# help with unit conversion --- candela, lumens & lxs and steradians

Discussion in 'Electronic Basics' started by frank, May 9, 2004.

1. ### frankGuest

Hello,

What EXACTLY is steradian ? I mean the EXACT definition.

I have the following dillema:

I am trying to figure out how many lx (illumunance units) does an area
"A" receive from a distance "d" away from a ligth source (more
precisily a LED).

The LED's luminous intensity is defined as 3 cd (candelas, luminous
intensity units).
A candelas is be defined as : [luminous flux]/ steradian.
Luminous flux is defined illuminance/area; lx/m^2.
http://www.convertit.com

I understand everything mentionned above EXCEPT for the following:
From what i remember of my calculus days, a steradian is suppose to be
a solid angle whose lengthwise cross-section makes an angle of 1 rad.
Hence, a steradian is suppose to resemble a cone of infinite
length/volume. It suppose to represent a percentage of space if one
perceive space as an expanding sphere starting from a point (in this
case, the ligth source), rather than a volume of finite dimenions.
This being said, I do not see how a ligth source's luminous intensity
can be constant at all distance from it -- this is what is interpreted
if one assumes a steradian to be a solid angle as described above.

I.e
Therefore, wherever the person is within the steradian, the LED should
seem just as brigth becaue the luminous flux remains the same. Doesn't
make sense.

However, if a steradian happened to be a cone of fixed volume, whose
lengthwise cross section makes 1 rad angle, it would make more sense.
Let's assume that the cone's sides are long by 1m.
Because in this case, the 3 [lx/m^2]/steradian would imply that at 1m
distance from the ligth source, the luminous flux is 3 lx/m^2. At 2m
distance the luminous flux would be 4 times less (i think), and so on.
Hence the closer one is, the brigther the ligth would appear. Now this
would make sense.

So getting back to the original question: What EXACTLY is steradian ?

Any help would be much appreciated.
Thx

-Frank

2. ### Sir Charles W. Shults IIIGuest

Imagine you have a sphere. The surface area of the sphere will be 4pi
steradians. The steradian is defined as the solid angle subtended by a
conic section that will cover an area of a sphere equal to the radius of the
sphere squared.
Think of a circle- it has 2pi circumference compared to its radius. For
a sphere, you square everything. Therefore, the circumference of the circle
is 2pi times the radius and the area of the sphere is 4pi times the radius
squared.
So in essence, the total area is all that matters- you could have a
bunch of little cones that add up to a steradian, but that would be clumsy.
In reality it is usually a single cone.
http://whatis.techtarget.com/definition/0,,sid9_gci528813,00.html

Cheers!

Chip Shults

3. ### LioNiNoiL_a t_NetscapE_D 0 T_NeTGuest

So getting back to the original question: What EXACTLY is steradian?

It's the solid angle that intersects R-squared units of area on a sphere

4. ### frankGuest

Got it.
Area of the "sperical base" of 1 steradian from distance D from the pt
of origin = D^2.

Then if the above true, how can the following be true also:
How can the datasheet of a LED, define its luminous intensity as
constant number of candelas. This would imply that a LED would remain
just as brigth as it is from 1 cm than it is from 1 km.
datasheet of LED:
http://www.omroncomponents.co.uk/Press/DR-LED.pdf

Reminder
Luminous intensity is defined in candelas (cd): [luminous flux]/
Luminous flux is defined as illuminance/area : lx/m^2.
Illuminace is defined in lx units.

Frank

5. ### Robert C MonsenGuest

Luminous flux is a measure of power. The power is spread over an area.
In the case of your 3 cd LED, you have 3/683 W for each solid angle of
1 sr. At some distance from the source, your eye subtends a solid
X*3/683 W. Clearly X depends on both the size of your eye, and the
distance.

If, for example, the size of your eye is 0.001 m^2, and its r meters
from the source, then the ratio of that solid angle to the total
sphere is 0.001/(4pi*r^2). The ratio of your eye to 1 steradian is
obviously 0.001/r^2. (thats why they like to work in steradians, I
bet.)

