 

[Xenon02]

No. The illuminance (in lux) of any area depends on how far the area is from the lamp. The values of 100lux and 300lux must refer to some specific region; here it is the table-top.

So ok the lamp has 300 lux.
The area is the Table. I know that the bulb provides 800 Lumens so the area is 2,6 m^2. This is the whole table or is this the distance table - lamp ?

All I know is that 800 Lumens must be distributed to the whole area.
Edit : But also this table is the whole area that this bult provides the light, walls or floor also contains some of lux so this equation for table 300 lux = 800 lumens/2,6m^2 isn't true because on the table there isn't 800 lumens which means the whole light distributed to whole table but part of 800 lumensOr here :

1m^2 is the surface or the distance ?

The key point is that using the lampshade increases the illuminance of the region below it. The figures are for illustration only. E.g. if the lampshade were a perfect reflector, maybe you would get 390lux instead of 300lux - it's not important.

Yes but those 300 lux when having a lampshade still uses the same equation
Lux = 800Lumens/(area)m^2 this indicates that whole 800 lumens are distributed to the whole table but some of it isn't reflected by the lampshade so the table doesn't receive 800 lumens.

The 'first picture' (in Post #1) shows an 800 lumen bulb and this is different to the diagram you refer to. So I can't tell what you mean. The shape of the box would depend very much on the distribution of light from the bulb. Draw a diagram showing your solution and we can comment on it.

I meant something like this :

 

[Steve4Physics]

So ok the lamp has 300 lux.

No. A lamp doesn't have a 'lux value'. A lamp's total light output is measured in lumens.

When the light from the lamp then reaches a region with a certain area, the amount of light per unit area is then measured in lux.

The area is the Table. I know that the bulb provides 800 Lumens so the area is 2,6 m^2. This is the whole table or is this the distance table - lamp ?

2.6m² is an area, not a distance (because m² is an unit of area, not distance).

I'm guessing that you have assumed that ALL the light from an 800 lumen bulb hits a table, giving the table an illuminance of 300lux. In that case the area of the illuminated part of the table is given by A = 800lumen/300lux = 2.7m².

It's impossible to find the table-lamp distance unless you also know the shape of the light beam (the angle in steradians) coming from the lamp+shade.

All I know is that 800 Lumens must be distributed to the whole area.

Or here :

View attachment 305700
1m^2 is the surface or the distance ?
See above.

Yes but those 300 lux when having a lampshade still uses the same equation
Lux = 800Lumens/(area)m^2 this indicates that whole 800 lumens are distributed to the whole table but some of it isn't reflected by the lampshade so the table doesn't receive 800 lumens.I meant something like this :

View attachment 305702

Sorry. You've lost me.

Can I suggest you take a step back and re-read post #6. Forget about light for a moment and think in terms of water because it's more familiar and easier to think about.

You have a nozzle spraying water downwards and delivering 800 litres per second. Suppose all of the water lands on the table below and there are 300 litres per second per m² landing on the table. Water hits all parts of the table - right up to the edges. That means the table's area is 800/300 = 2.7m² approx.

Once you fully understand that, understanding lumens and lux is easy; just:
- replace litres per second by lumens;
- replace litres per second per m² by lux.
The logic is the same for water and light.

If you wanted to construct a box so that the inner surface receives 800 litres per second per m² you can imagine the water being sprayed onto the inside of the box.

There are several issues/questions getting muddled-up with one another here. I suggest you post a new thread with a single specific question about what you are trying to design and your attempt at a solution. That will be much more manageable.

 

[Xenon02]

There are several issues/questions getting muddled-up with one another here. I suggest you post a new thread with a single specific question about what you are trying to design and your attempt at a solution. That will be much more manageable.

I'm just trying to understand how this lux is calculated. How it changes and what happens with the rest of lumens.

Let's say again that the table receives 300 lux.
the table is the size of 0,64 m^2 typical 80cmx80cm. To the amount of lumen that is distributed all over the table is 192 lumens ? Even if the bulb has 800 Lumens, the table receives 192 lumens. The rest of the lumens is distributed all over table for example walls, floors etc.

