This is an exercise that needed to be done, as I had no indication of the amount of light as a percentage of the total produced that will actually hit the surface of the water.

The above fixture has its 50W T8HO bulbs closely packed together, and in that configuration, it loses 49% (!)of its light.

What would happen if we removed the middle bulb?

What would be the difference in light output between 2 bulbs and 3 bulbs?

When light bulbs are closely packed in, light from the top half of the bulb never makes it to the aquarium, since there's an adjacent bulb to block the reflected light.

- It is assumed that all reflected light comes from light bouncing off of the adjacent light bulbs 50% attenuated.
- For the 2 bulb case, there's enough room to bounce around, so - with modern contoured and highly polished reflectors -- weâ€™ll assume that 75% of the reflected light makes it down to the aquarium top.

Fluorescent bulbs produce light 360 degrees around the tube. However, not all the light directly makes to the aquarium. The amount of light directly making it down to the aquarium top depends on the center to center spacing

of the bulbs. The angle which describes the amount of light directly striking the aquarium top can be estimated from

theta = 2 x ARCCOS(0.5 inches/D)

where:

theta is the angle of light directly hitting the aquarium top

0.5 inches -> comes from half the diameter of a T8 bulb.

D is the bulb center to center distance

ARCCOS() is the trigonometric inverse cosine function

As one can see only 1/2 the light (theta = 180 degrees) makes it directly to the aquarium top when the separation is infinity (very wide aquarium), and only 1/3 of the light (theta = 120 degrees) when D = 1 inch (packed-in bulbs).

Note: T8HO lamps can lose about 1/6 of its light output for each 40Â°F rise in temperature.

We'll assume that a two bulb fixture will produce a 40 degree rise when using 2 light bulbs.

Since we'll use the same fixture for the three bulb case, another 20 degree rise in temperature is expected for an internal hood temperature of 140Â°F.(The thermal impedance of the hood is the same for both cases, but the amount

of heat generated is 50% higher when using three bulbs.)

Anyways, let's see where this leads to ...

For the 2 bulb case, each bulb can be separated 2.5 inches center to center. Thus, theta ends up: 2 * ARCCOS(0.5/2.5) being about 157 degrees.

Which means about 43.6% (157/360) of the light makes it directly to the aquarium top, while 75% of the remaining 56.4% also makes it to the aquarium top.

Since the hood temperature rises by 40 degrees, the two bulbs only produce 5/6 (83.3%) of their light. (i.e. loss of about 1/6 of it light output for each 40Â°F rise in temperature.)

Total lumens available to the aquarium top two bulb case, with each bulb producing 3400 lumens at 80Â°F:

2 (lights) x 3400 lumens x ( 43.6% +75% x 56.4%) x 83.3% (@120Â°F) = 4867 lumens (71.6% !)

For the 3 bulb case,

theta = 2 * ARCCOS(0.5 inches/1.25)= 133Â°. Therefore 36.9% of the light is direct, while 63.1% is reflected off the adjacent bulbs and attenuated by 50% = 31.5%.

The hood temperature is 140 degrees F, so the light bulbs loses 3/12 of its light compared to when the bulb temperature is at 80Â°F.

Total lumens available to the aquarium top six bulb case:

3 (lights) * 3400 lumens * ( 36.9% + 31.5%) * 9/12 (@140Â°F) = 5200 lumens (51% !)

Well - what happened?

In the three bulb case, practically half of the lighting is lost by not having a separating gap between the bulbs. The added heat also reduce the amount of light.

There are a few things one could do to improve the design.

- At about 15 Watts each, adding fan(s) to reduce bulb temperatures seems reasonable.
- Switching to the thinner T5 bulbs would make the reflectors more efficient since they will not block as much reflected light.

The problem is that - in our light exercise - what we called "spatial losses" (=light that doesn't get to the water's surface), set at an average of 15%, is in fact much larger (Â±30% according to this example).