0.001 * 3 / (683 * r^2) W, or (4.4/r^2) uW

Thus, as you move farther away from the source, the amount of radiant
energy per second is decreased as the square of the distance, just as
expected.

So, the formula for the power received on an area A at distance r
being exposed to c candelas is

P(A,r,c) = A * c / (683 * r^2) Watts

The candela actually measures radiation of a particular frequency, so
your LED should be outputting a frequency of 540e12... A steradian is really a unit of solid angle measurement. Think about a
solid angle. You can measure it as a ratio of the total sphere. Given
a particular area A subtended at a particular distance r, that ratio
would be

A/(4 pi r^2)

ie, the ratio of the part to the whole. However, dropping the 4 pi
gives you the angle measured in steradians, since there are 4 pi
steradians in a sphere. So, given our area A at distance r, the solid
angle it cuts out is (in steradians)

A/r^2

which is more convienient.

Regards,
Bob Monsen

6. ### R. Steve WalzGuest

----------------
You know how a radian is the arc that subtends a length of one unit
of circumference for a unit radius circle? Well, a steradian is the
solid angle that subtends a one square unit area for a unit radius,
usually shown as a circular piece out of the surface of a sphere.

-------------
Not quite.

-Steve
--
-Steve Walz ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public

7. ### John FieldsGuest

---
Consider a circle where the length of an arc on its circumference is
equal to the length of its radius. Now, drop a line from each of the
ends of the arc to the center of the circle in order to form the legs
of an angle with its vertex at the center of the circle. The included
angle between the legs will be one radian.

Now, bisect the radian and sweep the legs around the line bisecting
the radian so they form a cone with the bisector as its axis.

The "solid angle" at the apex of the cone will be one steradian.

Now, replace the circle with a sphere of some diameter and negligible
wall thickness, place an LED at the center of the sphere and turn it
on. The LED will illuminate an area on the surface of the sphere upon
which will fall all of the power emitted by the LED. If the
illuminated area is measured and found to be one square meter and the
optical power of the LED measured and found to be one watt, then the
power falling on that area will be 1W/m².

If the beamwidth of the LED was ~57.3°, then the power incident on
that surface would be 1W/m²/sr, but for our example we don't care.

Now, If the diameter of the sphere is doubled and the power falling on
one square meter of its surface measured it will be found to be
250mW/m², since the area illuminated by the same total power output
from the LED will have increased by a factor of four and the power per
unit area thus diminished by a factor of four.

8. ### Sir Charles W. Shults IIIGuest

Got it.
Area of the "sperical base" of 1 steradian from distance D from the pt
of origin = D^2.

Then if the above true, how can the following be true also:
How can the datasheet of a LED, define its luminous intensity as
constant number of candelas. This would imply that a LED would remain
just as brigth as it is from 1 cm than it is from 1 km.
datasheet of LED:
http://www.omroncomponents.co.uk/Press/DR-LED.pdf[/QUOTE]

The light output is as intense as some other light source (the candle)
times some factor, and the area that light is spread out over is a
steradian. It decreases in intensity by the inverse square law.

9. ### LioNiNoiL_a t_NetscapE_D 0 T_NeTGuest

How can the datasheet of a LED, define its luminous intensity
No. The angle subtended by the detector (e.g., pupil of the eye) at one
kilometer is much smaller than at one centimeter.
No. Luminous flux is illuminance multiplied by area.

10. ### frankGuest

Yup. I messed the unit conversion.
Luminous flux IS illuminance multiplied by area, NOT divided by area.
It all makes sense now.

BTW, does anyone know what diffraction is ?

Thx
-Frank

11. ### Louis BybeeGuest

[/QUOTE]

As I understand it (not to any great degree), it is the control of the

Louis--
*********************************************
Remove the two fish in address to respond

12. ### John FieldsGuest  