I was also just curious just how the lumens and lux works when the light isn't just a circle but it is spread on the walls because light has 180 degrees or even more. Simple ones like flashlight have less than 90 degrees so it is easier.

I have seen this flash light with 200 lux and 800 lux. So I was wondering if I could do the same with a light bulb that his range is 180 degrees like here :

And the whole area he is lighting is 800 lux. Not the table but the whole area he is providing the light.
I mean here this equation 800 lux = 800 lumen/1m^2.
So if this light is giving light to walls and floor it is not a simple beam that has a shape of a circle so maybe if the floor and wall were summary of 1m^2 then the whole area he is giving the light would be 800 lux.

 

[Steve4Physics]

Let's say again that the table receives 300 lux.
the table is the size of 0,64 m^2 typical 80cmx80cm. To the amount of lumen that is distributed all over the table is 192 lumens ?

Yes.

Even if the bulb has 800 Lumens, the table receives 192 lumens. The rest of the lumens is distributed all over table for example walls, floors etc.

Not "all over table".
800 - 192 = 608 lumens.
192 lumens illuminates the table.
608 lumens illuminates the walls, the floor and whatever else is in the room apart from the table.

Of course it's not that simple. There will be reflections and absorptions at all surfaces. Some of the light reaching the table will have been reflected from the walls. But we have to simplify the problem to make it manageable.

I was also just curious just how the lumens and lux works when the light isn't just a circle but it is spread on the walls because light has 180 degrees or even more. Simple ones like flashlight have less than 90 degrees so it is easier.

Its not really an angle in degrees beause we are working in three dimensions, not a simple (2D) plane. We have to use 'solid ang;les' which are usually measured in units called steradians. This starts to make the maths a bit more omplicated.

If you are interested, look up 'solid angle'.

I have seen this flash light with 200 lux and 800 lux.

A flash light should not be rated in lux. I suppose you could use lux if you also gave a reference distance and area. E.g. you could rate a a flash light as '200lux at 10m over 0.8m^2'. This would mean that, when shone at a wall 10m away, the flashlight lights up a 0.8m² area of wall with an illuminance of 200lux.

So I was wondering if I could do the same with a light bulb that his range is 180 degrees like here :

View attachment 305708

And the whole area he is lighting is 800 lux. Not the table but the whole area he is providing the light.
I mean here this equation 800 lux = 800 lumen/1m^2.
So if this light is giving light to walls and floor it is not a simple beam that has a shape of a circle so maybe if the floor and wall were summary of 1m^2 then the whole area he is giving the light would be 800 lux.

It's more complicated.

1) Different parts of the room are at different distances from the bulb; e.g. the bottom corners of the room are further from the lamp than the floor below the lamp. Illuminance (in lux) depends on the distance.

2) The light given off by the lamp may be stronger in some directions than others (e.g. downwards stronger than sidewards); bulbs seldom produce 'evenly spread' light.

These sorts of problem can be dealt with, but quite a lot of maths is involved to do it accurately. A lighting engineer would typically use some ‘rules of thumb’ rather than accurate calculations.

However, I'm off to bed now!

 

[Xenon02]

Not "all over table".
800 - 192 = 608 lumens.
192 lumens illuminates the table.
608 lumens illuminates the walls, the floor and whatever else is in the room apart from the table.

Of course it's not that simple. There will be reflections and absorptions at all surfaces. Some of the light reaching the table will have been reflected from the walls. But we have to simplify the problem to make it manageable.

My english is pretty bad :D
I wanted to say that 192 lumen is table and 608 the rest :D

Yes.

And I thought all along that when I have 800 lumens and a place with 300 lux then it used all 800 lumens. But it seems that it is not 800 lumens but it is 192 lumens.

ts not really an angle in degrees beause we are working in three dimensions, not a simple (2D) plane. We have to use 'solid ang;les' which are usually measured in units called steradians. This starts to make the maths a bit more omplicated.

If you are interested, look up 'solid angle'.

