Steve4Physics said:
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
Steve4Physics said:
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.
Steve4Physics said:
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.
Steve4Physics said:
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.