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'[EE] Solid geometry'
2008\04\19@131847 by David VanHorn

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I'm having trouble remembering where to look for the pieces of this,
I'd appreciate any pointers.

The problem is to find the minimum LED output to illuminate the
surface of a sphere to 1 lux at a given distance, with the
complicating factor that the LED's pattern is not symmetrical.

1 Lux = 1 lumen per square meter, and at a 30' radius the area of the
sphere is 1050m^2, but I need to find the area of the sphere that
would be illuminated by the LED.  It's been a long time, and I'm
having trouble remembering where to look.

Any pointers?  (fwiw, I'm using Mathcad 11)

2008\04\19@162458 by Sean Breheny

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Hi Dave,

Try: http://en.wikipedia.org/wiki/Solid_angle

I think that you can probably get away with an approximation (a 30
foot radius gives a pretty flat spherical surface, so you can probably
approximate the area with a rectangle). Otherwise, you'll have to do
an integral involving the pattern function of the LED.

By the way, what kind of an LED do you have that is visible at 30 feet
away?! Must either be very focused or be a super-bright LED module.

Sean


On Sat, Apr 19, 2008 at 1:18 PM, David VanHorn <spam_OUTmicrobrixTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

>  --

2008\04\19@171909 by Philip Stubbs

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On 19/04/2008, David VanHorn <.....microbrixKILLspamspam@spam@gmail.com> wrote:
> I'm having trouble remembering where to look for the pieces of this,
>  I'd appreciate any pointers.
>
>  The problem is to find the minimum LED output to illuminate the
>  surface of a sphere to 1 lux at a given distance, with the
>  complicating factor that the LED's pattern is not symmetrical.
>
>  1 Lux = 1 lumen per square meter, and at a 30' radius the area of the
>  sphere is 1050m^2, but I need to find the area of the sphere that
>  would be illuminated by the LED.  It's been a long time, and I'm
>  having trouble remembering where to look.
>
>  Any pointers?  (fwiw, I'm using Mathcad 11)

http://www.lib.uwaterloo.ca/discipline/opt/documents/light-measurement.pdf

That has some stuff that may help.

--
Philip Stubbs

2008\04\19@172947 by David VanHorn

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> By the way, what kind of an LED do you have that is visible at 30 feet
> away?! Must either be very focused or be a super-bright LED module.

Osram Golden Dragons.  1A.   Very nice, needs thermalcore PCB or equivalent.

2008\04\19@173103 by David VanHorn

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On Sat, Apr 19, 2008 at 4:24 PM, Sean Breheny <shb7spamKILLspamcornell.edu> wrote:
> Hi Dave,
>
> Try: http://en.wikipedia.org/wiki/Solid_angle
>
> I think that you can probably get away with an approximation (a 30
> foot radius gives a pretty flat spherical surface, so you can probably
> approximate the area with a rectangle). Otherwise, you'll have to do
> an integral involving the pattern function of the LED.

I was figuring on simplifying, using the half-power angle and assuming
that the power was in fact flat. If I get more than I was figuring on,
that's ok.  I just need to show that I won't get less.

2008\04\19@174126 by Apptech

face
flavicon
face
> I'm having trouble remembering where to look for the
> pieces of this,
> I'd appreciate any pointers.
>
> The problem is to find the minimum LED output to
> illuminate the
> surface of a sphere to 1 lux at a given distance, with the
> complicating factor that the LED's pattern is not
> symmetrical.

The problem is probably easy enough once fully stated.
As defined it sounds ambiguous.
Can you describe what's needed a bit more completely?

Do you want to spread the LED energy evenly over a whole
sphere? / have a selected part at or above a certain level?
/ .... ?

       Russell



2008\04\19@192046 by David VanHorn

picon face
> The problem is probably easy enough once fully stated.
> As defined it sounds ambiguous.
> Can you describe what's needed a bit more completely?

That's probably why I'm having trouble.

> Do you want to spread the LED energy evenly over a whole
> sphere? / have a selected part at or above a certain level?

The swamp I'm trying to drain, is how much energy is a given LED
painting on the scene at 30' (or other distance)

My thought was to start with the sphere, cut away at the 50% power
points, and say that everything within that area is at least X.  The
led's pattern isn't symmetrical in X and Y though, so I can't use a
simple cone.

2008\04\19@200501 by Sean Breheny

face picon face
Hi Dave,

What is "the scene" you are talking about below?

Sean


On Sat, Apr 19, 2008 at 7:20 PM, David VanHorn <.....microbrixKILLspamspam.....gmail.com> wrote:
{Quote hidden}

> --

2008\04\19@212104 by David VanHorn

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On Sat, Apr 19, 2008 at 8:04 PM, Sean Breheny <EraseMEshb7spam_OUTspamTakeThisOuTcornell.edu> wrote:
> Hi Dave,
>
> What is "the scene" you are talking about below?

Dosen't really matter.. The inside of a sphere of 30' radius. :)

2008\04\19@213125 by Sean Breheny

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Hi Dave,

So you want the LED's total light output over the entire sphere? If
so, you can get that directly from the LED specs, you don't need to do
lots of fancy math.

Sean


On Sat, Apr 19, 2008 at 9:20 PM, David VanHorn <microbrixspamspam_OUTgmail.com> wrote:
> On Sat, Apr 19, 2008 at 8:04 PM, Sean Breheny <@spam@shb7KILLspamspamcornell.edu> wrote:
>  > Hi Dave,
>  >
>  > What is "the scene" you are talking about below?
>
>  Dosen't really matter.. The inside of a sphere of 30' radius. :)
>  --
>

2008\04\19@214138 by David VanHorn

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> http://www.lib.uwaterloo.ca/discipline/opt/documents/light-measurement.pdf
>
> That has some stuff that may help.