I'm not very smart when it comes to phisics.
I just used the degrees that some sites showed me. But I believe you undestand what do I mean.

A flash light should not be rated in lux. I suppose you could use lux if you also gave a reference distance and area. E.g. you could rate a a flash light as '200lux at 10m over 0.8m^2'. This would mean that, when shone at a wall 10m away, the flashlight lights up a 0.8m² area of wall with an illuminance of 200lux.

ayay me and my english.
What did I mean is that the area that this flash light is providing the light is 200 lux or 800 lux and the flashlight it self is 800 lumens. And I thought that when the area uses all lumens which for the flash light is 800 Lumens. Then I thought I could do the same with the light bulb that has 600 lumens. But the angles that both of them have are different.

For flashlight is easy because it's a simple circle. But for the lightbulb that has 180 degrees which is not just a simple circle (I know I should use other measures, but I believe you know what do I mean like in the picture from the 1st Post).
Then how close should the wall be to each other to make the whole surface (floor and the side walls) that I can use maximum of lumens which is 600 Lumens and 1m^2. then there would be 600 luxes. Like this equation 600 Lux = 600 lumens/1m^2 for this 180 degrees light bulb. Without using reflective material, just shortening the distance of the light bulb and the walls + floor.

Or is it more complicated than with the flashlight ? Of course I just saying theoretically that the light is evenly spread (let's just assume that) and the distance Light bulb - walls + floors are equal.

May I also ask you that when I use lampshade, and the lightbulb has for example 600 lumens. Theoretically the light should be reflected from the lampshades so that the lightbulb is illuminating on the area is still 600 lumens. But in reality some light still illuminates the lampshades not everything is reflected right ? So the lightbulb that has 600 lumens now has a beam that can maximally provide for example 550 lumens because some of them are already illuminating lampshade right ?

Like here :

On the red lightbult lumens.
On the black distributed lumens.
1st one doesn't have lampshade and the 2nd one has one. So I thought that some light is reflected and some is not so I thought that the lumen is reduced. Because not all light is received by the area it is illuminating, some of it is on the lampshade.

If something I said isn't understandable please let me know. I'll try to rewrite.
Thank you again for helping me, because I'm pretty bad at understanding simple things while watching how it works in real life. Especially when I see it everyday in my room.

 

[Steve4Physics]

And I thought all along that when I have 800 lumens and a place with 300 lux then it used all 800 lumens. But it seems that it is not 800 lumens but it is 192 lumens.

Please think about water if confused. E.g. you can have 800 litres/minute coming out of a nozzle. If one surface receives 192 litres/minute then other surfaces must receive 608 litres/minute.

I just used the degrees that some sites showed me. But I believe you undestand what do I mean.

I think I know what you mean. But you will need to use the correct mathematics if you want to improve your understanding. This requires some basic geometry in 3 dimensions.

For flashlight is easy because it's a simple circle.

A flashlight produces a cone-shaped beam of light. The angle (actually solid angle) of the cone is important. It the light then shines squarely on a wall, we get a disc ('simple circle') of light.

But for the lightbulb that has 180 degrees which is not just a simple circle

The lightbulb produces a hemisphere of light. (This is a solid angle which we call '2π steradians'.) If the light shines on a wall, the whole wall is illuminated.

However the centre of the wall is a lot nearer to the bulb than the edges of the wall. The centre of the wall will have more lux than the edges of the wall.

Then how close should the wall be to each other to make the whole surface (floor and the side walls) that I can use maximum of lumens which is 600 Lumens and 1m^2. then there would be 600 luxes.

The question is wrong. You can't just ask how to make the illuminance of the walls 600lux. Is that 600 lux at the centre of the wall? Or at the edges? Or some sort of average? See above.

Like this equation 600 Lux = 600 lumens/1m^2 for this 180 degrees light bulb. Without using reflective material, just shortening the distance of the light bulb and the walls + floor.

Or is it more complicated than with the flashlight ?

Yes - it is more complicated because different parts of the wall are different distances from the bulb as explained above.

