Post by thread:[EE] Solid geometry

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)

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_OUTmicrobrixTakeThisOuTgmail.com> wrote: {Quote hidden}> 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) > --

On 19/04/2008, David VanHorn <.....microbrixKILLspam@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

On Sat, Apr 19, 2008 at 4:24 PM, Sean Breheny <shb7KILLspamcornell.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.

> 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

> 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.

Hi Dave, What is "the scene" you are talking about below? Sean On Sat, Apr 19, 2008 at 7:20 PM, David VanHorn <.....microbrixKILLspam.....gmail.com> wrote: {Quote hidden}> > 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. > > > --

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 <microbrixspam_OUTgmail.com> wrote: > On Sat, Apr 19, 2008 at 8:04 PM, Sean Breheny <@spam@shb7KILLspamcornell.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. :) > -- >

On Sat, Apr 19, 2008 at 9:31 PM, Sean Breheny <KILLspamshb7KILLspamcornell.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.

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 <RemoveMEmicrobrixTakeThisOuTgmail.com> wrote: {Quote hidden}> On Sat, Apr 19, 2008 at 9:31 PM, Sean Breheny <spamBeGoneshb7spamBeGonecornell.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. > -- >

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 <TakeThisOuTmicrobrixEraseMEspam_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. > -- >

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

>> 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