Illuminating a
Room
Suppose we need to decide where to put light fixtures on the ceiling of a room measuring 10 meters by 4 meters by 3 meters high in order to best illuminate it. For aesthetic reasons, we are asked to use a small number of incandescent bulbs. We want the bulbs to total a maximum of 300 watts. For a given number of bulbs, how should they be placed to maximize the intensity of the light in the darkest part of the room? We also would like to see how much improvement there is in going from one 300 watt bulb to two 150 watt bulbs to three 100 watt bulbs, and so on. To keep things simple, we assume that there is no furniture in the room and that the light reflected from the walls is insignificant compared with the direct light from the bulbs.
One 300 Watt Bulb
If
there is only one bulb, then we want to put the bulb in the center of the
ceiling. Let's picture how well the
floor is illuminated. We introduce
coordinates x running from 0 to 10 in
the long direction of the room, and y
running from 0 to 4 in the short direction.
The intensity at a given point, measured in watts per square meter, is
the power of the bulb, 300, divided by 4p times the square of the distance from the
bulb. Since the bulb is 3 meters above
the point (5, 2) on the floor, at a point (x,
y) on the floor, we can express the
intensity as follows.
syms x y; illum = 300/(4*pi*((x - 5)^2 + (y - 2)^2 + 3^2))
75/pi/((x-5)^2+(y-2)^2+9)
We
can use ezcontourf to plot this expression over the entire floor. We use the option colormap to arrange for a color
gradation that helps us to see the illumination. (See the online help for more colormap options.)
ezcontourf(illum, [0 10 0 4]); colormap('gray'); axis equal tight
The
darkest parts of the floor are the corners.
Let us find the intensity of the light at the corners, and at the center
of the room.
subs(illum, {x, y}, {5, 2})
0.6282
ans =
2.6526
The
center of the room, at floor level, is about 4 times as bright as the corners
when there is only one bulb on the ceiling.
Our objective is to light the room more uniformly using more bulbs with
the same total amount of power. Before
proceeding to deal with multiple bulbs, we observe that the use of ezcontourf is
somewhat confining, as it does not allow us to control the number of contours
in our pictures. This will be helpful
in seeing the light intensity; therefore we shall plot numerically rather than
symbolically, that is we shall use contourf instead of ezcontourf.
Two 150 Watt
Bulbs
In
this case we need to decide where to put the two bulbs. Common sense tells us to arrange the bulbs
symmetrically along a line down the center of the room in the long direction;
that is, along the line y = 2. Define a function that gives the intensity
of light at a point (x, y) on the floor due to a 150 watt bulb
at a position (d, 2) on the ceiling.
light2 =
inline(vectorize('150/(4*pi*((x - d)^2 + (y - 2)^2 + 3^2))'), 'x', 'y', 'd')
Inline function:
light2(x,y,d) = 150./(4.*pi.*((x - d).^2 + (y - 2).^2 + 3.^2))
Let's
get an idea of the illumination pattern if we put one light at d = 3 and the other at d = 7.
We specify the drawing of 20 contours in this and the following plots.
The
floor is more evenly lit than with one bulb, but it looks as if the bulbs are
closer together than they should be. If
we move the bulbs further apart, the center of the room will get dimmer but the
corners will get brigher. Let's try
changing the location of the lights to d
= 2 and d = 8.
contourf(light2(X, Y, 2) + light2(X, Y, 8), 20); colormap('gray'); axis equal tight
This
is an improvement. The corners are
still the darkest spots of the room, though the light intensity along the walls
toward the middle of the room (near x
= 5) is diminishing as we move the bulbs further apart. Still, to better illuminate the darkest
spots we should keep moving the bulbs apart.
Let's try lights at d = 1 and d = 9.
contourf(light2(X, Y, 1) + light2(X, Y, 9), 20); colormap('gray'); axis equal tight
Looking
along the long walls, the room is now darker toward the middle than at the
corners. This indicates that we have
spread the lights too far apart.
We
could proceed with further contour plots, but instead let's be systematic about
finding the best position for the lights.
In general, we can put one light at x
= d and the other symmetrically at x = 10 - d for d between 0 and
5. Judging from the examples above, the
darkest spots will be either at the corners or at the midpoints of the two long
walls. By symmetry, the intensity will be
the same at all four corners, so let's graph the intensity at one of the
corners (0,0) as a function of d.
d = 0:0.1:5; plot(d, light2(0, 0, d) + light2(0, 0, 10 - d))
As
expected, the smaller d is, the
brighter the corners are. In contrast,
the graph for the intensity at the midpoint (5, 0) of a long wall (again by
symmetry it does not matter which of the two long walls we choose) should grow
as d increases toward 5.
d = 0:0.1:5; plot(d, light2(5, 0, d) + light2(5, 0, 10 - d))
We
are after the value of d for which
the lower of the two numbers on the above graphs (corresponding to the darkest
spot in the room) is as high as possible.
We can find this value by showing both curves on one graph.
hold on; plot(d, light2(0, 0, d) + light2(0, 0, 10 - d)); hold off
The
optimal value of d is at the point of
intersection, near 1.4, with minimum intensity a little under 1. To get the optimum value of d, we find exactly where the two curves
intersect.
Inline function:
eqn(d) = 75/2/pi/(d^2+13)+75/2/pi/((-10+d)^2+13)-75/2/pi/((5-d)^2+13)-75/2/pi/((-5+d)^2+13)
1.4410
So
the lights should be placed about 1.44 meters from the short walls. For this configuration, the approximate
intensity at the darkest spots on the floor is as follows.
light2(0, 0, 1.441) + light2(0, 0, 10 - 1.441)
0.9301
The
darkest spots in the room have intensity around 0.93, as opposed to 0.63 for a
single bulb. This represents an
improvement of about 50%.
Three 100 Watt
Bulbs
We
refedine the intensity function for 100 watt bulbs.
light3 = inline(vectorize('100/(4*pi*((x - d)^2 + (y - 2)^2 + 3^2))'), 'x', 'y', 'd')
Inline function:
light3(x,y,d) = 100./(4.*pi.*((x - d).^2 + (y - 2).^2 + 3.^2))
Assume
we put one bulb at the center of the room and place the other two symmetrically
as before. Here we show the
illumination of the floor when the off-center bulbs are one meter from the short
walls.
It
appears that we should put the bulbs even closer to the walls. (This may not please everyone's
aesthetics!) Let d be the distance of the bulbs from the short walls. We define a function giving the intensity at
position x along a long wall and then
graph the intensity as a function of d
for several values of x.
plot(d, light3(x, 0, d) +
light3(x, 0, 5) + light3(x, 0, 10 - d))
hold on
end
hold off
We
know that for d near 5, the intensity
will be increasing as x increases
from 0 to 5, so the bottom curve corresponds to x = 0 and the top curve to x
= 5. Notice that the x = 0 curve is the lowest one for all d, and it rises as d decreases. Thus d = 0 maximizes the intensity of the
darkest spots in the room, which are the corners (corresponding to x = 0).
There the intensity is as follows.
light3(0, 0, 0) + light3(0, 0, 5) + light3(0, 0, 10)
0.8920
This
is surprising; we do worse than with two bulbs. In going from two bulbs to three, with a decrease in wattage per
bulb, we are forced to move wattage away from the ends of the room and bring it
back to the center. We could probably
improve on the two-bulb scenario if we used brighter bulbs at the ends of the
room and a dimmer bulb in the center, or if we used four 75-watt bulbs. But our results so far indicate that the
amount to be gained in going to more than two bulbs is likely to be small
compared with the amount we gained by going from one bulb to two.