Of course I just saying theoretically that the light is evenly spread (let's just assume that) and the distance Light bulb - walls + floors are equal.

No. You can't say these things are equal. It makes no sense. Different parts of the wall are different distances from the bulb. You can never get evenly spread light on flat walls.

May I also ask you that when I use lampshade, and the lightbulb has for example 600 lumens. Theoretically the light should be reflected from the lampshades so that the lightbulb is illuminating on the area is still 600 lumens. But in reality some light still illuminates the lampshades not everything is reflected right ?

Ok. Some of the light hitting the lampshade will be absorbed - it gets permanently removed by the lampshade.

So the lightbulb that has 600 lumens now has a beam that can maximally provide for example 550 lumens because some of them are already illuminating lampshade right ?

Like here :

View attachment 305714

On the red lightbult lumens.
On the black distributed lumens.
1st one doesn't have lampshade and the 2nd one has one. So I thought that some light is reflected and some is not so I thought that the lumen is reduced. Because not all light is received by the area it is illuminating, some of it is on the lampshade.

No sure I understand. The lampshade does two things - it absorbs some of the light and it redirects (reflects) some of the light. You must take both of these into account.
_________________

There are too many different/overlapping questions, and it is becoming too difficult for me to answer you.

If you are still struggling, you need to reviiew what has alaready been said. If necessary post a new single question.

 

[malawi_glenn]

If you take a piece from the sphere with a certain area - that piece is curved.
So when you compare that area with the area projected on a flat surface, you have to be careful.

Consider the analogy with a circle segment and a straight line

##L_1 \neq L_2##
But the larger the radius, the better the approximation ##L_1 \approx L_2## becomes.

 

[Xenon02]

Please think about water if confused. E.g. you can have 800 litres/minute coming out of a nozzle. If one surface receives 192 litres/minute then other surfaces must receive 608 litres/minute.I think I know what you mean. But you will need to use the correct mathematics if you want to improve your understanding. This requires some basic geometry in 3 dimensions.A flashlight produces a cone-shaped beam of light. The angle (actually solid angle) of the cone is important. It the light then shines squarely on a wall, we get a disc ('simple circle') of light.The lightbulb produces a hemisphere of light. (This is a solid angle which we call '2π steradians'.) If the light shines on a wall, the whole wall is illuminated.

However the centre of the wall is a lot nearer to the bulb than the edges of the wall. The centre of the wall will have more lux than the edges of the wall.The question is wrong. You can't just ask how to make the illuminance of the walls 600lux. Is that 600 lux at the centre of the wall? Or at the edges? Or some sort of average? See above.Yes - it is more complicated because different parts of the wall are different distances from the bulb as explained above.No. You can't say these things are equal. It makes no sense. Different parts of the wall are different distances from the bulb. You can never get evenly spread light on flat walls.Ok. Some of the light hitting the lampshade will be absorbed - it gets permanently removed by the lampshade.No sure I understand. The lampshade does two things - it absorbs some of the light and it redirects (reflects) some of the light. You must take both of these into account.
_________________

There are too many different/overlapping questions, and it is becoming too difficult for me to answer you.

If you are still struggling, you need to reviiew what has alaready been said. If necessary post a new single question.

I think I understand now how it works.
I'll try to summarize it :

1. if there is some amount of lux in a certain area like 300 lux on a table it doesn't mean there is full 800 lumens of lightbulb in this area. It can be part of it like 192 lumens.
2. In a flashlight when I have 200 lux on the 4 m^2 wall contains 800 lumens of flashlight because the whole light is focused in that 4m^2 area so that's why there is 800 lumens and not part of it like in the table example.
3. If I want to have 800 lux using full 800 lumens from lightbulb in a room might be quite difficult to calculate because the light isn't cone-shaped beam like in the flashlight. So it is not focused in one place this light.
4. Shadelamp might receive some of this light from the lightbulb like 20 lumens out of 600 lumens but the rest is focused like the cone-shaped beam so I can use the rest which was 600-20 = 580 lumens and calculate the lux on a specific area using full 580 lumens for example Lux = 580lumens/2m^2 = 240 lux.

