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

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

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  1. frank

    frank Guest

    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
    Luminous intensity: 3 [lx/m^2]/steradian
    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. 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.
    Here is a good link.
    http://whatis.techtarget.com/definition/0,,sid9_gci528813,00.html

    Cheers!

    Chip Shults
     
  3. 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
    of radius "R".
     
  4. frank

    frank Guest

    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]/
    steradian.
    Luminous flux is defined as illuminance/area : lx/m^2.
    Illuminace is defined in lx units.

    Frank
     
  5. 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
    angle, say its X, measured in steradians. Then, your eye will receive
    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.)

    Thus, given your 3 cd source, your eye will receive

    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. ----------------
    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 Fields

    John Fields Guest

    ---
    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. 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. 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. frank

    frank Guest

    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 Bybee

    Louis Bybee Guest

    [/QUOTE]

    As I understand it (not to any great degree), it is the control of the
    pattern or direction of radiation.

    Google is your friend. :-]

    http://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q="diffraction"&btn
    G=Google+Search

    Louis--
    *********************************************
    Remove the two fish in address to respond
     
  12. John Fields

    John Fields Guest

     
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