Thanks, that's a very good doc!

2008\04\19@214225 by David VanHorn

picon face
On Sat, Apr 19, 2008 at 9:31 PM, Sean Breheny <KILLspamshb7KILLspamspamcornell.edu> wrote:
> Hi Dave,
>
> So you want the LED's total light output over the entire sphere? If
> so, you can get that directly from the LED specs, you don't need to do
> lots of fancy math.

The dragons have a 60 degree half-angle, and I'm going to simplify and
assume that power across that cone is equal to power at the edges.  If
I have more in the middle, that's fine.

2008\04\19@221148 by Sean Breheny

face picon face
I'm still confused. If you just want total light output, for higher
power LEDs, the total luminous output (lumens) is often spec'd. Is it
not for this LED?

If you want to guarantee a minimum illumination intensity on the
surface of the sphere, then clearly you need to specify some area of
the sphere as a region of interest because the parts of the sphere
directly behind the LED will not be lit much at all.

Sean


On Sat, Apr 19, 2008 at 9:42 PM, David VanHorn <RemoveMEmicrobrixTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

2008\04\19@230312 by David VanHorn

picon face
> If you want to guarantee a minimum illumination intensity on the
> surface of the sphere, then clearly you need to specify some area of
> the sphere as a region of interest because the parts of the sphere
> directly behind the LED will not be lit much at all.

That's why I was referring to the cone. Figure a cone whose apex is at the LED.

2008\04\20@002047 by Sean Breheny

face picon face
I get the feeling that we are talking in circles here. :)

Let's say you have the LED at the center of an actual 30 foot radius
spherical shell. If you were also at the center and took a look
around, you would see some kind of a light pattern on the inside of
the spherical shell. There would most likely be an area of high
brightness within which there would be several local maxima, and then
the edges of this area would fade to lower and lower levels of
brightness, perhaps with some local maxima and minima along the way
(sidelobes and nulls).

You said that the pattern does not have symmetry around the radius of
the sphere (axis of radiation from the LED). This means that the
"bright spot" on the inside of the shell would be an ellipsoid rather
than a circle.

Now, can you describe what part of the inside of the sphere you need
to consider for your light intensity measurement? In general, this
area of interest will NOT be the same as the "bright spot" from the
LED, especially since the actual bright spot will vary somewhat from
LED to LED and especially across different models of LEDs. The area of
interest would usually be dictated by your application and NOT the
device being used to do the illumination.

It sounds like you are saying that you want to find the minimum
illumination level that occurs within the main lobe of the LED, where
you are defining the boundaries of the main lobe by the half-power
beamwidths (which are different in the two angular coordinates since
it is not a cone). Is this what you want? If so, beware that this may
not actually be spec'd very well. I'd suspect that at 30 feet it would
actually vary a fair amount within the spot due to imperfections in
the LED lens.

Sean


On Sat, Apr 19, 2008 at 11:02 PM, David VanHorn <TakeThisOuTmicrobrixEraseMEspamspam_OUTgmail.com> wrote:
> > If you want to guarantee a minimum illumination intensity on the
>  > surface of the sphere, then clearly you need to specify some area of
>  > the sphere as a region of interest because the parts of the sphere
>  > directly behind the LED will not be lit much at all.
>
>  That's why I was referring to the cone. Figure a cone whose apex is at the LED.
>  --
>

2008\04\20@004727 by William \Chops\ Westfield

face picon face
On Apr 19, 2008, at 4:20 PM, David VanHorn wrote:

> The swamp I'm trying to drain, is how much energy is a given LED
> painting on the scene at 30' (or other distance)

Are you making things more difficult than they need to be?  I mean,  
ALL of the energy is painting the scene at 30 feet, and any  
particular spot has intensity falling off proportionally to  1/d^2,  
right?  You just need the particular constants associated with the  
"usual" equation...

BillW

2008\04\20@105305 by Apptech

face
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face
>> The swamp I'm trying to drain, is how much energy is a
>> given LED
>> painting on the scene at 30' (or other distance)

Do you have specified candella at beam centre and
distribution patterns?
Or just gross lumens and distribution patterns?

If the former then it's very easy [tm].
And probably wrong :-).

Given what you said about guaranteed minimum and half power
points.

- Select beam orientation with NARROWEST angle.
- Note angle for half power or power at desired angle.
- Candella are lumens/m^2 at 1m. so
If radius = 30 feet = 9.144m.
At 30 feet brightness will be down 1/9.144^2 over 1m =
1.196% of level at 1m.

Minimum guaranteed irradiance =

Cd x k_a x 0.01196

   k_a = 0.5 for half power point
             or whatever for selected angle in step 2 above


QED ???

if you have gross lumens and no other info then you will
need to integrate the lumens with angle. 'Untidy' for a non
rotationally symmetrical pattern. Easy enough with squared
paper for a rotationally symmetric pattern.

We still really don't know exactly what you have available
and what you are trying to achieve.
Does your Golden Dragon have a custom lens on it or is it an
OTS unit? If the latter can you point us to a data sheet. If
the former is the lens spec and LED spec available? Or the
combined radiation pattern?


       Russell McMahon



       Russell



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