I think I slowly get the idea how it works. It is far more complicated. Also I thought that when there was some lux value that means it uses full lumen value. That's what every side showed using flashlights :D But now I know it doesn't have to be like that. I believe that thsi summarizing is quite good ?

 

[Steve4Physics]

1. if there is some amount of lux in a certain area like 300 lux on a table it doesn't mean there is full 800 lumens of lightbulb in this area. It can be part of it like 192 lumens.

Yes.

2. In a flashlight when I have 200 lux on the 4 m^2 wall contains 800 lumens of flashlight because the whole light is focused in that 4m^2 area so that's why there is 800 lumens and not part of it like in the table example.

Yes.

3. If I want to have 800 lux using full 800 lumens from lightbulb in a room might be quite difficult to calculate because the light isn't cone-shaped beam like in the flashlight. So it is not focused in one place this light.

Yes. It's not just a difficult calculation. It's difficult to achieve in reality. (For example, a lighting-engineer might use many lamps and diffusers to achieve an approximately uniform light distribution.)

4. Shadelamp might receive some of this light from the lightbulb like 20 lumens out of 600 lumens but the rest is focused like the cone-shaped beam so I can use the rest which was 600-20 = 580 lumens and calculate the lux on a specific area using full 580 lumens for example Lux = 580lumens/2m^2 = 240 lux.

Ok, but 580/2 = 290, not 240!

I think I slowly get the idea how it works. It is far more complicated. Also I thought that when there was some lux value that means it uses full lumen value. That's what every side showed using flashlights :D But now I know it doesn't have to be like that.

Yes, I think you 'get it'.

I believe that thsi summarizing is quite good ?

Yes - well done! You understand much better than you did at Post #1!

 

[Xenon02]

Ah yes can I ask one more thing ?

For a flash light it's easy to calculate lux because it's a simple equation. Lux = max Lumens/ m^2.
For a light bulb that has this wide range like in the picture with 180 degrees. If I use max Lumens from this lightbulb then I can't use this simple equation to calculate lux which is Lux = max Lumens/ m^2 right ? This equation works only for cone-shaped beam and for a certain area like this table. But for wide range of light this equation doesn't work to calculate lux yea ?

 

[Steve4Physics]

Ah yes can I ask one more thing ?

For a flash light it's easy to calculate lux because it's a simple equation. Lux = max Lumens/ m^2.

Ok. First, here’s a short exercise for you to check your basic understanding!

All the light from a 100 lumen torch-beam falls on a circular 50cm diameter region on a wall.
a) Find the average lux value of the disc of light.
b) Why is the word 'average' important'?
Post your answer!

For a light bulb that has this wide range like in the picture with 180 degrees. If I use max Lumens from this lightbulb then I can't use this simple equation to calculate lux which is Lux = max Lumens/ m^2 right ? This equation works only for cone-shaped beam and for a certain area like this table. But for wide range of light this equation doesn't work to calculate lux yea ?

You need some additional mathematics. You can’t use degrees. You have to learn to use ‘solid angles’ (measured in units of steradians) because the light spreads out in three dimensions.

It isn’t practical to explain here. I suggest doing some background reading/viewing. E.g. a YouTube search gives this:

Edit: in the video, the presenter gives solid angle the symbol dΩ. But he should really use just Ω.

 

[Xenon02]

Ok. First, here’s a short exercise for you to check your basic understanding!

All the light from a 100 lumen torch-beam falls on a circular 50cm diameter region on a wall.
a) Find the average lux value of the disc of light.
b) Why is the word 'average' important'?
Post your answer!

For A it would be 31.8 lux ?
For B I can try to answear. That maybe this is estimated amount and not exact amount on 50 cm area.

You need some additional mathematics. You can’t use degrees. You have to learn to use ‘solid angles’ (measured in units of steradians) because the light spreads out in three dimensions.

It isn’t practical to explain here. I suggest doing some background reading/viewing. E.g. a YouTube search gives this:

Edit: in the video, the presenter gives solid angle the symbol dΩ. But he should really use just Ω.

