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NC-v4 SSc6:Light pollution reduction

Uplight under semi-opaque canopy

Profile picture for user emily.purcell
Emily Purcell
Sustainable Design Lead
CannonDesign
LEEDuser Expert
379 thumbs up

Hi,

I'm working on a project that is evaluating some different facade lighting and canopy options. We are in LZ2, so we can't use the facade lighting uplight exemption. We are using the BUG method (unless we can't resolve this without photometrics...).

The lighting designer has proposed a fixture like this: https://brucklighting.com/products/outdoor-cylinder-4in-up-and-down.html on the facade, with a brighter light pointing down for walkway safety lighting and a softer uplight for the facade. There is a roof-level canopy (2 stories above the fixture).

The canopy may be used for PV, in which case we'd have an opaque surface blocking the uplight from the sky and would show it as exempt from the U-2 limit. Another option would be a fritted glass canopy (frit level TBD). My gut feeling says this (at least at a higher frit %) would meet the intent of the credit, as it would block some of the light output and diffuse the light that does get through. If you looked at it from above, probably no different from downlight reflecting off concrete elsewhere on the site. But I'd like to be more sure on it meeting the credit requirement before giving the team the go-ahead. 

 

I guess ultimately my question is - is there any semi-opaque surface we could treat as opaque for the purpose of blocking uplight? A minimum frit % it should meet? Or any review experience with demonstrating that uplight is shielded? Thanks! 

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Mon, 07/27/2020 - 16:32

Emily, The canopy, reagrdless of its opacity is going to have to block enough of the light to give you an "effective" U rating of U2. BUG ratings are based on absolute lumens emitted in different solid angles. So you would have to do some sort of analysis to show how many lumens from each the fixture going to miss or go thorugh the canopy and then make the argument that the fixture is effectively U2 or lower
Profile picture for user emily.purcell
Emily Purcell
Sustainable Design Lead
CannonDesign
LEEDuser Expert
379 thumbs up

Mon, 07/27/2020 - 17:01

Thanks Glenn, that makes sense. 

Tue, 07/02/2024 - 15:01

Hello Emily, Could you please confirm if you installed these lighting fixtures on facades for your project? Also, could you share how you ensured compliance with the credit requirements? I am working on a similar project that proposes to use the same type of fixture. Thank you!
Profile picture for user emily.purcell
Emily Purcell
Sustainable Design Lead
CannonDesign
LEEDuser Expert
379 thumbs up

Tue, 07/02/2024 - 15:15

Hi Sofia, I'm no longer working on that project unfortunately! Hopefully someone else here has done a study similar to what Glenn suggested and can share some wisdom...

Tue, 07/02/2024 - 15:22

Thank you very much Emily for your quick response. Meanwhile, I will continue to research possible ways to demonstrate the intent of the credit. 
Profile picture for user swanson.bill
Bill Swanson
Sr. Electrical Engineer
Integrated Design Solutions
LEEDuser Expert
516 thumbs up

Tue, 07/02/2024 - 16:18

Per the IES table, U2 rating allows a max of 50 lumens between 90-100 degrees, and 50 lumens between 100-180 degrees.

Putting that example light into the Photometric Tool, the output says 0.7 lumens between 90-100 degrees, and 705.5 lumens between 100-180 degrees.  The problem for this fixture is 100-180 degrees.  A maximum of 50 lumens compared to the 705.5 lumen output. You might be able to argue that the back half of the light output hits the wall and shield direct output.  I'd suggest 40% uplight shielded by the wall since the light source sticks out from the wall.  It might also be a column and not a wall.  705.5 lumens - 40% = 423.3 lumens.  If all of the remaining light hits the canopy, the canopy would need to block most of the light.  The opacity needs to be 88.2%.
423.2 lumens * 11.8% < 50 lumens.  It is unlikely the canopy will cover all of the light between 100-180 degrees. Most likely there is an angle between the horizontal and the edge of the canopy above.  This is the hard part. The Photometric tool does break down the light output for every 5 degrees.  Hypothetical, let's say there is only 30 degrees.  We need the light output between 100-120.  105 degrees = 18 lumens (- 40% wall shield) = 10.8 lumens
110 degrees = 27 lumens (- 40% wall shield) = 16.2 lumens
115 degrees = 45 lumens (- 40% wall shield) = 27.0 lumens
120 degrees = 65 lumens (- 40% wall shield) = 39.0 lumens
                  Total between 100-120 degrees = 93.0 lumens This fixture would fail in this installation since 93 lumens are emitted when there is a max of only 50 lumens allowed. The opacity of the canopy is irrelevant in this situation since too much light has already escaped below the canopy.

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