I know I should use solid angles but I didn't know how to change this 180 degrees from the picture so I had to use this as my reference. I know it is not correct. I just wanted to show which light did I mean.

I wanted to know that calculating the lux from this type of light with wide range and not a cone shaped beam or a simple plain like table, that it would be more complicated than just Lux = Lumens/m^2. Because it is not a simple plain it is a three dimension and not a two dimension like a simple table or a beam that is a cone shaped beam. But I wanted to know that If I want to calculate the lux using max Lumens from this light bulb with this wide range then this equation Lux = max Lumens/ m^2 wouldn't work yes ? It requires more complex calculation to calculate the lux in a room using maximum value of Lumens with a light bulb with wide range.

 

[Steve4Physics]

For A it would be 31.8 lux ?

No. Please post your working. Unless you can handle the simple problems, you won't be able to handle the harder ones.

For B I can try to answear. That maybe this is estimated amount and not exact amount on 50 cm area.

No. And 50cm is not an area; it is a length!

Note: If you make a 5 hour car journey and your average speed is 50 km/h, does that mean your speed was exactly 50km/h for whole journey?

If you have a 5m² surface with an average illuminance of 50 lux (=50lumens/m²), do you think that each square metre receives exactly 50 lux?

I know I should use solid angles but I didn't know how to change this 180 degrees from the picture so I had to use this as my reference. I know it is not correct. I just wanted to show which light did I mean.

I wanted to know that calculating the lux from this type of light with wide range and not a cone shaped beam or a simple plain like table, that it would be more complicated than just Lux = Lumens/m^2. Because it is not a simple plain it is a three dimension and not a two dimension like a simple table or a beam that is a cone shaped beam. But I wanted to know that If I want to calculate the lux using max Lumens from this light bulb with this wide range then this equation Lux = max Lumens/ m^2 wouldn't work yes ?

It requires more complex calculation to calculate the lux in a room using maximum value of Lumens with a light bulb with wide range.

Yes - more complex calculations are needed .

You have to learn, understand and use the relevant mathematics, e.g. solid angles, area of a sphere, inverse square law.

 

[Xenon02]

No. Please post your working. Unless you can handle the simple problems, you won't be able to handle the harder ones.

Ok so I'll make some corrections.
50 cm diameter region so I have to calculate the surface area which will be 0,5m = d so the radius will be r = 0,25m
A = Pi * (0,25)^2 let's say it is 0,20 m ok ?

So now lux = 100 Lumen (maximum value)/0,20 = 500 Lux.

Yes - more complex calculations are needed .

You have to learn, understand and use the relevant mathematics, e.g. solid angles, area of a sphere, inverse square law.

As I thought. This simple equation Lux = Max Lumens/ m^2 wouldn't work for wide light beams.

So this equation Lux = Lumens/m^2. Is for specific area (table, wall, object) or if I want to use max Lumens for example flash light then I can use Lux = Max Lumens/m^2. But this doesn't work for wide light beams.

 

[Steve4Physics]

Ok so I'll make some corrections.
50 cm diameter region so I have to calculate the surface area which will be 0,5m = d so the radius will be r = 0,25m
A = Pi * (0,25)^2 let's say it is 0,20 m ok ?

Not quite:
1) You have the wrong units. The unit for area is m², not m. In an examination, you lose 1 mark!
2) It is not usually a good idea to round intermediate values too much. I would have used A=0.1963m².

So now lux = 100 Lumen (maximum value)/0,20 = 500 Lux.

100lumen /0.1963m² = 509 lux = 510 lux (rounded to 2 significant figures)
You need to be more careful.

As I thought. This simple equation Lux = Max Lumens/ m^2 wouldn't work for wide light beams.

So this equation Lux = Lumens/m^2. Is for specific area (table, wall, object) or if I want to use max Lumens for example flash light then I can use Lux = Max Lumens/m^2. But this doesn't work for wide light beams.

It can work if you only want an average lux value.

E.g. a light bulb in an emppty room emits 100 lumens downwards.
The walls and floor have a total area of 200m².
Some light goes on the walls and some goes on the floor. No light from the bulb goes directly onto the ceiling. (We will ignore reflections.)

The average illuminance of the walls and floor = 100 lumens/200m² = 0.5lux.

However - this is an average. For example the walls could be dim and the floor bright in the middle and dim at the edges. We found the average but it is not usefu herel.

 

[Xenon02]

Not quite:
1) You have the wrong units. The unit for area is m², not m. In an examination, you lose 1 mark!
2) It is not usually a good idea to round intermediate values too much. I would have used A=0.1963m².

For the first one I completely forgot to add ^2. For the second one, I thought I coudl use round it to 0,2 m²

100lumen /0.1963m² = 509 lux = 510 lux (rounded to 2 significant figures)
You need to be more careful.

Sorry.

It can work if you only want an average lux value.

E.g. a light bulb in an emppty room emits 100 lumens downwards.
The walls and floor have a total area of 200m².
Some light goes on the walls and some goes on the floor. No light from the bulb goes directly onto the ceiling. (We will ignore reflections.)

The average illuminance of the walls and floor = 100 lumens/200m² = 0.5lux.

However - this is an average. For example the walls could be dim and the floor bright in the middle and dim at the edges. We found the average but it is not usefu herel.

And I thought that I can't use this equation for wide light beam. Because I thought it would be more complex. But here the thing I had to do is add up all the surface areas (I mean here floor : 80 m^2 + 1st wall : 40m^2 + ...). I thought that this equation can be used only for for light that falls perpendicularly on some surface, and it was easy for flash light, but the wide light beam they go in many diretion.

So for the flashlight those luxes are also average amount as I know.
In your example with torch where you calculated 50 lux for 5m². Some square meters receives more and some less luxes. For flash lights it works the same as I know. I thought that it is more of an exact number. Many sites didn't mention this is average amount.

 

[pinball1970]

Great thread and why pf is so good. This particular problem raised its ugly head in a problem for me at work. I am going to read this from the beginning to the end as my misconceptions were @Xenon02 also.

 

[Xenon02]

So I wasn't alone here :D Good to know because it was a very interesting topic for me.

 

[Steve4Physics]

Many sites didn't mention this is average amount.

Yes. At an introductory teaching level, the lux values are usually average ones; and this is not usually made clear.

But in the real world, the average lux value for a large area is often not useful. However, for fun, here a short problem...

The square floor of a room measures 4m x 4m.
A 300 lumen bulb hangs 2m above the centre of the floor.
The bulb emits light equally in all directions.
What is the average illuminance (lux) of the floor?

(There is a simple answer because of the values I’ve used.)

And, if you are interested, the lux-level at the centre of the floor will be more than three times bigger than the lux level near a corner of the floor. So knowing the average value is of limited use.

 

[hutchphd]

Having made much of my fabulous wealth by designing precison LED based measurement devices, it may be some solace to @Xenon02 that I always kept a "cheat sheet" of the various photometric and radiometric quantities and their definition (and associated units) on the wall next to me in my office. And I still get it wrong occasionally...

 

[Xenon02]

Yes. At an introductory teaching level, the lux values are usually average ones; and this is not usually made clear.

Good to know :D So I was also right with adding every surface together ? Floor + walls + ... like in the example I gave ? To use this funny equation ?

But in the real world, the average lux value for a large area is often not useful. However, for fun, here a short problem...

The square floor of a room measures 4m x 4m.
A 300 lumen bulb hangs 2m above the centre of the floor.
The bulb emits light equally in all directions.
What is the average illuminance (lux) of the floor?

From those 3 dots I can say that I am quite tough to teach me :D

The light bulbt emits the light in all directions which is not only the floor but also walls ect ?
If the walls to then it's hard for me to tell because I don't know how much lumen receives the floor.
But I'll tell that the floor receives all 300 lumens. Then surface is 16 m².

Lux = 300/16 = 18,75 lux per m².

And, if you are interested, the lux-level at the centre of the floor will be more than three times bigger than the lux level near a corner of the floor. So knowing the average value is of limited use.

I can see it now that Those lux are only average value.

I think it also works with the flash lights.
If I look for example here :

The one with 200 Lux. It would be also average because the center of this area is more brighter and the edges. Soo 200 Lux is also average.

 

[Steve4Physics]

Good to know :D So I was also right with adding every surface together ? Floor + walls + ... like in the example I gave ? To use this funny equation ?

Yes, you were right - providing you really want the average lux value for the different surfaces. There is usually no point in calculating this average. Try answering this question:

The average lux value for a room is 100 lux. What is the lux value for the floor?
a) More than 100lux
b) 100lux
c) Less than 100 lux
d) Can't tell

Click the 'Spoiler' to check your answer:

Spoiler

The correct answer is d). You can't tell the floor's lux level. Knowing the average value for the whole room doesn't help. It is useless in this situation.

The light bulbt emits the light in all directions which is not only the floor but also walls ect ?

Yes. In my problem, the light hits the ceiling, walls and floor. For example, if you used a simple old-fashioned filament bulb.

If the walls to then it's hard for me to tell because I don't know how much lumen receives the floor.

There is a way to work out how much light reaches the floor. For a hint, click the spoiler:

Spoiler

The 300 lumen bulb emit light equally in all directions. Imagine the bulb is at the centre of a cube. How much light passes through each face of the cube?

But I'll tell that the floor receives all 300 lumens. Then surface is 16 m².

Lux = 300/16 = 18,75 lux per m².

So that's not the answer we want. Can you do it using the hint given above?

I can see it now that Those lux are only average value.

I think it also works with the flash lights.

Yes. For a simple flashlight beam, falling squarely on a wall, the average is useful.

 

[Xenon02]

Yes, you were right - providing you really want the average lux value for the different surfaces. There is usually no point in calculating this average. Try answering this question:

The average lux value for a room is 100 lux. What is the lux value for the floor?
a) More than 100lux
b) 100lux
c) Less than 100 lux
d) Can't tell

Click the 'Spoiler' to check your answer:

I'm fully aware that average lux value has no point here. It was just curiosity.
Also it's logical that I can't say what is the lux value on the floor. Because on the floor it could be 300-500 lux. but average of every lux value on every surface is 100 lux. Some can have 50 lux some even 2 lux (maybe).

There is a way to work out how much light reaches the floor. For a hint, click the spoiler:

Spoiler

The 300 lumen bulb emit light equally in all directions. Imagine the bulb is at the centre of a cube. How much light passes through each face of the cube?

thinking of the cube then 300/6 (4 walls, 1 floor, 1 ceiling).
So 50 lumen on the floor. So looking at it this way then. 50lumens/16m² = 3,125 lux average.

Yes. For a simple flashlight beam, falling squarely on a wall, the average is useful.

it's easier more practical because all lumens are focused in this one area, instead of part of lumens in one area.

 

[Steve4Physics]

I'm fully aware that average lux value has no point here. It was just curiosity.

Sorry. I wasn't sure if you had realized.

thinking of the cube then 300/6 (4 walls, 1 floor, 1 ceiling).
So 50 lumen on the floor. So looking at it this way then. 50lumens/16m² = 3,125 lux average.

Yes. Well done!

I'd think about it like this. A cube has 6 equal-area faces. If the 300 lumen (evenly spread) bulb is at the centre of any cube, then, by symmetry, exactly the same amount of light must fall on each face of the cube. So each face recives 300/6 = 50 lumen.

For the problem in Post #29, the floor measures 4m and 4m and the bulb is 2m above the centre of the floor. The floor must receive 1/6 of the total light, i.e. 50 lumens.

(Note, if the bulb were at a different height, the problem gets *much* more complicated!)

 

[Xenon02]

(Note, if the bulb were at a different height, the problem gets *much* more complicated!)

Because the light then isn't evenly spread as I can assume. And it will be more complicated to calculate the lumens that the floor receives.